Galois Theory Tom Leinster, University of Edinburgh Version of 13 May 2021 Note to the Reader2 1 Overview of Galois Theory4 1.1 the View of C from Q

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Galois Theory Tom Leinster, University of Edinburgh Version of 13 May 2021 Note to the reader2 1 Overview of Galois theory4 1.1 The view of C from Q ...... 4 1.2 Every polynomial has a symmetry group...... 9 1.3 . . . which determines whether it can be solved...... 11 2 Rings and fields 14 2.1 Rings ...... 14 2.2 Fields ...... 19 3 Polynomials 24 3.1 The ring of polynomials...... 24 3.2 Factorizing polynomials ...... 28 3.3 Irreducible polynomials...... 33 4 Field extensions 38 4.1 Definition and examples ...... 38 4.2 Algebraic and transcendental elements...... 43 4.3 Simple extensions...... 47 5 Degree 51 5.1 Degrees of extensions and polynomials...... 51 5.2 The tower law...... 56 5.3 Algebraic extensions ...... 58 5.4 Ruler and compass constructions...... 60 6 Splitting fields 67 6.1 Extending homomorphisms...... 68 6.2 Existence and uniqueness of splitting fields ...... 70 6.3 The Galois group...... 76 7 Preparation for the fundamental theorem 82 7.1 Normality...... 83 7.2 Separability...... 91 7.3 Fixed fields ...... 96 8 The fundamental theorem of Galois theory 100 8.1 Introducing the Galois correspondence ...... 100 8.2 The theorem...... 104 8.3 A specific example ...... 109 9 Solvability by radicals 115 9.1 Radicals...... 116 9.2 Solvable polynomials have solvable groups ...... 119 9.3 An unsolvable polynomial ...... 126 10 Finite fields 130 10.1 ?th roots in characteristic ? ...... 131 10.2 Classification of finite fields ...... 133 10.3 Multiplicative structure...... 135 10.4 Galois groups for finite fields...... 136

1 Note to the reader

These are the 2020–21 course notes for Galois Theory (MATH10080).

Structure Each chapter corresponds to one week of the semester. You should read Chapter 1 in Week 1, Chapter 2 in Week 2, and so on. I’m writing the notes as we go along, so the chapters will appear one by one: keep your eye on Learn.

Exercises looking like this are sprinkled through the notes. The idea is that you try them immediately, before you continue reading. Most of them are meant to be quick and easy, much easier than assignment or workshop questions. If you can do them, you can take it as a sign that you’re following. For those that defeat you, talk with your group or ask on Piazza; or if you’re really stuck, ask me. I promise you that if you make a habit of trying every exercise, you’ll enjoy the course more and understand it better than if you don’t bother.

Digressions like this are optional and not examinable, but might interest you. They’re usually on points that I ﬁnd interesting, and often describe connections between Galois theory and other parts of mathematics. Here you’ll see titles of relevant References to theorem numbers, page numbers, etc., are clickable links. videos What to prioritize You know by now that the most important things in almost any course are the deﬁnitions and the results called Theorem. But I also want to emphasize the proofs. This course presents a wonderful body of theory, and the idea is that you learn it all, including the proofs that are its beating heart. A closed-book exam would test that by asking you to reproduce some proofs. Your exam will be open book, so it can’t ask you to reproduce proofs, but it will test something arguably harder: that you understand them. So the proofs will need your attention and energy.

2 Prerequisites You are required to have taken these two courses:

• Honours Algebra: We’ll need some linear algebra, corresponding to Chap- ter 1 of that course. For example, you should be able to convince yourself that an endomorphism of a ﬁnite-dimensional vector space is injective if and only if it is surjective. We’ll also need everything from Honours Algebra about rings and polynomials (Chapter 3 there), including ideals, quotient rings (factor rings), the universal property of quotient rings, and the ﬁrst isomorphism theorem for rings.

• Group Theory: From that course, we’ll need fundamentals such as normal subgroups, quotient groups, the universal property of quotient groups, and the ﬁrst isomorphism theorem for groups. You should know lots about the symmetric groups (=, alternating groups =, cyclic groups = and dihedral groups =, and I hope you can list all of the groups of order < 8 without having to think too hard. Chapter 8 of Group Theory, on solvable groups, will be crucial for us! If you skipped it, you’ll need to go back and ﬁx that. For example, you’ll need to understand the statement that (4 is solvable but 5 is not. We won’t need anything on free groups, the Sylow theorems, or the Jordan– Hölder theorem.

If you’re a visiting student and didn’t take those courses, please get in touch so we can decide whether your background is suitable.

Mistakes I’ll be grateful to hear of any mistakes ([email protected]), even if it’s something very small and even if you’re not sure.

3 Chapter 1

Overview of Galois theory

This chapter stands apart from all the others, Modern treatments of Galois theory take advantage of several well-developed branches of algebra: the theories of groups, rings, ﬁelds, and vector spaces. This is as it should be! However, assembling all the algebraic apparatus will take us some time, during which it’s easy to lose sight of what it’s all for. Introduction to Galois theory came from two basic insights: Week 1 • every polynomial has a symmetry group; • this group determines whether the polynomial can be solved by radicals (in a sense I’ll deﬁne). In this chapter, I’ll explain these two ideas in as short and low-tech a way as I can manage. In Chapter2 we’ll start again, beginning the modern approach that will take up the rest of the course. But I hope that all through that long build-up, you’ll keep in mind the fundamental ideas that you learn in this chapter.

1.1 The view of C from Q

Imagine you lived several centuries ago, before the discovery of complex numbers. Your whole mathematical world is the real numbers, and there is no square root of −1. This situation frustrates you, and you decide to do something about it. So, you invent a new symbol 8 (for ‘imaginary’) and decree that 82 = −1. You still want to be able to do all the usual arithmetic operations (+, ×, etc.), and you want to keep all the rules that govern them (associativity, commutativity, etc.). So you’re also forced to introduce new numbers such as 2 + 3 × 8, and you end up with what today we call the complex numbers. So far, so good. But then you notice something strange. When you invented the complex numbers, you only intended to introduce one square root of −1. But

4 accidentally, you introduced a second one at the same time: −8. (You wait centuries for a square root of −1, then two arrive at once.) Maybe that’s not so strange in itself; after all, positive reals have two square roots too. But then you realize something genuinely weird:

There’s nothing you can do to distinguish 8 from −8.

Try as you might, you can’t find any reasonable statement that’s true for 8 but not −8. For example, you notice that 8 is a solution of 17 I3 − I2 − I − = , 3 16 3 I but then you realize that −8 is too. Of course, there are unreasonable statements that are true for 8 but not −8, such as ‘I = 8’. We should restrict to statements that only refer to the known world of real numbers. More precisely, let’s consider statements of the form ? (I) ? (I) 1 = 3 , (1.1) ?2(I) ?4(I) where ?1,?2,?3,?4 are polynomials with real coefficients. Any such equation can be rearranged to give ?(I) = 0, where again ? is a polynomial with real coefficients, so we might as well just consider statements of that form. The point is that if ?(8) = 0 then ?(−8) = 0. Let’s make this formal.

Deﬁnition 1.1.1 Two complex numbers I and I0 are indistinguishable when seen from R, or conjugate over R, if for all polynomials ? with coeﬃcients in R,

?(I) = 0 ⇐⇒ ?(I0) = 0.

Warning 1.1.2 The standard term is ‘conjugate over R’. ‘Indistin- guishable’ is a term I invented for this chapter only, to express the idea of not being able to tell the two numbers apart.

For example, 8 and −8 are indistinguishable when seen from R. This follows from a more general result:

Lemma 1.1.3 Let I,I0 ∈ C. Then I and I0 are indistinguishable when seen from R if and only if I0 = I or I0 = I.

5 Proof ‘Only if’: suppose that I and I0 are indistinguishable when seen from R. Write I = G + 8H with G,H ∈ R. Then (I − G)2 = −H2. Since G and H are real, indistinguishability implies that (I0 − G)2 = −H2, so I0 − G = ±8H, so I0 = G ± 8H. ‘If’: obviously I is indistinguishable from itself, so it’s enough to prove that I is indinguishable from I. I’ll give two proofs. Each one teaches us a lesson that will be valuable later. First proof: recall that complex conjugation satisﬁes

F1 + F2 = F1 + F2, F1 · F2 = F1 · F2 for all F1,F2 ∈ C, and 0 = 0 for all 0 ∈ R. It follows by induction that for any polynomial ? over R, ?(F) = ?(F) for all F ∈ C. So

?(I) = 0 ⇐⇒ ?(I) = 0 ⇐⇒ ?(I) = 0.

Second proof: write I = G + 8H with G,H ∈ R. Let ? be a polynomial over R such that ?(I) = 0. We will prove that ?(I) = 0. This is trivial if H = 0, so suppose that H ≠ 0. Consider the polynomial <(C) = (C − G)2 + H2. Then <(I) = 0. You know from Honours Algebra that

?(C) = <(C)@(C) + A(C) (1.2) for some polynomials @ and A with deg(A) < deg(<) = 2 (so A is either a constant or of degree 1). Putting C = I in (1.2) gives A(I) = 0. But it’s easy to see that this is impossible unless A is the zero polynomial (using the assumption that H ≠ 0). So ?(C) = <(C)@(C). But <(I) = 0, so ?(I) = 0, as required. We have just shown that for all polynomials ? over R, if ?(I) = 0 then ?(I) = 0. Exchanging the roles of I and I proves the converse. Hence I and I are indistinguishable when seen from R.

Exercise 1.1.4 Both proofs of ‘if’ contain little gaps: ‘It follows by induction’ in the ﬁrst proof, and ‘it’s easy to see’ in the second. Fill them.

Digression 1.1.5 With complex analysis in mind, we could imagine a stricter deﬁnition of indistinguishability in which polynomials are replaced by arbitrary convergent power series (still with coeﬃcients in R). This would allow functions such as exp, cos and sin, and equations such as exp(8c) = −1.

6 But this apparently diﬀerent deﬁnition of indistinguishability is, in fact, equivalent. A complex number is still indistinguishable from exactly itself and its complex conjugate. (For example, exp(−8c) = −1 too.) Do you see why?

Lemma 1.1.3 tells us that indistinguishability over R is rather simple. But the same idea becomes much more interesting if we replace R by Q. And in this course, we will mainly focus on polynomials over Q. Define indistinguishability seen from Q, or officially conjugacy over Q, by replacing R by Q in Definition 1.1.1. From now on, I will usually just say ‘indistinguishable’, dropping the ‘seen from Q’. √ √ Example 1.1.6 I claim that 2 and − 2 are indistinguishable. And I’ll give you two different proofs, closely analogous to the two proofs of the ‘if’ part of Lemma 1.1.3. First proof: write √ √ Q( 2) = {0 + 1 2 : 0, 1 ∈ Q}. √ √ √ For F ∈ Q( 2), there are unique 0, 1 ∈ Q such that F = 0 + 1 2, because 2 is irrational. So it is logically valid to define Example 1.1.6 √ √ F 0 1 . e = − 2 ∈ Q( 2) It is straightforward to check that F F F F , F F F F 1 + 2 = f1 + f2 1 · 2 = f1 · f2 √ F ,F 0 0 0 for all 1 2 ∈ Q( 2), and that e = for all ∈ Q. Just as in the proof√ of F F F Lemma 1.1.3,√ it follows that and e are indistinguishable√ for every ∈ Q( 2). In particular, 2 is indistinguishable from − 2. ? ? C √Second proof: let = ( ) be a polynomial with coefficients in Q such that ?( 2) = 0. You know from Honours Algebra that

?(C) = (C2 − 2)@(C) + A(C) √ @ C A C A < C for√ some polynomials√ ( ) and ( ) over Q with deg 2. Putting = 2 gives A A C 0C 1 0, 1 A ( 2) = 0. But 2 is irrational and ( ) is of the form + with √ ∈ Q, so ? C C2 @ C ? must be the zero polynomial. Hence ( ) = ( − 2) ( ), giving (−√ 2) = 0. ? ? We√ have just shown that for all polynomials √over Q, if√ ( 2) = 0 then ? (− 2) = 0. The same√ argument with the roles of 2 and − 2 reversed proves the converse. Hence ± 2 are indistinguishable.

7 l l2

l3 l4

Figure 1.1: The 5th roots of unity.

Exercise 1.1.7 Let I ∈ Q. Show that I is distinguishable from I0 for any complex number I0 ≠ I.

One thing that makes indistinguishability more subtle over Q than over R is that over Q, more than two numbers can be indistinguishable:

Example 1.1.8 The 5th roots of unity are

1, l, l2, l3, l4, where l = 42c8/5 (Figure 1.1). Now 1 is distinguishable from the rest, since it is a root of the polynomial C − 1 and the others are not. (See also Exercise 1.1.7.) But it turns out that l, l2, l3, l4 are all indistinguishable from each other. Complex conjugates are indistinguishable when seen from R, so they’re certainly indistinguishable when seen from Q. Since l4 = 1/l = l, it follows that l and l4 are indistinguishable when seen from Q. By the same argument, l2 and l3 are indistinguishable. What’s not so obvious is that l and l2 are indistinguishable. I know two proofs, which are like the two proofs of Lemma 1.1.3 and Example 1.1.6. But we’re not equipped to do either yet.

Example 1.1.9 More generally, let ? be any prime and put l = 42c8/?. Then l, l2, . . . , l?−1 are indistinguishable.

So far, we have asked when one complex number can be distinguished from another, using only polynomials over Q. But what about more than one? : I ,...,I I0 ,...,I0 : Deﬁnition 1.1.10 Let ≥ 0 and let ( 1 : ) and ( 1 : ) be -tuples of I ,...,I I0 ,...,I0 complex numbers. We say that ( 1 : ) and ( 1 : ) are indistinguishable if for all polynomials ?(C1,...,C: ) over Q in : variables, ? I ,...,I ? I0 ,...,I0 . ( 1 : ) = 0 ⇐⇒ ( 1 : ) = 0

8 When : = 1, this is just the previous deﬁnition of indistinguishability.

I ,...,I I0 ,...,I0 Exercise 1.1.11 Suppose that ( 1 : ) and ( 1 : ) are in- I I0 distinguishable. Show that 8 and 8 are indistinguishable, for each 8 ∈ {1,...,:}.

Example 1.1.12 For any I1,...,I: ∈ C, the :-tuples (I1,...,I: ) and (I1,..., I: ) are indistinguishable. For let ?(C1,...,C: ) be a polynomial over Q. Then

?(I1,...,I: ) = ?(I1,..., I: ) since the coeﬃcients of ? are real, by a similar argument to the one in the ﬁrst proof of Lemma 1.1.3. Hence

?(I1,...,I: ) = 0 ⇐⇒ ?(I1,..., I: ) = 0, which is what we had to prove.

Example 1.1.13 Let l = 42c8/5, as in Example 1.1.8. Then

(l, l2, l3, l4) and (l4, l3, l2, l) are indistinguishable, by Example 1.1.12. It can also be shown that

(l, l2, l3, l4) and (l2, l4, l, l3) are indistinguishable, although the proof is beyond us for now. But

(l, l2, l3, l4) and (l2, l, l3, l4)

? C ,C ,C ,C C C2 are distinguishable, since if we put ( 1 2 3 4) = 2 − 1 then

?(l, l2, l3, l4) = 0 ≠ ?(l2, l, l3, l4).

I I0 So the converse of Exercise 1.1.11 is false: just because 8 and 8 are indistinguish- 8 I ,...,I I0 ,...,I0 able for all , it doesn’t follow that ( 1 : ) and ( 1 : ) are indistinguishable.

1.2 Every polynomial has a symmetry group. . .

We are now ready to describe the ﬁrst main idea of Galois theory: every polynomial has a symmetry group.

9 Deﬁnition 1.2.1 Let 5 be a polynomial with coeﬃcients in Q. Write U1,...,U: for its distinct roots in C. The Galois group of 5 is

Gal( 5 ) = {f ∈ (: : (U1,...,U: ) and (Uf(1),...,Uf(:)) are indistinguishable}. ‘Distinct roots’ means that we ignore any repetition of roots: e.g. if 5 (C) = 5 9 C (C − 1) then : = 2 and {U1,U2} = {0, 1}.

Exercise 1.2.2 Show that Gal( 5 ) is a subgroup of (: . (This one is maybe a bit tricky notationally. Stick to : = 3 if you like.)

Exercise 1.2.2 Digression 1.2.3 I brushed something under the carpet. The definition of Gal( 5 ) depends on the order in which the roots are listed. Different orderings gives different subgroups of (: . However, these subgroups are all conjugate to each other, and therefore isomorphic as abstract groups. So Gal( 5 ) is well-defined as an abstract group, independently of the choice of ordering.

Example 1.2.4 Let 5 be a polynomial over Q all of whose complex roots U1,...,U: are rational. If f ∈ Gal( 5 ) then Uf(8) and U8 are indistinguishable for each 8, by Exercise 1.1.11. But since they are rational, that forces Uf(8) = U8 (by Exercise 1.1.7), and since U1,...,U: are distinct, f(8) = 8. Hence f = id. So the Galois group of 5 is trivial. Example 1.2.5 Let 5 be a quadratic over Q. If 5 has rational roots then as we have just seen, Gal( 5 ) is trivial. If 5 has two non-real roots then they are complex conjugate, so Gal( 5 ) = (2 by Example 1.1.12. The remaining case is where 5 has two distinct roots that are real but not rational, and it can be shown that in that case too, Gal( 5 ) = (2.

Warning 1.2.6 On terminology: note that just now I said ‘non-real’. Sometimes people casually say ‘complex’ to mean ‘not real’. But try not to do this yourself. It makes as little sense as saying ‘real’ to mean ‘irrational’, or ‘rational’ to mean ‘not an integer’.

Example 1.2.7 Let 5 (C) = C4 + C3 + C2 + C + 1. Then (C − 1) 5 (C) = C5 − 1, so 5 has roots l, l2, l3, l4 where l = 42c8/5. We saw in Example 1.1.13 that 1 2 3 4 1 2 3 4 1 2 3 4 , ∈ Gal( 5 ), ∉ Gal( 5 ). 4 3 2 1 2 4 1 3 2 1 3 4 In fact, it can be shown that 1 2 3 4 Gal( 5 ) = . 2 4 1 3 4

10 Example 1.2.8 Let 5 (C) = C3 + 1C2 + 2C + 3 be a cubic over Q with no rational roots. Then √ ( 2 3 3 2 2 3 if −273 + 18123 − 42 − 41 3 + 1 2 ∈ Q, Gal( 5 ) (3 otherwise. This appears as Proposition 22.4 in Stewart, but is way beyond us for now: calculating Galois groups is hard. Galois groups, informally 1.3 . . . which determines whether it can be solved

Here we meet the second main idea of Galois theory: the Galois group of a polynomial determines whether it can be solved. More exactly, it determines whether the polynomial can be ‘solved by radicals’. To explain what this means, let’s begin with the quadratic formula. The roots of a quadratic 0C2 + 1C + 2 are √ −1 ± 12 − 02 4 . 20 After much struggling, it was discovered that there is a similar formula for cubics 0C3 + 1C2 + 2C + 3: the roots are given by

q3 √ q3 √ −27023+9012−213+30 3(270232−180123+4023+4133−1222) + −27023+9012−213−30 3(270232−180123+4023+4133−1222) √ . 3 3 20 (No, you don’t need to memorize that!) This is a complicated formula, and there’s also something strange about it. Any nonzero complex number√ has three cube roots, √ 3 and there are two 3 signs in the formula (ignoring the 2 in the denominator), so it looks as if the formula gives nine roots for the cubic. But a cubic can only have three roots. What’s going on? It turns out that some of the nine aren’t roots of the cubic at all. You have to choose your cube roots carefully. Section 1.4 of Stewart’s book has much more on this point, as well as an explanation of how the cubic formula was obtained. (We won’t be going into this ourselves.) As Stewart also explains, there is a similar but even more complicated formula for quartics (polynomials of degree 4).

Digression 1.3.1 Stewart doesn’t actually write out the explicit formula for the cubic, let alone the much worse one for the quartic. He just describes algorithms by which they can be solved. But if you unwind the algorithm for the cubic, you get the formula above. I have done this exercise once and do not recommend it.

11 Once mathematicians discovered how to solve quartics, they naturally looked for a formula for quintics (polynomials of degree 5). But it was eventually proved by Abel and Ruﬃni, in the early 19th century, that there is no formula like the quadratic, cubic or quartic formula for polynomials of degree ≥ 5. A bit more precisely, there is no formula for the roots in terms of the coeﬃcients that uses only the usual arithmetic operations (+, −, ×, ÷) and :th roots (for integers :). Spectacular as this result was, Galois went further, and so will we. Informally, let us say that a complex number is radical if it can be obtained from the rationals using only the usual arithmetic operations and :th roots. For example, √ √ 1 p3 7 2 2 + 2 − 7 r q 4 5 2 6 + 3 is radical, whichever square root, cube root, etc., we choose. A polynomial over Q is solvable (or soluble) by radicals if all of its complex roots are radical.

Example 1.3.2 Every quadratic√ over Q is solvable by radicals. This follows from the quadratic formula: (−1 ± 12 − 402)/20 is visibly a radical number.

Example 1.3.3 Similarly, the cubic formula shows that every cubic over Q is solvable by radicals. The same goes for quartics.

Example 1.3.4 Some quintics are solvable by radicals. For instance,

(C − 1)(C − 2)(C − 3)(C − 4)(C − 5) is solvable by radicals, since all its roots are rational and, therefore, radical. A bit C 5 less trivially, ( − 123) √+ 456 is solvable by radicals, since its roots are the ﬁve complex numbers 123 + 5 −456, which are all radical.

What determines whether a polynomial is solvable by radicals? Galois’s amazing achievement was to answer this question completely:

Theorem 1.3.5 (Galois) Let 5 be a polynomial over Q. Then

5 is solvable by radicals ⇐⇒ Gal( 5 ) is a solvable group.

Example 1.3.6 Deﬁnition 1.2.1 implies that if 5 has degree = then Gal( 5 ) is isomorphic to a subgroup of (=. You saw in Group Theory that (4 is solvable, and that every subgroup of a solvable group is solvable. Hence the Galois group of any polynomial of degree ≤ 4 is solvable. It follows from Theorem 1.3.5 that every polynomial of degree ≤ 4 is solvable by radicals.

12 5 Example 1.3.7 Put 5 (C) = C −6C +3. Later we’ll show that Gal( 5 ) = (5. You saw in Group Theory that (5 is not solvable (or at least, you saw that 5 isn’t solvable, which implies that (5 isn’t either, as (5 contains 5 as a subgroup). Hence 5 is not solvable by radicals.

If there was a quintic formula then all quintics would be solvable by radicals, for the same reason as in Examples 1.3.2 and 1.3.3. But since this is not the case, there is no quintic formula. Galois’s result is much sharper than Abel and Ruﬃni’s. They proved that there is no formula providing a solution by radicals of every quintic, whereas Galois found a way of determining which quintics (and higher) can be solved by radicals and which cannot.

Digression 1.3.8 From the point of view of modern numerical computation, this is all a bit odd. Computationally speaking, there is probably not much 5 difference√ between solving C + 3 = 0 to 100 decimal places (that is, finding 5 −3) and solving C5 − 6C + 3 = 0 to 100 decimal places (that is, solving a polynomial that isn’t solvable by radicals). Numerical computation and abstract algebra have different ideas about what is easy and what is hard!

∗ ∗ ∗

This completes our overview of Galois theory. What’s next? Mathematics increasingly emphasizes abstraction over calculation. Individual mathematicians’ tastes vary, but the historical trend is clear. In the case of Galois theory, this means dealing with abstract algebraic structures, principally fields, instead of manipulating explicit algebraic expressions such as polynomials. The cubic formula already gave you a taste of how hairy that can get. Developing Galois theory using abstract algebraic structures helps us to see its connections to other parts of mathematics, and also has some fringe benefits. For example, we’ll solve some notorious geometry problems that perplexed the ancient Greeks and remained unsolved for millennia. For that and many other things, we’ll need some ring theory and some field theory—and that’s what’s next.

13 Chapter 2

Rings and ﬁelds

We now start again. This chapter is a mixture of revision and material that is likely to be new to you. The revision is from Honours Algebra and Introduction to Number Theory (if you took it, which I won’t assume). Introduction to Week 2 2.1 Rings

We’ll begin with some stuﬀ you know—but with a twist. In this course, the word ring means commutative ring with 1 (multiplicative identity). Noncommutative rings and rings without 1 are important in some parts of mathematics, but since we’ll be focusing on commutative rings with 1, it will be easier to just call them ‘rings’. Given rings ' and (, a homomorphism from ' to ( is a function i : ' → ( satisfying the equations i(A + A0) = i(A) + i(A0), i(0) = 0, i(−A) = −i(A), i(AA0) = i(A)i(A0), i(1) = 1 (note this!) for all A,A0 ∈ '. For example, complex conjugation is a homomorphism C → C. It is a very useful lemma that if i(A + A0) = i(A) + i(A0), i(AA0) = i(A)i(A0), i(1) = 1 for all A,A0 ∈ ' then i is a homomorphism. In other words, to show that i is a homomorphism, you only need to check it preserves +, · and 1; preservation of 0 and negatives then comes for free. But you do need to check it preserves 1. That doesn’t follow from the other conditions. A subring of a ring ' is a subset ( ⊆ ' that contains 0 and 1 and is closed under addition, multiplication and negatives. Whenever ( is a subring of ', the inclusion ]: ( → ' (deﬁned by ](B) = B) is a homomorphism.

14 Warning 2.1.1 In Honours Algebra, rings had 1s but homomorphisms were not required to preserve 1. Similarly, subrings of ' had to have a 1, but it was not required to be the same as the 1 of '. For example, take the ring C of complex numbers, the noncommutative ring " of 2 × 2 matrices over C, and the function i : C → " deﬁned by I 0 i(I) = . 0 0 In the terminology of Honours Algebra, i is a homomorphism and its image im i is a subring of ". But in our terminology, i is not a homomorphism (as i(1) ≠ ) and im i is not a subring of " (as ∉ im i).

Exercise 2.1.2 Let ' be a ring and let S be any set (perhaps inﬁnite) ' Ñ ( of subrings of . Prove that their intersection (∈S is also a subring of '. (In contrast, in the Honours Algebra setup, even the intersection of two subrings need not be a subring.)

Example 2.1.3 For any ring ', there is exactly one homomorphism Z → '. Here is a sketch of the proof. To show there is at least one homomorphism j : Z → ', we will construct one. Deﬁne j on nonnegative integers = inductively by j(0) = 0 and j(=+1) = j(=)+1. (Thus, j(=) = 1' + · · · + 1'.) Deﬁne j on negative integers = by j(=) = −j(−=). A series of tedious checks shows that j is indeed a ring homomorphism. To show there is only one homomorphism Z → ', let i be any homomorphism Z → '; we have to prove that i = j. Certainly i(0) = 0 = j(0). Next prove by induction on = that i(=) = j(=) for nonnegative integers =. I leave the details to you, but the crucial point is that because homomorphisms preserve 1, we must have i(= + 1) = i(=) + i(1) = i(=) + 1 for all = ≥ 0. Once we have shown that i and j agree on the nonnegative integers, it follows that for negative =,

i(=) = −i(−=) = −j(−=) = j(=).

Hence i(=) = j(=) for all = ∈ Z; that is, i = j. j(=) = · = The meaning of Usually we write as 1', or simply as if it is clear from the context ‘= · 1’, and that = is to be interpreted as an element of '. Exercise 2.2.7 15 Every ring homomorphism i : ' → ( has an image im i, which is a subring of (, and a kernel ker i, which is an ideal of '.

Warning 2.1.4 Subrings in ring theory are analogous to subgroups in group theory, and ideals in ring theory are analogous to normal subgroups in group theory. But whereas normal subgroups are a special kind of subgroup, ideals are not a special kind of subring! Subrings must contain 1, but most ideals don’t.

Exercise 2.1.5 Prove that the only subring of a ring ' that is also an ideal is ' itself.

Given a ring ' and an ideal ', we obtain the quotient ring or factor ring '/ and the canonical homomorphismP c : ' → '/, which is surjective and has kernel . Quotient rings As explained in Honours Algebra, the quotient ring together with the canonical homomorphism has a ‘universal property’: given any ring ( and any homomorphism i : ' → ( satisfying ker i ⊇ , there is exactly one homomorphism i¯ : '/ → ( such that this diagram commutes:

c ' / '/

i¯ i (.

(For a diagram to commute means that whenever there are two diﬀerent paths from one object to another, the composites along the two paths are equal. Here, it means that i = i¯ ◦ c.) The ﬁrst isomorphism theorem says that if i is surjective and has kernel equal to then i¯ is an isomorphism. So c : ' → '/ is essentially the only surjective homomorphism out of ' with kernel .

Digression 2.1.6 Loosely, the ideals of a ring ' correspond one-to-one with the surjective homomorphisms out of '. This means four things:

• given an ideal ', we get a surjective homomorphism out of ' P (namely, c : ' → '/); • given a surjective homomorphism i out of ', we get an ideal of ' (namely, ker i); • if we start with an ideal of ', take its associated surjective homomorphism c : ' → '/, then take its associated ideal, we end up where we started (that is, ker(c ) = );

16 • if we start with a surjective homomorphism i : ' → (, take its associated ideal ker i, then take its associated surjective homomorphism cker i : ' → '/ker i, we end up where we started (at least ‘up to isomorphism’, in that we have the isomorphism i¯ : '/ker i → ( making the triangle commute). This is the ﬁrst isomorphism theorem.

Analogous stories can be told for groups and for modules.

0 An integral domain is a ring ' such that 0' ≠ 1' and for A,A ∈ ',

AA0 = 0 ⇒ A = 0 or A0 = 0.

Exercise 2.1.7 The trivial ring or zero ring is the one-element set with its only possible ring structure. Show that the only ring in which 0 = 1 is the trivial ring.

Equivalently, an integral domain is a nontrivial ring in which cancellation is valid: AB = A0B implies A = A0 or B = 0.

Digression 2.1.8 Why is the condition 0 ≠ 1 in the deﬁnition of integral domain? My answer begins with a useful general point: the sum of no things should always be interpreted as 0. (The amount you pay in a shop is the sum of the prices of the individual things. If you buy no things, you pay £0.) Similarly, the product of no things should be interpreted as 1. Now consider the following condition on a ring ': for all = ≥ 0 and A1,...,A= ∈ ',

A1A2 ··· A= = 0 ⇒ there exists 8 ∈ {1, . . . , =} such that A8 = 0. (2.1)

In the case = = 0, this says: if 1 = 0 then there exists 8 ∈ such that A8 = 0. But any statement beginning ‘there exists 8 ∈ ’ is false! So in the case = = 0, condition (2.1) states that 1 ≠ 0. And for = = 2, it’s the main condition in the deﬁnition of integral domain. So ‘1 ≠ 0’ is the 0-fold analogue of the main condition. On the other hand, if (2.1) holds for = = 0 and = = 2 then a simple induction shows that it holds for all = ≥ 0. Conclusion: an integral domain can equivalently be deﬁned as a ring in which (2.1) holds for all = ≥ 0.

Let ) be a subset of a ring '. The ideal h)i generated by ) is the intersection of all the ideals of ' containing ). You can show that any intersection of ideals is an ideal (much as you did for subrings in Exercise 2.1.2), so h)i is an ideal. It is

17 the smallest ideal of ' containing ). That is, h)i is an ideal containing ), and if is another ideal containing ) then h)i ⊆ . When ) is a ﬁnite set {A1,...,A=}, we write h)i as hA1,...,A=i, and it satisﬁes

hA1,...,A=i = {01A1 + · · · + 0=A= : 01, . . . , 0= ∈ '}. (2.2) In particular, when = = 1, hAi = {0A : 0 ∈ '}. Ideals of the form hAi are called principal ideals.A principal ideal domain is an integral domain in which every ideal is principal. Example 2.1.9 Z is a principal ideal domain. Indeed, if Z then either = {0}, in which case = h0i, or contains some positive integer,P in which case we can deﬁne = to be the least positive integer in and use the division algorithm to show that = h=i.

Exercise 2.1.10 Fill in the details of Example 2.1.9.

Let A and B be elements of a ring '. We say that A divides B, and write A | B, if there exists 0 ∈ ' such that B = 0A. This condition is equivalent to B ∈ hAi, and to hBi ⊆ hAi. An element D ∈ ' is a unit if it has a multiplicative inverse, or equivalently if hDi = '. The units form a group '× under multiplication. For instance, Z× = {1, −1}.

Exercise 2.1.11 Let A and B be elements of an integral domain. Show that A | B | A ⇐⇒ hAi = hBi ⇐⇒ B = DA for some unit D.

Elements A and B of a ring are coprime if for 0 ∈ ', 0 | A and 0 | B ⇒ 0 is a unit. Proposition 2.1.12 Let ' be a principal ideal domain and A,B ∈ '. Then A and B are coprime ⇐⇒ 0A + 1B = 1 for some 0, 1 ∈ '. Proof ⇒: suppose that A and B are coprime. Since ' is a principal ideal domain, hA,Bi = hDi for some D ∈ '. Since A ∈ hA,Bi = hDi, we must have D | A, and similarly D | B. But A and B are coprime, so D is a unit. Hence 1 ∈ hDi = hA,Bi. But by equation (2.2), hA,Bi = {0A + 1B : 0, 1 ∈ '}, and the result follows. ⇐: suppose that 0A + 1B = 1 for some 0, 1 ∈ '. If D ∈ ' with D | A and D | B then D | (0A + 1B) = 1, so D is a unit. Hence A and B are coprime.

18 2.2 Fields

A ﬁeld is a ring in which 0 ≠ 1 and every nonzero element is a unit. Equivalently, it is a ring such that × = \{0}. Every ﬁeld is an integral domain.

Exercise 2.2.1 Write down all the examples of ﬁelds that you know.

A ﬁeld has exactly two ideals: {0} and . For if {0} ≠ then D ∈ for some D ≠ 0; but then D is a unit, so hDi = , so = . P

Lemma 2.2.2 Every homomorphism between ﬁelds is injective.

Proof Let i : → ! be a homomorphism between fields. Then ker i , so ker i is either {0} or . If ker i = then i(1) = 0; but i(1) = 1 by definitionP of homomorphism, so 0 = 1 in !, contradicting the assumption that ! is a field. Hence ker i = {0}, that is, i is injective.

Warning 2.2.3 With the Honours Algebra deﬁnition of homomorphism, Lemma 2.2.2 would be false, since the map with constant value 0 would be a homomorphism.

Let ' be any ring. By Example 2.1.3, there is a unique homomorphism j : Z → '. Its kernel is an ideal of the principal ideal domain Z. Hence ker j = h=i for a unique integer = ≥ 0. This = is called the characteristic of ', and written as char '. Explicitly, ( the least = > 0 such that = · 1 = 0 , if such an = exists; char ' = ' ' (2.3) 0, otherwise.

Another way to say it: for < ∈ Z, we have < · 1' = 0 if and only if < is a multiple of char '. The concept of characteristic is mostly used in the case of ﬁelds.

Examples 2.2.4 i. Q, R and C all have characteristic 0.

ii. For a prime number ?, we write F? for the ﬁeld Z/h?i of integers modulo ?. Then char F? = ?.

Lemma 2.2.5 The characteristic of an integral domain is 0 or a prime number.

19 Proof Let ' be an integral domain and write = = char '. Suppose that = > 0; we must prove that = is prime. Since 1 ≠ 0 in an integral domain, = ≠ 1. (Remember that 1 is not a prime! So that step was necessary.) Now let :, < > 0 with :< = =. Writing j for the unique homomorphism Z → ', we have

j(:) j(<) = j(:<) = j(=) = 0,

and ' is an integral domain, so j(:) = 0 or j(<) = 0. WLOG, j(:) = 0. But ker j = h=i, so = | :, so : = =. Hence = is prime.

Examples 2.2.4 show that there exist fields of every possible characteristic. But there is no way of mapping between fields of different characteristics: Lemma 2.2.6 Let i : → ! be a homomorphism of fields. Then char = char !.

Proof Write j and j! for the unique homomorphisms from Z to and !, respectively. Since j! is the unique homomorphism Z → !, the triangle Z j j!

! i /

commutes. (Concretely, this says that i(= · 1 ) = = · 1! for all = ∈ Z.) Hence ker(i ◦ j ) = ker j!. But i is injective by Lemma 2.2.2, so ker(i ◦ j ) = ker j . Hence ker j = ker j!, or equivalently, char = char !. For example, the inclusion Q → R is a homomorphism of ﬁelds, and both have characteristic 0.

Exercise 2.2.7 This proof of Lemma 2.2.6 is quite abstract. Find a more concrete proof, taking equation (2.3) as your definition of characteristic. (You will still need the fact that i is injective.) The meaning of ‘= · 1’, and A subfield of a field is a subring that is a field. The prime subfield of Exercise 2.2.7 is the intersection of all the subfields of . It is straightforward to show that any intersection of subfields is a subfield (just as you showed in Exercise 2.1.2 that any intersection of subrings is a subring). Hence the prime subfield is a subfield. It is the smallest subfield of , in the sense that any other subfield of contains it. Concretely, the prime subfield of is < · 1 : <, = ∈ Z with = · 1 ≠ 0 . = · 1

20 To see this, first note that this set is a subfield of . It is the smallest subfield of : for if ! is a subfield of then 1 ∈ ! by definition of subfield, so < · 1 ∈ ! for all integers <, so (< · 1 )/(= · 1 ) ∈ ! for all integers < and = such that = · 1 ≠ 0. Examples 2.2.8 i. The field Q has no proper subfields, so the prime subfield of Q is Q itself.

ii. Let ? be a prime. The field F? has no proper subfields, so the prime subfield of F? is F? itself.

Exercise 2.2.9 What is the prime subﬁeld of R? Of C?

The prime subﬁelds appearing in Examples 2.2.8 were Q and F?. In fact, these are the only prime subﬁelds of anything:

Lemma 2.2.10 Let be a ﬁeld.

i. If char = 0 then the prime subﬁeld of is Q.

ii. If char = ? > 0 then the prime subﬁeld of is F?.

In the statement of this lemma, as so often in mathematics, the word ‘is’ means ‘is isomorphic to’. I hope you’re comfortable with that by now.

Proof For (i), suppose that char = 0. By definition of characteristic, = · 1 ≠ 0 for all integers = ≥ 0. One can check that there is a well-defined homomorphism i : Q → defined by 0. By Lemma 2.2.5, ? is prime. The unique homomorphism j : Z → has kernel h?i, by definition. By the first isomorphism theorem, im j Z/h?i = F?. But im j is a subring of , and in fact a subfield since is it isomorphic to F?. It is the prime subfield, since F? has no proper subfields.

Lemma 2.2.11 Every ﬁnite ﬁeld has positive characteristic.

Proof By Lemma 2.2.10, a ﬁeld of characteristic 0 contains a copy of Q and is therefore inﬁnite.

21 Warning 2.2.12 There are also inﬁnite ﬁelds of positive characteristic! We haven’t met one yet, but we will soon.

So far, we have rather few examples of ﬁelds. The following construction will allow us to manufacture many, many more. An element A of a ring ' is irreducible if A is not 0 or a unit, and if for 0, 1 ∈ ', Building blocks A = 01 ⇒ 0 or 1 is a unit.

For example, the irreducibles in Z are ±2, ±3, ±5,.... An element of a ring is reducible if it is not 0, a unit, or irreducible. So 0 and units count as neither reducible nor irreducible.

Exercise 2.2.13 What are the irreducible elements of a ﬁeld?

Proposition 2.2.14 Let ' be a principal ideal domain and 0 ≠ A ∈ '. Then

A is irreducible ⇐⇒ '/hAi is a ﬁeld.

Proof Write c for the canonical homomorphism ' → '/hAi. ⇒: suppose that A is irreducible. To show that 1'/hAi ≠ 0'/hAi, note that since A is not a unit, 1' ∉ hAi = ker c, so

1'/hAi = c(1') ≠ 0'/hAi. Next we have to show that every nonzero element of '/hAi is a unit, or equivalently that c(B) is a unit whenever B ∈ ' with B ∉ hAi. We have A - B, and A is irreducible, so A and B are coprime. Hence by Proposition 2.1.12 (and the assumption that ' is a principal ideal domain), we can choose 0, 1 ∈ ' such that

0A + 1B = 1'. Applying c to each side gives

c(0)c(A) + c(1)c(B) = 1'/hAi. But c(A) = 0, so c(1)c(B) = 1, so c(B) is a unit. ⇐: suppose that '/hAi is a ﬁeld. Then 1'/hAi ≠ 0'/hAi, that is, c(1') ∉ ker c = hAi, that is, A - 1'. Hence A is not a unit. Next we have to show that if 0, 1 ∈ ' with A = 01 then 0 or 1 is a unit. We have 0 = c(A) = c(0)c(1)

22 and '/hAi is an integral domain, so WLOG c(0) = 0. Then 0 ∈ ker c = hAi, so 0 = A10 for some 10 ∈ '. This gives

A = 01 = A101.

But A ≠ 0 by hypothesis, and ' is an integral domain, so 101 = 1. Hence 1 is a unit.

Example 2.2.15 When = is an integer, Z/h=i is a ﬁeld if and only if = is irreducible (that is, ± a prime number).

Proposition 2.2.14 enables us to construct ﬁelds from irreducible elements. . . but irreducible elements of a principal ideal domain. Right now that’s not much help, because we don’t have many examples of principal ideal domains. But we will soon.

23 Chapter 3

Polynomials

This chapter revisits and develops some themes you met in Honours Algebra. Although it’s long, it contains material you’ve seen before. Before you begin, it may help you to reread Section 3.3 (Polynomials) of the Honours Algebra notes. Introduction to Week 3 3.1 The ring of polynomials

You already know the definition of polynomial, but I want to make a point by phrasing it in an unfamiliar way. Definition 3.1.1 Let ' be a ring. A polynomial over ' is an infinite sequence (00, 01, 02,...) of elements of ' such that {8 : 08 ≠ 0} is finite. The set of polynomials over ' forms a ring as follows:

(00, 01,...) + (10, 11,...) = (00 + 10, 01 + 11,...), (3.1) (00, 01,...)·(10, 11,...) = (20, 21,...) (3.2) Õ where 2: = 081 9 , (3.3) 8, 9 : 8+ 9=: the zero of the ring is (0, 0,...), and the multiplicative identity is (1, 0, 0,...). 2 Of course, we almost always write (00, 01, 02,...) as 00 + 01C + 02C + · · · , or the same with some other symbol in place of C. In that notation, formulas (3.1) and (3.2) look like the usual formulas for addition and multiplication of polynomials. Nevertheless:

Warning 3.1.2 A polynomial is not a function! A polynomial gives rise to a function, as we’ll recall in a moment. But a polynomial itself is a purely formal object.

24 The set of polynomials over ' is written as '[C] (or '[D], '[G], etc.). Since ( = '[C] is itself a ring, we can consider the ring ([D] = ('[C])[D], usually written as '[C,D], and similarly '[C,D,E] = ('[C,D])[E], etc. Why study We use 5 , 6, ℎ, . . . and 5 (C), 6(C), ℎ(C),... interchangeably to denote elements polynomials? of '[C]. A polynomial 5 = (00, 01,...) over ' gives rise to a function ' → ' 2 A 7→ 00 + 01A + 02A + · · · .

(The sum on the right-hand side makes sense because only ﬁnitely many 08s are nonzero.) This function is usually called 5 too. But calling it that is slightly dangerous, because:

Warning 3.1.3 Different polynomials can give rise to the same func- 2 tion. For example, consider C,C ∈ F2 [C]. They are different polynomials: going back to Definition 3.1.1, they’re alternative notation for the sequences

(0, 1, 0, 0,...) and (0, 0, 1, 0,...),

which are plainly not the same. On the other hand, they induce the 2 same function F2 → F2, because 0 = 0 for all (both) 0 ∈ F2.

Exercise 3.1.4 Show that whenever ' is a finite nontrivial ring, it is possible to find distinct polynomials over ' that induce the same function ' → '. (Hint: are there finitely or infinitely many polynomials over '? Functions ' → '?)

The ring of polynomials has a universal property: a homomorphism from '[C] to some other ring is determined by its eﬀect on constant polynomials and on C itself, in the following sense. The universal property of '[C] Lemma 3.1.5 (Universal property of the polynomial ring) Let ' and be rings. For every homomorphism i : ' → and every 1 ∈ , there is exactly one homomorphism \ : '[C] → such that

\(0) = i(0) for all 0 ∈ ', (3.4) \(C) = 1. (3.5)

On the left-hand side of (3.4), the ‘0’ means the polynomial 0 + 0C + 0C2 + · · · .

25 Proof To show there is at most one such \, take any homomorphism \ : '[C] → Í 0 C8 ' satisfying (3.4) and (3.5). Then for every polynomial 8 8 over ,

Õ 8 Õ 8 \ 08C = \(08)\(C) since \ is a homomorphism 8 8 Õ 8 = i(08)1 by (3.4) and (3.5). 8 So \ is uniquely determined. To show there is at least one such \, deﬁne a function \ : '[C] → by

Õ 8 Õ 8 \ 08C = i(08)1 8 8 Í 0 C8 ' C \ ( 8 8 ∈ [ ]). Then clearly satisﬁes conditions (3.4) and (3.5). It remains to check that \ is a homomorphism. I will do the worst part of this, which is to check that \ preserves multiplication, and leave the rest to you. 5 C Í 0 C8 6 C Í 1 C 9 5 C 6 C So, take polynomials ( ) = 8 8 and ( ) = 9 9 . Then ( ) ( ) = Í 2 C: 2 : : , where : is as deﬁned in equation (3.3). We have

Õ : \( 5 6) = \ 2: C : Õ : = i(2: )1 by definition of \ : Õ Õ : = i 081 9 1 by definition of 2: : 8, 9 : 8+ 9=: Õ Õ : = i(08)i(1 9 )1 since i is a homomorphism : 8, 9 : 8+ 9=: Õ 8+ 9 = i(08)i(1 9 )1 8, 9 Õ 8 Õ 9 = i(08)1 i(1 9 )1 8 9 = \( 5 )\(6) by definition of \.

Here are three uses for the universal property of the ring of polynomials. First: Deﬁnition 3.1.6 Let i : ' → ( be a ring homomorphism. The induced homomorphism i∗ : '[C] → ([C] is the unique homomorphism '[C] → ([C] such that i∗(0) = i(0) for all 0 ∈ ' and i∗(C) = C.

26 The universal property guarantees that there is one and only one homomorphism i∗ with these properties. Concretely,

Õ 8 Õ 8 i∗ 08C = i(08)C 8 8 Í 0 C8 ' C for all 8 8 ∈ [ ]. Second, let ' be a ring and A ∈ '. By the universal property, there is a unique homomorphism evA : '[C] → ' such that evA (0) = 0 for all 0 ∈ ' and evA (C) = A. Concretely, Õ 8 Õ 8 evA 08C = 08A 8 8 Í 0 C8 ' C A for all 8 8 ∈ [ ]. This map evA is called evaluation at . Í 8 (The notation 08C for what is oﬃcially (00, 01,...) makes it look obvious that we can evaluate a polynomial at an element and that this gives a homomorphism: of course ( 5 · 6)(A) = 5 (A)6(A), for instance! But that’s only because of the notation: there was actually something to prove here.) Third, let ' be a ring and 2 ∈ '. For any 5 (C) ∈ '[C], we can ‘substitute C = D + 2’ to get a polynomial in D. What exactly does this mean? Formally, there is a unique homomorphism \ : '[C] → '[D] such that \(0) = 0 for all 0 ∈ ' and \(C) = D + 2. Concretely,

Õ 8 Õ 8 \ 08C = 08 (D + 2) . 8 8 This particular substitution is invertible. Informally, the inverse is ‘substitute D = C − 2’. Formally, there is a unique homomorphism \0 : '[D] → '[C] such that \0(0) = 0 for all 0 ∈ ' and \0(D) = C − 2. These maps \ and \0 carrying out the substitutions are inverse to each other, as you can deduce this from either the universal property or the concrete descriptions. So, the substitution maps

\ / '[C] o '[D] (3.6) \0 deﬁne an isomorphism between '[C] and '[D]. For example, since isomorphism preserve irreducibility (and everything else that matters!), 5 (C) is irreducible if and only if 5 (C − 2) is irreducible.

Exercise 3.1.7 What happens to everything in the previous paragraph if we substitute C = D2 + 2 instead?

The rest of this section is about degree.

27 Í 8 Deﬁnition 3.1.8 The degree deg( 5 ) of a nonzero polynomial 5 (C) = 08C is the largest = ≥ 0 such that 0= ≠ 0. By convention, deg(0) = −∞, where −∞ is a formal symbol which we give the properties −∞ < =, (−∞) + = = −∞, (−∞) + (−∞) = −∞ for all integers =.

Digression 3.1.9 Deﬁning deg(0) like this is helpful because it allows us to make statements about all polynomials without having to make annoying exceptions for the zero polynomial (e.g. Lemma 3.1.10(i)). But putting deg(0) = −∞ also makes intuitive sense. At least for polynomials over R, the degree of a nonzero polynomial tells us how fast it grows: when C is large, 5 (C) behaves roughly like Cdeg( 5 ) . What about the zero polynomial? Well, whether or not C is large, 0(C) = 0, and C−∞ can sensibly be interpreted A as limA→−∞ C = 0. So it makes sense to put deg(0) = −∞.

Lemma 3.1.10 Let ' be an integral domain. Then: i. deg( 5 6) = deg( 5 ) + deg(6) for all 5 , 6 ∈ '[C]; ii. '[C] is an integral domain; iii. when ' is a field, the units in '[C] are the polynomials of degree 0 (that is, the nonzero constants); iv. when ' is a field, 5 (C) ∈ '[C] is irreducible if and only if deg( 5 ) > 0 and 5 cannot be expressed as a product of two polynomials of degree > 0. Proof At least parts (i)–(iii) were in Honours Algebra (Section 3.3). Part (iv) follows from the general definition of irreducible element of a ring.

3.2 Factorizing polynomials

Every nonzero integer can be expressed as a product of primes in an essentially unique way. But the analogous statement is not true in all rings, or even all integral domains. Some rings have elements that can’t be expressed as a product of irreducibles at all. In other rings, factorizations into irreducibles exist but are not unique. (By ‘not unique’ I mean more than just changing the order of the factors or multiplying them by units.) The big theorem of this section is that, happily, every polynomial over a ﬁeld can be factorized, essentially uniquely, into irreducibles. We begin with a result on division of polynomials from Section 3.3 of Honours Algebra.

28 Proposition 3.2.1 Let be a ﬁeld and 5 , 6 ∈ [C]. Then there is exactly one pair of polynomials @,A ∈ [C] such that 5 = @6 + A and deg(A) < deg(6).

We use this to prove an extremely useful fact: Proposition 3.2.2 Let be a ﬁeld. Then [C] is a principal ideal domain.

Proof First, [C] is an integral domain, by Lemma 3.1.10(ii). Now let [C]. If = {0} then = h0i. Otherwise, put 3 = min{deg( 5 ) : 0 ≠ 5 ∈ } andP choose 6 ∈ such that deg(6) = 3. I claim that = h6i. To prove this, let 5 ∈ ; we must show that 6 | 5 . By Proposition 3.2.1, 5 = @6 + A for some @,A ∈ [C] with deg(A) < 3. Now A = 5 − @6 ∈ since 5 , 6 ∈ , so the minimality of 3 implies that A = 0. Hence 5 = @6, as required.

If you struggled with Exercise 2.1.10, that proof should give you a clue.

Warning 3.2.3 Lemma 3.1.10(ii) implies that [C1,...,C=] is an integral domain for all = ≥ 2, but it is not a principal ideal domain. For example, the ideal

hC1,C2i = { 5 (C1,C2) ∈ Q[C1,C2] : 5 has constant term 0}

is not principal. Also, Proposition 3.2.2 really needed the hypothesis that is a ﬁeld; it’s not enough for it to be a principal ideal domain. For example, Z is a principal ideal domain, but in Z[C], the ideal

h2,Ci = { 5 (C) ∈ Z[C] : the constant term of 5 is even}

is not principal.

Exercise 3.2.4 Prove that the ideals in Warning 3.2.3 are indeed not principal.

Exercise 3.2.4: a At the end of Chapter2, I promised I’d give you a way of manufacturing lots non-principal ideal of new ﬁelds. Here it is! Corollary 3.2.5 Let be a ﬁeld and let 0 ≠ 5 ∈ [C]. Then

5 is irreducible ⇐⇒ [C]/h 5 i is a ﬁeld.

Proof This follows from Propositions 2.2.14 and 3.2.2.

29 To manufacture new ﬁelds using Corollary 3.2.5, we’ll need a way of knowing which polynomials are irreducible. That’s the topic of Section 3.3, but for now let’s stick to our mission: proving that every polynomial factorizes into irreducibles in an essentially unique way. To achieve this mission, we’ll need two more lemmas.

Lemma 3.2.6 Let be a ﬁeld and let 5 (C) ∈ [C] be a nonconstant polynomial. Then 5 (C) is divisible by some irreducible in [C].

The word nonconstant just means ‘of degree > 0’.

Proof Let 6 be a nonconstant polynomial of smallest possible degree such that 6 | 5 . (For this to make sense, there must be at least one nonconstant polynomial dividing 5 , and there is: 5 .) I claim that 6 is irreducible. Proof: if 6 = 6162 then each 68 divides 5 , so by the minimality of deg(6), each 68 has degree 0 or deg(6). They cannot both have degree deg(6), since deg(61) + deg(62) = deg(6) > 0. So at least one has degree 0, i.e., is a unit.

Lemma 3.2.7 Let be a ﬁeld and 5 , 6, ℎ ∈ [C]. Suppose that 5 is irreducible and 5 | 6ℎ. Then 5 | 6 or 5 | ℎ.

This behaviour is familiar in the integers: if a prime ? divides some product 01, then ? | 0 or ? | 1.

Proof Suppose that 5 - 6. Since 5 is irreducible, 5 and 6 are coprime. Since [C] is a principal ideal domain, Proposition 2.1.12 implies that there are ?,@ ∈ [C] such that ? 5 + @6 = 1. Multiplying both sides by ℎ gives

? 5 ℎ + @6ℎ = ℎ.

But 5 | ? 5 ℎ and 5 | 6ℎ, so 5 | ℎ.

Theorem 3.2.8 Let be a ﬁeld and 0 ≠ 5 ∈ [C]. Then

5 = 0 51 52 ··· 5=

for some = ≥ 0, 0 ∈ and monic irreducibles 51, . . . , 5= ∈ [C]. Moreover, = and 0 are uniquely determined by 5 , and 51, . . . , 5= are uniquely determined up to reordering.

30 In the case = = 0, the product 51 ··· 5= should be interpreted as 1 (as in Digression 2.1.8). Monic means that the leading coeﬃcient is 1. Proof First we prove that such a factorization exists, by induction on deg( 5 ). If deg( 5 ) = 0 then 5 is a constant 0 and we take = = 0. Now suppose that deg( 5 ) > 0 and assume the result for polynomials of smaller degree. By Lemma 3.2.6, there is an irreducible 6 dividing 5 , and we can assume that 6 is monic by dividing by a constant if necessary. Then 5 /6 is a nonzero polynomial of smaller degree than 5 , so by inductive hypothesis,

5 /6 = 0ℎ1 ··· ℎ< for some 0 ∈ and monic irreducibles ℎ1, . . . , ℎ<. Rearranging gives

5 = 0ℎ1 ··· ℎ<6, completing the induction. Now we prove uniqueness, again by induction on deg( 5 ). If deg( 5 ) = 0 then 5 is a constant 0 and the only possible factorization is the one with = = 0. Now suppose that deg( 5 ) > 0, and take two factorizations

0 51 ··· 5= = 5 = 161 ··· 6< (3.7) where 0, 1 ∈ and 58, 6 9 are monic irreducible. Since deg( 5 ) > 0, we have =, < ≥ 1. Now 5= | 161 ··· 6<, so by Lemma 3.2.7, 5= | 6 9 for some 9. By rearranging, we can assume that 9 = <. But 6< is also irreducible, so 5= = 26< for some nonzero 2 ∈ , and both 5= and 6< are monic, so 2 = 1. Hence 5= = 6<. Cancelling in (3.7) (which we can do as [C] is an integral domain) gives

0 51 ··· 5=−1 = 161 ··· 6<−1.

By inductive hypothesis, = = <, 0 = 1, and the lists of irreducibles 51, . . . , 5=−1 and 61, . . . , 6<−1 are the same up to reordering. This completes the induction. One way to find an irreducible factor of a polynomial 5 (C) ∈ [C] is to find a root (an element 0 ∈ such that 5 (0) = 0): Lemma 3.2.9 Let be a field, 5 (C) ∈ [C] and 0 ∈ . Then 5 (0) = 0 ⇐⇒ (C − 0) | 5 (C). Proof ⇒: suppose that 5 (0) = 0. By Proposition 3.2.1, 5 (C) = (C − 0)@(C) + A(C) (3.8) for some @,A ∈ [C] with deg(A) < 1. Then A is a constant, so putting C = 0 in (3.8) gives A = 0. ⇐: if 5 (C) = (C − 0)@(C) for some polynomial @ then 5 (0) = 0.

31 Deﬁnition 3.2.10 Let be a ﬁeld, let 0 ≠ 5 ∈ [C], and let 0 ∈ be a root of 5 . The multiplicity of 0 is the unique integer < ≥ 1 such that (C − 0)< | 5 (C) but (C − 0)<+1 - 5 (C).

Exercise 3.2.11 This deﬁnition assumes that there is a unique < with these properties (that is, there is one and only one such <). Prove it.

Example 3.2.12 Over any ﬁeld , the polynomial C has a root at 0 with multiplicity 1, and the polynomial C2 has a root at 0 with multiplicity 2. This is true even when 2 = F2, in which case C and C induce the same function → (Warning 3.1.2).

Proposition 3.2.13 Let be a ﬁeld and let 0 ≠ 5 ∈ [C]. Write 01, . . . , 0: for the distinct roots of 5 in , and <1, . . . , <: for their multiplicities. Then

<1 <: 5 (C) = (C − 01) ···(C − 0: ) 6(C) for some 6(C) ∈ [C] that has no roots. Proof By induction on :. If : = 0, this is immediate (again interpreting an empty < product as 1). Now suppose that : ≥ 1. By deﬁnition, (C − 0: ) : | 5 (C), so we can put 5 (C) 5 (C) = ∈ [C]. e < (C − 0: ) :

Any root of e5 is a root of 5 , so it is one of 01, . . . , 0: . But e5 (0: ) ≠ 0: for < +1 if e5 (0: ) = 0 then (C − 0: ) | e5 (C) by Lemma 3.2.9, so (C − 0: ) : | 5 (C), a contradiction. Hence any root of e5 is one of 01, . . . , 0:−1. These are indeed roots of e5 , with multiplicities <1, . . . , <:−1. So by inductive hypothesis,

<1 <:−1 e5 (C) = (C − 01) ···(C − 0:−1) 6(C) for some 6(C) ∈ [C] with no roots, completing the induction. Corollary 3.2.14 Let be a field and 5 ∈ [C]. Suppose that 5 has distinct roots 01, . . . , 0: ∈ with multiplicities <1, . . . , <: . Then <1 + · · · + <: ≤ deg( 5 ). In other words, a polynomial of degree = has no more than = roots, even when you count the roots with multiplicities (e.g. count a double root twice). A field is algebraically closed if every nonconstant polynomial has at least one root. For example, C is algebraically closed (the fundamental theorem of algebra). Proposition 3.2.13 implies: Corollary 3.2.15 Let be an algebraically closed field and let 0 ≠ 5 ∈ [C]. Write 01, . . . , 0: for the distinct roots of 5 in , and <1, . . . , <: for their multiplicities. Then <1 <: 5 (C) = 2(C − 01) ···(C − 0: ) , where 2 is the leading coefficient of 5 .

32 3.3 Irreducible polynomials

Determining whether an integer is prime is generally hard, and determining whether a polynomial is irreducible is hard too. This section presents a few techniques for doing so. Let’s begin with the simplest cases. Recall Lemma 3.1.10(iv): a polynomial is irreducible if and only if it is nonconstant (has degree > 0) and cannot be expressed as a product of two nonconstant polynomials. Lemma 3.3.1 Let be a ﬁeld and 5 ∈ [C]. i. If 5 is constant then 5 is not irreducible.

ii. If deg( 5 ) = 1 then 5 is irreducible. iii. If deg( 5 ) ≥ 2 and 5 has a root then 5 is reducible. iv. If deg( 5 ) ∈ {2, 3} and 5 has no root then 5 is irreducible.

Proof Parts (i) and (ii) follow from what we just recalled, and (iii) follows from Lemma 3.2.9. For (iv), suppose for a contradiction that 5 = 6ℎ with deg(6), deg(ℎ) ≥ 1. We have deg(6) + deg(ℎ) ∈ {2, 3}, so without loss of generality, deg(6) = 1. Also without loss of generality, 6 is monic, say 6(C) = C + 0; but then 5 (−0) = 0, a contradiction.

Warning 3.3.2 The converse of (iii) is false! To show a polynomial is irreducible, it’s not enough to show it has no root (unless it has degree 2 or 3). For instance, (C2 + 1)2 ∈ Q[C] has no root but is reducible.

?−1 Examples 3.3.3 i. Let ? be a prime. Then 5 (C) = 1 + C + · · · + C ∈ F? [C] is reducible, since 5 (1) = 0. ii. Let 5 (C) = C3 − 10 ∈ Q[C]. Then deg( 5 ) = 3 and 5 has no root in Q, so 5 is irreducible by part (iv) of the lemma. iii. Over C or any other algebraically closed ﬁeld, the irreducibles are exactly the polynomials of degree 1.

Exercise 3.3.4 If I gave you a quadratic over Q, how would you decide whether it is reducible or irreducible?

From now on we focus on = Q. Any polynomial over Q can be multiplied by a nonzero rational constant to get a polynomial over Z, and that’s often a helpful move, so we’ll look at Z[C] too.

33 Definition 3.3.5 A polynomial over Z is primitive if its coefficients have no common divisor except for ±1. For example, 15 + 6C + 10C2 is primitive but 15 + 6C + 30C2 is not. Lemma 3.3.6 Let 5 (C) ∈ Q[C]. Then there exist a primitive polynomial (C) ∈ Z[C] and U ∈ Q such that 5 = U. 5 C Í 0 1 C8 0 1 Proof Write ( ) = 8 ( 8/ 8) , where 8 ∈ Z and 0 ≠ 8 ∈ Z. Take any common Í 8 multiple 1 of the 18s; then writing 28 = 081/18 ∈ Z, we have 5 (C) = (1/1) 28C . Now let 2 be the greatest common divisor of the 28s, put 38 = 28/2 ∈ Z, and put Í 8 (C) = 38C . Then (C) is primitive and 5 (C) = (2/1)(C). If the coefficients of a polynomial 5 (C) ∈ Q[C] happen to all be integers, the word ‘irreducible’ could mean two things: irreducibility in the ring Q[C] or in the ring Z[C]. We say that 5 is irreducible over Q or Z to distinguish between the two. Suppose we have a polynomial over Z that’s irreducible over Z. In principle it could still be reducible over Q: although there’s no nontrivial way of factorizing it over Z, perhaps it can be factorized when you give yourself the freedom of non-integer coefficients. But the next result tells us that you can’t. Lemma 3.3.7 (Gauss) i. The product of two primitive polynomials over Z is primitive. ii. If a polynomial over Z is irreducible over Z, it is irreducible over Q. Proof For (i), let 5 and 6 be primitive polynomials over Z. Let ? be a prime number. (We’re going to show that ? doesn’t divide all the coefficients of 5 6.) Write c : Z → Z/?Z = F? for the canonical homomorphism, which induces a homomorphism c∗ : Z[C] → F? [C] as in Definition 3.1.6. Since 5 is primitive, ? does not divide all the coefficients of 5 . Equivalently, c∗( 5 ) ≠ 0. Similarly, c∗(6) ≠ 0. But F? [C] is an integral domain, so

c∗( 5 6) = c∗( 5 )c∗(6) ≠ 0, so ? does not divide all the coeﬃcients of 5 6. This holds for all primes ?, so 5 6 is primitive. For (ii), let 5 ∈ Z[C] be a polynomial irreducible over Z. Let 6, ℎ ∈ Q[C] with 5 = 6ℎ. By Lemma 3.3.6, 6 = U and ℎ = V for some U,V ∈ Q and primitive , ∈ Z[C]. Then UV =

34 Gauss’s lemma quickly leads to a test for irreducibility. It involves taking a polynomial over Z and reducing it mod ?, for some prime ?. This means applying the map c∗ : Z[C] → F? [C] from the last proof. As we saw after Deﬁnition 3.1.6, Í 8 Í 8 if 5 (C) = 08C then c∗( 5 )(C) = c(08)C , where c(08) is the congruence class of 08 mod ?. I’ll write 0 = c(0) and 5 = c∗( 5 ). That is, 5 is ‘ 5 mod ?’. = Proposition 3.3.8 (Mod p method) Let 5 (C) = 00 + 01C + · · · + 0=C ∈ Z[C]. If there is some prime ? such that ? - 0= and 5 ∈ F? [C] is irreducible, then 5 is irreducible over Q.

I’ll give some examples ﬁrst, then the proof. Examples 3.3.9 i. Let’s use the mod ? method to show that 5 (C) = 9+14C−8C3 3 is irreducible over Q. Take ? = 7: then 5 (C) = 2 − C ∈ F7 [C], so it’s enough 3 to show that 2−C is irreducible over F7. Since this has degree 3, it’s enough 3 to show that C = 2 has no solution in F7 (by Lemma 3.3.1(iv)). And you can easily check this by computing 03, (±1)3, (±2)3 and (±3)3 mod 7.

ii. The condition in Proposition 3.3.8 that ? - 0= can’t be dropped. For instance, consider 5 (C) = 6C2 + C and ? = 2.

Warning 3.3.10 Take 5 (C) as in Example 3.3.9(i), but this time take 3 ? = 3. Then 5 (C) = −C + C ∈ F3 [C], which is reducible. But that doesn’t mean 5 is reducible! The mod ? method only ever lets you show that a polynomial is irreducible over Q, not reducible.

Proof of Proposition 3.3.8 Take a prime ? satisfying the conditions in the Propo- sition. By Gauss’s lemma, it is enough to prove that 5 is irreducible over Z. Since 5 is irreducible, deg( 5 ) > 0, so deg( 5 ) > 0. Now let 5 = 6ℎ in Z[C]. We have 5 = 6ℎ and 5 is irreducible, so without loss of generality, 6 is constant. The leading coefficient of 5 is the product of the leading coefficients of 6 and ℎ, and is not divisible by ?, so the leading coefficient of 6 is not divisible by ?. Hence deg(6) = deg(6) = 0.

We ﬁnish with an irreducibility test that turns out to be surprisingly powerful.

= Proposition 3.3.11 (Eisenstein’s criterion) Let 5 (C) = 00 + · · · + 0=C ∈ Z[C], with = ≥ 1. Suppose there exists a prime ? such that:

• ? - 0=;

• ? | 08 for all 8 ∈ {0, . . . , = − 1};

2 • ? - 00.

35 Then 5 is irreducible over Q.

To prove this, we will use the concept of the codegree codeg( 5 ) of a polynomial 5 C Í 0 C8 8 0 5 ( ) = 8 8 , which is deﬁned to be the least such that 8 ≠ 0 (if ≠ 0), or as the formal symbol ∞ if 5 = 0. For polynomials 5 and 6 over an integral domain,

codeg( 5 6) = codeg( 5 ) + codeg(6).

Clearly codeg( 5 ) ≤ deg( 5 ) unless 5 = 0. Proof By Gauss’s lemma, it is enough to show 5 is irreducible over Z. Let 6, ℎ ∈ Z[C] with 5 = 6ℎ. Continue to write 5 (C) ∈ F? [C] for 5 reduced mod ?; then 5 = 6ℎ. Since 2 ? - 00 = 5 (0) = 6(0)ℎ(0), we may assume without loss of generality that ? - 6(0). Hence codeg(6) = 0. Also, codeg( 5 ) = =, since ? divides each of 00, . . . , 0=−1 but not 0=. So

= = codeg( 5 ) = codeg(6) + codeg(ℎ) = codeg(ℎ) ≤ deg(ℎ) ≤ deg(ℎ), (3.9)

giving = ≤ deg(ℎ). But 5 = 6ℎ with deg( 5 ) = =, so deg(ℎ) = = and deg(6) = 0.

Exercise 3.3.12 The last step in (3.9) was ‘deg(ℎ) ≤ deg(ℎ)’. Why is that true? And when does equality hold?

Example 3.3.13 Let 2 5 1 6(C) = C5 − C4 + C3 + ∈ Q[C]. 9 3 3 Then 6 is irreducible over Q if and only if

96(C) = 2C5 − 15C4 + 9C3 + 3

is irreducible over Q, which it is by Eisenstein’s criterion with ? = 3. Testing for irreducibility Exercise 3.3.14 Use Eisenstein’s criterion to show that for every = ≥ 1, there is an irreducible polynomial over Q of degree =.

I’ll give you one more example, and it’s not just any old polynomial: it’s an important one that we’ll need when we come to think about solvability by radicals. It needs a lemma. ? < 8 < ? ? ? Lemma 3.3.15 Let be a prime and 0 . Then | 8 .

36 For example, the 7th row of Pascal’s triangle is 1, 7, 21, 35, 35, 21, 7, 1, and 7 divides all of these numbers apart from the ﬁrst and last. 8 ? 8 ? ? ? ? 8 ? 8 ? Proof We have !( − )! 8 = !, and divides ! but not ! or ( − )! (since < 8 < ? ? ? is prime and 0 ), so must divide 8 . Example 3.3.16 Let ? be a prime. The ?th cyclotomic polynomial is C ? − 1 Φ (C) = 1 + C + · · · + C ?−1 = . (3.10) ? C − 1 I claim that Φ? is irreducible. We can’t apply Eisenstein to Φ? as it stands, because whichever prime we choose (whether it’s ? or another one) doesn’t divide any of the coeﬃcients. However, we saw on p. 27 that Φ? (C) is irreducible if and only if Φ? (C − 2) is irreducible, for any 2 ∈ Q. We’ll take 2 = −1. We have (C + 1) ? − 1 Φ (C + 1) = ? (C + 1) − 1 ? 1 Õ C = C8 C 8 8=1 ? ? = ? + C + · · · + C ?−2 + C ?−1. 2 ? − 1

So Φ? (C + 1) is irreducible by Eisenstein’s criterion and Lemma 3.3.15, hence Φ? (C) is irreducible too.

Digression 3.3.17 I deﬁned the ?th cyclotomic polynomial Φ? only when ? is prime. The deﬁnition of Φ= for general = ≥ 1 is not the obvious generalization of (3.10). It’s this: Ö Φ= (C) = (C − Z), Z

where the product runs over all primitive =th roots of unity Z. (In this context, ‘primitive’ means that = is the smallest number satisfying Z = = 1; it’s a diﬀerent usage from ‘primitive polynomial’.)

Many surprising things are true. It’s not obvious that the coeﬃcients of Φ= are real, but they are. Even given that they’re real, it’s not obvious that they’re rational, but they are. Even given that they’re rational, it’s not obvious that they’re integers, but they are. The degree of Φ= is i(=), the number of integers between 1 and = that are coprime with = (Euler’s function). It’s also true that the polynomial Φ= is irreducible for all =, not just primes. Some of these things are quite hard to prove, and results from Galois theory help. We probably won’t get into all of this, but you can read more here.

37 Chapter 4

Field extensions

Roughly speaking, an ‘extension’ of a field is a field " that contains as a subfield. It’s not much of an exaggeration to say that field extensions are the central objects of Galois theory, in much the same way that vector spaces are the central objects of linear algebra. Introduction to It will be a while before it becomes truly clear why field extensions are so Week 4 important, but here are a couple of indications:

• For any polynomial 5 over Q, we can take the smallest subﬁeld " of C that contains all the complex roots of 5 , and that’s an extension of Q.

• For any irreducible polynomial 5 over a field , the quotient ring " = [C]/h 5 i is a field. The constant polynomials form a subfield of " isomorphic to , so " is an extension of .

It’s important to distinguish between these two types of example. The ﬁrst extends Q by all the roots of 5 , whereas the second extends by just one root of 5 —as we’ll see. But let’s begin at the beginning.

4.1 Deﬁnition and examples

Definition 4.1.1 A field extension consists of a field , a field ", and a homomorphism ]: → ".

Since homomorphisms between fields are injective (Lemma 2.2.2), is isomorphic to the subfield im(]) of ". It is usually safe to identify ](0) with 0 for each 0 ∈ —in other words, pretend that is actually a subfield of " and ](0) = 0. We then say that " is an extension of and write " : .

38 It’s worth taking a minute to make sure you understand the relationship between subsets and injections. This is a fundamental point about sets, not fields. Given a set and a subset ⊆ , there’s an inclusion function ]: → , defined by ](1) = 1 for all 1 ∈ . (Remember that any function has a specified domain and codomain, so this isn’t the same thing as the identity on .) This inclusion function is an injective. On the other hand, given any injective function between sets, say i : - → , the image im i is a subset of , and there’s a bijection i0 : - → im i given by i0(G) = i(G) (G ∈ -). Hence - is ‘isomorphic to’ (in bijection with) the subset im i of . This back-and-forth process means that subsets and injections are more or less the same thing.

Digression 4.1.2 If a field extension is basically the same thing as a field " together with a subfield , you might wonder why we bother with the more general Definition 4.1.1, involving an arbitrary homomorphism ]. It turns out that in the long run, it makes things easier. There are two factors at play. The first is purely set-theoretic. The concept of subset isn’t actually as simple as it appears, at least when you look at what mathematicians do rather than what we claim we do. For example, (1) it’s common to define the set C as R2, (2) everyone treats R as a subset of C, but (3) almost no one would say that R is a subset of R2 (if you wrote ‘the point c of R2’ on an exam, you’d be marked wrong). In truth, the common conventions are inconsistent. A good way to make everything respectable is to do everything in terms of injections rather than subsets. It would take up too much space to go into this here, but Definition 4.1.1 is one example of this approach in action. The second factor is algebraic. According to Definition 4.1.1, a field extension is simply a homomorphism between fields, so it includes examples such as the conjugation map : C → C. You might feel this example obeys the letter but not the spirit of Definition 4.1.1. But again, it turns out to be useful to include such examples. When we come to count isomorphisms between fields in a few weeks, you’ll see why.

Examples 4.1.3 i. The inclusion ]: Q → C is a ﬁeld extension, usually just written as C : Q. Similarly, there are ﬁeld extensions C : R and R : Q.

ii. Let √ √ Q( 2) = {0 + 1 2 : 0, 1 ∈ Q}. √ Then Q( 2) is a subring of C (easily), and in fact it’s a subﬁeld: for if 0, 1 , ( ) ≠ (0 0) then √ 1 0 − 1 2 √ = 0 + 1 2 02 − 212

39 √ (noting that the denominators√ are not 0 because 2 √is irrational). So we have an extension√ C : Q( 2). Also, because Q ⊆ Q( 2), we have another extension Q( 2) : Q. iii. By direct calculation or later theory (which will make it much easier), √ √ √ Q( 2, 8) = {0 + 1 2 + 28 + 3 28 : 0, 1, 2, 3 ∈ Q} √ is also a subﬁeld of C, so we have an extension Q( 2, 8) : Q.

iv. Let be a ﬁeld. A rational expression over is a ratio of two polynomials 5 (C) , 6(C)

where 5 (C), 6(C) ∈ [C] with 6 ≠ 0. Two such expressions, 51/61 and 52/62, are regarded as equal if 5162 = 5261 in [C]. So formally, a rational expression is an equivalence class of pairs ( 5 , 6) under the equivalence relation in the last sentence. The set of rational expressions over is denoted by (C). Rational expressions are added, subtracted and multiplied in the ways you’d expect, making (C) into a ﬁeld. There is a homomorphism ]: → (C) given by ](0) = 0/1 (0 ∈ ). In other words, (C) contains a copy of as the constant rational expressions. So, we have a ﬁeld extension (C) : .

v. In particular, when = F? for some prime ?, we have the field extension F? (C) : F?. Note that F? (C) is an infinite field of characteristic ?! Fields of positive characteristic don’t have to be finite. So I’ve now fulfilled the promise I made in Warning 2.2.12.

Warning 4.1.4 People sometimes say ‘rational function’ to mean ‘rational expression’. But just as for polynomials (Warnings 3.1.2 and 3.1.3), I want to emphasize that rational expressions are not functions. For instance, 1/(C − 1) is a totally respectable element of (C). You don’t have to—and shouldn’t—worry about what happens when C = 1, because C is just a formal symbol (a mark on a piece of paper) rather than a variable, and 1/(C − 1) is just a formal expression, not a function. If this puzzles you, I suggest going back to those warnings about polynomials, which make the same point in a simpler setting.

Exercise√ 4.1.5 Find two examples of ﬁelds such that Q( ( Q( 2, 8). (The symbol ( means proper subset.)

For any kind of algebraic structure, there is a notion of the ‘substructure generated by’ a given subset. For example, when - is a subset of a group , you know what the subgroup generated by - is: it’s the intersection of all the subgroups containing -. Similarly, when - is a subset of a vector space +, the linear subspace generated by - (or ‘spanned by -’, as one usually says) is the intersection of all the linear subspaces containing -. We now make a similar deﬁnition for ﬁelds.

Definition 4.1.6 Let be a field and - a subset of . The subfield of generated by - is the intersection of all subfields of containing -.

You can check that any intersection of subfields of is a subfield of (even if it’s an uncountably infinite intersection). In fact, you already did most of the work for this in Exercise 2.1.2. So, the subfield of generated by - really is a subfield of . It contains - itself. By definition of intersection, is the smallest subfield of containing -, in the sense that any subfield of containing - contains .

Exercise 4.1.7 Check the truth of all the statements in the previous paragraph.

Examples 4.1.8 i. The subﬁeld of generated by is the prime subﬁeld of .

ii. Let ! be the subﬁeld of C generated by {8}. I claim that

! = {0 + 18 : 0, 1 ∈ Q}.

To prove this, we have to show that ! is the smallest subfield of C containing 8. First, it is a subfield of C (by an argument similar to Example 4.1.3(ii)) and it contains 0 + 18 = 8. Now let !0 be any subfield of C containing 8. Then !0 contains the prime subfield of C (by definition of prime subfield), which is Q. So whenever 0, 1 ∈ Q, we have 0, 1, 8 ∈ !0 and so 0 + 18 ∈ !0. Hence ! ⊆ !0, as required. √ iii. A very similar argument shows√ that the subfield of C generated by 2 is what we have been calling Q( 2).

41 Exercise 4.1.9 What is the subﬁeld of C generated by {7/8}? By {2 + 38}? By R ∪ {8}?

We will be very interested in chains of ﬁelds

⊆ ! ⊆ " in which and " are regarded as ﬁxed and ! as variable. You can think of as the ﬂoor, " as the ceiling, and ! as varying in between.

Definition 4.1.10 Let " : be a field extension and . ⊆ ". We write (.) for the subfield of " generated by ∪ ., and call it with . adjoined.

So, (.) is the smallest subfield of " containing both and .. When . is a finite set {U1,...,U=}, we write ({U1,...,U=}) as (U1,...,U=). √ " . . Examples 4.1.11 i. Take : to be C√: Q and = { 2}. Then ( ) is the smallest subfield of C containing Q∪ { 2}. But every subfield of C contains . Q: that’s what it means for Q to be√ the prime subfield of C. So, ( ) is the smallest subfield of C containing√ 2. By Example 4.1.8√(iii), that’s exactly√ what we’ve been calling Q( 2) all along. We refer to Q( 2) as ‘Q with 2 adjoined’.

ii. Similarly, Q with 8 adjoined is

Q(8) = {0 + 18 : 0, 1 ∈ Q} √ , 8 (Example√ 4.1.8(ii)), and Q with { 2 } adjoined is the subﬁeld denoted by Q( 2, 8) in Example 4.1.3(iii).

iii. Let " be a field and - ⊆ ". Write for the prime subfield of ". Then (-) is the smallest subfield of " containing and -. But every subfield of " contains , by definition of prime subfield. So (-) is the smallest subfield of " containing -; that is, it’s the subfield of " generated by -. √ We already saw this argument in (i), in the case " = C and - = { 2}.

iv. Let be any field and let " be the field (C) of rational expressions over , which is an extension of . You might worry that there’s some ambiguity in the notation: (C) could either mean the field of rational expressions over (as defined in Example 4.1.3(iv)) or the subfield of (C) obtained by adjoining the element C of (C) to (as in Definition 4.1.10).

42 In fact, they’re the same. In other words, the smallest subfield of (C) containing and C is (C) itself. Or equivalently, the only subfield of (C) containing and C is (C) itself. To see this, let ! be any such subfield. For Í 8 any polynomial 5 (C) = 08C over , we have 5 (C) ∈ !, since 08,C ∈ ! and ! is closed under multiplication and addition. Hence for any polynomials 5 (C), 6(C) over with 6(C) ≠ 0, we have 5 (C), 6(C) ∈ !, so 5 (C)/6(C) ∈ ! as ! is closed under division by nonzero elements. So ! = (C).

Warning 4.1.12 It is not true in general that

(U) = {0 + 1U : 0, 1 ∈ } (false!) (4.1) √ 8 Examples√ like Q( 2) and Q( ) do satisfy this, but that’s only because 2 and 8 satisfy quadratic equations. Certainly the right-hand side is a subset of (U), but in general it’s much smaller, and isn’t a subﬁeld. You’ve just seen an example: the ﬁeld (C) of rational expressions is much bigger than the set {0 + 1C : 0, 1 ∈ } of polynomials of degree ≤ 1. And that set of polynomials isn’t closed under multiplication. Another example: let U be the real cube root of 2. You can show that U2 cannot be expressed as 0 + 1U for any 0, 1 ∈ Q (a fact we’ll come back to in Example 4.2.9(ii)). But U ∈ Q(U), so U2 ∈ Q(U), so (4.1) fails in this case. In fact,

Q(U) = {0 + 1U + 2U2 : 0, 1, 2 ∈ Q}.

We’ll see why next week.

Exercise 4.1.13 Let " : be a ﬁeld extension. Show that (. ∪/) = ( (.))(/) whenever .,/ ⊆ ". (For example, (U,V) = ( (U))(V) whenever U,V ∈ ".)

4.2 Algebraic and transcendental elements

A complex number U is said to be ‘algebraic’ if

= 00 + 01U + · · · + 0=U = 0 for some rational numbers 08, not all zero. (You may have seen this definition with ‘integer’ instead of ‘rational number’; it makes no difference, as you can always clear the denominators.) This concept generalizes to arbitrary field extensions:

43 Deﬁnition 4.2.1 Let " : be a ﬁeld extension and U ∈ ". Then U is algebraic over if there exists 5 ∈ [C] such that 5 (U) = 0 but 5 ≠ 0, and transcendental otherwise.

Exercise 4.2.2 Show that every element of is algebraic over .

Examples 4.2.3 i. Let = ≥ 1. Then 42c8/= ∈ C is algebraic over Q, since 5 (C) = C= − 1 is a nonzero polynomial such that 5 (42c8/=) = 0.

ii. The numbers c and 4 are both transcendental over Q. Both statements are hard to prove (and we won’t). By Exercise 4.2.2, any complex number transcendental over Q is irrational. Proving the irrationality of c and 4 is already a challenge; proving they’re transcendental is even harder.

iii. Although c is transcendental over Q, it is algebraic over R, since it’s an element of R. (Again, we’re using Exercise 4.2.2.) Moral: you shouldn’t say an element of a ﬁeld is just ‘algebraic’ or ‘transcendental’; you should say it’s ‘algebraic/transcendental over ’, specifying your . Or at least, you should do this when there’s any danger of confusion.

iv. Take the ﬁeld (C) of rational expressions over a ﬁeld . Then C ∈ (C) is transcendental over , since 5 (C) = 0 ⇐⇒ 5 = 0.

The set of complex numbers algebraic over Q is written as Q. It’s a fact that Q is a subﬁeld of C, but this is extremely hard to prove by elementary means. Next week I’ll show you that with a surprisingly small amount of abstract algebra, you can transform this from a very hard problem into an easy one. So that you appreciate the miracle later, I give you this unusual exercise now.

Exercise 4.2.4 Attempt to prove any part of the statement that Q is a subfield of C. For example, try to show that Q is closed under addition, or multiplication, or reciprocals. I have no idea how to do any of these using only our current tools, but it’s definitely worth a few minutes of doomed effort to get a sense of the difficulties.

Let " : be a ﬁeld extension and U ∈ ". An annihilating polynomial of U is a polynomial 5 ∈ [C] such that 5 (U) = 0. So, U is algebraic if and only if it has some nonzero annihilating polynomial. It is natural to ask not only whether U is annihilated by some nonzero polynomial, but which polynomials annihilate it. The situation is pleasantly simple:

44 Lemma 4.2.5 Let " : be a ﬁeld extension and U ∈ ". Then there is a polynomial <(C) ∈ [C] such that

If U is transcendental over then < = 0. If U is algebraic over then there is a unique monic polynomial < satisfying (4.2).

Proof By the universal property of polynomial rings (Lemma 3.1.5), there is a unique homomorphism \ : [C] → " such that \(0) = 0 for all 0 ∈ and \(C) = U. (Here we’re taking the ‘i’ of Lemma 3.1.5 to be the inclusion → ".) Then

Õ 8 Õ 8 \ 08C = 08U

Í 8 for all 08C ∈ [C], so

ker \ = {annihilating polynomials of U over }.

Now ker \ is an ideal of the principal ideal domain [C] (using Proposition 3.2.2), so ker \ = h

Exercise 4.2.7 What is the minimal polynomial of an element 0 of ?

This is an important deﬁnition, so we give some equivalent ways of stating it.

Lemma 4.2.8 Let " : be a ﬁeld extension, let U ∈ " be algebraic over , and let < ∈ [C] be a monic polynomial. The following are equivalent:

i. < is the minimal polynomial of U over ;

45 ii. <(U) = 0, and < | 5 for all annihilating polynomials 5 of U over ; iii. <(U) = 0, and deg(<) ≤ deg( 5 ) for all nonzero annihilating polynomials 5 of U over ; iv. <(U) = 0 and < is irreducible over . Part (iii) says the minimal polynomial is a monic annihilating polynomial of least degree. Proof (i)⇒(ii) follows from the deﬁnition of minimal polynomial. (ii)⇒(iii) because if < | 5 ≠ 0 then deg(<) ≤ deg( 5 ). (iii)⇒(iv): assume (iii). First, < is not constant: for if < is constant then < = 1 (since < is monic); but <(U) = 0, so 1 = 0 in , a contradiction. Next, suppose that < = 5 6 for some 5 , 6 ∈ [C]. Then 0 = <(U) = 5 (U)6(U), so without loss of generality, 5 (U) = 0. By (iii), deg( 5 ) ≥ deg(<), so deg( 5 ) = deg(<) and deg(6) = 0. This proves (iv). (iv)⇒(i): assume (iv), and write quadratic equation over Q, i.e. that the equation √ 2 √ 3 2 = 0 3 2 + 1 has no solution for 0, 1 ∈ Q. It’s not impossible, but it’s a mess. (You naturally begin by cubing both sides, but look what happens next. . . ) So the theory really gets us something here. Two traps 46 iii. Let ? be a prime number, and put l = 42c8/? ∈ C. Then l is a root of C ? − 1, but that is not the minimal polynomial of l, since it is reducible:

C ? − 1 = (C − 1)<(C)

where <(C) = C ?−1 + · · · + C + 1. Since l? −1 = 0 but l−1 ≠ 0, we must have <(l) = 0. By Example 3.3.16, < is irreducible over Q. Hence < is the minimal polynomial of l over Q.

4.3 Simple extensions

We can say a lot about extensions generated by a single element.

Deﬁnition 4.3.1 A ﬁeld extension " : is simple if there exists U ∈ " such that " = (U).

Examples 4.3.2 √ i. √Surprisingly many algebraic extensions are simple. For , instance, Q√( √2 3) : Q√is a simple√ extension (despite appearances), because in fact Q( 2, 3) = Q( 2 + 3).

ii. (C) : is simple, where (C) is the ﬁeld of rational expressions over .

√ √ √ √ , Exercise√ 4.3.3√Prove that Q( 2 3) = Q( 2 + 3). Hint: begin by ﬁnding ( 2 + 3)3.

Given a simple extension " : and an element U that generates it, we can take the minimal polynomial of U, which is irreducible over . But in the opposite direction, if you hand me a ﬁeld and an irreducible polynomial < over , I can cook up for you a simple extension " : and an element U ∈ " whose minimal polynomial is the < you gave me. This works as follows. Whenever < is an irreducible polynomial over a ﬁeld , the quotient [C]/h

c → [C] −→ [C]/h

47 Lemma 4.3.4 Let < be a monic irreducible polynomial over a ﬁeld . Then [C]/h

i " / "0 ` >

] ]0

commutes. If i is invertible then we call i an isomorphism over .

Exercise 4.3.6 In this deﬁnition, show that if i is invertible then i−1 is also a homomorphism over .

How to understand The next result not only classifies the simple extensions by an algebraic element simple algebraic (part (i)), but also those by a transcendental element (part (ii)). extensions Theorem 4.3.7 (Classification of simple extensions) Let be a field.

i. Let < ∈ [C] be a monic irreducible polynomial. Then there exist an extension " : and an algebraic element U ∈ " such that " = (U) and U has minimal polynomial < over . Moreover, if (",U) and ("0,U0) are two such pairs, there is an isomorphism i : " → "0 over such that i(U) = U0.

ii. There exist an extension " : and a transcendental element U ∈ " such that " = (U). Moreover, if (",U) and ("0,U0) are two such pairs, there is an isomor-

48 phism i : " → "0 over such that i(U) = U0.

Proof The ﬁrst part of (i) follows from Lemma 4.3.4. For ‘Moreover’, we may as well take " = [C]/h

\ : [C] → "0 Í 8 Í 08 (4.3) 08C 7→ 08U has kernel h

c \ % [C] / [C]/h

(If you’re losing the thread, it may help to go back to the review of the universal property of quotients and the ﬁrst isomorphism theorem on p. 16.) The isomorphism \ : [C]/h

49 Conclusion: given any ﬁeld (not necessarily Q!) and any monic irreducible <(C) ∈ [C], we can say the words ‘adjoin to a root U of <’, and this unambiguously deﬁnes an extension (U) : . (At least, unambiguously up to isomorphism over —but who could want more?) Similarly, we can unambiguously adjoin to a transcendental element.

Examples 4.3.8 i. Let be any ﬁeld not containing a square root of 2. Then C2 C2 − 2 is irreducible√ over . So we can adjoin to a root of − 2, giving an extension ( 2) : .

We√ have already seen this example many times when = Q, in which case

( 2) can be seen as a subﬁeld of C. But the construction works for √any . For instance, 2 has no square root in F3, so there is an extension F3( 2) 2 of F3. It can be constructed as F3 [C]/hC − 2i.

2 ii. The polynomial <(C) = 1 + C + C is irreducible over F2, so we may adjoin to 2 F2 a root U of <. Then F2(U) = F2 [C]/h1 + C + C i. √ Exercise 4.3.9 How many elements does the ﬁeld F3( 2) have? What about F2(U)?

Warning 4.3.10 Take = Q and <(C) = C3 − 2, which is irreducible. Write U1,U2,U3 for the roots of < in C. Then Q(U1), Q(U2) and Q(U3) are all different as subsets of C. For example, one of the U8 is the real cube root of 2 (say U1), which implies that Q(U1) ⊆ R, whereas the other two are not real, so Q(U8) *R for 8 ≠ 1. However, Q(U1) : Q, Q(U2) : Q and Q(U3) : Q are all isomorphic as abstract field extensions of Q. This follows from Theorem 4.3.7, since all the U8 have the same minimal polynomial, <. You’re already very familiar with this kind of situation in other branches of algebra, whether you realize it or not. For instance, in lin- 2 ear algebra, take three vectors v1, v2, v3 in R , none a scalar multiple of any other. Then span(v1), span(v2) and span(v3) are all different as subsets of R2, but they are all isomorphic as abstract vector spaces (since they’re all 1-dimensional). A similar example could be given with a group containing several subgroups that are all isomorphic.

You’ve seen that Galois theory involves aspects of group theory and ring theory. In the next chapter, you’ll see how linear algebra enters the picture too.

50 Chapter 5

Degree

We’ve already seen that if you adjoin to Q a square root of 2, then each element of the resulting ﬁeld can be speciﬁed using two rational numbers, 0 and 1: √ √ Q( 2) = 0 + 1 2 : 0, 1 ∈ Q .

We’ve also seen that if you adjoin to Q a cube root of 2, then it takes three rational numbers to specify each element of the resulting field: Introduction to √ √ √ Week 5 3 n 3 3 2 o Q( 2) = 0 + 1 2 + 2 2 : 0, 1, 2 ∈ Q √ 3 (Warning 4.1.12). This might√ lead us to suspect that Q( 2) : Q is in some sense a ‘bigger’ extension than Q( 2) : Q. The first thing we’ll do in this chapter is to make√ this intuition rigorous.√ We’ll define the ‘degree’ of an extension and see that Q( 2) : Q and Q( 3 2) : Q have degrees 2 and 3, respectively. The concept of degree is incredibly useful, and not only in Galois theory. In fact, I’ll show you how it can be used to solve three problems that remained unsolved for literally millennia, since the time of the ancient Greeks.

5.1 Degrees of extensions and polynomials

The concept of degree is an excellent illustration of a powerful mathematical technique: forgetting. Let " : be a ﬁeld extension. What happens if we forget how to multiply together elements of " that aren’t in ? We still have the ﬁeld . What remains of " is its underlying additive abelian group (", +, 0), and because we haven’t forgotten how to multiply elements of

51 with elements of ", we still have the multiplication function × " → ". So, we have a field , an abelian group ", and an action of on ". In other words, whenever " : is a field extension, " is a vector space over in a natural way. Definition 5.1.1 The degree [" : ] of a field extension " : is the dimension of " as a vector space over .

If " is a ﬁnite-dimensional vector space over , it’s clear what this means. If " is inﬁnite-dimensional over , we write [" : ] = ∞, where ∞ is a formal symbol which we give the properties

= < ∞, = · ∞ = ∞ (= ≥ 1), ∞ · ∞ = ∞

(where = is an integer). An extension " : is ﬁnite if [" : ] < ∞.

Digression 5.1.2 You know that whenever + is a finite-dimensional vector space, (i) there exists a basis of +, and (ii) there is a bijection between any two bases. This makes it possible to define the dimension of a vector space as the number of elements in a basis. In fact, both (i) and (ii) are true for every vector space, not just the finite-dimensional ones. So we can define the dimension of an arbitrary vector space as the ‘number’ of elements in a basis, where now ‘number’ means cardinal, i.e. isomorphism class of sets. We could interpret Definition 5.1.1 using this general definition of dimension. For instance, suppose we had one field extension " : such that " had a countably infinite basis over , and another, " 0 : , such that " 0 had an uncountably infinite basis over . Then [" : ] and [" 0 : ] would be different. However, we’ll lump all the infinite-dimensional extensions together and say that their degrees are all ∞. We’ll mostly be dealing with finite extensions anyway, and won’t need to distinguish between sizes of ∞. It’s a bit like the difference between a house that costs a million pounds and a house that costs ten million: although the difference in cost is huge, most of us would lump them together in a single category called ‘unaffordable’.

Examples 5.1.3 i. Every ﬁeld " contains at least one nonzero element, namely, 1. So [" : ] ≥ 1 for every ﬁeld extension " : . If " = then {1} is a basis, so [" : ] = 1. On the other hand, if [" : ] = 1 then the one-element linearly independent set {1} must be a basis, which implies that every element of " is equal to 0 · 1 = 0 for some 0 ∈ , and so " = . Hence

[" : ] = 1 ⇐⇒ " = .

52 ii. Every element of C is equal to G + H8 for a unique pair (G,H) of elements of R. That is, {1, 8} is a basis of C over R. Hence [C : R] = 2.

iii. Let be a ﬁeld and (C) the ﬁeld of rational expressions over . Then 1,C,C2,... are linearly independent over , so [ (C) : ] = ∞.

Warning 5.1.4 The degree [ : ] of over itself is 1, not 0. Degrees of extensions are never 0. See Example 5.1.3(i).

Theorem 5.1.5 Let (U) : be a simple extension, with U algebraic over . Write < ∈ [C] for the minimal polynomial of U and = = deg(<). Then

1,U,...,U=−1

is a basis of (U) over . In particular, [ (U) : ] = deg(<).

Proof By Lemma 4.3.4 and Theorem 4.3.7(i), we might as well take (U) = [C]/h

Equivalently, there are unique 00, . . . , 0=−1 ∈ such that

=−1 c( 5 ) = c 00 + 01C + · · · + 0=−1C .

Equivalently (since c(C) = U), there are unique 00, . . . , 0=−1 ∈ such that

=−1 c( 5 ) = 00 + 01U + · · · 0=−1U .

We have now shown that every element of (U) can be expressed as a -linear combination of 1,U,...,U=−1 in a unique way. In other words, 1,U,...,U=−1 is a basis of (U) over .

Exercise 5.1.6 For a prime ?, ﬁnd [Q(42c8/?) : Q].

53 Theorem 5.1.5 implies that when U ∈ " is algebraic over , with minimal =−1 polynomial of degree =, the subset {00 + 01U + · · · + 0=−1U : 08 ∈ } is a subﬁeld of ". This isn’t particularly obvious: for instance, why is it closed under taking reciprocals? But it’s true. Corollary 5.1.7 Let " : be a ﬁeld extension and U ∈ ". Then

(U) : is ﬁnite ⇐⇒ U is algebraic over .

Proof For ⇒, we prove the contrapositive. If U is transcendental over then (U) is isomorphic to (C) over (by Theorem 4.3.7(ii)), and [ (C) : ] = ∞ by Example 5.1.3(iii). Theorem 5.1.5 gives ⇐.

For a ﬁeld extension " : and U ∈ ", the degree of U over is [ (U) : ]. U U < U We write it as deg ( ). So, deg ( ) ∞ if and only if is algebraic over , and U U in that case, deg ( ) is the degree of the minimal polynomial of over . Examples 5.1.8 i. Let U ∈ C be an algebraic number over Q whose minimal polynomial is quadratic. Then by Theorem 5.1.5,

Q(U) = {0 + 1U : 0, 1 ∈ Q}. √ We’ve already seen this in many examples, such as U = 2 and U = 8. ii. Let U be the real cube root of 2. By Example 4.2.9(ii), the minimal U C3 U polynomial of over Q is − 2, so degQ( ) = 3. It follows that Q(U) ≠ {0 + 1U : 0, 1 ∈ Q}, since otherwise the two-element set {1,U} would span the three-dimensional vector space Q(U). So we have another proof that 22/3 cannot be written as a Q-linear combination of 1 and 21/3. As observed in Example 4.2.9(ii), this is messy to prove directly.

Theorem 5.1.5 is quite powerful. Here are two more corollaries of it. Corollary 5.1.9 i. Let " : ! : be ﬁeld extensions and V ∈ ". Then [!(V) : !] ≤ [ (V) : ]. ii. Let " : be a ﬁeld extension and U,V ∈ ". Then [ (U,V) : (U)] ≤ [ (V) : ].

Proof For (i): if [ (V) : ] = ∞ then the inequality is clear. Otherwise, V is algebraic over (by Corollary 5.1.7), with minimal polynomial < ∈ [C], say. Then < is an annihilating polynomial for V over !, so the minimal polynomial of V over ! has degree ≤ deg(<). The result follows from Theorem 5.1.5. Part (ii) follows by taking ! = (U).

54 V " [ (V): ] [!(V):!] !

Figure 5.1: Visualization of Corollary 5.1.9(i) (not to be taken too seriously).

Informally, I think of part (i) as in Figure 5.1. The degree of V over measures how far V is from being in . Since ! contains , it might be that V is closer to ! than to (i.e. [!(V) : !] < [ (V) : ]), and it’s certainly no further away.

Exercise 5.1.10 Give an example to show that the inequality in Corol- lary 5.1.9(ii) can be strict. Your example can be as trivial as you like.

Corollary 5.1.11 Let " : be a ﬁeld extension and U1,...,U= ∈ ". Sup- pose that each U8 is algebraic over of degree 38. Then every element U ∈ (U1,...,U=) can be represented as a polynomial in U1,...,U= over : Õ U = 2 UA1 ··· UA= A1,...,A= 1 = A1,...,A= 2 A , . . . , 3 for some A1,...,A= ∈ , where 8 ranges over 0 8 − 1. Proof By induction on =. When = = 0, this is trivial. Now let = ≥ 1 and suppose the result holds for = − 1. Let U ∈ (U1,...,U=) = (U1,...,U=−1) (U=).

By Theorem 5.1.5 applied to the extension ( (U1,...,U=−1))(U=) : (U ,...,U ) (U ) ≤ (U ) = 3 1 =−1 , noting that deg (U1,...,U=−1) = deg = =, we have

3=−1 U Õ 2 UA = A = (5.1) A=0 2 , . . . , 2 U ,...,U A for some 0 3=−1 ∈ ( 1 =−1). By inductive hypothesis, for each we have Õ 2 = 2 UA1 ··· UA=−1 A A1,...,A=−1,A 1 =−1 (5.2) A1,...,A=−1 2 A , . . . , 3 for some A1,...,A=−1,A ∈ , where 8 ranges over 0 8 − 1. Substituting (5.2) into (5.1) completes the induction.

55 Example 5.1.12 Back in Example 4.1.11(ii), I claimed that √ √ √ Q( 2, 8) = {0 + 1 2 + 28 + 3 28 : 0, 1, 2, 3 ∈ Q}. √ √ , 8 8 Corollary 5.1.11 applied to Q( 2 ) : Q proves this, since degQ( 2) = degQ( ) = 2.

Exercise 5.1.13 Let " : be a ﬁeld extension and U a transcendental element of ". Can every element of (U) be represented as a polynomial in U over ?

5.2 The tower law

We now know about the degrees of simple extensions—those obtained by adjoining a single element. What about extensions obtained by adjoining several elements? The following result is invaluable.

Theorem 5.2.1 (Tower law) Let " : ! : be ﬁeld extensions.

i. If (U8)8∈ is a basis of ! over and (V 9 )9∈ is a basis of " over !, then (U8 V 9 )(8, 9)∈× is a basis of " over . ii. " : is ﬁnite ⇐⇒ " : ! and ! : are ﬁnite.

iii. [" : ] = [" : !][! : ].

The sets and here could be infinite. I’ll say that a family (08)8∈ of elements of a field is finitely supported if the set {8 ∈ : 08 ≠ 0} is finite.

Proof To prove (i), we show that (U8 V 9 )(8, 9)∈× is a linearly independent spanning set of " over . For linear independence, let (28 9 )(8, 9)∈× be a finitely supported family of ele- Í 2 U V Í Í 2 U V Í 2 U ! ments of such that 8, 9 8 9 8 9 = 0. Then 9 ( 8 8 9 8) 9 = 0, with 8 8 9 8 ∈ 9 V ! Í 2 U for each ∈ . Since ( 9 )9∈ is linearly independent over , we have 8 8 9 8 = 0 for each 9 ∈ . But (U8)8∈ is linearly independent over , so 28 9 = 0 for each 8 ∈ and 9 ∈ . To show (U8 V 9 )(8, 9)∈× spans " over , let 4 ∈ ". Since (V 9 )9∈ spans " over ! 4 Í 3 V 3 , we have = 9 9 9 for some finitely supported family ( 9 )9∈ of elements of ! U ! 9 3 Í 2 U . Since ( 8)8∈ spans over , for each ∈ we have 9 = 8 8 9 8 for some 2 4 Í 2 U V finitely supported family ( 8 9 )8∈ of . Hence = 8, 9 8 9 8 9 , as required. Parts (ii) and (iii) follow.

56 √ √ Example 5.2.2 What is [Q( 2, 3) : Q]? The tower law gives √ √ √ √ √ √ Q( 2, 3) : Q = Q( 2, 3) : Q( 2) Q( 2) : Q √ √ √ = 2Q( 2, 3) : Q( 2).

We have √ √ √ √ Q( 2, 3) : Q( 2) ≤ Q( 3) : Q = 2 √ √ √ √ √ , by Corollary√ √ 5.1.9(ii√). On the other hand, 3 ∉ Q( 2), so√Q(√2 3) ≠√ Q( 2), , > , so [Q( 2 3) : Q( 2)]√ √1 by Example 5.1.3(i). So [Q( 2 3) : Q( 2)] = 2, , giving the answer: [Q( 2 3) : Q] = 4. √ √ √ √ √ , , , , By the same argument as in Example √5.1.12√ , {1 2 3 6} spans Q( 2 3) , over Q. But we have just shown that Q( 2 3) has dimension√ 4√over Q. Hence this spanning set is a basis. That is, for every element U ∈ Q( 2, 3), there is one and only one 4-tuple (0, 1, 2, 3) of rational numbers such that √ √ √ U = 0 + 1 2 + 2 3 + 3 6. √ √ Exercise 5.2.3 In that example, I claimed that 3 ∉ Q( 2). Prove it.

Corollary 5.2.4 Let " : !0 : ! : be ﬁeld extensions. If " : is ﬁnite then [!0 : !] divides [" : ]. Proof Apply the tower law to " : !0 : ! then " : ! : . That result might remind you of Lagrange’s theorem on group orders. The resemblance is no coincidence, as we’ll see.

Exercise 5.2.5 Show that a ﬁeld extension whose degree is a prime number must be simple.

That result might remind you of the fact that a group of prime order must be cyclic, and that’s no coincidence either! A second corollary of the tower law:

Corollary 5.2.6 Let " : be a ﬁeld extension and U1,...,U= ∈ ". Then

[ (U1,...,U=) : ] ≤ [ (U1) : ]···[ (U=) : ]. Proof By the tower law and then Corollary 5.1.9(ii),

[ (U1,...,U=) : ] = [ (U1,...,U=) : (U1,...,U=−1)]···[ (U1,U2) : (U1)][ (U1) : ] ≤ [ (U=) : ]···[ (U2) : ][ (U1) : ].

57 ﬁnitely algebraic generated ﬁnite

Figure 5.2: Finiteness conditions on a ﬁeld extension

Example 5.2.7 What is [Q(121/4, 61/15) : Q]? You can check (hint, hint) that 1/4 1/15 1/4, 1/15 degQ(12 ) = 4 and degQ(6 ) = 15. So by Corollary 5.2.4, [Q(12 6 ) : Q] is divisible by 4 and 15. But also, Corollary 5.2.6 implies that [Q(121/4, 61/15) : Q] ≤ 4 × 15 = 60. Since 4 and 15 are coprime, the answer is 60.

Exercise 5.2.8 Generalize Example 5.2.7. In other words, what general result does the argument of Example 5.2.7 prove, not involving the particular numbers chosen there?

5.3 Algebraic extensions

We defined a field extension " : to be finite if [" : ] < ∞, that is, " is finite-dimensional as a vector space over . Here are two related conditions.

Definition 5.3.1 A field extension " : is finitely generated if " = (.) for some finite subset . ⊆ ".

Deﬁnition 5.3.2 A ﬁeld extension " : is algebraic if every element of " is algebraic over .

Recall from Corollary 5.1.7 that U is algebraic over if and only if (U) : is finite. So for a field extension to be algebraic is also a kind of finiteness condition.

Examples 5.3.3 i. For any field , the extension (C) : is finitely generated (take the ‘.’ above to be {C}) but not finite, by Corollary 5.1.7.

ii. In Section 4.2 you met the set Q of complex numbers algebraic over Q. We’ll very soon prove that it’s a subfield of C. It is algebraic over Q, by definition. But you’ll show in Workshop 3 that it is not finite over Q.

Our three ﬁniteness conditions are related as follows (Figure 5.2).

58 Proposition 5.3.4 The following conditions on a ﬁeld extension " : are equivalent:

i. " : is ﬁnite;

ii. " : is ﬁnitely generated and algebraic;

iii. " = (U1,...,U=) for some ﬁnite set {U1,...,U=} of elements of " algebraic over .

Proof (i)⇒(ii): suppose that " : is finite. To show that " : is finitely generated, take a basis U1,...,U= of " over . Every subfield ! of " containing is a -linear subspace of ", so if U1,...,U= ∈ ! then ! = ". This proves that the only subfield of " containing ∪ {U1,...,U=} is " itself; that is, " = (U1,...,U=). So " : is finitely generated. To show that " : is algebraic, let U ∈ ". Then by part (ii) of the tower law (Theorem 5.2.1), (U) : is finite, so by Corollary 5.1.7, U is algebraic over . (ii)⇒(iii) is immediate from the definitions. (iii)⇒(i): suppose that " = (U1,...,U=) for some U8 ∈ " algebraic over . Then [" : ] ≤ [ (U1) : ]···[ (U=) : ] by Corollary 5.2.6. For each 8, we have [ (U8) : ] < ∞ since U8 is algebraic over (using Corollary 5.1.7 again). So [" : ] < ∞.

We already saw that when " = (U1,...,U=) with each U8 algebraic, every element of " is a polynomial in U1,...,U= (Corollary 5.1.11). So for any ﬁnite extension " : , there is some ﬁnite set of elements such that everything in " can be expressed as a polynomial over in these elements.

Exercise 5.3.5 Let " : be a ﬁeld extension and ⊆ ! ⊆ ". In the proof of Proposition 5.3.4, I said that if ! is a subﬁeld of " then ! is a -linear subspace of ". Why is that true? And is the converse also true? Give a proof or a counterexample.

Corollary 5.3.6 Let (U) : be a simple extension. The following are equivalent:

i. (U) : is ﬁnite;

ii. (U) : is algebraic;

iii. U is algebraic over .

59 Proof (i)⇒(ii) follows from (i)⇒(ii) of Proposition 5.3.4. (ii)⇒(iii) is immediate from the deﬁnitions. (iii)⇒(i) follows from (iii)⇒(i) of Proposition 5.3.4.

Here’s a spectacular application of Corollary 5.3.6. Proposition 5.3.7 Q is a subﬁeld of C.

Proof By Corollary 5.3.6,

Q = {U ∈ C : [Q(U) : Q] < ∞}.

For all U,V ∈ Q,

[Q(U,V) : Q] ≤ [Q(U) : Q][Q(V) : Q] < ∞ by Corollary 5.2.6. Hence

[Q(U + V) : Q] ≤ [Q(U,V) : Q] < ∞, giving U + V ∈ Q. Similarly, U · V ∈ Q. For all U ∈ Q,

[Q(−U) : Q] = [Q(U) : Q] < ∞, giving −U ∈ Q. Similarly, 1/U ∈ Q (if U ≠ 0). And clearly 0, 1 ∈ Q.

If you did Exercise 4.2.4, you’ll appreciate how hard that result is to prove from ﬁrst principles, and how amazing it is that the proof above is so clean and simple.

5.4 Ruler and compass constructions

The ancient Greeks developed planar geometry to an extraordinary degree, dis- covering how to perform a very wide range of constructions using only ruler and compasses. But there were three particular constructions that they couldn’t ﬁgure out how to do using only these instruments: • Trisect the angle: given an angle \, construct the angle \/3. • Duplicate the cube: given a length, construct a new length whose cube is ! twice the cube of the original.√ That is, given two points distance apart, construct two points distance 3 2! apart. • Square the circle: given a circle, construct a square with the same area. ! √That is, given two points distance apart, construct two points distance c! apart.

60 The challenge of finding constructions lay unanswered for thousands of years. And it wasn’t for lack of attention: mathematicians kept on looking. My Galois theory lecture notes from when I was an undergraduate contain the following words: Thomas Hobbes claimed to have solved these. John Wallis disagreed. A 17th century pamphlet war ensued. Twitter users may conclude that human nature has not changed. It turns out that the reason why no one could find a way to do these constructions is that they’re impossible. We’ll prove it using field theory. In order to prove that you can’t do these things using ruler and compasses, it’s necessary to know that you can do certain other things using ruler and compasses. I’ve made a video showing how to do the various constructions we’ll need. Ruler and compass constructions Digression 5.4.1 The standard phrase is ‘ruler and compass constructions’, but it’s slightly misleading. A ruler has distance markings on it, whereas for the problems of ancient Greece, you’re supposed to use only a ‘straight edge’: a ruler without markings (and no, you’re not allowed to mark it). As Stewart explains (Section 7.1), with a marked or markable straight edge, you can solve all three problems. Also, for what it’s worth, an instrument for drawing circles is strictly speaking a pair of compasses. But like everyone else, we’ll say ‘ruler and compass’—

—when we really mean ‘straight edge and compasses’—

The problems as stated above are maybe not quite precise; let’s formalize them. Starting from a subset Σ of the plane, our instruments allow the following constructions: • given two distinct points , of Σ, draw the (inﬁnite) line through and ; • given two distinct points , of Σ, draw the circle with centre passing through .

61 A point in the plane is immediately constructible from Σ if it is a point of intersection between two distinct lines, or two distinct circles, or a line and a circle, of the form above. A point in the plane is constructible from Σ if there is a finite sequence 1, . . . , = = of points such that 8 is immediately constructible from Σ ∪ {1, . . . , 8−1} for each 8. So far I have written in the Greek spirit by saying ‘the plane’ rather than R2. But now fix a coordinate system. For Σ ⊆ R2, write Σ = Q {U ∈ R : U is a coordinate of some point in Σ} , which is a subfield of R. The condition on U means that (U,H) ∈ Σ for some H ∈ R or (G,U) ∈ Σ for some G ∈ R. The key to the impossibility proofs is the following definition. For subfields ⊆ " ⊆ R, let us say that " : is an iterated quadratic extension if there is some finite sequence of subfields

= 0 ⊆ 1 ⊆ · · · ⊆ = = " such that [ 8 : 8−1] = 2 for all 8 ∈ {1, . . . , =}.

Theorem 5.4.2 Let Σ ⊆ R2 and (G,H) ∈ R2. If (G,H) is constructible from Σ then there is an iterated quadratic extension of Σ containing G and H.

Before I show you the proof, I’ll state a corollary that reveals how this theorem leads to proofs of impossibility.

Corollary 5.4.3 Let Σ ⊆ R2 and (G,H) ∈ R2. If (G,H) is constructible from Σ then G and H are algebraic over Σ, and their degrees over Σ are powers of 2.

Proof Take an iterated quadratic extension " of Σ with G ∈ ". Then [" : ] = = = ≥ (G) = [ (G) ] Σ 2 for some 0, by the tower law. But then deg Σ Σ : Σ divides 2= by Corollary 5.2.4, and is therefore a power of 2. And similarly for H.

We will show that if (for instance) we could trisect angles, we would be able to construct a point whose coordinates do not have degree a power of 2, giving a contradiction. To prove Theorem 5.4.2, we need a lemma.

Lemma 5.4.4 Let be a subﬁeld of R and U,V ∈ R. Suppose that U and V are each contained in some iterated quadratic extension of . Then there is some iterated quadratic extension of containing both U and V.

62 Proof Take subﬁelds

= 0 ⊆ 1 ⊆ · · · ⊆ = ⊆ R, = !0 ⊆ !1 ⊆ · · · ⊆ !< ⊆ R with U ∈ =, V ∈ !<, and [ 8 : 8−1] = 2 = [! 9 : ! 9−1] for all 8, 9. For each 9 ∈ {1, . . . , <}, choose some V 9 ∈ ! 9 \ ! 9−1; then ! 9 = ! 9−1(V 9 ). Hence ! 9 = (V1,...,V 9 ) for each 9. Now consider the chain of subﬁelds

= 0 ⊆ 1 ⊆ · · · ⊆ = ⊆ = (V1) ⊆ = (V1,V2) ⊆ · · · ⊆ = (V1,...,V<). (5.3)

For each 9 ∈ {1, . . . , <}, Corollary 5.1.9(i) gives the inequality

[ = (V1,...,V 9 ) : = (V1,...,V 9−1)] ≤ [ (V1,...,V 9 ) : (V1,...,V 9−1)] = [! 9 : ! 9−1] = 2.

So in the chain of subﬁelds (5.3), each successive extension has degree 1 or 2. An extension of degree 1 is an equality, so by ignoring repeats, we see that = (V1,...,V<) is an iterated quadratic extension of . Finally,

U ∈ = ⊆ = (V1,...,V<),V ∈ !< = (V1,...,V<) ⊆ = (V1,...,V<), so U,V ∈ = (V1,...,V<), as required.

Exercise 5.4.5 In the second paragraph of the proof, I claimed that ! 9 = ! 9−1(V 9 ). The general principle here is that if " : is a ﬁeld extension of degree 2 and W ∈ " \ then " = (W). Prove this.

Proof of Theorem 5.4.2 Suppose that (G,H) is constructible from Σ in = steps. If = = 0 then (G,H) ∈ Σ, so G,H ∈ Σ, which is trivially an iterated quadratic extension of Σ. Now let = ≥ 1, and suppose inductively that each coordinate of each point of R2 constructible from Σ in < = steps lies in some iterated quadratic extension of Σ. By deﬁnition, (G,H) is an intersection point of two distinct lines/circles through points constructible in < = steps. By inductive hypothesis, each coordinate of each of those points lies in some iterated quadratic extension of Σ, so by Lemma 5.4.4, there is an iterated quadratic extension ! of Σ containing all the points’ coordinates. The coeﬃcients in the equations of the lines/circles then also lie in !. G , We now show that deg! ( ) ∈ {1 2}.

63 If (G,H) is the intersection point of two distinct lines, then G and H satisfy two linearly independent equations 0G + 1H + 2 = 0, 00G + 10H + 20 = 0 with 0, 1, 2, 00, 10, 20 ∈ !. Solving gives G ∈ !. If (G,H) is an intersection point of a line and a circle, then 0G + 1H + 2 = 0, G2 + H2 + 3G + 4H + 5 = 0 with 0,1,2,3,4, 5 ∈ !. If 1 = 0 then 0 ≠ 0 and G = −2/0 ∈ !. Otherwise, we H ! G G , can eliminate to give a quadratic over satisﬁed by , so that deg! ( ) ∈ {1 2}. If (G,H) is an intersection point of two circles, then G2 + H2 + 3G + 4H + 5 = 0, G2 + H2 + 30G + 40H + 5 0 = 0 with 3, 4, 5 , 30, 40, 5 0 ∈ !. Subtracting, we reduce to the case of a line and a circle, G , again giving deg! ( ) ∈ {1 2}. G , G G ! Hence deg! ( ) ∈ {1 2}. If deg! ( ) = 1 then ∈ , which is an iterated G ! G ! ! G quadratic extension of Σ. If deg! ( ) = 2, i.e. [ ( ) : ] = 2, then ( ) is an iterated quadratic extension of Σ. In either case, G lies in some iterated quadratic extension of Σ. The same is true of H. Hence by Lemma 5.4.4, there is an iterated quadratic extension of Σ containing G and H. This completes the induction. Now we solve the problems of ancient Greece. In all cases, we take Σ = {(0, 0), (1, 0)}. Then Σ = Q. Proposition 5.4.6 The angle cannot be trisected by ruler and compass. Proof Suppose it can be. Construct an equilateral triangle with (0, 0) and (1, 0) as two of its vertices (which can be done by ruler and compass; Figure 5.3). Trisect the angle of the triangle at (0, 0). Plot the point (G,H) where the trisector meets the circle with centre (0, 0) through (1, 0). Then G = cos(c/9), so by Corollary 5.4.3, c degQ(cos( /9)) is a power of 2. Now we use the trigonometric formula cos 3\ = 4(cos \)3 − 3 cos \. Taking \ = c/9 and using cos(c/3) = 1/2, we get 8G3 − 6G − 1 = 0. Reduced mod 5, this cubic has no roots and is therefore irreducible (by Lemma 3.3.1(iv)). So by the mod ? test, 8C3 − 6C − 1 is irreducible over Q. Hence (8C3 − 6C − 1)/8 is the G G minimal polynomial of over Q, giving degQ( ) = 3. Since 3 is not a power of 2, this is a contradiction.

64 (G,H) (0, 0)(1, 0)

Figure 5.3: The impossibility of trisecting 60◦.

Proposition 5.4.7 The cube cannot be duplicated by ruler and compass. , , Proof Suppose it can be. Since (0 0) and (1 0√) are distance 1 apart, we can 3 construct from them two points and distance√ 2 apart. From and we can 3 , construct, using ruler and compass, the point√ ( 2 0). (The video shows how to ‘transport distances’ like this.) So deg ( 3 2) is a power of 2, by Corollary 5.4.3. √ Q 3 But degQ( 2) = 3 by Example 5.1.8(ii), a contradiction. Proposition 5.4.8 The circle cannot be squared by ruler and compass. This one is the most outrageously false, yet the hardest to prove. , , Proof Suppose it can be. Since the circle with centre (0 0) through (1 0) √has c c area , we can construct by ruler and compass a square with side-length√ , c, and from that, we√ can construct by ruler and compass the point ( 0). So by Corollary 5.4.3, c is algebraic over Q with degree a power of 2. Since Q is a subﬁeld of C, it follows that c is algebraic over Q. But it is a (hard) theorem that c is transcendental over Q.

Digression 5.4.9 Stewart has a nice alternative approach to all this, in his Chapter 7. He treats the plane as the complex plane, and he shows that the set of all points in C constructible from 0 and 1 is a subﬁeld. In fact, it is the smallest subﬁeld of C closed under taking square roots. He calls it Qpy, the ‘Pythagorean closure’ of Q. It can also be described as the set of complex numbers contained in some iterated quadratic extension of Q.

There is one more famous ruler and compass problem: for which integers = is the regular =-sided polygon constructible, starting from just a pair of points in the plane? The answer has to do with Fermat primes, which are prime numbers of the form 2D + 1 for some D ≥ 1. A little exercise in number theory shows that if 2D + 1 is prime then D must itself be a power of 2. The only known Fermat primes are

0 1 2 3 4 22 + 1 = 3, 22 + 1 = 5, 22 + 1 = 17, 22 + 1 = 257, 22 + 1 = 65537.

65 Whether there are any others is a longstanding open question. In any case, it can be shown that the regular =-sided polygon is constructible if and only if

A = = 2 ?1 ··· ?: for some A,: ≥ 0 and distinct Fermat primes ?1,...,?: . We will not do the proof, but it involves cyclotomic polynomials. A glimpse of the connection: let ? be a prime such that the regular ?-sided polygon is constructible. Consider the regular ?-sided polygon inscribed in the unit circle in C, with one of its vertices at 1. Then another vertex is at 42c8/?, and from 42c8/? constructibility, one can deduce that degQ( ) is a power of 2. But we saw in 42c8/? ? ? ? Example 3.3.16 that degQ( ) = − 1. So − 1 is a power of 2, that is, is a Fermat prime. Galois theory, number theory and Euclidean geometry come together!

66 Chapter 6

Splitting ﬁelds

In Chapter1, we met a definition of the symmetry group of a polynomial over Q. It was phrased in terms of indistinguishable tuples, it was possibly a little mysterious, and it was definitely difficult to work with (e.g. we couldn’t compute the symmetry group of 1 + C + C2 + C3 + C4). Introduction to In this chapter, we’re going to give a different but equivalent definition of the Week 6 symmetry group of a polynomial. It’s a two-step process: 1. We show how every polynomial 5 over gives rise to an extension of , called the ‘splitting field’ of 5 .

2. We show how every field extension has a symmetry group. The symmetry group, or ‘Galois group’, of a polynomial is then defined to be the symmetry group of its splitting field extension. How does these two steps work? 1. When = Q, the splitting field of 5 is the smallest subfield of C containing all the complex roots of 5 . For a general field , it’s constructed by adding the roots of 5 one at a time, using simple extensions, until we obtain an extension of in which 5 splits into linear factors.

2. The symmetry group of a field extension " : is defined as the group of automorphisms of " over . This is the same idea you’ve seen many times before, for symmetry groups of other mathematical objects. Why bother? Why not define the symmetry group of 5 directly, as in Chapter1? • Because this strategy works over every field , not just Q.

• Because there are ﬁeld extensions that do not arise from a polynomial, and their symmetry groups are sometimes important. For example, an important

67 structure in number theory, somewhat mysterious to this day, is the symmetry group of the algebraic numbers Q over Q.

• Because using abstract algebra means you can cut down on explicit calculations with polynomials. (By way of analogy, you’ve seen how abstract linear algebra with vector spaces and linear maps allows you to cut down on calculations with matrices.) It also makes connections with other parts of mathematics more apparent.

6.1 Extending homomorphisms

In your degree so far, you’ll have picked up the general principle that for many kinds of mathematical object (such as groups, rings, fields, vector spaces, modules, metric spaces, topological spaces, measure spaces, . . . ), it’s important to consider the appropriate notion of mapping between them (such as homomorphisms, linear maps, continuous maps, . . . ). And since Chapter4, you know that the basic objects of Galois theory are field extensions. So it’s no surprise that sooner or later, we’ll have to think about mappings from one field extension to another. That moment is now: we’ll need what’s in this section in order to establish the basic facts about splitting fields. When we think about a field extension " : , we generally regard the field as our starting point and " as a field that extends it. (This doesn’t mean anything rigorous.) Similarly, we might start with a homomorphism k : → 0 between fields, together with extensions " of and "0 of 0, and look for a homomorphism " → "0 that extends k. The language is as follows.

Definition 6.1.1 Let ]: → " and ]0 : 0 → "0 be field extensions. Let k : → 0 be a homomorphism of fields. A homomorphism i : " → "0 extends k if the square i Extension problems " / "0 O O ] ]0 / 0 k commutes (that is, i ◦ ] = ]0 ◦ k).

Here I’ve used the definition of a field extension as a homomorphism ] of fields (Definition 4.1.1). Most of the time we view as a subset of " and 0 as a subset of "0, with ] and ]0 being the inclusions. In that case, for i to extend k just means that i(0) = k(0) for all 0 ∈ .

68 The basic questions are: given the two field extensions and the homomorphism k, is there some i that extends k? If so, how many? We’ll consider these questions later. This section simply gathers together three general results about extensions of field homomorphisms. Recall that any ring homomorphism k : ' → ( induces a homomorphism k∗ : '[C] → ([C] (Definition 3.1.6). Lemma 6.1.2 Let " : and "0 : 0 be field extensions, let k : → 0 be a homomorphism, and let i : " → "0 be a homomorphism extending k. Let Explanation of U ∈ " and 5 (C) ∈ [C]. Then Lemma 6.1.2 5 (U) = 0 ⇐⇒ k∗( 5 ) i(U) = 0. 5 C Í 0 C8 0 k 5 C Í k 0 C8 Proof Write ( ) = 8 8 , where 8 ∈ . Then ( ∗( ))( ) = 8 ( 8) ∈ 0[C], so

Õ 8 Õ 8 (k∗( 5 ))(i(U)) = k(08)i(U) = i(08)i(U) = i( 5 (U)), 8 8 where the second equality holds because i extends k. Since i is injective (Lemma 2.2.2), the result follows.

Exercise 6.1.3 Show that if k is injective then so is k∗, and if k is an isomorphism then so is k∗.

Lemma 6.1.4 Let " : and "0 : 0 be ﬁeld extensions, let k : → 0 be an isomorphism, and let i : " → "0 be a homomorphism extending k. Let U ∈ " be algebraic over with minimal polynomial <. Then i(U) is algebraic over 0 with minimal polynomial k∗(<).

0 Proof By Lemma 6.1.2, k∗(<) is an annihilating polynomial of i(U) over . 0 Also k∗(<) ≠ 0, since < ≠ 0 and k∗ is injective. So i(U) is algebraic over . (Recall that an element is algebraic if it has a nonzero annihilating polynomial.) 0 Since k∗ : [C] → [C] is an isomorphism and < ∈ [C] is irreducible, 0 k∗(<) ∈ [C] is irreducible. It is also monic. Hence k∗(<) is a monic irreducible annihilating polynomial of i(U), so it is the minimal polynomial of i(U). An isomorphism between fields, rings, groups, vector spaces, etc., can be understood as simply a renaming of the elements. For example, if I tell you that the ring ' is left Noetherian but not right Artinian, and that ( is isomorphic to ', then you can deduce that ( is left Noetherian but not right Artinian without having the slightest idea what those terms mean. Just as long as they don’t depend on the names of the elements of the ring concerned (which such definitions never do), you’re fine.

69 Proposition 6.1.5 Let k : → 0 be an isomorphism of ﬁelds, let (U) : be a simple extension where U has minimal polynomial < over , and let 0(U0) : 0 0 0 be a simple extension where U has minimal polynomial k∗(<) over . Then there is a unique isomorphism i : (U) → 0(U0) that extends k and satisﬁes i(U) = U0.

Diagram: i (U) / 0(U0) O O

 / 0 k We often use a dotted arrow to denote a map whose existence is part of the conclusion of a theorem, rather than a hypothesis.

k Proof View 0(U0) as an extension of via the composite homomorphism −→ 0 → 0(U0). Then the minimal polynomial of U0 over is <. (If this isn’t intuitively clear to you, think of the isomorphism k as renaming.) Hence by the classification of simple extensions, Theorem 4.3.7, there is an isomorphism i : (U) → 0(U0) over such that i(U) = U0. Then i extends k. i U It only remains to prove uniqueness. Let e be any homomorphism ( ) → 0 U0 k i U U0 i 0 k 0 i 0 ( ) that extends and satisfies e( ) = . Then e( ) = ( ) = ( ) for all 0 i U U0 i U U U ∈ and e( ) = = ( ). Since every element of ( ) is a polynomial in i i with coefficients in , it follows that e = .

6.2 Existence and uniqueness of splitting ﬁelds

Let 5 be a polynomial over a ﬁeld . Informally, a splitting ﬁeld for 5 is an extension of where 5 has all its roots, and which is no bigger than it needs to be.

Warning 6.2.1 If 5 is irreducible, we know how to create an extension of where 5 has at least one root: take the simple extension [C]/h 5 i, in which the equivalence class of C is a root of 5 (Lemma 4.3.4). But [C]/h 5 i is not usually a splitting ﬁeld for 5 . For example, take = Q and 5 (C) = C3 − 2, as in Warning 4.3.10. Write U for the real cube root of 2. (Half the counterexamples in Galois theory involve the real cube root of 2.) Then Q[C]/h 5 i is isomorphic to the subﬁeld Q(U) of R, which only contains one root of 5 : the other two are non-real, hence not in Q(U).

70 Deﬁnition 6.2.2 Let 5 be a polynomial over a ﬁeld ". Then 5 splits in " if

5 (C) = V(C − U1)···(C − U=) for some = ≥ 0 and V,U1,...,U= ∈ ".

Equivalently, 5 splits in " if all its irreducible factors in " [C] are linear.

Examples 6.2.3 i. A ﬁeld " is algebraically closed if and only if every polynomial over " splits in ". √ ii. Let 5 (C) = C4 − 4C2 − 5. Then 5 splits in Q(8, 5), since

5 (C) = (C2 + 1)(C2 − 5) √ √ = (C − 8)(C + 8)(C − 5)(C + 5).

But 5 does not split in Q(8), as its factorization into irreducibles in Q(8)[C] is 5 (C) = (C − 8)(C + 8)(C2 − 5), which contains a nonlinear factor. Moral: 5 may have one root or even several roots in ", but still not split in ".

2 iii. Let " = F2(U), where U is a root of 5 (C) = 1+C +C , as in Example 4.3.8(ii). We have

5 (1 + U) = 1 + (1 + U) + (1 + 2U + U2) = 1 + U + U2 = 0,

so 5 has two distinct roots in ", giving

5 (C) = (C − U)(C − (1 + U))

in " [C]. Hence 5 splits in ". In this example, adjoining one root of 5 gave us a second root for free. But this doesn’t typically happen (Warning 6.2.1).

Definition 6.2.4 Let 5 be a polynomial over a field .A splitting field of 5 over is an extension " of such that:

i. 5 splits in ";

ii. " = (U1,...,U=), where U1,...,U= are the roots of 5 in ".

71 Exercise 6.2.5 Show that (ii) can equivalently be replaced by: ‘if ! is a subﬁeld of " containing , and 5 splits in !, then ! = "’.

Examples 6.2.6 i. Let 5 ∈ Q[C]. Write U1,...,U= for the complex roots of 5 . Then Q(U1,...,U=) (the smallest subﬁeld of C containing U1,...,U=) is a splitting ﬁeld of 5 over Q.

ii. Let 5 (C) = C3 − 2 ∈ Q[C]. Its complex roots are U, lU and l2U, where U is the real cube root of 2 and l = 42c8/3. Hence a splitting ﬁeld of 5 over Q is

Q(U, lU, l2U) = Q(U, l).

U 5 l l Now degQ( ) = 3 as is irreducible, and degQ( ) = 2 as has minimal polynomial 1 + C + C2. By an argument like that in Example 5.2.7, it follows that [Q(U, l) : Q] = 6. On the other hand, [Q(U) : Q] = 3. So again, the extension we get by adjoining all the roots of 5 is bigger than the one we get by adjoining just one root of 5 .

2 iii. Take 5 (C) = 1 + C + C ∈ F2 [C], as in Example 6.2.3(iii). By Theorem 5.1.5, {1,U} is a basis of F2(U) over F2, so

F2(U) = {0, 1,U, 1 + U} = F2 ∪ {the roots of 5 in F2(U)}.

Hence F2(U) is a splitting ﬁeld of 5 over F2.

Exercise 6.2.7 In Example 6.2.6(ii), I said that Q(U, lU, l2U) = Q(U, l). Why is that true?

Digression 6.2.8 Splitting fields over Q are easy, as Example 6.2.6(i) shows. That’s because we have a ready-made algebraically closed field containing Q, namely, C. In fact, for any field , it’s possible to build an algebraically closed field containing . And there is a unique ‘smallest’ algebraically closed field containing , called its algebraic closure . For example, the algebraic closure of Q is Q, the subfield of C consisting of the algebraic numbers. (For a proof that Q is indeed algebraically closed, see Workshop 3.) Many texts on Galois theory include constructions of the algebraic closure of a field, but we won’t do it.

72 Our mission for the rest of this section is to show that every polynomial 5 has exactly one splitting field. So that’s actually two tasks: first, show that 5 has at least one splitting field, then, show that 5 has only one splitting field. The first task is easy, and in fact we prove a little bit more: Lemma 6.2.9 Let 5 be a polynomial over a field . Then there exists a splitting field " of 5 over such that [" : ] ≤ deg( 5 )!. (So that this lemma holds for 5 = 0, we had better define (−∞)! = 1.) Proof If 5 is constant then is a splitting field of 5 over , and the result holds trivially. Now suppose inductively that deg( 5 ) ≥ 1. We may choose an irreducible factor < of 5 . By Theorem 4.3.7, there is an extension (U) of with <(U) = 0. Then (C − U) | 5 (C) in (U)[C], giving a polynomial 6(C) = 5 (C)/(C − U) over (U). We have deg(6) = deg( 5 ) − 1, so by inductive hypothesis, there is a splitting field " of 6 over (U) with [" : (U)] ≤ deg(6)!. Then " is a splitting field of 5 over . (Check that you understand why.) Also, by the tower law, [" : ] = [" : (U)][ (U) : ] ≤ (deg( 5 ) − 1)! · deg(<) ≤ deg( 5 )!, completing the induction. Proving that every polynomial has only one splitting field is harder. As ever, ‘only one’ has to be understood up to isomorphism: after all, if you’re given a splitting field, you can always rename its elements to get an isomorphic copy that’s not literally identical to the original one. But isomorphism is all that matters. Our proof of the uniqueness of splitting fields depends on the following result, which will also be useful for other purposes as we head towards the fundamental theorem of Galois theory. Proposition 6.2.10 Let k : → 0 be an isomorphism of fields, let 5 ∈ [C], let 0 0 " be a splitting field of 5 over , and let " be a splitting field of k∗( 5 ) over . Then: i. there exists an isomorphism i : " → "0 extending k; ii. there are at most [" : ] such extensions i. Diagram: i " / "0 O O

/ 0 k

73 Proof We prove both statements by induction on deg( 5 ). If 5 is constant then both ﬁeld extensions are trivial, so there is exactly one isomorphism i extending k. Now suppose that deg( 5 ) ≥ 1. We can choose a monic irreducible factor < of 5 . Then < splits in " since 5 does and < | 5 ; choose a root U ∈ " of <. We have 5 (U) = 0, so (C − U) | 5 (C) in (U)[C], giving a polynomial 6(C) = 5 (C)/(C − U) over (U). Then " is a splitting ﬁeld of 6 over (U), and deg(6) = deg( 5 ) − 1. 0 Also, k∗(<) splits in " since k∗( 5 ) does and k∗(<) | k∗( 5 ). Write U0 ,...,U0 k < "0 1 B for the distinct roots of ∗( ) in . Note that

1 ≤ B ≤ deg(k∗(<)) = deg(<). (6.1)

Counting Since k∗ is an isomorphism, k∗(<) is monic and irreducible, and is therefore the isomorphisms: the U0 9 ,...,B minimal polynomial of 9 for each ∈ {1 }. Hence by Proposition 6.1.5, proof of 9 \ (U) → 0(U0 ) k Proposition 6.2.10 for each , there is a unique isomorphism 9 : 9 that extends and \ U U0 satisﬁes 9 ( ) = 9 . (See diagram below.) For each 9 ∈ {1,...,B}, we have a polynomial

\ ∗( 5 ) k ( 5 ) \ (6) = 9 = ∗ 9∗ \ C U C U0 9∗( − ) − 9 0 U0 "0 k 5 0 "0 over ( 9 ), and is a splitting ﬁeld of ∗( ) over , so is also a splitting \ 6 0 U0 ﬁeld of 9∗( ) over ( 9 ). To prove that there is at least one isomorphism i extending k, choose any 9 ∈ {1,...,B} (as we may since B ≥ 1). By applying the inductive hypothesis to 6 and \ 9 , there is an isomorphism i extending \ 9 :

i " / "0 O O

U \ 9 0 U0 ( ) / ( 9 ) O O

/ 0 k

But then i also extends k, as required. To prove there are at most [" : ] isomorphisms i : " → "0 extending k, first note that any such i satisfies (k∗( 5 ))(i(U)) = 0 (by Lemma 6.1.2), so i U U0 9 ,...,B U ( ) = 9 for some ∈ {1 }. Every element of ( ) is a polynomial in U over , and i(0) = k(0) ∈ 0 for all 0 ∈ , so i maps (U) into 0 U0 i U 0 k U0 ( 9 ). Now ( ( )) contains (since is an isomorphism) and 9 , so i U 0 U0 i ( ( )) = ( 9 ). Since homomorphisms of fields are injective, restricts to U 0 U0 U U0 an isomorphism ( ) → ( 9 ) satisfying 7→ 9 . By the uniqueness part of

74 Proposition 6.1.5, this restricted isomorphism must be \ 9 . Thus, i extends \ 9 for a unique 9 ∈ {1,...,B}, giving

B Õ (number of isos i extending k) = (number of isos i extending \ 9 ). 9=1

For each 9, the number of isomorphisms i extending \ 9 is ≤ [" : (U)], by inductive hypothesis. So, using the tower law and (6.1),

[" : ] (number of isos i extending k) ≤ B ·[" : (U)] = B · ≤ [" : ], deg(<) completing the induction.

Exercise 6.2.11 Why does the proof of Proposition 6.2.10 not show that there are exactly [" : ] isomorphisms i extending k? How could you strengthen the hypotheses in order to obtain that conclusion? (The second question is a bit harder, and we’ll see the answer next week.)

This brings us to the foundational result on splitting ﬁelds. Recall that an automorphism of an object - is an isomorphism - → -.

Theorem 6.2.12 Let 5 be a polynomial over a ﬁeld . Then:

i. there exists a splitting ﬁeld of 5 over ;

ii. any two splitting ﬁelds of 5 are isomorphic over ;

iii. when " is a splitting ﬁeld of 5 over ,

(number of automorphisms of " over ) ≤ [" : ] ≤ deg( 5 )!.

Proof Part (i) is immediate from Lemma 6.2.9, and part (ii) follows from Propo- 0 sition 6.2.10 by taking = and k = id . The ﬁrst inequality in (iii) follows 0 0 from Proposition 6.2.10 by taking = , " = " and k = id , and the second follows from Lemma 6.2.9.

Up to now we have been saying ‘a’ splitting field. Theorem 6.2.12 gives us the right to speak of the splitting field of a given polynomial 5 over a given field . We write it as SF ( 5 ). We finish with a left over lemma that will be useful later.

75 Lemma 6.2.13 i. Let " : ( : be field extensions, 5 ∈ [C], and . ⊆ ". Suppose that ( is the splitting field of 5 over . Then ((.) is the splitting field of 5 over (.).

ii. Let 5 be a polynomial over a field , and let ! be a subfield of SF ( 5 ) containing (so that SF ( 5 ) : ! : ). Then SF ( 5 ) is the splitting field of 5 over !.

Proof For (i), 5 splits in (, hence in ((.). Writing - for the set of roots of 5 in (, we have ( = (-) and so ((.) = (-)(.) = (- ∪ .) = (.)(-); that is, ((.) is generated over (.) by -. This proves (i), and (ii) follows by taking " = SF ( 5 ) and . = !.

6.3 The Galois group

What gives Galois theory its special flavour is the use of groups to study fields and polynomials. Here is the central definition.

Deﬁnition 6.3.1 The Galois group Gal(" : ) of a ﬁeld extension is the group of automorphisms of " over , with composition as the group operation.

Exercise 6.3.2 Check that this really does deﬁne a group. You’ll need the result of Exercise 4.3.6, for instance.

In other words, an element of Gal(" : ) is an isomorphism \ : " → " such that \(0) = 0 for all 0 ∈ .

Examples 6.3.3 i. What is Gal(C : R)? Certainly the identity is an automorphism of C over R. So is complex conjugation ^, as implicitly shown in the ﬁrst proof of Lemma 1.1.3. So {id, ^} ⊆ Gal(C : R). I claim that Gal(C : R) has no other elements. For let \ ∈ Gal(C : R). Then

(\(8))2 = \(82) = \(−1) = −\(1) = −1

as \ is a homomorphism, so \(8) = ±8. If \(8) = 8 then

\(0 + 18) = \(0) + \(1)\(8) = 0 + 18

for all 0, 1 ∈ R (since \ is an automorphism over R), giving \ = id. Similarly, if \(8) = −8 then \ = ^. So Gal(C : R) = {id, ^} 2.

76 ii. Let U be the real cube root of 2. For each \ ∈ Gal(Q(U) : Q), we have

(\(U))3 = \(U3) = \(2) = 2

and \(U) ∈ Q(U) ⊆ R, so \(U) = U. Every element of Q(U) can be expressed as a polynomial in U over Q (by Theorem 5.1.5), so \ = id. Hence Gal(Q(U) : Q) is trivial.

Exercise 6.3.4 Prove that Gal(Q(42c8/3) : Q) = {id, ^}, where ^(I) = I. (Hint: imitate Example 6.3.3(i).)

The Galois group of a polynomial is deﬁned to be the Galois group of its splitting ﬁeld extension:

Deﬁnition 6.3.5 Let 5 be a polynomial over a ﬁeld . The Galois group Gal ( 5 ) of 5 over is Gal(SF ( 5 ) : ).

So the deﬁnitions ﬁt together like this:

polynomial 7−→ ﬁeld extension 7−→ group.

We will soon prove that Deﬁnition 6.3.5 is equivalent to the deﬁnition of Galois group in Chapter1, where we went straight from polynomials to groups. Theorem 6.2.12(iii) says that

| Gal ( 5 )| ≤ [SF ( 5 ) : ] ≤ deg( 5 )!. (6.2)

In particular, Gal ( 5 ) is always a ﬁnite group. 2 Examples 6.3.6 i. GalQ(C + 1) = Gal(Q(8) : Q) = {id, ^} 2, where ^ is complex conjugation on Q(8). The second equality is proved by the same argument as in Example 6.3.3(i), replacing C : R by Q(8) : Q. Calculating the 5 C C2 C2 5 Galois group with ii. Let ( √) = ( + 1)( − 2). Then GalQ( ) is the group of automorphisms bare hands, part 1 8, of Q( 2) over Q. Similar arguments to those in Examples√ √ 6.3.3 show that every \ ∈ GalQ( 5 ) must satisfy \(8) = ±8 and \( 2) = ± 2, and that the two choices of sign determine \ completely. And one can show that all four choices are possible, so that | GalQ( 5 )| = 4. There are two groups of order Calculating the 5 Galois group with four, 4 and 2 × 2. But each element of GalQ( ) has order 1 or 2, so bare hands, part 2 GalQ( 5 ) is not 4, so GalQ( 5 ) 2 × 2. I’ve been sketchy with the details here, because it’s not really sensible to try to calculate Galois groups until we have a few more tools at our disposal. We start to assemble them now.

77 Figure 6.1: The Galois group = Gal ( 5 ) permutes the roots U8 of 5 . (Image adapted from @rowvector.)

In the examples so far, we’ve seen that if U is a root of 5 then so is \(U) for every \ ∈ Gal ( 5 ). This is true in general, and is best expressed in terms of group actions (Figure 6.1). In a slogan: the Galois group permutes the roots. Lemma 6.3.7 Let 5 be a polynomial over a ﬁeld . Let - be the set of roots of 5 in SF ( 5 ). Then there is an action of Gal ( 5 ) on - deﬁned by Gal ( 5 ) × - → - (6.3) (\,U) 7→ \(U).

Proof First, if \ ∈ Gal ( 5 ) and U ∈ - then \(U) ∈ -, by Lemma 6.1.2 with = 0, " = "0 and k = id. For the function (6.3) to be an action means that (i ◦ \)(U) = i(\(U)) and id(U) = U for all i, \ ∈ Gal ( 5 ) and U ∈ -, which are true by deﬁnition. The action of the An action of a group on a set - is essentially the same thing as a homomor- Galois group phism from to the group Sym(-) of bijections from - to -. (If we write 6G as f6 (G), then the homomorphism → Sym(-) is 6 7→ f6.) We now adopt this viewpoint in the case of Galois groups. Let 5 be a polynomial over a ﬁeld , and write U1,...,U: for the distinct roots of 5 in SF ( 5 ). For each 8 ∈ {1,...,:}, Lemma 6.3.7 implies that \(U8) = U 9 for a unique 9. Write 9 as f\ (8), so that \ U U . ( 8) = f\ (8) Then we have a function Γ: Gal ( 5 ) → ( : (6.4) \ 7→ f\, and it is straightforward to check that Γ is a homomorphism.

78 Exercise 6.3.8 What is the kernel of Γ, in concrete terms?

If you remember the definition of Galois group in Chapter1 (Definition 1.2.1), the mention of (: should have set your antennae tingling. There, we defined the Galois group as a certain subgroup of (: , namely, the one consisting of those permutations f for which the tuples

(U1,...,U: ), (Uf(1),...,Uf(:)) are indistinguishable. Let’s now make the definition of indistinguishability official, switch to the standard name (recall Warning 1.1.2), and generalize from Q to an arbitrary field.

Deﬁnition 6.3.9 Let " : be a ﬁeld extension, let : ≥ 0, and let (U1,...,U: ) and U0 ,...,U0 : " U ,...,U U0 ,...,U0 ( 1 : ) be -tuples of elements of . Then ( 1 : ) and ( 1 : ) are conjugate over if for all ? ∈ [C1,...,C: ], ? U ,...,U ? U0 ,...,U0 . ( 1 : ) = 0 ⇐⇒ ( 1 : ) = 0 In the case : = 1, we omit the brackets and say that U and U0 are conjugate to mean that (U) and (U0) are.

We show now that the two deﬁnitions of the Galois group of 5 are equivalent.

Proposition 6.3.10 Let 5 be a polynomial over a ﬁeld , with distinct roots U1,...,U: in SF ( 5 ). Deﬁne the group homomorphism Γ: Gal ( 5 ) → (: as in (6.4). Then Γ is injective, and its image is

{f ∈ (: : (U1,...,U: ) and (Uf(1),...,Uf(:)) are conjugate over }. (6.5)

In particular, (6.5) is a subgroup of (: isomorphic to Gal ( 5 ).

Proof To show that Γ is injective, let \ ∈ ker Γ. Then \(U8) = U8 for all 8. Now SF ( 5 ) = (U1,...,U: ), with each U8 algebraic over , so every element of SF ( 5 ) can be expressed as a polynomial in U1,...,U: over (by Corol- lary 5.1.11). Since \ : SF ( 5 ) → SF ( 5 ) ﬁxes each element of and each U8, it is the identity on all of SF ( 5 ). Thus, ker Γ is trivial, so Γ is injective. Now we prove that im Γ is the set (6.5). In one direction, let f ∈ im Γ. Then f = f\ for some \ ∈ Gal ( 5 ) (writing Γ(\) = f\, as before). For every ? ∈ [C1,...,C: ], ? U ,...,U ? \ U ,...,\ U \ ? U ,...,U , ( f\ (1) f\ (:)) = ( ( 1) ( : )) = ( ( 1 : ))

79 where the ﬁrst equality is by deﬁnition of f\ and the second is because \ is a homomorphism over . But \ is an isomorphism, so it follows that ? U ,...,U ? U ,...,U . ( f\ (1) f\ (:)) = 0 ⇐⇒ ( 1 : ) = 0 Hence f belongs to the set (6.5). In the other direction, let f be a permutation in (6.5). By Corollary 5.1.11, every element of SF ( 5 ) can be expressed as ?(U1,...,U: ) for some ? ∈ [C1,...,C: ]. Now for ?,@ ∈ [C1,...,C: ], we have

?(U1,...,U: ) = @(U1,...,U: ) ⇐⇒ ?(Uf(1),...,Uf(:)) = @(Uf(1),...,Uf(:)) (by applying Deﬁnition 6.3.9 of conjugacy with ? − @ as the ‘?’). So there is a well-deﬁned, injective function \ : SF ( 5 ) → SF ( 5 ) satisfying

\(?(U1,...,U: )) = ?(Uf(1),...,Uf(:)) (6.6) for all ? ∈ [C1,...,C: ]. Moreover, \ is surjective because f is a permutation, \(0) = 0 for all 0 ∈ (by taking ? = 0 in (6.6)), and \(U8) = Uf(8) for all 8 (by taking ? = C8 in (6.6)). You can check that \ is a homomorphism of ﬁelds. Hence \ ∈ Gal ( 5 ) with Γ(\) = f, proving that f ∈ im Γ. The ﬁnal sentence of the proposition follows because every injective group homomorphism W : → induces an isomorphism between and the subgroup im W of .

Exercise 6.3.11 I skipped two small bits in that proof: ‘\ is surjective because f is a permutation’ (why?), and ‘You can check that \ is a homomorphism of ﬁelds’. Fill the gaps.

Digression 6.3.12 As you may know, an action of a group on a set - is called faithful if the corresponding homomorphism → Sym(-) is injective. A more concrete way to say that is that the only element 6 ∈ that ﬁxes everything (6G = G for all G ∈ -) is the identity. Equivalently, if 6, ℎ ∈ and 6G = ℎG for all G then 6 = ℎ. Most actions that one meets in practice are faithful; those that aren’t involve a kind of redundancy.

It’s important in Galois theory to be able to move easily between fields. For example, you might start with a polynomial whose coefficients belong to one field , but later decide to consider the coefficients as belonging to some larger field !. Here’s what happens to the Galois group when you do that.

Corollary 6.3.13 Let ! : be a ﬁeld extension and 5 ∈ [C]. Then Gal! ( 5 ) embeds naturally as a subgroup of Gal ( 5 ).

80 The phrasing here is slightly vague: it means there is an injective homomorphism Gal! ( 5 ) → Gal ( 5 ), and there is such an obvious choice of homomorphism that we tend to regard Gal! ( 5 ) as being a subgroup of Gal ( 5 ). Proof This follows from Proposition 6.3.10 together with the observation that if two :-tuples are conjugate over !, they are conjugate over .

Example 6.3.14 Consider the Galois group of 5 (C) = (C2 + 1)(C2 − 2) over Q, R and C. In Example 6.3.6(ii), we saw that GalQ( 5 ) 2 × 2. Over R, the Galois group of 5 is the same as that of C2 + 1, since both 2 roots of C − 2 lie in R and are therefore preserved by elements of GalR( 5 ). So 2 GalR( 5 ) = GalR(C + 1) 2. Finally, GalC( 5 ) is trivial. Indeed, every polynomial 6 ∈ C[C] has trivial Galois group over C: for 6 splits in C, so SFC(6) = C, so SFC(6) : C is a trivial extension and so has trivial Galois group.

Corollary 6.3.15 Let 5 be a polynomial over a ﬁeld , with : distinct roots in SF ( 5 ). Then | Gal ( 5 )| divides :!.

Proof By Proposition 6.3.10, Gal ( 5 ) is isomorphic to a subgroup of (: , which has :! elements. The result follows from Lagrange’s theorem.

The inequalities (6.2) already gave us | Gal ( 5 )| ≤ deg( 5 )!. Corollary 6.3.15 improves on this in two respects. First, it gives us | Gal ( 5 )| ≤ :!, where : ≤ deg( 5 ) in all cases, and : < deg( 5 ) if 5 has repeated roots in its splitting field. A trivial example: if 5 (C) = C2 then : = 1 and deg( 5 ) = 2. Second, it tells us that | Gal ( 5 )| is not only less than or equal to :!, but a factor of it. Galois theory is about the interplay between field extensions and groups. In the next chapter, we’ll see that just as every field extension giving rise to a group of automorphisms (its Galois group), every group of automorphisms gives rise to a field extension. We’ll also go deeper into the different types of field extension: normal extensions (the mirror image of normal subgroups) and separable extensions (which have to do with repeated roots). All that will lead us towards the fundamental theorem of Galois theory.

81 Chapter 7

Preparation for the fundamental theorem

Very roughly, the fundamental theorem of Galois theory says that you can tell a lot about a field extension by looking at its Galois group. A bit more specifically, Introduction to it says that the subgroups and quotients of Gal(" : ), and their orders, give Week 7 us information about the subfields of " containing , and their degrees. For example, one part of the fundamental theorem is that [" : ] = | Gal(" : )|. The theorem doesn’t hold for all extensions, just those that are ‘nice enough’. Crucially, this includes splitting field extensions SFQ( 5 ) : Q of polynomials 5 over Q—the starting point of classical Galois theory. Let’s dip our toes into the water by thinking about why it might be true that [" : ] = | Gal(" : )|, at least for extensions that are nice enough. The simplest nontrivial extensions are the simple algebraic extensions, " = (U). Write < for the minimal polynomial of U over and U1,U2,...,UB for the distinct roots of < in ". For every element i of Gal(" : ), we have <(i(U)) = 0 by Lemma 6.1.2, and so i(U) = U 9 for some 9 ∈ {1,...,B}. On the other hand, for each 9 ∈ {1,...,B}, there is exactly one i ∈ Gal(" : ) such that i(U) = U 9 , by Proposition 6.1.5. So | Gal(" : )| = B. We know that [" : ] = deg(<). So [" : ] = | Gal(" : )| if and only if deg(<) is equal to B, the number of distinct roots of < in ". Certainly B ≤ deg(<). But are B and deg(<) equal? < " There are two reasons why√ they might not be. First, might not split in . U 3 < C C3 For√ instance, if =√Q and = 2 then ( ) = − 2, which has only one root in Q( 3 2), so | Gal(Q( 3 2) : Q)| = 1 < 3 = deg(<). An algebraic extension is called ‘normal’ if this problem doesn’t occur, that is, if the minimal polynomial of every element does split. That’s what Section 7.1 is about.

82 Second, we might have B < deg(<) because some of the roots of < in " are repeated. If they are, the number B of distinct roots will be less then deg(<). An algebraic extension is called ‘separable’ if this problem doesn’t occur, that is, if the minimal polynomial of every element has no repeated roots in its splitting field. That’s what Section 7.2 is about. If we take any finite extension " : (not necessarily simple) that is both normal and separable, then it is indeed true that | Gal(" : )| = [" : ]. And in fact, these conditions are enough to make the whole fundamental theorem work, as we’ll see next week. I hesitated before putting normality and separability into the same chapter, because you should think of them in quite different ways:

• Normality has a clear conceptual meaning (as I explain in a video), and its importance was recognized by Galois himself. Despite the name, most ﬁeld extensions aren’t normal; normality isn’t something to be taken for granted.

• In contrast, Galois never considered separability, because it holds automat- ically over Q (his focus), and in fact over any field of characteristic 0, as well as any finite field. It takes some work to find an extension that isn’t separable. You can view separability as more of a technicality.

There’s one more concept in this chapter: the ‘fixed field’ of a group of automorphisms (Section 7.3). Every Galois theory text I’ve seen contains at least one proof that makes you ask ‘how did anyone think of that?’ I would argue that the proof of Theorem 7.3.6 is the one and only truly ingenious argument in this course: maybe not the hardest, but the most ingenious. Your opinion may differ!

7.1 Normality

Deﬁnition 7.1.1 An algebraic ﬁeld extension " : is normal if for all U ∈ ", the minimal polynomial of U splits in ".

We also say " is normal over to mean that " : is normal.

Lemma 7.1.2 Let " : be an algebraic extension. Then " : is normal if and only if every irreducible polynomial over either has no roots in " or splits in ".

Put another way, normality means that any irreducible polynomial over with at least one root in " has all its roots in ".

Proof Suppose that " : is normal, and let 5 be an irreducible polynomial over . If 5 has a root U in " then the minimal polynomial of U is 5 /2, where 2 ∈

83 is the leading coeﬃcient of 5 . Since " : is normal, 5 /2 splits in ", so 5 does too. Conversely, suppose that every irreducible polynomial over either has no roots in " or splits in ". Let U ∈ ". Then the minimal polynomial of U has at least one root in " (namely, U), so it splits in ". √ √ 3 3 Examples 7.1.3 i. Consider Q( 2) √: Q. The minimal√ polynomial√ of 2 over C3 3 l 3 , l2 3 Q is − 2, whose roots√ in C are 2 ∈ R and 2 2 ∈ C \ R, where l 42c8/3 3 C3 = √ . Since Q(√2) ⊆ R, the minimal polynomial − 2 does not split in Q( 3 2). Hence Q( 3 2) is not normal over Q. √ Alternatively, using the equivalent condition in Lemma 7.1.2, Q( 3 2) : Q is C3 not normal√ because − 2 is an irreducible polynomial over Q that has a root in Q( 3 2) but does not split there. √ One way to think about the non-normality of Q( 3 2) : Q is as follows. The three roots of C3 −2 are indistinguishable (conjugate) over Q, since they have the same minimal polynomial. But if they’re indistinguishable, it would What does it mean be strange for an extension to contain some but not all of them, since that to be normal? would be making a distinction between√ elements that are supposed to be indistinguishable. In that sense, Q( 3 2) is ‘abnormal’.

ii. Let 5 be a polynomial over a ﬁeld . Then SF ( 5 ) : is always normal, as we shall see (Theorem 7.1.5).

iii. Every extension of degree 2 is normal (just as, in group theory, every subgroup of index 2 is normal). You’ll be asked to show this in Workshop 4, but you also know enough to do it now.

Exercise 7.1.4 What happens if you drop the word ‘irreducible’ from Lemma 7.1.2? Is it still true?

Normality of field extensions is intimately related to normality of subgroups, and conjugacy in field extensions is also related to conjugacy in groups. The video ‘What does it mean to be normal?’ explains both kinds of normality and conjugacy in intuitive terms. Here’s the first of our two theorems about normal extensions. It describes which extensions arise as splitting field extensions.

Theorem 7.1.5 Let " : be a ﬁeld extension. Then

" = SF ( 5 ) for some 5 ∈ [C] ⇐⇒ " : is ﬁnite and normal.

84 SF" (<) O

i " = (U1,...,U=) / (U1,...,U=,Y) O O

(X) / (Y) g \ 7

Figure 7.1: Maps used in the proof that splitting ﬁeld extensions are normal.

Proof For ⇐, suppose that " : is finite and normal. By finiteness, there is a basis U1,...,U= of " over , and each U8 is algebraic over (by Proposition 5.3.4). For each 8, let <8 be the minimal polynomial of U8 over ; then by normality, <8 splits in ". Hence 5 = <1<2 ··· <= ∈ [C] splits in ". The set of roots of 5 in " contains {U1,...,U=}, and " = (U1,...,U=), so " is generated over by the set of roots of 5 in ". Thus, " is a splitting field of 5 over . For ⇒, take 5 ∈ [C] such that " = SF ( 5 ). We may assume that 5 ≠ 0, since if 5 = 0 then " = , which is certainly finite and normal over . Write U1,...,U= for the roots of 5 in ". Then " = (U1,...,U=). Each U8 is algebraic over (since 5 ≠ 0), so by Proposition 5.3.4, " : is finite. We now show that " : is normal, which is the most substantial part of the proof (Figure 7.1). Let X ∈ ", with minimal polynomial < ∈ [C]. Certainly < splits in SF" (<), so to show that < splits in ", it is enough to show that every Y < (<) " Splitting field root of in SF" lies in . extensions are Since < is a monic irreducible annihilating polynomial of Y over , it is the normal minimal polynomial of Y over . Hence by Theorem 4.3.7, there is an isomorphism \ : (X) → (Y) over . Now observe that:

• " = SF (X) ( 5 ), by Lemma 6.2.13(ii);

• (U1,...,U=,Y) is a splitting ﬁeld of 5 over (Y), by Lemma 6.2.13(i) (taking the ‘"’, ‘(’ and ‘.’ there to be SF" ( 5 ), " and {Y});

• \∗( 5 ) = 5 , since 5 ∈ [C] and \ is a homomorphism over .

So by Proposition 6.2.10, there is some isomorphism i : " → (U1,...,U=,Y) extending \. Then i is an isomorphism over , since \ is.

85 Since X is in (U1,...,U=), it can be expressed as a polynomial in U1,...,U= over (by Corollary 5.1.11). Since i is a map over , it follows that i(X) is a polynomial in i(U1), . . . , i(U=) over . But i(X) = \(X) = Y; moreover, for each 8 we have 5 (i(U8)) = 0 (by Lemma 6.1.2 with k = id ) and so i(U8) ∈ {U1,...,U=}. Hence Y is a polynomial in U1,...,U= over , giving Y ∈ (U1,...,U=) = ", as required.

Corollary 7.1.6 Let " : ! : be ﬁeld extensions. If " : is ﬁnite and normal then so is " : !.

Proof Follows from Theorem 7.1.5 and Lemma 6.2.13(ii).

! Warning 7.1.7√ It does not follow√ that : is normal. For instance, 3 , 42c8/3 3 consider Q( 2 ) : Q( 2) : Q. The first field is the√ splitting field of C3 − 2 over Q, and therefore normal over Q, but Q( 3 2) is not (Example 7.1.3(i)).

Theorem 7.1.5 is the ﬁrst of two theorems about normality. The second is to do with the action of the Galois group of an extension.

Warning 7.1.8 By definition, the Galois group Gal(" : ) of an extension " : acts on ". But if " is the splitting field of some polynomial 5 over then the action of Gal(" : ) on " restricts to an action on the roots of 5 (a finite set), as we saw in Section 6.3. So there are two actions of the Galois group under consideration, one the restriction of the other. Both are important.

When a group acts on a set, a basic question is: what are the orbits? For Gal(" : ) acting on ", the answer is: the conjugacy classes of " over . Or at least, that’s the case when " : is ﬁnite and normal:

Proposition 7.1.9 Let " : be a ﬁnite normal extension and U,U0 ∈ ". Then

U and U0 are conjugate over ⇐⇒ U0 = i(U) for some i ∈ Gal(" : ).

Proof For ⇐, let i ∈ Gal(" : ) with U0 = i(U). Then U and U0 are conjugate 0 0 over , by Lemma 6.1.2 (with " = ", = and k = id ). For ⇒, suppose that U and U0 are conjugate over . Since " : is ﬁnite, both are algebraic over , and since they are conjugate over , they have the same minimal polynomial < ∈ [C]. By Theorem 4.3.7, there is an isomorphism \ : (U) → (U0) over such that \(U) = U0 (see diagram below).

86 By Theorem 7.1.5, " is the splitting ﬁeld of some polynomial 5 over . Hence " is also the splitting ﬁeld of 5 over both (U) and (U0), by Lemma 6.2.13(ii). Moreover, \∗( 5 ) = 5 since \ is a homomorphism over and 5 is a polynomial over . So by Proposition 6.2.10(i), there is an automorphism i of " extending \: i " / " O O

(U) / (U0) b \ <

Then i ∈ Gal(" : ) with i(U) = \(U) = U0, as required.

That result was about the action of Gal(" : ) on the ﬁeld ", but it has a powerful corollary involving the action of the Galois group on the roots of an irreducible polynomial 5 , when " = SF ( 5 ): Corollary 7.1.10 Let 5 be an irreducible polynomial over a ﬁeld . Then the action of Gal ( 5 ) on the roots of 5 in SF ( 5 ) is transitive.

Recall what transitive means, for an action of a group on a set -: for all G,G0 ∈ -, there exists 6 ∈ such that 6G = G0.

Proof Since 5 is irreducible, the roots of 5 in SF ( 5 ) all have the same minimal polynomial, namely, 5 divided by its leading coeﬃcient. So they are all conjugate over . Since SF ( 5 ) : is ﬁnite and normal (by Theorem 7.1.5), the result follows from Proposition 7.1.9.

Exercise 7.1.11 Show by example that Corollary 7.1.10 becomes false if you drop the word ‘irreducible’.

Example 7.1.12 Let 5 (C) = 1 + C + · · · + C ?−1 ∈ Q[C], where ? is prime. Since (1 − C) 5 (C) = 1 − C ?, the roots of 5 in C are l, l2, . . . , l?−1, where l = 42c8/?. By Example 3.3.16, 5 is irreducible over Q. Hence by Corollary 7.1.10, for each 8 8 ∈ {1,...,? − 1}, there is some i ∈ GalQ( 5 ) such that i(l) = l . This is spectacular! Until now, we’ve been unable to prove such things without a huge amount of explicit checking, which, moreover, only works on a case-by-case basis. For example, if you watched the video ‘Calculating Galois groups with bare hands, part 2’, you’ll have seen how much tedious calculation went into the single

87 case ? = 5, 8 = 2:

But the theorems we’ve proved make all this unnecessary. In fact, for each 8 ∈ {1,...,? − 1}, there’s exactly one element i8 of GalQ( 5 ) 8 ?−1 such that i8 (l) = l . For since SFQ( 5 ) = Q(l, . . . , l ) = Q(l), two elements of GalQ( 5 ) that take the same value on l must be equal. Hence

GalQ( 5 ) = {i1, . . . , i?−1}.

We’ll see later that GalQ( 5 ) ?−1.

3 3 Example 7.1.13 Let’s calculate = GalQ(C − 2). Since C − 2 has 3 distinct roots in C, it has 3 distinct roots in its splitting ﬁeld. By Proposition 6.3.10, is isomorphic to a subgroup of (3. Now acts transitively on the 3 roots, so it has at least 3 elements, so it is isomorphic to either 3 or (3. Since two of the roots are non-real complex conjugates, one of the elements of is complex conjugation, which has order 2. Hence 2 divides ||, forcing (3.

We now show how a normal ﬁeld extension gives rise to a normal subgroup. Whenever you meet a normal subgroup, you should immediately want to form the resulting quotient, so we do that too.

Theorem 7.1.14 Let " : ! : be ﬁeld extensions with " : ﬁnite and normal.

i. ! : is a normal extension ⇐⇒ i! = ! for all i ∈ Gal(" : ).

ii. If ! : is a normal extension then Gal(" : !) is a normal subgroup of Gal(" : ) and Gal(" : ) Gal(! : ). Gal(" : !)

88 Before the proof, here’s some context and explanation. Part (i) answers the question implicit in Warning 7.1.7: we know from Corol- lary 7.1.6 that " : ! is normal, but when is ! : normal? In part (i), i! means {i(U) : U ∈ !}. For i! to be equal to ! means that i ﬁxes ! as a set (in other words, permutes it within itself), not that i ﬁxes each element of !. In part (ii), it’s true for all " : ! : that Gal(" : !) is a subset of Gal(" : ), since

Gal(" : !) = {automorphisms i of " such that i(U) = U for all U ∈ !} ⊆ {automorphisms i of " such that i(U) = U for all U ∈ } = Gal(" : ).

It’s also always true that Gal(" : !) is a subgroup of Gal(" : ), as you can easily check. But part (ii) tells us something much more substantial: it’s a normal subgroup when ! : is a normal extension.

Proof of Theorem 7.1.14 For (i), ﬁrst suppose that ! is normal over , and let i ∈ Gal(" : ). For all U ∈ !, Proposition 7.1.9 implies that U and i(U) are conjugate over , so they have the same minimal polynomial, so i(U) ∈ ! by normality. Hence i! ⊆ !. The same argument with i−1 in place of i gives i−1! ⊆ !, and applying i to each side then gives ! ⊆ i!. So i! = !. Conversely, suppose that i! = ! for all i ∈ Gal(" : ). Let U ∈ ! with minimal polynomial <. Since " : is normal, < splits in ". Each root U0 of < in " is conjugate to U over , so by Proposition 7.1.9, U0 = i(U) for some i ∈ Gal(" : ), giving U0 ∈ i! = !. Hence < splits in ! and ! : is normal. For (ii), suppose that ! : is normal. To prove that Gal(" : !) is a normal subgroup of Gal(" : ), let i ∈ Gal(" : ) and \ ∈ Gal(" : !). We show that i−1\i ∈ Gal(" : !), or equivalently,

i−1\i(U) = U for all U ∈ !, or equivalently, \i(U) = i(U) for all U ∈ !. But by (i), i(U) ∈ ! for all U ∈ !, so \(i(U)) = i(U) since \ ∈ Gal(" : !). This completes the proof that Gal(" : !) Gal(" : ). Finally, we prove the statement on quotientsP (still supposing that ! : is a normal extension). Every automorphism i of " over satisﬁes i! = ! (by (i)), and therefore restricts to an automorphism iˆ of !. The function

a : Gal(" : ) → Gal(! : ) i 7→ iˆ

89 is a group homomorphism, since it preserves composition. Its kernel is Gal(" : !), by deﬁnition. If we can prove that a is surjective then the last part of the theorem will follow from the ﬁrst isomorphism theorem. To prove that a is surjective, we must show that each automorphism k of ! over extends to an automorphism i of ":

i " / " O O

! / ! ` k >

The argument is similar to the second half of the proof of Proposition 7.1.9. By Theorem 7.1.5, " is the splitting ﬁeld of some 5 ∈ [C]. Then " is also the splittting ﬁeld of 5 over !. Also, k∗( 5 ) = 5 since k is a homomorphism over and 5 is a polynomial over . So by Proposition 6.2.10(i), there is an automorphism i of " extending k, as required.

Example 7.1.15 Take " : ! : to be √ Q 3 2, l : Q(l) : Q, √ where l = 42c8/3. As you will recognize by now, Q( 3 2, l) is the splitting field of C3 − 2 over Q, so it is a finite normal extension of Q by Theorem 7.1.5. Also, Q(l) is the splitting field of C2+C+1 over Q, so it too is a normal extension 3 of Q. Part (i) of Theorem 7.1.14 implies that every element of GalQ(C −2) restricts to an automorphism of Q(l). Part (ii) implies that √ √ GalQ 3 2, l : Q(l) GalQ 3 2, l : Q P and that √ GalQ 3 2, l : Q √ Gal(Q(l) : Q). (7.1) GalQ 3 2, l : Q(l) √ What does this say explicitly? We showed in Example 7.1.13 that Gal(Q( 3 2, l) : Q) (3. That is, each element of the Galois group permutes the three roots √ √ √ 3 2, l 3 2, l2 3 2

90 C3 of − 2, and all six permutations√ are realized by some element of the Galois 3 , l l group. An element√ of Gal(Q( 2 ) : Q) that√ ﬁxes is determined by which 3 3 of the three roots 2 is mapped to, so Gal(Q( 2, l) : Q(l)) 3. Finally, Gal(Q(l) : Q) 2 by Example 7.1.12. So in this case, the isomorphism (7.1) states that (3 2. 3

C3 Exercise 7.1.16 Draw a diagram√ showing the three roots of − 2 and 3 , l l the elements of = Gal(Q( 2 ) : Q( )) acting on them.√ There is a simple geometric description of the elements of Gal(Q( 3 2, l) : Q) that belong to the subgroup . What is it?

7.2 Separability

Theorem 6.2.12 implies that | Gal(" : )| ≤ [" : ] whenever " : is a splitting field extension. Why is this an inequality, not an equality? The answer can be traced back to the proof of Proposition 6.2.10 on extension of isomorphisms. There, we had an irreducible polynomial called k∗(<), and we wrote B for the number of distinct roots of k∗(<) in its splitting field. Ultimately, the source of the inequality | Gal(" : )| ≤ [" : ] was the fact that B ≤ deg(k∗(<)). But is this last inequality actually an equality? That is, does an irreducible polynomial of degree 3 always have 3 distinct roots in its splitting field? Certainly it has 3 roots when counted with multiplicity. But there will be fewer than 3 distinct roots if any of the roots are repeated (have multiplicity ≥ 2, as defined in Definition 3.2.10). The question is whether this can ever happen.

Exercise 7.2.1 Try to ﬁnd an example of an irreducible polynomial of degree 3 with fewer than 3 distinct roots in its splitting ﬁeld. Or if you can’t, see if you can prove that this is impossible over Q—that is, an irreducible over Q has no repeated roots in C. Both are quite hard, but ten minutes spent trying may help you to appreciate what’s to come.

Definition 7.2.2 An irreducible polynomial over a field is separable if it has no repeated roots in its splitting field.

Equivalently, an irreducible polynomial 5 ∈ [C] is separable if it splits into distinct linear factors in SF ( 5 ):

5 (C) = 0(C − U1)···(C − U=)

91 for some 0 ∈ and distinct U1,...,U= ∈ SF ( 5 ). Put another way, an irreducible 5 is separable if and only if it has deg( 5 ) distinct roots in its splitting ﬁeld.

Example 7.2.3 C3 − 2 ∈ Q[C] is separable, since it has 3 distinct roots in C, hence in its splitting ﬁeld.

Example 7.2.4 This is an example of an irreducible polynomial that’s inseparable. It’s a little bit complicated, although in fact it’s the simplest example there is. Let ? be a prime, let be the field F? (D) of rational expressions over F? (where D is the variable symbol), and let 5 (C) = C ? − D. By definition, 5 has at least one root U in its splitting field. We have

? Õ ? (C − U) ? = C8 (−U) ?−8 = C ? − U? = 5 (C), 8 8=0 where the second equality follows from Lemma 3.3.15. So 5 (C) = (C − U) ? in SF ( 5 ), which means that U is the only root of 5 in its splitting ﬁeld—despite 5 having degree ? > 1. We now show that 5 is irreducible over . The unique factorization of 5 ? into irreducible polynomials over SF ( 5 ) is 5 (C) = (C − U) , so any nontrivial factorization of 5 in [C] is of the form

5 (C) = (C − U)@ (C − U) ?−@ where 0 < @ < ? and both factors belong to [C]. The coeﬃcient of C@−1 in (C − U)@ is −@U, so −@U ∈ . But @ is invertible in , so U ∈ , contradicting Exercise 7.2.5.

Exercise 7.2.5 Show that D has no ?th root in F? (D); that is, there is no ? U ∈ F? (D) with U = D. (Hint: consider the degree of polynomials.)

Warning 7.2.6 Definition 7.2.2 is only a definition of separability for irreducible polynomials. There is a definition of separability for arbitrary polynomials, but it’s not simply Definition 7.2.2 with the word ‘irreducible’ deleted. We won’t need it, but here it is: an arbitrary polynomial is called separable if each of its irreducible factors is separable. So C2 is separable, even though it has a repeated root.

In real analysis, we can test whether a root is repeated by asking whether the

92 derivative is 0 there:

Over an arbitrary field, there’s no general definition of differentiation, as there is no sense of what a ‘limit’ might be. But even without limits, we can differentiate polynomials in the following sense. 5 C Í= 0 C8 C Definition 7.2.7 Let be a field and let ( ) = 8=0 8 ∈ [ ]. The formal derivative of 5 is = Õ 8−1 ( 5 )(C) = 808C ∈ [C]. 8=1 We use 5 rather than 5 0 to remind ourselves not to take the familiar properties of differentiation for granted. Nevertheless, the usual basic laws hold:

Lemma 7.2.8 Let be a ﬁeld. Then

( 5 + 6) = 5 + 6, ( 5 6) = 5 · 6 + 5 · 6, 0 = 0 for all 5 , 6 ∈ [C] and 0 ∈ .

Exercise 7.2.9 Check a couple of the properties in Lemma 7.2.8.

The real analysis test for repetition of roots has an algebraic analogue:

Lemma 7.2.10 Let 5 be a nonzero polynomial over a ﬁeld . The following are equivalent:

i. 5 has a repeated root in SF ( 5 );

ii. 5 and 5 have a common root in SF ( 5 ); iii. 5 and 5 have a nonconstant common factor in [C].

Proof (i)⇒(ii): suppose that 5 has a repeated root U in SF ( 5 ). Then 5 (C) = 2 (C − U) 6(C) for some 6(C) ∈ (SF ( 5 ))[C]. Hence

( 5 )(C) = (C − U)26(C) + (C − U)·(6)(C) ,

93 so U is a common root of 5 and 5 in SF ( 5 ). (ii)⇒(iii): suppose that 5 and 5 have a common root U in SF ( 5 ). Then U is algebraic over (since 5 ≠ 0), and the minimal polynomial of U over is then a nonconstant common factor of 5 and 5 . (iii)⇒(ii): if 5 and 5 have a nonconstant common factor 6 then 6 splits in SF ( 5 ), and any root of 6 in SF ( 5 ) is a common root of 5 and 5 . (ii)⇒(i): suppose that 5 and 5 have a common root U ∈ SF ( 5 ). Then 5 (C) = (C − U)6(C) for some 6 ∈ (SF ( 5 ))[C], giving ( 5 )(C) = 6(C) + (C − U)·(6)(C).

But ( 5 )(U) = 0, so 6(U) = 0, so 6(C) = (C − U)ℎ(C) for some ℎ ∈ (SF ( 5 ))[C]. Hence 5 (C) = (C − U)2ℎ(C), and U is a repeated root of 5 in its splitting ﬁeld.

The point of Lemma 7.2.10 is that condition (iii) allows us to test for repetition of roots in SF ( 5 ) without ever leaving [C], or even knowing what SF ( 5 ) is. Proposition 7.2.11 Let 5 be an irreducible polynomial over a ﬁeld. Then 5 is inseparable if and only if 5 = 0.

Proof This follows from (i) ⇐⇒ (iii) in Lemma 7.2.10. Since 5 is irreducible, 5 and 5 have a nonconstant common factor if and only if 5 divides 5 ; but deg( 5 ) < deg( 5 ), so 5 | 5 if and only if 5 = 0.

Corollary 7.2.12 Let be a ﬁeld. i. If char = 0 then every irreducible polynomial over is separable. ii. If char = ? > 0 then an irreducible polynomial 5 ∈ [C] is inseparable if and only if ? A? 5 (C) = 10 + 11C + · · · + 1AC

for some 10, . . . , 1A ∈ .

In other words, the only irreducible polynomials that are inseparable are the polynomials in C ? in characteristic ?. Inevitably, Example 7.2.4 is of this form. Í 8 Proof Let 5 (C) = 08C be an irreducible polynomial. Then 5 is inseparable if and only if 5 = 0, if and only if 808 = 0 for all 8 ≥ 1. If char = 0, this implies that 08 = 0 for all 8 ≥ 1, so 5 is constant, which contradicts 5 being irreducible. If char = ?, then 808 = 0 for all 8 ≥ 1 is equivalent to 08 = 0 whenever ? - 8.

Remark 7.2.13 In the final chapter we will show that every irreducible polynomial over a finite field is separable. So, it is only over infinite fields of characteristic ? that you have to worry about inseparability.

94 We now build up to showing that | Gal(" : )| = [" : ] whenever " : is a finite normal extension in which the minimal polynomial of every element of " is separable. First, some terminology: Definition 7.2.14 Let " : be an algebraic extension. An element of " is separable over if its minimal polynomial over is separable. The extension " : is separable if every element of " is separable over . Examples 7.2.15 i. Every algebraic extension of fields of characteristic 0 is separable, by Corollary 7.2.12. ii. Every algebraic extension of a finite field is separable, by Remark 7.2.13.

? iii. The splitting ﬁeld of C − D over F? (D) is inseparable: indeed, the element denoted by U in Example 7.2.4 is inseparable over F? (D), since its minimal polynomial is the inseparable polynomial C ? − D.

Exercise 7.2.16 Let " : ! : be ﬁeld extensions. Show that if " : is algebraic then so are " : ! and ! : .

Lemma 7.2.17 Let " : ! : be field extensions, with " : algebraic. If " : is separable then so are " : ! and ! : . Proof Both " : ! and ! : are algebraic by Exercise 7.2.16, so it does make sense to ask whether they are separable. (We only defined what it means for an algebraic extension to be separable.) That ! : is separable is immediate from the definition. To show that " : ! is separable, let U ∈ ". Write

95 Theorem 7.2.19 | Gal(" : )| = [" : ] for every ﬁnite normal separable extension " : .

Proof By Theorem 7.1.5, " = SF ( 5 ) for some 5 ∈ [C]. The result follows 0 0 from Proposition 7.2.18, taking " = ", = , and k = id .

Examples 7.2.20 i. | Gal ( 5 )| = [SF ( 5 ) : ] for any polynomial 5 over a ﬁeld of characteristic 0. 3 For instance, [SFQ(C − 2) : Q] = 6 by a similar argument to Example 5.2.7, 3 using that SFQ(C − 2) contains elements of degree 2 and 3 over Q. Hence | GalQ( 5 )| = 6. But GalQ( 5 ) embeds into (3 by Proposition 6.3.10, so GalQ( 5 ) (3. We already proved this in Example 7.1.15, using a diﬀerent argument.

? ii. Consider = F? (D) and " = SF (C − D). With notation as in Exam- " U " U ? ple 7.2.4, we have = ( ), so [ : ] = deg ( ) = . On the other hand, | Gal(" : )| = 1 by Corollary 6.3.15. So Theorem 7.2.19 fails if we drop the separability hypothesis.

Digression 7.2.21 With some effort, one can show that in any algebraic extension " : , the separable elements form a subfield of ". (See Stewart, Theorem 17.22.) It follows that a finite extension (U1,...,U=) : is separable if and only if each U8 is. Hence a splitting field extension SF ( 5 ) : is separable if and only if every root of 5 is separable in SF ( 5 ), which itself is equivalent to 5 being separable in the sense of Warning 7.2.6.

So: SF ( 5 ) is separable over if and only if 5 is separable over . Thus, the diﬀerent meanings of ‘separable’ interact nicely.

Digression 7.2.22 It’s a stunning fact that every ﬁnite separable extension is simple. This is called the theorem of the primitive element. For instance, whenever U1,...,U= are complex numbers algebraic over Q, there is some U ∈ C (a ‘primitive element’) such that√ Q√(U1,...,U√ =) =√Q(U). We saw one case of this in Example 4.3.2(i): Q( 2, 3) = Q( 2 + 3). The theorem of the primitive element was at the heart of most early accounts of Galois theory, and appears in many modern treatments too, but we will not prove it.

7.3 Fixed ﬁelds

When a group acts on a set -, we usually want to know which elements of - are ﬁxed by every element of . Here, we’ll ask this question for groups of

96 automorphisms of a ﬁeld. The following preliminary deﬁnition will come in handy.

Deﬁnition 7.3.1 Let - and . be sets, and let ⊆ {functions - → . }. The equalizer of is

Eq() = {G ∈ - : 5 (G) = 6(G) for all 5 , 6 ∈ }.

So the equalizer of is the part of - on which all the functions in are equal.

Lemma 7.3.2 Let " and "0 be ﬁelds, and let ⊆ {homomorphisms " → "0}. Then Eq() is a subﬁeld of ".

Proof We must show that 0, 1 ∈ Eq(), that if U ∈ Eq() then −U ∈ Eq() and 1/U ∈ Eq() (when U ≠ 0), and that if U,V ∈ Eq() then U + V,UV ∈ Eq().I will show just the last of these, leaving the rest to you. Suppose that U,V ∈ Eq(). For all i, \ ∈ , we have

i(UV) = i(U)i(V) = \(U)\(V) = \(UV), so UV ∈ Eq().

Write Aut(") for the group of automorphisms of a ﬁeld ".

Definition 7.3.3 Let " be a field and let a subgroup of Aut("). The fixed field of is Fix() = {U ∈ " : i(U) = U for all i ∈ }.

Lemma 7.3.4 Fix() is a subﬁeld of ".

Proof In fact, Fix() = Eq(), since id" is an element of . The result follows from Lemma 7.3.2.

Exercise 7.3.5 Using Lemma 7.3.4, show that every automorphism of a field is an automorphism over its prime subfield. In other words, Aut(") = Gal(" : ) whenever " is a field with prime subfield .

Here’s the big, ingenious, result about ﬁxed ﬁelds. It will play a crucial part in the proof of the fundamental theorem of Galois theory.

97 Theorem 7.3.6 Let " be a ﬁeld and a ﬁnite subgroup of Aut("). Then [" : Fix()] ≤ ||.

As Fix() gets bigger, [" : Fix()] gets smaller. Theorem 7.3.6 tells us that the smaller || is, the more of " must be fixed by . For some intuition about both the statement and the proof, watch the video ‘The size of fixed fields’.

Proof Write = = ||. It is enough to prove that any = + 1 elements U0,...,U= of " are linearly dependent over Fix(). Let

 =+1 , = (G0,...,G=) ∈ " : G0\(U0) + · · · + G=\(U=) = 0 for all \ ∈ .

Then , is defined by = homogeneous linear equations in = + 1 variables, so it is a The size of fixed nontrivial "-linear subspace of "=+1. fields Claim: let (G0,...,G=) ∈ , and i ∈ . Then (i(G0), . . . , i(G=)) ∈ ,. Proof: For all \ ∈ , we have

−1 −1 G0(i ◦ \)(U0) + · · · + G= (i ◦ \)(U=) = 0

since i−1 ◦ \ ∈ . Applying i to both sides gives that for all \ ∈ ,

i(G0)\(U0) + · · · + i(G=)\(U=) = 0,

proving the claim. Since , is nontrivial, there is a least : ≥ 0 such that , contains some nonzero vector x = (G0,...,G=) with G8 = 0 for all 8 > :. Then G: ≠ 0, and since , is Remarks on the " G proof of closed under scalar multiplication by , we may assume that : = 1. Theorem 7.3.6 We now show that G8 ∈ Fix() for all 8. Let i ∈ . By the claim, (actually a PDF, (i(G0), . . . , i(G=)) ∈ ,. Since , is a linear subspace, not a video) (i(G0) − G0, . . . , i(G=) − G=) ∈ ,.

But i(G: ) − G: = i(1) − 1 = 0 and i(G8) − G8 = 0 − 0 = 0 for all 8 > :, so by minimality of :, also i(G8) − G8 = 0 for all 8 < :. Hence G8 ∈ Fix() for all 8. We have shown that , contains a nonzero vector x ∈ Fix()=+1. But taking Í \ = id in the deﬁnition of , gives G8U8 = 0. Hence U0,...,U= are linearly dependent over Fix().

Example 7.3.7 Let ^ : C → C be complex conjugation. Then = {id, ^} is a subgroup of Aut(C), and Theorem 7.3.6 predicts that [C : Fix()] ≤ 2. Since Fix() = R, this is true.

98 Exercise 7.3.8 Find another example of Theorem 7.3.6.

Digression 7.3.9 In fact, Theorem 7.3.6 is an equality: [" : Fix()] = ||. This is proved directly in many Galois theory books (e.g. Stewart, Theorem 10.5). In our approach it will be a consequence of the fundamental theorem of Galois theory (rather than a step on the way to proving it), at least under certain hypotheses on " and . The reverse inequality, [" : Fix()] ≥ ||, is closely related to the result called ‘linear independence of characters’. (A good reference is Lang, Alge- bra, 3rd edition, Theorem 4.1.) Another instance of linear independence of characters is that the functions I 7→ 42c8=I (= ∈ Z) are linearly independent, a fundamental fact in the theory of Fourier series.

We finish by adding a further connecting strand between the concepts of normal extension and normal subgroup, complementary to the strands in Theorem 7.1.14. It needs a lemma that fundamentally has nothing to do with fields, and is really a general result about groups acting on sets, as the proof shows. Lemma 7.3.10 Let " be a field, a subgroup of Aut("), and i ∈ Aut("). Then Fix(ii−1) = i Fix().

Proof Let U ∈ ". Then

U ∈ Fix(ii−1) ⇐⇒ i\i−1(U) = U for all \ ∈ ⇐⇒ \i−1(U) = i−1(U) for all \ ∈ ⇐⇒ i−1(U) ∈ Fix() ⇐⇒ U ∈ i Fix().

Proposition 7.3.11 Let " : be a ﬁnite normal extension and a normal subgroup of Gal(" : ). Then Fix() is a normal extension of .

Proof Since every element of is an automorphism over , the subﬁeld Fix() of " contains . For each i ∈ Gal(" : ), we have

i Fix() = Fix(ii−1) = Fix(), where the ﬁrst equality holds by Lemma 7.3.10 and the second because is normal in Gal(" : ). Hence by Theorem 7.1.14(i), Fix() : is a normal extension.

The stage is now set for the central result of the course: the fundamental theorem of Galois theory.

99 Chapter 8

The fundamental theorem of Galois theory

We’ve been building up to this moment all semester. Let’s do it! Introduction to Week 8 8.1 Introducing the Galois correspondence

Let " : be a field extension, with viewed as a subfield of ", as usual. An intermediate field of " : is a subfield of " containing . Write F = {intermediate fields of " : }. For ! ∈ F , we draw diagrams like this: "

, with the bigger ﬁelds higher up. Also write G = {subgroups of Gal(" : )}. For ∈ G , we draw diagrams like this: 1

Gal(" : ),

100 where 1 denotes the trivial subgroup and the bigger groups are lower down. It will become clear soon why we’re using opposite conventions for the field and group diagrams. For ! ∈ F , the group Gal(" : !) consists of all automorphisms i of " that fix each element of !. Since ⊆ !, any such i certainly fixes each element of . Hence Gal(" : !) is a subgroup of Gal(" : ). This process defines a function Gal(" : −) : F → G ! 7→ Gal(" : !). In the expression Gal(" : −), the symbol − should be seen as a blank space into which arguments can be inserted.

Warning 8.1.1 The group we’re associating with ! is Gal(" : !), not Gal(! : )! Both groups matter, but only one is a subgroup of Gal(" : ), which is what we’re interested in here. We showed just now that Gal(" : !) is a subgroup of Gal(" : ). If you wanted to show that Gal(! : ) is (isomorphic to) a subgroup of Gal(" : )—which it isn’t—then you’d probably do it by trying to prove that every automorphism of ! over extends uniquely to ". And that’s false. For instance, when ! = , the identity on ! typically has many extensions to ": they’re the elements of Gal(" : ). Although Gal(! : ) isn’t a subgroup of Gal(" : ), it is a quotient of it, at least when both extensions are ﬁnite and normal. We saw this in Theorem 7.1.14, and we’ll come back to it in Section 8.2.

In the other direction, for ∈ G , the subfield Fix() of " contains . Indeed, ⊆ Gal(" : ), and by definition, every element of Gal(" : ) fixes every element of , so Fix() ⊇ . Hence Fix() is an intermediate field of " : . This process defines a function Fix: G → F 7→ Fix(). We have now defined functions Gal(":−) / F o G . Fix The fundamental theorem of Galois theory tells us how these functions behave: how the concepts of Galois group and fixed field interact. The proof will bring together most of the big results we’ve proved so far, and assumes that the extension is finite, normal and separable. But first, let’s say the simple things that are true for all extensions:

101 " 1

!2 Gal(" : !2)

!1 Gal(" : !1)

Gal(" : )

Figure 8.1: The function ! 7→ Gal(" : !) is order-reversing (Lemma 8.1.2(i)).

Lemma 8.1.2 Let " : be a ﬁeld extension, and deﬁne F and G as above.

i. For !1,!2 ∈ F ,

!1 ⊆ !2 ⇒ Gal(" : !1) ⊇ Gal(" : !2)

(Figure 8.1). For 1, 2 ∈ G ,

1 ⊆ 2 ⇒ Fix(1) ⊇ Fix(2).

ii. For ! ∈ F and ∈ G ,

! ⊆ Fix() ⇐⇒ ⊆ Gal(" : !).

iii. For all ! ∈ F , ! ⊆ Fix(Gal(" : !)). For all ∈ G , ⊆ Gal(" : Fix()).

Warning 8.1.3 In part (i), the functions Gal(" : −) and Fix reverse inclusions. The bigger you make !, the smaller you make Gal(" : !), because it gets harder for an automorphism to fix everything in !. And the bigger you make , the smaller you make Fix(), because it gets harder for an element of " to be fixed by everything in . That’s why the field and group diagrams are opposite ways up.

102 Proof (i): I leave the ﬁrst half as an exercise. For the second, suppose that 1 ⊆ 2, and let U ∈ Fix(2). Then \(U) = U for all \ ∈ 2, so \(U) = U for all \ ∈ 1, so U ∈ Fix(1). (ii): both sides are equivalent to the statement that \(U) = U for all \ ∈ and U ∈ !. (iii): the ﬁrst statement follows from the ⇐ direction of (ii) by taking = Gal(" : !), and the second follows from the ⇒ direction of (ii) by taking ! = Fix(). (Or, they can be proved directly.)

Exercise 8.1.4 Prove the ﬁrst half of Lemma 8.1.2(i).

Exercise 8.1.5 Draw a diagram like Figure 8.1 for the second half of Lemma 8.1.2(i).

Digression 8.1.6 If you’ve done some algebraic geometry, the formal structure of Lemma 8.1.2 might seem familiar. Given a ﬁeld and a natural number =, we can form the set F of subsets of = and the set G of ideals of [C1,...,C=], and there are functions F G deﬁned by taking the annihi- = lating ideal of a subset of and the zero-set of an ideal of [C1,...,C=]. The analogue of Lemma 8.1.2 holds. In general, a pair of ordered sets F and G equipped with functions F G satisfying the properties above is called a Galois connection. This in turn is a special case of the category-theoretic notion of adjoint functors.

The functions Gal(":−) / F o G . Fix are called the Galois correspondence for " : . This terminology is mostly used in the case where the functions are mutually inverse, meaning that

! = Fix(Gal(" : !)), = Gal(" : Fix()) for all ! ∈ F and ∈ G . We saw in Lemma 8.1.2(iii) that in both cases, the left-hand side is a subset of the right-hand side. But they are not always equal: √ Example 8.1.7 Let " : be Q( 3 2) : Q. Since [" : ] is 3, which is a prime number, the tower law implies that there are no nontrivial intermediate ﬁelds: F = {", }. We saw in Example 6.3.3(ii) that Gal(" : ) is trivial, so G = {Gal(" : )}. Hence F has two elements and G has only one. This

103 makes it impossible for there to be mutually inverse functions between F and G . Speciﬁcally, what goes wrong is that √ √ 3 √ 3 Q Q = 3 = Q ≠ Q. Fix Gal 2 : Fix idQ( 2) 2

Exercise 8.1.8 Let ? be a prime number, let = F? (D), and let " be the splitting ﬁeld of C ? − D over , as in Examples 7.2.4 and 7.2.20(ii). Prove that Gal(" : −) and Fix are not mutually inverse.

If Gal(" : −) and Fix are mutually inverse then they set up a one-to-one correspondence between the set F of intermediate ﬁelds of " : and the set G of subgroups of Gal(" : ). The fundamental theorem of Galois theory tells us that this dream comes true when " : is ﬁnite, normal and separable. And it tells us more besides.

8.2 The theorem

The moment has come.

Theorem 8.2.1 (Fundamental theorem of Galois theory) Let " : be a ﬁnite normal separable extension. Write

F = {intermediate ﬁelds of " : }, G = {subgroups of Gal(" : )}.

Gal(":−) / i. The functions F o G are mutually inverse. Fix

ii. | Gal(" : !)| = [" : !] for all ! ∈ F , and [" : Fix()] = || for all ∈ G .

iii. Let ! ∈ F . Then

! is a normal extension of ⇐⇒ Gal(" : !) is a normal subgroup of Gal(" : ),

and in that case, Gal(" : ) Gal(! : ). Gal(" : !)

104 Proof First note that for each ! ∈ F , the extension " : ! is finite and normal (by Corollary 7.1.6) and separable (by Lemma 7.2.17). Also, the group Gal(" : ) is finite (by Theorem 7.2.19), so every subgroup is finite too. We prove (i) and (ii) together. First let ∈ G . We have

|| ≤ | Gal(" : Fix())| = [" : Fix()] ≤ ||, (8.1) where the first inequality holds because ⊆ Gal(" : Fix()) (Lemma 8.1.2(iii)), the equality follows from Theorem 7.2.19 (since " : Fix() is finite, normal and separable), and the second inequality follows from Theorem 7.3.6 (since is finite). So equality holds throughout (8.1), giving

 = Gal(" : Fix()), [" : Fix()] = ||.

Now let ! ∈ F . We have

[" : Fix(Gal(" : !))] = | Gal(" : !)| = [" : !], where the ﬁrst equality follows from the previous paragraph by taking = Gal(" : !), and the second follows from Theorem 7.2.19. But ! ⊆ Fix(Gal(" : !)) by Lemma 8.1.2(iii), so ! = Fix(Gal(" : !)) by Workshop 3, q. 1. This completes the proof of (i) and (ii). We have already proved most of (iii) as Theorem 7.1.14(ii). It only remains to show that whenever ! is an intermediate ﬁeld such that Gal(" : !) is a normal subgroup of Gal(" : ), then ! is a normal extension of . By Proposition 7.3.11, Fix(Gal(" : !)) : is normal. But by (i), Fix(Gal(" : !)) = !, so ! : is normal, as required.

The fundamental theorem of Galois theory is about field extensions that are finite, normal and separable. Let’s take a moment to think about what those conditions mean. An extension " : is finite and normal if and only if " is the splitting field of some polynomial over (Theorem 7.1.5). So, the theorem can be understood as a result about splitting fields of polynomials. Not every splitting field extension is separable (Example 7.2.15(iii)). However, we know of two settings where separability is guaranteed. The first is fields of characteristic zero (Example 7.2.15(i)). The most important of these is Q, which is our focus in this chapter: we’ll consider examples in which " : is the splitting field extension of a polynomial over Q. The second is where the fields are finite (Example 7.2.15(ii)). We’ll come to finite fields in the final chapter.

105 Digression 8.2.2 Normality and separability are core requirements of Galois theory, but there are extensions of the fundamental theorem (well beyond this course) in which the finiteness condition on " : is relaxed. The first level of relaxation replaces ‘finite’ by ‘algebraic’. Then Gal(" : ) is no longer a finite group, but it does acquire an interesting topology. One example is where " is the algebraic closure of , and Gal( : ) is called the absolute Galois group of . It contains all splitting fields of polynomials over , so to study it is to study all polynomials over at once. Going further, we can even drop the condition that the extension is algebraic. In this realm, we need the notion of ‘transcendence degree’, which counts how many algebraically independent elements can be found in the extension.

You’ll want to see some examples! Section 8.3 is devoted to a single example of the fundamental theorem, showing every aspect of the theorem in all its glory. I’ll give a couple of simpler examples in a moment, but before that, it’s helpful to review some of what we did earlier: Remark 8.2.3 When working out the details of the Galois correspondence for a polynomial 5 ∈ [C], it’s not only the fundamental theorem that’s useful. Some of our earlier results also come in handy, such as the following.

i. Lemma 6.3.7 states that Gal ( 5 ) acts on the set of roots of 5 in SF ( 5 ), and the injectivity part of Proposition 6.3.10 states that the action is faithful: if i, \ ∈ Gal ( 5 ) and i(U) = \(U) for every root U of 5 in SF ( 5 ), then i = \. In words, an element of the Galois group is entirely determined by what it does to the roots.

ii. Corollary 6.3.15 states that | Gal ( 5 )| divides :!, where : is the number of distinct roots of 5 in its splitting ﬁeld.

iii. Let U and V be roots of 5 in SF ( 5 ). Then there is an element of the Galois group mapping U to V if and only if U and V are conjugate over (have the same minimal polynomial). This follows from Proposition 7.1.9. In particular, if 5 is irreducible then there is always an element of the Galois group that maps U to V (Corollary 7.1.10).

iv. Let ! be an intermediate ﬁeld of SF ( 5 ) : such that ! : is normal. Then every automorphism of ! over can be extended to an automorphism of SF ( 5 ) over . This was shown in the last paragraph of the proof of Theorem 7.1.14.

Examples 8.2.4 i. Let " : be a normal separable extension of prime degree ?. By the fundamental theorem, | Gal(" : )| = [" : ] = ?.

106 Every group of prime order is cyclic, so Gal(" : ) ?. By the tower law, " : has no nontrivial intermediate ﬁelds, and by Lagrange’s theorem, Gal(" : ) has no nontrivial subgroups. So F = {", } and G = {1, Gal(" : )}: " 1

Gal(" : )

Both " and are normal extensions of , and both 1 and Gal(" : ) are normal subgroups of Gal(" : ). √ 2 2 ii. Let 5 (C) = (C + 1)(C − 2) ∈ Q[C]. Put " = SFQ( 5 ) = Q( 2, 8) and = Gal(" : ) = GalQ( 5 ). Then " : is a ﬁnite normal separable extension, so the fundamental theorem applies. We already calculated in a sketchy way in Example 6.3.6(ii). Let’s do it again in full, using what we now know. First, √ √ √ [" : ] = Q 2, 8 : Q 2 Q 2 : Q = 2 × 2 = 4

(much as in Example 5.2.2). √ , 8 5 Now consider√ how√ acts√ on the set {± 2 ± } of roots of . The√ conjugacy√ class of 2 is { 2, − 2}, so for each i ∈ we have i( 2) = ± 2. Similarly, i(8) = ±8 for each i ∈ . The two choices of sign determine i entirely, so || ≤ 4. But by the fundamental theorem, || = [" : ] = 4, so each of the four possibilities does in fact occur. So = {id, i+−, i−+, i−−}, where √ √ √ √ √ √ i+− 2 = 2, i−+ 2 = − 2, i−− 2 = − 2, i+−(8) = −8, i−+(8) = 8, i−−(8) = −8.

The only two groups of order 4 are 4 and 2 × 2, and each element of has order 1 or 2, so 2 × 2. The subgroups of are

hi+−i hi−+i hi−−i (8.2)

107 where lines indicate inclusions. Here hi+−i is the subgroup generated by i+−, which is {id, i+−}, and similarly for i−+ and i−−. What are the fixed fields of these subgroups?√ √ The fundamental√ theorem implies that Fix() = Q. Also, i+−( 2) = 2, so Q( 2) ⊆ Fix(hi+−i). But √ √ √ Q 2, 8 : Q 2 = 2 = |hi+−i| = Q 2, 8 : Fix(hi+−i) √ (where the last step is by the fundamental theorem), so Q( 2) = Fix(hi+−i). Similar arguments apply to i−+ and i−−, so the fixed fields of the groups in diagram (8.2) are √ Q 2, 8

√ √ Q 2 Q(8) Q 28 (8.3)

Q √ , 8 Equivalently, the groups in (8.2) are√ the Galois groups of Q( 2 ) over the ﬁelds in (8.3). For instance, Gal(Q( 2, 8) : Q(8)) = hi−+i.

Since the overall Galois group 2 × 2 is abelian, every subgroup is normal. Hence all the extensions in diagram (8.3) are normal too.

Exercise 8.2.5 In this particular example, one can also see directly that all the extensions in (8.3) are normal. How?

Like any big theorem, the fundamental theorem of Galois theory has some important corollaries. Here’s one.

Corollary 8.2.6 Let " : be a ﬁnite normal separable extension. Then for every U ∈ " \ , there is some automorphism i of " over such that i(U) ≠ U.

Proof Theorem 8.2.1(i) implies that Fix(Gal(" : )) = . Now U ∉ , so U ∉ Fix(Gal(" : )), which is what had to be proved.

Example 8.2.7 For any 5 ∈ Q[C] and irrational U ∈ SFQ( 5 ), there is some i ∈ GalQ( 5 ) that does not ﬁx U. This is clear if U ∉ R, as we can take i to be complex conjugation restricted to SFQ( 5 ). But it is not so obvious otherwise.

108 d b8

−b b ^

−b8

Figure 8.2: The roots of 5 , and the eﬀects on them of d, ^ ∈ GalQ( 5 ).

8.3 A speciﬁc example

Chapter 13 of Stewart’s book opens with these words:

The extension that we discuss is a favourite with writers on Galois theory, because of its archetypal quality. A simpler example would be too small to illustrate the theory adequately, and anything more complicated would be unwieldy. The example is the Galois group of the splitting ﬁeld of C4 − 2 over Q.

We go through the same example here. My presentation of it is diﬀerent from Stewart’s, so you can consult his book if anything that follows is unclear. Write 5 (C) = C4 − 2 ∈ Q[C], which is irreducible by Eisenstein’s criterion. Write = GalQ( 5 ).

Splitting ﬁeld Write b for the unique real positive root of 5 . Then the roots of 5 are ±b and ±b8 (Figure 8.2). So SFQ( 5 ) = Q(b, b8) = Q(b, 8). We have

[Q(b, 8) : Q] = [Q(b, 8) : Q(b)][Q(b) : Q] = 2 × 4 = 8, where the ﬁrst factor is 2 because Q(b) ⊆ R and the second factor is 4 because 5 is the minimal polynomial of b over Q (being irreducible) and deg( 5 ) = 4. By the fundamental theorem, || = 8.

Galois group We now look for the 8 elements of the Galois group. We’ll use the principle that if i, \ ∈ with i(b) = \(b) and i(8) = \(8) then i = \. To see this, note that i(U) = \(U) whenever U is a root of 5 , and since the action of on the roots of 5 is faithful, i = \. Complex conjugation on C restricts to an automorphism ^ of Q(b, 8) over Q, giving an element ^ ∈ of order 2.

109 I now claim that has an element d satisfying d(b) = b8 and d(8) = 8. In that case, d will act on the roots of 5 as follows:

b 7→ b8 7→ −b 7→ −b8 7→ b

(Figure 8.2). This element d will have order 4. Proof of claim: since 5 is irreducible, acts transitively on the roots of 5 in SFQ( 5 ), so there is some i ∈ such that i(b) = b8. The conjugacy class of 8 over Q is {±8}, so i(8) = ±8. If i(8) = 8 then we can take d = i. If i(8) = −8 then

(i ◦ ^)(b) = i(b) = b8, (i ◦ ^)(8) = i(−8) = −i(8) = 8, so we can take d = i ◦ ^. (From now on, I will usually omit the ◦ sign and write things like i^ instead. Of course, juxtaposition is also used to mean multiplication, as in b8. But confusion shouldn’t arise: automorphisms are composed and numbers are multiplied.) Figure 8.2 suggests that is the dihedral group 4, the symmetry group of the square.

Warning 8.3.1 The symmetry group of a regular =-sided polygon has 2= elements: = rotations and = reﬂections. Some authors call it = and others call it 2=. I will call it =, as in the Group Theory course.

If this is right, we should have ^d = d−1^. (This is one of the deﬁning equations of the dihedral group; you saw it in Example 3.2.12 of Group Theory.) Let’s check this algebraically:

^d(b) = b8 = −b8, d−1^(b) = d−1(b) = −b8, ^d(8) = ^(8) = −8, d−1^(8) = d−1(−8) = −8, so ^d and d−1^ are equal on b and 8, so ^d = d−1^. It follows that ^dA = d−A ^ for all A ∈ Z. Figure 8.3 shows the eﬀect of 8 elements of on b, 8 and b8. Since no two of them have the same eﬀect on both b and 8, they are all distinct elements of . Since || = 8, they are the only elements of . So 4.

Warning 8.3.2 The ‘geometric description’ in Figure 8.3 applies only to the roots, not the whole of the splitting ﬁeld Q(b, 8). For example, d2 is rotation by c on the set of roots, but it is not rotation by c on the rest of Q(b, 8): it ﬁxes each element of Q, for instance.

110 i ∈ i(b) i(8) i(b8) order geometric description (see Warning 8.3.2) id b 8 b8 1 identity d b8 8 −b 4 rotation by c/2 d2 −b 8 −b8 2 rotation by c d3 = d−1 −b8 8 b 4 rotation by −c/2 ^ b −8 −b8 2 reflection in real axis ^d = d−1^ −b8 −8 −b 2 reflection in axis through 1 − 8 ^d2 = d2^ −b −8 b8 2 reflection in imaginary axis ^d−1 = d^ b8 −8 b 2 reflection in axis through 1 + 8

Figure 8.3: The Galois group of C4 − 2 over Q.

Subgroups of the Galois group Since || = 8, any nontrivial proper subgroup of has order 2 or 4. Let’s look in turn at subgroups of order 2 and 4, also determining which ones are normal. This is pure group theory, with no mention of ﬁelds.

• The subgroups of order 2 are of the form hii = {id, i} where i ∈ has order 2. So, they are hd2i, h^i, h^di, h^d2i, h^d−1i.

If you watched the video ‘What does it mean to be normal?’, you may be able to guess which of these subgroups are normal in , the symmetry group of the square. It should be those that can be speciﬁed without referring to particular vertices or edges of the square. So, just the ﬁrst should be normal. Let’s check. We know that ^d2 = d2^, so d2 commutes with both ^ and d, which generate . Hence d2 is in the centre of (commutes with everything in ). It follows that hd2i is a normal subgroup of . On the other hand, for each A ∈ Z, the subgroup h^dA i is not normal, since

d(^dA )d−1 = (d^)dA−1 = (^d−1)dA−1 = ^dA−2 ∉ h^dA i.

• The subgroups of of order 4 are isomorphic to either 4 or 2 × 2, since these are the only groups of order 4. The only elements of of order 4 are d±1, so the only subgroup of 2 3 −1 isomorphic to 4 is hdi = {id, d, d , d = d }.

Now consider subgroups of isomorphic to 2 × 2.

111 Exercise 8.3.3 Show that every such must contain d2. (Hint: think geometrically.)

We have d2 ∈ , and both other nonidentity elements of have order 2, so they are of the form ^dA for some A ∈ Z. The two such subgroups are

h^, d2i = {id, ^, d2, ^d2}, h^d, d2i = {id, ^d, d2, ^d−1}.

Finally, any subgroup of index 2 of any group is normal, so all the subgroups of of order 4 are normal.

Hence the subgroup structure of 4 is as follows, where a box around a subgroup means that it is normal in .

1 order 1

h^i h^d2i hd2i h^di h^d−1i order 2

2 d 2 h^, d i 2 × 2 h i 4 h^d, d i 2 × 2 order 4

 = h^, di 4 order 8

Fixed ﬁelds We now ﬁnd Fix() for each ∈ G , again considering the subgroups of orders 2 and 4 in turn.

• Order 2: take Fixhî (officially Fix(hî), but let’s drop the brackets). We have ^(b) = b, so b ∈ Fixhî, so Q(b) ⊆ Fixhî. But [Q(b, 8) : Q(b)] = 2, and by the fundamental theorem, [Q(b, 8) : Fixhî] = |hî| = 2, so Fixhî = Q(b). The same argument shows that for any i ∈ of order 2, if we can spot some U ∈ Q(b, 8) such that i(U) = U and [Q(b, 8) : Q(U)] ≤ 2, then 2 Finding fixed fields Fixhii = Q(U). For i = ^d , we can take U = b8 (by Figure 8.3). We have b8 b8 5 b8 degQ( ) = 4 since is a root of , so [Q( ) : Q] = 4, or equivalently, [Q(b, 8) : Q(b8)] = 2. Hence Fixh^d2i = Q(b8).

112 Exercise 8.3.4 I took a small liberty in the sentence beginning ‘The same argument’, because it included an inequality but the previous argument didn’t. Prove the statement made in that sentence.

It is maybe not so easy to spot an U for ^d, but the geometric description in Figure 8.3 suggests taking U = b(1 − 8). And indeed, one can check that ^d fixes b(1 − 8). One can also check that b(1 − 8) is not the root of any b 8 nonzero quadratic over Q, so degQ( (1 − )) is ≥ 4 (since it divides 8), so [Q(b, 8) : Q(b(1 −8))] ≤ 8/4 = 2. Hence Fixh^di = Q(b(1 −8)). Similarly, Fixh^d−1i = Q(b(1 + 8)). Finally, d2(b2) = (d2(b))2 = (−b)2 = b2, d2(8) = 8, so Q(b2, 8) ⊆ Fixhd2i. But [Q(b, 8) : Q(b2, 8)] = 2, so Fixhd2i = Q(b2, 8). • Order 4: for = h^, d2i, note that b2 is fixed by both ^ and d2, so b2 ∈ Fix(), so Q(b2) ⊆ Fix(). But b2 ∉ Q, so [Q(b2) : Q] ≥ 2, so [Q(b, 8) : Q(b2)] ≤ 4. The fundamental theorem guarantees that [Q(b, 8) : Fix()] = || = 4, so Fix() = Q(b2). The same argument applies to the other two subgroups of order 4: if we can spot an element U ∈ Q(b, 8)\ Q fixed by the generators of , then Fix() = Q(U). This gives Fixhdi = Q(8) and Fixh^d, d2i = Q(b28). In summary, the fixed fields of the subgroups of are as follows. Q(b, 8) degree 1

Q(b) Q(b8) Q(b2, 8) Q(b(1 − 8)) Q(b(1 + 8)) degree 2

Q(b2) Q(8) Q(b28) degree 4

Q degree 8 In the right-hand column, ‘degree’ means the degree of Q(b, 8) over the subfield concerned. The fundamental theorem implies that the Galois group of Q(b, 8) over each intermediate field is the corresponding subgroup of in the earlier diagram. For example, Gal(Q(b, 8) : Q(b2, 8)) = hd2i. The fundamental theorem also implies that the intermediate fields that are normal over Q are the boxed ones.

113 Quotients Finally, the fundamental theorem tells us that

Gal(Q(b, 8) : Q) Gal(! : Q) Gal(Q(b, 8) : !) whenever ! is an intermediate ﬁeld normal over Q. For ! = Q(b2, 8), this gives

/hd2i Gal(Q(b2, 8) : Q). (8.4)

The left-hand side is the quotient of 4 by a subgroup isomorphic to 2. It has order 4, but it has no element of order 4: for the only elements of of order 4 ±1 2 2 are d , whose images in /hd i have order 2. Hence /hd i 2 × 2. On the other hand, Q(b2, 8) is the splitting field over Q of (C2 − 2)(C2 + 1), which by Example 8.2.4(ii) has Galois group 2 ×2. This confirms the isomorphism (8.4). Normal subgroups The other three intermediate fields normal over Q, I leave to you: and normal extensions Exercise 8.3.5 Choose one of Q(b2), Q(8) or Q(b28), and do the same for it as I just did for Q(b2, 8).

As you’ve now seen, it can take quite some time to work through a particular example of the Galois correspondence. You’ll get practice at doing this in workshop questions. Beyond examples, there are at least two things we can do with the fundamental theorem of Galois theory. The first is to resolve the old question on solvability of polynomials by radicals, which we met back in Chapter1. The second is to work out the structure of finite fields. We will carry out these two missions in the remaining two weeks.

114 Chapter 9

Solvability by radicals

We began this course with a notorious old problem: can every polynomial be solved by radicals? Theorem 1.3.5 gave the answer and more: not only is it impossible to find a general formula that does it, but we can tell which specific polynomials can be solved by radicals. Introduction to Theorem 1.3.5 states that a polynomial over Q is solvable by radicals if and only Week 9 if it has the right kind of Galois group—a solvable one. In degree 5 and higher, there are polynomials that have the wrong kind of group. These polynomials are not, therefore, solvable by radicals. We’ll prove one half of this ‘if and only if’ statement: if 5 is solvable by radicals then GalQ( 5 ) is solvable. This is the half that’s needed to show that some polynomials are not solvable by radicals. The proof of the other direction is in Chapter 18 of Stewart’s book, but we won’t do it. If you’re taking Algebraic Topology, you’ll already be familiar with the idea that groups can be used to solve problems that seem to have nothing to do with groups. You have a problem about some objects (such as topological spaces or field extensions), you associate groups with those objects (maybe their fundamental groups or their Galois groups), you translate your original problem into a problem about groups, and you solve that instead. For example, the question of whether R2 and R3 are homeomorphic is quite difficult using only general topology; but using algebraic topology, we can answer ‘no’ by noticing that the fundamental group of R2 with a point removed is not isomorphic to the fundamental group of R3 with a point removed. In much the same way, we’ll answer a difficult question about field extensions by converting it into a question about groups. For this chapter, you’ll need to remember something about solvable groups. At a minimum, you’ll need the definition, the fact that any quotient of a solvable group is solvable, and the fact that (5 is not solvable.

115 9.1 Radicals

We speak of square roots, cube roots, and so on, but we also speak about roots of polynomials. To distinguish between these two related usages, we will use the word ‘radical’ for square roots etc. (Radical comes from the Latin word for root. A radish is a root, and a change or policy is radical if it gets right down to the roots of the matter.) Back in Chapter1, I said that a complex number is called radical if ‘it can be obtained from the rationals using only the usual arithmetic operations [addition, subtraction, multiplication and division] and :th roots [for : ≥ 1]’. As an example, I said that

p3 √ √ 1 + 7 2 − 2 7 2 (9.1) r q 4 5 2 6 + 3

is radical, whichever square root, cube root, etc., we choose (p. 12). Let’s now make this deﬁnition precise. √ The ﬁrst point is that the notation = I or I1/= is highly dangerous:

I = ≥ Warning 9.1.1 Let be a complex√ number and 2. Then there = I I1/= = U is no single number called or √ . There are elements of C such that U= = I. So, the notation = I or I1/= makes no sense if it is intended to denote a single complex number. It is simply invalid. I + When√ belongs to the set R of nonnegative reals, the convention is that = I or I1/= denotes the unique U ∈ R+ such that U= = I. There is I = also a√ widespread convention that when is a negative real and is odd, = I or I1/= denotes the unique real U such that U= = I. In these cases, there is a sensible and systematic way of choosing one of the =th roots of I. But for a general I and =, there is not. Complex analysis has a lot to say about diﬀerent choices of =th roots. But we don’t need to go into that. We simply treat all the =th roots of I on an equal footing, not attempting to pick out any of them as special.

With√ this warning in mind, we define the radical numbers without using notation like = I or I1/=. Definition 9.1.2 Let Qrad be the smallest subfield of C such that for U ∈ C, The definition of radical number U= ∈ Qrad for some = ≥ 1 ⇒ U ∈ Qrad. (9.2)

A complex number is radical if it belongs to Qrad.

116 So any rational number is radical; the sum, product, difference or quotient of radical numbers is radical; any =th root of a radical number is radical; and there are no more radical numbers than can be obtained by those rules. For the definition of Qrad to make sense, we need there to be a smallest subfield of C with the property (9.2). This will be true as long as the intersection of any family of subfields of C satisfying (9.2) is again a subfield of C satisfying (9.2): for then Qrad is the intersection of all subfields of C satisfying (9.2).

Exercise 9.1.3 Check that the intersection of any family of subfields of C satisfying (9.2) is again a subfield of C satisfying (9.2). (That any intersection of subfields is a subfield is a fact we met back on p. 20; the new aspect is (9.2).)

Example 9.1.4 Consider again the expression (9.1). It’s not quite as random as it looks. I chose it so that the various radicals are covered by one of the two conventions mentioned in Warning 9.1.1: they’re all =th roots of positive reals p3 √ √ except for 7 2 − 2 7, which is an odd root of a negative real. Let I be the number (9.1), choosing the radicals according to those conventions. I claim that I is radical, or equivalently that I belongs to every subfield of C satisfying (9.2). First, Q ⊆ since Q is the prime subfield of C. So 2/3 ∈ , and so p5 2/3 ∈ by (9.2). Also, 6 ∈ and is a field, so 6 + p5 2/3 ∈ . But then by (9.2) again, the denominator of (9.1) is in . A similar argument shows that the numerator is in . Hence I ∈ . Definition 9.1.5 A polynomial over Q is solvable by radicals if all of its complex roots are radical. The simplest nontrivial example of a polynomial solvable by radicals is something of the form C= − 0, where 0 ∈ Q. The theorem we’re heading for is that any polynomial solvable by radicals has solvable Galois group, and if that’s true = then the group GalQ(C − 0) must be solvable. Let’s consider that group now. The results we prove about it will form part of the proof of the big theorem. We begin with the case 0 = 1. = Lemma 9.1.6 For all = ≥ 1, the group GalQ(C − 1) is abelian. Proof Write l = 42c8/=. The complex roots of C= − 1 are 1, l, . . . , l=−1, so = SFQ(C − 1) = Q(l). = = Let i, \ ∈ GalQ(C − 1). Since i permutes the roots of C − 1, we have i(l) = l8 for some 8 ∈ Z. Similarly, \(l) = l 9 for some 9 ∈ Z. Hence (i ◦ \)(l) = i(l 9 ) = i(l) 9 = l8 9 ,

117 8 9 = and similarly (\ ◦ i)(l) = l . So (i ◦ \)(l) = (\ ◦ i)(l). Since SFQ(C − 1) = Q(l), it follows that \ ◦ i = i ◦ \.

Exercise 9.1.7 In the last sentence of that proof, how exactly does it ‘follow’?

Much more can be said about the Galois group of C= − 1, and you’ll see a bit more at Workshop 5. But this is all we need for our purposes. Now that we’ve considered C= − 1, let’s do C= − 0 for an arbitrary 0. Lemma 9.1.8 Let be a ﬁeld and = ≥ 1. Suppose that C= − 1 splits in . Then = Gal (C − 0) is abelian for all 0 ∈ . The hypothesis that C= − 1 splits in might seem so restrictive as to make this lemma useless. For instance, it doesn’t hold in Q or even R (for = > 2). Nevertheless, this turns out to be the key lemma in the whole story of solvability by radicals. = Proof If 0 = 0 then Gal (C − 0) is trivial; suppose otherwise. = = Choose a root b of C − 0 in SF (C − 0). For any other root a, we have (a/b)= = 0/0 = 1 (valid since 0 ≠ 0), and C= − 1 splits in , so a/b ∈ . = = It follows that SF (C − 0) = (b). Moreover, given i, \ ∈ Gal (C − 0), we have i(b)/b ∈ (since i(b) is a root of C= − 0), so i(b) i(b) i(b)\(b) (\ ◦ i)(b) = \ · b = · \(b) = . b b b

= Similarly, (i◦\)(b) = i(b)\(b)/b, so (\◦i)(b) = (i◦\)(b). Since SF (C −0) = (b), it follows that i ◦ \ = \ ◦ i.

Warning 9.1.9 For 0 ∈ Q, the Galois group of C= − 0 over Q is not usually abelian. For instance, you saw on Assignment 4 that 3 GalQ(C − 2) is the nonabelian group (3.

Exercise 9.1.10 What does the proof of Lemma 9.1.8 tell you about = the eigenvectors and eigenvalues of the elements of Gal (C − 0)?

= Exercise 9.1.11 Use Lemmas 9.1.6 and 9.1.8 to show that GalQ(C −0) is solvable for all 0 ∈ Q. This is harder than most of these exercises, but I recommend it as a way of getting into the right frame of mind for the theory that’s coming in Section 9.2.

118 Digression 9.1.12 We’re only going to do the theory of solvability by radicals over Q. It can be done over any field, but Q has two special features. First, Q can be embedded into an algebraically closed field that we know very well, namely, C. This makes some things easier. Second, char Q = 0. For fields of characteristic ?, the proof that any polynomial with a solvable Galois group is solvable by radicals has some extra complications.

9.2 Solvable polynomials have solvable groups

Here we’ll prove that every polynomial over Q that is solvable by radicals has solvable Galois group. You know by now that in Galois theory, we tend not to jump straight from polynomials to groups. We go via the intermediate stage of ﬁeld extensions, as in the diagram polynomial 7−→ ﬁeld extension 7−→ group

that I first drew after the definition of Gal ( 5 ) (page 77). That is, we understand polynomials through their splitting field extensions. So it shouldn’t be a surprise that we do the same here, defining a notion of ‘solvable extension’ and showing (roughly speaking) that

solvable polynomial 7−→ solvable extension 7−→ solvable group.

In other words, we’ll define ‘solvable extension’ in such a way that (i) if 5 ∈ Q[C] is a polynomial solvable by radicals then SFQ( 5 ) : Q is a solvable extension, and (ii) if " : is a solvable extension then Gal(" : ) is a solvable group. Hence Solvable if 5 is solvable by radicals then GalQ( 5 ) is solvable—the result we’re aiming for. polynomials have (I glossed over some details in that paragraph; we’ll get to those.) solvable groups: a map Definition 9.2.1 Let " : be a finite normal separable extension. Then " : is solvable (or " is solvable over ) if there exist A ≥ 0 and intermediate fields

= !0 ⊆ !1 ⊆ · · · ⊆ !A = "

such that !8 : !8−1 is normal and Gal(!8 : !8−1) is abelian for each 8 ∈ {1,...,A}.

Exercise 9.2.2 Let # : " : be extensions, with # : ", " : and # : all ﬁnite, normal and separable. Show that if # : " and " : are solvable then so is # : .

We will focus on subﬁelds of C, where separability is automatic (Exam- ple 7.2.15(i)).

119 = Example 9.2.3 Let 0 ∈ Q and = ≥ 1. Then SFQ(C − 0) : Q is a finite normal separable extension, being a splitting field extension over Q. I claim that it is solvable. = Proof: if 0 = 0 then SFQ(C − 0) = Q, and Q : Q is solvable (taking A = 0 and !0 = Q in Definition 9.2.1). Now assume that 0 ≠ 0. Choose a complex root b of C= − 0 and write l = 42c8/=. Then the complex roots of C= − 0 are b, lb, . . . , l=−1b.

= 8 8 = = So SFQ(C − 0) contains (l b)/b = l for all 8, and so C − 1 splits in SFQ(C − 0). Hence = = Q ⊆ SFQ(C − 1) ⊆ SFQ(C − 0). = Now SFQ(C − 1) : Q is normal (being a splitting field extension) and has abelian = = Galois group by Lemma 9.1.6. Also SFQ(C −0) : SFQ(C −1) is normal (being the = = splitting field extension of C − 0 over SFQ(C − 1), by Lemma 6.2.13(ii)), and has = abelian Galois group by Lemma 9.1.8. So SFQ(C − 0) : Q is a solvable extension, as claimed. The definition of solvable extension bears a striking resemblance to the definition of solvable group. Indeed: Lemma 9.2.4 Let " : be a finite normal separable extension. Then " : is solvable ⇐⇒ Gal(" : ) is solvable. Proof We will only need the ⇒ direction, and that is all I prove here. For the converse, see Workshop 5. Suppose that " : is solvable. Take intermediate fields

= !0 ⊆ !1 ⊆ · · · ⊆ !A = " as in Deﬁnition 9.2.1. For each 8 ∈ {1,...,A}, the extension " : !8−1 is ﬁnite, normal and separable (by Corollary 7.1.6 and Lemma 7.2.17), so we can apply the fundamental theorem of Galois theory to it. Since !8 : !8−1 is a normal extension, Gal(" : !8) is a normal subgroup of Gal(" : !8−1) and

Gal(" : !8−1) Gal(!8 : !8−1). Gal(" : !8) By hypothesis, the right-hand side is abelian, so the left-hand side is too. So the sequence of subgroups

Gal(" : ) = Gal(" : !0) ⊇ Gal(" : !1) ⊇ · · · ⊇ Gal(" : !A ) = 1 exhibits Gal(" : ) as a solvable group.

120 Exercise 9.2.5 Prove the ⇐ direction of Lemma 9.2.4. It’s a very similar argument to the proof of ⇒.

According to the story I’m telling, solvability by radicals of a polynomial should correspond to solvability of its splitting field extension. Thus, the subfields of C that are solvable over Q should be exactly the splitting fields SFQ( 5 ) of polynomials 5 that are solvable by radicals. (This is indeed true, though we won’t entirely prove it.) Now if 5 , 6 ∈ Q[C] are both solvable by radicals then so is 5 6, and SFQ( 5 6) is a solvable extension of Q containing both SFQ( 5 ) and SFQ(6). So it should be the case that for any two subfields of C solvable over Q, there is some larger subfield, also solvable over Q, containing both. We now prove this.

Lemma 9.2.6 Let ! and " be subfields of C such that the extensions ! : Q and " : Q are finite, normal and solvable. Then there is some subfield # of C such that ! ∪ " ⊆ # and # : Q is also finite, normal and solvable.

Proof Take subﬁelds

Q = !0 ⊆ · · · ⊆ !A = !, Q = "0 ⊆ · · · ⊆ "B = " such that !8 : !8−1 is normal with abelian Galois group for each 8, and similarly for "9 . There is a chain of subﬁelds

Q = !0 ⊆ · · · ⊆ !A = ! = "0(!) ⊆ · · · ⊆ "B (!) = "(!) (9.3) of C, where "9 (!) is the subfield of C generated by "9 ∪ ! (an instance of the notation (.) of Definition 4.1.10). Put # = "(!). Certainly ! ∪ " ⊆ #. We show that # : Q is finite, normal and solvable. Since ! : Q is finite and normal, ! = SFQ( 5 ) for some 5 ∈ Q[C], and similarly, " = SFQ(6). Now # is the subfield of C generated by ! ∪ ", so it is generated by the roots of 5 and the roots of 6, or equivalently the roots of 5 6. So # = SFQ( 5 6), which is finite and normal over Q. To see that # : Q is solvable, we show that each successive extension in (9.3) is normal with abelian Galois group. For those to the left of !, this is immediate. For those to the right, let 9 ∈ {1,...,B}. Since "9 : "9−1 is fi- " ℎ ℎ " C " ! nite and normal, 9 = SF" 9−1 ( ) for some ∈ 9−1 [ ]. Then 9 ( ) = ℎ " ! " ! SF" 9−1 (!) ( ) by Lemma 6.2.13(i), so 9 ( ) : 9−1( ) is normal. Its Ga- ℎ lois group is Gal" 9−1 (!) ( ), which by Corollary 6.3.13 is isomorphic to a sub- ℎ ℎ " " group of Gal" 9−1 ( ). But Gal" 9−1 ( ) is Gal( 9 : 9−1), which is abelian, so Gal("9 (!) : "9−1(!)) is abelian.

121 Digression 9.2.7 This proof, and some others in this section, can be unclut- tered slightly using the notion of compositum. By definition, the compositum of subfields !," of a field is the smallest subfield containing ! and ". For example, the # of the proof is the compositum of ! and ". In our usual notation, the compositum of ! and " is "(!) or !("). The standard notation for the compositum is !", which has the advantage of being symmetric and the disadvantage of being misleading: it is not the set {UV : U ∈ !,V ∈ "} (although it is the smallest subfield of containing that set). The various fields used in the proof of Lemma 9.2.6, together with the fields !8 (") = !8 ", can be drawn like this:

!"B !A "

......

!"0 !0 " ! "

!A "B

......

!0 "0 Q

I have chosen not to use the compositum explicitly in this course, but once you know about it, you’ll notice how often it appears implicitly in Galois theory proofs.

The heart of the proof that solvable polynomials have solvable Galois groups is the following lemma (which in turn depends on Lemmas 9.1.6 and 9.1.8 on the Galois groups of C= −1 and C= −0). Loosely, it says that the set of complex numbers that can be reached from Q by solvable extensions is closed under taking =th roots. Write

Qsol = {U ∈ C : U ∈ ! for some subﬁeld ! ⊆ C that is ﬁnite, normal and solvable over Q}.

In fact, Qsol = Qrad, but we don’t know that yet.

Lemma 9.2.8 Let U ∈ C and = ≥ 1. If U= ∈ Qsol then U ∈ Qsol.

122 The proof (below) is slightly subtle. Here’s why. Let ! be a subfield of C that’s finite, normal and solvable over Q, and take U ∈ C and = ≥ 1 such that U= ∈ !. To find some larger " that contains U = = itself and is also solvable over Q, we could try putting " = SF! (C − U ). Now " : Q is indeed solvable (as can be shown using Exercises 9.1.11 and 9.2.2), but the trouble is that " : Q is not in general normal. And normality is part of the definition of Qsol, ultimately because it’s an essential requirement if we want to use the fundamental theorem of Galois theory. An example should clarify. √ √ √ U 4 = U2 Example√ 9.2.9 Put = 2 and take = 2. We have = 2 ∈ Q( 2), and Q( 2) : Q is finite, normal and solvable (since its Galois group is√ the abelian 2 sol 4 group 2), so U ∈ Q . Hence, according to Lemma 9.2.8, U = 2 should be " contained in some finite normal solvable extension of Q. √ How can we find such an "? We can’t take " = SF √ (C2 − 2), since this √ Q( 2) 4 is Q( 2), which√ is not√ normal over Q. (You may have already contemplated the extensions Q( 4 2) : Q( 2) : Q in Workshop 4, q. 4.) " To find a bigger , still finite√ and solvable over Q but also normal,√ we have to adjoin a square root not just of 2 but also of its conjugate, − 2. This is the crucial point: the whole idea of normality is that conjugates are treated equally. (Normal behaviour means that anything√ you do for one element, you do for all 4 , 8 C4 its conjugates.) The result is Q( 2 ) = SFQ(√ − 2), which is indeed a finite, solvable and normal extension of Q containing 4 2.

Proof of Lemma 9.2.8 Write 0 = U= ∈ Qsol. Choose a subﬁeld of C such that 0 ∈ and : Q is ﬁnite, normal and solvable.

= = Step 1: enlarge to a field in which C − 1 splits. Put ! = SF (C − 1) ⊆ C. Since : Q is finite and normal, = SFQ( 5 ) for some 5 ∈ Q[C], and then = ! = SFQ (C − 1) 5 (C) . Hence ! : Q is finite and normal. It follows from = Corollary 7.1.6 that ! : is normal. Its Galois group is Gal (C − 1), which is = isomorphic to a subgroup of GalQ(C − 1) (by Corollary 6.3.13), which is abelian (by Lemma 9.1.6). Hence Gal(! : ) is abelian. Also : Q is solvable, so ! : Q is solvable. In summary, ! is a subfield of C such that 0 ∈ ! and ! : Q is finite, normal and solvable, and, moreover, C= − 1 splits in !. We now forget about .

Step 2: adjoin the =th roots of the conjugates of 0. Write < ∈ Q[C] for the = minimal polynomial of 0 over Q, and put " = SF! (<(C )) ⊆ C. Then U ∈ ", as <(U=) = <(0) = 0. We show that " : Q is ﬁnite, normal and solvable.

123 Since ! : Q is finite and normal, ! = SFQ(6) for some 6 ∈ Q[C]. Then = " = SFQ(6(C)<(C )), so " : Q is finite and normal. It follows that " : ! is finite and normal too (by Corollary 7.1.6). To show that " : Q is solvable, it is enough to show that " : ! is solvable (by Exercise 9.2.2). Since ! : Q is normal and < ∈ Q[C] is the minimal polynomial of 0 ∈ !, it follows by definition of normality that < splits in !, say B Ö <(C) = (C − 08) 8=1

(08 ∈ !). Deﬁne subﬁelds !0 ⊆ · · · ⊆ !B of C by

!0 = ! ! C= 0 1 = SF!0 ( − 1) ! C= 0 2 = SF!1 ( − 2) . . ! C= 0 . B = SF!B−1 ( − B) Then = !8 = ! V ∈ " : V ∈ {01, . . . , 08} .

In particular, !B = ". For each 8 ∈ {1,...,B}, the extension !8 : !8−1 is finite and normal (being a splitting field extension), and its Galois group is abelian (by = Lemma 9.1.8 and the fact that C − 1 splits in ! ⊆ !8−1). So " : ! is solvable. Now we can relate the set Qrad of radical numbers, defined in terms of basic arithmetic operations, to the set Qsol, defined in terms of field extensions. Proposition 9.2.10 Qrad ⊆ Qsol. That is, every radical number is contained in some subfield of C that is a finite, normal, solvable extension of Q. As I mentioned, Qrad and Qsol are actually equal, although we won’t prove this. Proof By definition of Qrad, it is enough to show that Qsol is a subfield of C with the property that U= ∈ Qsol ⇒ U ∈ Qsol. We have just proved that property (Lemma 9.2.8), so it only remains to show that Qsol is a subfield of C. The argument is similar to the proof that the algebraic numbers form a field (Proposition 5.3.7). Let U,V ∈ Qsol. Then U ∈ ! and V ∈ " for some !," that are finite, normal and solvable over Q. By Lemma 9.2.6, U,V ∈ # for some # that is finite, normal and solvable over Q. Then U+V ∈ #, so U+V ∈ Qsol, and similarly U · V ∈ Qsol. This shows that Qsol is closed under addition and multiplication. The other parts of the proof (negatives, reciprocals, 0 and 1) are straightforward. This brings us to the main result of this chapter. Notice that it doesn’t mention field extensions: it goes straight from polynomials to groups.

124 Theorem 9.2.11 Let 5 ∈ Q[C]. If the polynomial 5 is solvable by radicals then the group GalQ( 5 ) is solvable.

Proof Suppose that 5 is solvable by radicals. Write U1,...,U= ∈ C for its roots. rad For each 8, we have U8 ∈ Q (by definition of solvability by radicals), hence sol U8 ∈ Q (by Proposition 9.2.10). So each of U1,...,U= is contained in some subfield of C that is finite, normal and solvable over Q. By Lemma 9.2.6, there is some subfield " of C that is finite, normal and solvable over Q and contains all of U1,...,U=. Thus, SFQ( 5 ) = Q(U1,...,U=) ⊆ ".

By Lemma 9.2.4, Gal(" : Q) is solvable. Now SFQ( 5 ) : Q is normal, so by the fundamental theorem of Galois theory, Gal(SFQ( 5 ) : Q) is a quotient of Gal(" : Q). But Gal(SFQ( 5 ) : Q) = GalQ( 5 ), and a quotient of a solvable group is solvable, so GalQ( 5 ) is solvable.

Examples 9.2.12 i. For 0 ∈ Q and = ≥ 1, the polynomial C= − 0 is solvable by = radicals, so the group GalQ(C −0) is solvable. You may already have proved this in Exercise 9.1.11. It also follows from Example 9.2.3 and Lemma 9.2.4.

= ii. Let 01, . . . , 0: ∈ Q and =1, . . . , =: ≥ 1. Each of the polynomials C 8 − 08 is Î C=8 0 solvable by radicals, so their product is too. Hence GalQ 8 ( − 8) is a solvable group.

Theorem 9.2.11 is most sensational in its contrapositive form: if GalQ( 5 ) is not solvable then 5 is not solvable by radicals. That’s the subject of the next section.

Digression 9.2.13 The converse of Theorem 9.2.11 is also true: if GalQ ( 5 ) is solvable then 5 is solvable by radicals. You can even unwind the proof to obtain an explicit formula for the solving the quartic by radicals (Stewart, Chapter 18). To prove this converse statement, we have to deduce properties of a field extension from assumptions about its Galois group. A solvable group is built up from abelian groups, and every finite abelian group is a direct sum of cyclic groups. The key step in the proof of the converse has come to be known as ‘Hilbert’s Theorem 90’ (Stewart’s Theorem 18.18), which gives information about any field extension whose Galois group is cyclic.

Digression 9.2.14 The proof of Theorem 9.2.11 might not have ended quite how you expected. Given my explanations earlier in the chapter, you might justiﬁably have imagined we were going to show that when the polynomial 5 is solvable by radicals, the extension SFQ ( 5 ) : Q is solvable. That’s not

125 what we did. We showed that SFQ ( 5 ) is contained in some larger subﬁeld " such that " : Q is solvable, then used that to prove the solvability of the group GalQ ( 5 ).

But all is right with the world: SFQ ( 5 ) : Q is a solvable extension. Indeed, its Galois group GalQ ( 5 ) is solvable, so Lemma 9.2.4 implies that SFQ ( 5 ) : Q is solvable too.

9.3 An unsolvable polynomial

Here we give a speciﬁc example of a polynomial over Q that is not solvable by radicals. By Theorem 9.2.11, our task is to construct a polynomial whose Galois group is not solvable. The smallest non-solvable group is 5 (of order 60). Our polynomial has Galois group (5 (of order 120), which is also non-solvable. Finding Galois groups is hard, and we will use a whole box of tools and tricks, from Cauchy’s theorem on groups to Rolle’s theorem on diﬀerentiable functions. First we establish a useful fact on the order of Galois groups.

Lemma 9.3.1 Let 5 be an irreducible polynomial over a ﬁeld , with SF ( 5 ) : separable. Then deg( 5 ) divides | Gal ( 5 )|.

Proof Let U be a root of 5 in SF ( 5 ). By irreducibility, deg( 5 ) = [ (U) : ], which divides [SF ( 5 ) : ] by the tower law, which is equal to | Gal ( 5 )| by Theorem 7.2.19 (using separability).

Next, we need some results about the symmetric group (=. I assume you know that (= is generated by the ‘adjacent transpositions’ (12), (23),..., (= − 1 =). This may have been proved in Fundamentals of Pure Mathematics, and as the Group Theory notes say (p. 58):

This is intuitively clear: suppose you have = people lined up and you want them to switch into a diﬀerent order. To put them in the order you want them, it’s clearly enough to have people move up and down the line; and each time a person moves one place, they switch places with the person next to them.

Here’s a diﬀerent way of generating (=.

Lemma 9.3.2 For = ≥ 2, the symmetric group (= is generated by (12) and (12 ··· =).

126 Proof We have (12 ··· =)(12)(12 ··· =)−1 = (23), −1 either by direct calculation or the general fact that f(01 ··· 0: )f = (f(01)··· f(0: )) for any f ∈ (= and cycle (01 ··· 0: ). So any subgroup of (= containing (12) and (12 ··· =) also contains (23). By the same argument, also contains (34),..., (= − 1 =). But the adjacent transpositions generate (=, so = (=.

Lemma 9.3.3 Let ? be a prime number, and let 5 ∈ Q[C] be an irreducible polynomial of degree ? with exactly ? − 2 real roots. Then GalQ( 5 ) (?.

Proof Since char Q = 0 and 5 is irreducible, 5 is separable and therefore has ? distinct roots in C. By Proposition 6.3.10, the action of GalQ( 5 ) on the roots of 5 in C deﬁnes an isomorphism between GalQ( 5 ) and a subgroup of (?. Since 5 is irreducible, ? divides | GalQ( 5 )| = || (by Lemma 9.3.1). So by Cauchy’s theorem, has an element f of order ?. Then f is a ?-cycle, since these are the only elements of (? of order ?. The complex conjugate of any root of 5 is also a root of 5 , so complex conjugation restricts to an automorphism of SFQ( 5 ) over Q. Exactly two of the roots of 5 are non-real; complex conjugation transposes them and ﬁxes the rest. So contains a transposition g. Without loss of generality, g = (12). Since f is a ?-cycle, fA (1) = 2 for some A ∈ {1,...,? − 1}. Since ? is prime, fA also has order ?, so it is a ?-cycle. Now without loss of generality, fA = (123 ··· ?). So (12), (12 ··· ?) ∈ , forcing = (? by Lemma 9.3.2. Hence GalQ( 5 ) (?.

Exercise 9.3.4 Explain why, in the last paragraph, fA has order ?.

Theorem 9.3.5 Not every polynomial over Q of degree 5 is solvable by radicals.

Proof We show that 5 (C) = C5 − 6C + 3 satisﬁes the conditions of Lemma 9.3.3. Then GalQ( 5 ) is (5, which is not solvable, so by Theorem 9.2.11, 5 is not solvable by radicals. Evidently deg( 5 ) is the prime number 5, and 5 is irreducible by Eisenstein’s criterion with prime 3. It remains to prove that 5 has exactly 3 real roots. This is where we use some analysis, considering 5 as a function R → R (Figure 9.1). We have

lim 5 (G) = −∞, 5 (0) > 0, 5 (1) < 0, lim 5 (G) = ∞, G→−∞ G→∞

127 Figure 9.1: The function G 7→ G5 − 6G + 3. and 5 is continuous on R, so by the intermediate value theorem, 5 has at least 3 real roots. On the other hand, 5 0(G) = 5G4 − 6 has only 2 real roots (±p4 6/5), so by Rolle’s theorem, 5 has at most 3 real roots. Hence 5 has exactly 3 real roots, as required.

Exercise 9.3.6 Prove that for every = ≥ 5, there is some polynomial of degree = that is not solvable by radicals.

Digression 9.3.7 We now know that some polynomials 5 over Q are not solvable by radicals, which means that not all their complex roots are radical. Could it be that some of the roots are radical and others are not? Yes: simply take a polynomial 6 that is not solvable by radicals and put 5 (C) = C6(C). Then the roots of 5 are 0 (which is radical) together with the roots of 6 (which are not all radical). But what if 5 is irreducible? In that case, either all the roots of 5 are radical or none of them are. This follows from the fact that the extension Qrad : Q is normal, which we will not prove.

Digression 9.3.8 There are many similarities between the theory of constructibility of points by ruler and compass and the theory of solvability of polynomials by radicals. In both cases, the challenge is to construct some things (points in the plane or roots of polynomials) using only certain tools (ruler and compass or a machine for taking =th roots). In both cases, there were diﬃcult questions of constructibility that remained open for a very long time, and in both cases, they were solved by Galois theory. The solutions have something in common too. For the geometry problem, we used iterated quadratic extensions, and for the polynomial problem, we used solvable extensions, which could reasonably be called iterated abelian extensions. For the geometry problem, we showed that the coordinates of any

128 point constructible by ruler and compass satisfy a certain condition on their degree over Q (Corollary 5.4.3); for the polynomial problem, we showed that any polynomial solvable by radicals satisﬁes a certain condition on its Galois group. There are other similarities: compare Lemmas 5.4.4 and 9.2.6, for example, and maybe you can ﬁnd more similarities still.

We have now used the fundamental theorem of Galois theory to solve a major problem about Q. What else can we do with it? The fundamental theorem is about separable extensions. Our two main sources of separable extensions are:

• ﬁelds of characteristic 0 such as Q (Example 7.2.15(i)), which we’ve explored extensively already;

• ﬁnite ﬁelds (Example 7.2.15(ii)), which we’ve barely touched.

In the next and final chapter, we’ll use the fundamental theorem and other results we’ve proved to explore the world of finite fields. In contrast to the complicated world of finite groups, finite fields are almost shockingly simple.

129 Chapter 10

Finite ﬁelds

This chapter is dessert. Through this semester, we’ve developed a lot of sophisti- cated theory for general fields. All of it works for finite fields, but becomes much simpler there. It’s a miniature world in which life is sweet. For example: Introduction to Week 10 • If we want to apply the fundamental theorem of Galois theory to a field extension " : , we first have to ask whether it is finite, and whether it is normal, and whether it is separable. When " and are finite, all three conditions are automatic (Lemma 10.4.2).

• There are many fields of different kinds, and to classify them all would be a near-impossible task. But for finite fields, the classificiation is very simple, as we’ll see. We know exactly what finite fields there are and how many elements they have.

• The Galois correspondence for arbitrary field extensions can also be complicated. But again, it’s simple for finite fields. Their Galois groups are very easy (they’re all cyclic), we know what their subgroups are, and it’s easy to describe all the subfields of any finite field.

So although the world of finite fields is not trivial, there’s a lot about it that’s surprisingly straightforward. Two aspects of finite fields may seem counterintuitive. First, they always have positive characteristic, which means they satisfy some equation like 1+· · ·+1 = 0. Second (and relatedly), ?th powers and ?th roots behave strangely in fields of characteristic ?—at least, ‘strange’ if, like most of us, characteristic 0 is what you’re most familiar with. But the behaviour of ?th roots and powers is fundamental to all the nice properties of finite fields.

130 10.1 ?th roots in characteristic ?

Recall from Lemma 2.2.11 that every ﬁnite ﬁeld has positive characteristic, which, by Lemma 2.2.5, must be a prime number ?. Square roots usually√ come in pairs: how many times in your life have you written a ± sign before a ? But in characteristic 2, plus and minus are the same, so the two square roots become one. We’ll see that this pattern persists.

Proposition 10.1.1 Let ? be a prime number and ' a ring of characteristic ?.

i. The function \ : ' → ' A 7→ A ? is a homomorphism.

ii. If ' is a ﬁeld then \ is injective.

iii. If ' is a ﬁnite ﬁeld then \ is an automorphism of '.

Proof For (i), certainly \ preserves multiplication and 1. To show that \ preserves addition, let A,B ∈ ': then by Lemma 3.3.15 and the hypothesis that char ' = ?,

? Õ ? \(A + B) = (A + B) ? = A8 B?−8 = A ? + B? = \(A) + \(B). 8 8=0 Now (ii) follows since every homomorphism between ﬁelds is injective, and (iii) since every injection from a ﬁnite set to itself is bijective.

The homomorphism \ : A 7→ A ? is called the Frobenius map, or, in the case of ﬁnite ﬁelds, the Frobenius automorphism. That \ is a homomorphism is a shocker. Writing ‘(G + H)= = G= + H=’ is a classic mistake in school-level algebra. But here, it’s true!

Examples 10.1.2 i. The Frobenius automorphism of F? = Z/h?i is not very interesting. When is a ﬁnite group of order =, Lagrange’s theorem implies that 6= = 1 for all 6 ∈ . Applying this to the multiplicative group × 0?−1 0 0? 0 F? = F? \{0} gives = 1 whenever 0 ≠ ∈ F?. It follows that = for all 0 ∈ F?. That is, \ is the identity. Everything is its own ?th root! ii. The proof that the Frobenius map preserves addition may seem familiar. We essentially did it in Example 7.2.4, in the case of the ring (F? (D))[C] of polynomials over the ﬁeld F? (D) of rational expressions over F?.

131 iii. We don’t have many examples of ﬁnite ﬁelds yet (but we will soon). Apart from those of the form F?, the simplest is F2(U), where U is a root of the irre- 2 ducible polynomial 1+C +C over F2 (Example 4.3.8(ii)). By Theorem 5.1.5,

F2(U) = {0 + 1U : 0, 1 ∈ F2} = {0, 1,U, 1 + U}.

2 Since 1 + U + U = 0 and F2(U) has characteristic 2,

U2 = 1 + U, (1 + U)2 = U.

So the Frobenius automorphism of F2(U) interchanges U and 1 + U. Like all automorphisms, it ﬁxes 0 and 1.

Exercise 10.1.3 Write out the addition and multiplication tables of F2(U).

Corollary 10.1.4 Let ? be a prime number.

i. In a ﬁeld of characteristic ?, every element has at most one ?th root.

ii. In a ﬁnite ﬁeld of characteristic ?, every element has exactly one ?th root.

Proof Part (i) says that the Frobenius map is injective, and part (ii) says that it is bijective, as Proposition 10.1.1 states.

Examples 10.1.5 i. In a ﬁeld of characteristic 2, every element has at most one square root.

ii. In C, there are ? diﬀerent ?th roots of unity. But in a ﬁeld of characteristic ?, there is only one: 1 itself.

iii. Let be a ﬁeld of characteristic ? and 0 ∈ . Corollary 10.1.4(i) says that 0 has at most one ?th root. It may have none. For instance, Exercise 7.2.5 asked you to show that in F? (D), the element D has no ?th root.

iv. In the 4-element ﬁeld F2(U) of Example 10.1.2(iii), the only square root of U is 1 + U, and the only square root of 1 + U is U. Each is the square root of the other.

Exercise 10.1.6√Work out the values of the Frobenius automorphism on the ﬁeld F3( 2), which you ﬁrst met in Exercise 4.3.9.

132 10.2 Classification of finite fields

If you try to write down a formula for the number of groups or rings with a given number of elements, you’ll find that it’s hard and the results are quite strange. For instance, more than 99% of the first 50 billion groups have order 1024. But fields turn out to be much, much easier. We’ll obtain a complete classification of finite fields in the next two pages. The order of a finite field " is its cardinality, or number of elements, |"|.

Warning 10.2.1 Order and degree mean different things. For instance, if the order of a field is 9, then its degree over its prime subfield F3 is 2.

Lemma 10.2.2 Let " be a ﬁnite ﬁeld. Then char " is a prime number ?, and = |"| = ? where = = [" : F?] ≥ 1.

In particular, the order of a finite field is a prime power. Proof By Lemmas 2.2.5 and 2.2.11, char " is a prime number ?. By Lemma 2.2.10, " has prime subfield F?. Since " is finite, 1 ≤ [" : F?] < ∞; write = = [" : F?]. As a vector space over F?, then, " is =-dimensional and so = = = ?= " ?= isomorphic to F?. But |F? | = |F? | = , so | | = . Example 10.2.3 There is no field of order 6, since 6 is not a prime power.

Lemma 10.2.2 prompts two questions: given a prime power ?=, is there some field of order ?=? And if so, how many are there? The answer to the first question is yes: Lemma 10.2.4 Let ? be a prime number and = ≥ 1. Then the splitting field of ?= = C − C over F? has order ? . 5 C C ?= C C " 5 5 Proof Put ( ) = − ∈ F? [ ] and = SFF? ( ). Then = −1 (since = ≥ 1), so by (i)⇒(ii) of Lemma 7.2.10, 5 has no repeated roots in ". Hence " has at least ?= elements. Write \ for the Frobenius map of ". The set ! of roots of 5 in " is the set of = = fixed points of \ = \ ◦ · · · ◦ \, which is Eq{\ , id" }. Since \ is a homomorphism, ! is a subfield of " (by Lemma 7.3.2), and contains all the roots of 5 in ". Hence by definition of splitting field, ! = "; that is, every element of " is a root of 5 . Since deg( 5 ) = ?=, this implies that " has at most ?= elements.

As for the second question, there is exactly one ﬁeld of each prime power order. To show this, we need a lemma.

133 Lemma 10.2.5 Let " be a ﬁnite ﬁeld of order @. Then U@ = U for all U ∈ ". The proof uses the same argument as in Example 10.1.2(i). Proof The multiplicative group "× has order @−1, so Lagrange’s theorem implies that U@−1 = 1 for all U ∈ "× = " \{0}. Hence U@ = U whenever 0 ≠ U ∈ ", and clearly the equation holds for U = 0 too.

Exercise 10.2.6 Verify that V4 = V for all V in the 4-element ﬁeld F2(U) of Examples 10.1.2(iii) and 10.1.5(iv),

@ Lemma 10.2.7 Every finite field of order @ is a splitting field of C − C over F?. Proof Let " be a field of order @. By Lemma 10.2.2, @ = ?= for some prime ? and = ≥ 1, and char " = ?. Hence " has prime subfield F?. By Lemma 10.2.5, = every element of " is a root of 5 (C) = C ? − C. So 5 has |"| = ?= = deg( 5 ) distinct roots in ", and therefore splits in ". The set of roots of 5 in " generates ", since it is equal to ". Hence " is a splitting field of 5 . Together, these results completely classify the finite fields. Theorem 10.2.8 (Classification of finite fields) i. Every finite field has order ?= for some prime ? and integer = ≥ 1.

ii. For each prime ? and integer = ≥ 1, there is exactly one field of order ?=, up to isomorphism. It has characteristic ? and is a splitting field for ?= C − C over F?. Proof This is immediate from the results above together with the uniqueness of splitting fields (Theorem 6.2.12(ii)).

When @ > 1 is a prime power, we write F@ for the one and only ﬁeld of order @.

Warning 10.2.9 F@ is not Z/h@i unless @ is a prime. It can’t be, because Z/h@i is not a ﬁeld (Example 2.2.15). To my knowledge, there is no description of F@ simpler than the splitting ﬁeld description.

We now know exactly how many finite fields there are of each order. But in algebra, it’s important to think not just about the objects (such as vector spaces, groups, modules, rings, fields, . . . ), but also the maps (homomorphisms) between objects. So now that we’ve counted the finite fields, it’s natural to try to count the homomorphisms between finite fields. Field homomorphisms are injective, so this boils down to counting subfields and automorphisms. Galois theory is very well equipped to do that! We’ll come to this in the final section. But first, we consider another way in which finite fields are very simple.

134 10.3 Multiplicative structure

The multiplicative group × of a ﬁnite ﬁeld is as easy as can be:

Proposition 10.3.1 For an arbitrary field , every finite subgroup of × is cyclic. In particular, if is finite then × is cyclic. The multiplicative group of a finite Proof This was Theorem 5.1.13 and Corollary 5.1.14 of Group Theory. field is cyclic Example 10.3.2 In examples earlier in the course, we frequently used the =th root of unity l = 42c8/= ∈ C, which has the property that every other =th root of unity is a power of l. Can we find an analogue of l in an arbitrary field ? It’s not obvious how to generalize the formula 42c8/=, since the exponential is a concept from complex analysis. But Proposition 10.3.1 solves our problem. For = ≥ 1, put

= *= ( ) = {U ∈ : U = 1}.

× = Then *= ( ) is a subgroup of , and is ﬁnite since its elements are roots of C − 1. So by Proposition 10.3.1, *= ( ) is cyclic. Let l be a generator of *= ( ). Then every =th root of unity in is a power of l, which is what we were aiming for. Note, however, that *= ( ) may have fewer than = elements, or equivalently, the order of l may be less than =. For instance, if char = ? then *? ( ) is trivial and l = 1, by Example 10.1.5(ii).

Exercise 10.3.3 Let be a ﬁeld and let be a ﬁnite subgroup of × of order =. Prove that ⊆ *= ( ).

× ? Example 10.3.4 The group F? is cyclic, for any prime . This means that there is some l ∈ {1,...,? − 1} such that l, l2,... runs through all elements of {1,...,? − 1} when taken mod ?. In number theory, such an l is called a primitive root mod ? (another usage of the word ‘primitive’). For instance, you can check that 3 is a primitive root mod 7, but 2 is not, since 23 ≡ 1 (mod 7).

Corollary 10.3.5 Every extension of one ﬁnite ﬁeld over another is simple.

Proof Let " : be an extension with " ﬁnite. By Proposition 10.3.1, the group "× is generated by some element U ∈ "×. Then " = (U).

This is yet another pleasant aspect of ﬁnite ﬁelds.

135 Exercise 10.3.6 In the proof of Corollary 10.3.5, once we know that the group "× is generated by U, how does it follow that " = (U)?

Digression 10.3.7 In Digression 7.2.22, I mentioned the theorem of the primitive element: every finite separable extension " : is simple. One of the standard proofs involves splitting into two cases, according to whether " is finite or infinite. We’ve just done the finite case.

Corollary 10.3.8 For every prime number ? and integer = ≥ 1, there exists an irreducible polynomial over F? of degree =.

Proof The ﬁeld F?= has prime subﬁeld F?. By Corollary 10.3.5, the extension F?= : F? is simple, say F?= = F? (U). The minimal polynomial of U over F? is irreducible of degree [F? (U) : F?] = [F?= : F?] = =.

This is not obvious. For example, can you ﬁnd an irreducible polynomial of degree 100 over F31?

10.4 Galois groups for ﬁnite ﬁelds

We now work out the Galois correspondence for any extension of one ﬁnite ﬁeld over another.

Warning 10.4.1 The term ‘finite field extension’ means an extension " : that’s finite in the sense defined on p. 52: " is finite- dimensional as a vector space over . It doesn’t mean that " and are finite fields. But the safest policy is to avoid this term entirely.

The three hypotheses of the fundamental theorem of Galois theory are always satisfied when both fields in the extension are finite:

Lemma 10.4.2 Let " : be a ﬁeld extension.

i. If is ﬁnite then " : is separable.

ii. If " is also ﬁnite then " : is ﬁnite and normal.

Proof For (i), we show that every irreducible polynomial 5 over is separable. Write char = ? > 0, and suppose for a contradiction that 5 is inseparable. By Corollary 7.2.12, ? A? 5 (C) = 10 + 11C + · · · + 1AC

136 for some 10, . . . , 1A ∈ . For each 8, there is a (unique) ?th root 28 of 18 in , by Corollary 10.1.4(ii). Then 5 (C) 2? + 2?C ? + · · · + 2?CA?. = 0 1 A But by Proposition 10.1.1(i), the function 6 7→ 6? is a homomorphism [C] → [C], so A ? 5 (C) = (20 + 21C + · · · + 2AC ) . This contradicts 5 being irreducible. For (ii), suppose that " is finite. Write char " = ? > 0. By Theorem 10.2.8, " is a splitting field over F?, so by Lemma 6.2.13(ii), it is also a splitting field over . Hence " : is finite and normal, by Theorem 7.1.5.

Part (i) fulfils the promise made in Remark 7.2.13 and Example 7.2.15(ii), and the lemma as a whole lets us use the fundamental theorem freely in the world of finite fields. We now work out the Galois correspondence for the extension F?= : F? of an arbitrary finite field over its prime subfield.

Proposition 10.4.3 Let ? be a prime and = ≥ 1. Then Gal(F?= : F?) is cyclic of order =, generated by the Frobenius automorphism of F?= .

By Workshop 4, q. 7, Gal(F?= : F?) is the group of all automorphisms of F?= .

Proof Write \ for the Frobenius automorphism of F?= ; then \ ∈ Gal(F?= : F?). ?= First we calculate the order of \. By Lemma 10.2.5, U = U for all U ∈ F?= , or < equivalently, \= = id. If < is a positive integer such that \< = id then U? = U for ?< = = < all U ∈ F?= , so the polynomial C − C has ? roots in F?= , so ? ≤ ? , so = ≤ <. Hence \ has order =. On the other hand, [F?= : F?] = =, so by the fundamental theorem of Galois theory, | Gal(F?= : F?)| = =. The result follows.

Exercise 10.4.4 What is the ﬁxed ﬁeld of h\i ⊆ Gal(F?= : F?)?

In Fundamentals of Pure Mathematics or Group Theory, you presumably saw that the cyclic group of order = has exactly one subgroup of order : for each divisor : of =. (And by Lagrange’s theorem, there are no subgroups of other orders.)

Exercise 10.4.5 Refresh your memory by proving this fact about subgroups of cyclic groups.

In the case at hand, Gal(F?= : F?) = h\i =, and when : | =, the unique subgroup of order : is h\=/: i.

137 Proposition 10.4.6 Let ? be a prime and = ≥ 1. Then F?= has exactly one subﬁeld of order ?< for each divisor < of =, and no others. It is

 ?< U ∈ F?= : U = U .

Proof The subfields of F?= are the intermediate fields of F?= : F?, which by the fundamental theorem of Galois theory are precisely the fixed fields Fix() of =/: subgroups of Gal(F?= : F?). Any such is of the form h\ i with : | =, and

=/: ?=/: Fixh\ i = U ∈ F?= : U = U . The tower law and the fundamental theorem give

[F?= : F?] = = [Fixh\=/: i : F ] = = = , ? =/: =/: [F?= : Fixh\ i] |h\ i| : so | Fixh\=/: i| = ?=/: . As : runs through the divisors of =, the quotient =/: also runs through the divisors of =, so putting < = =/: gives the result.

Warning 10.4.7 The subfields of F?= are of the form F?< where < divides =, not < ≤ =. For instance, F8 has no subfield isomorphic to 3 2 F4 (that is, no 4-element subfield), since 8 = 2 , 4 = 2 , and 2 - 3.

Let < be a divisor of =. By Proposition 10.4.6, F?= has exactly one subﬁeld isomorphic to F?< . We can therefore speak of the extension F?= : F?< without < < ambiguity. Since F?< = Fixh\ i and h\ i =/<, it follows from the fundamental theorem that

Gal(F?= : F?< ) =/<. (10.1)

So in working out the Galois correspondence for F?= : F?, we have accidentally derived the Galois group of a completely arbitrary extension of ﬁnite ﬁelds. In the Galois correspondence for F?= : F?, all the extensions and subgroups involved are normal, either by Lemma 10.4.2 or because cyclic groups are abelian. For < | =, the isomorphism

Gal(F?= : F?) Gal(F?< : F?) Gal(F?= : F?< ) supplied by the fundamental theorem amounts to

= <. =/<

Alternatively, substituting : = =/<, this is =/: =/: .

138 ? Example 10.4.8 Consider the Galois correspondence for F?12 : F?, where is \ any prime. Writing for the Frobenius automorphism of F?12 , the subgroups of = Gal(F?12 : F?) are

12 h\ i 1 1 order 1

6 h\ i 2 order 2 4 h\ i 3 order 3 3 h\ i 4 order 4 2 h\ i 6 order 6

 = h\i 12 order 12 Their ﬁxed ﬁelds are

F?12 degree 1

F?6 degree 2

F?4 degree 3

F?3 degree 4

F?2 degree 6

F? degree 12

Here, ‘degree’ means the degree of F?12 over the subfield, and (for instance) the subfield of F?12 called F?4 is U U?4 U . ∈ F?12 : = F?4 \4 The Galois group Gal(F?12 : F?4 ) is h i 3, and similarly for the other subfields.

Exercise 10.4.9 What do the diagrams of Example 10.4.8 look like for ?8 in place of ?12? What about ?432? (Be systematic!)

Ordered sets In Workshop 5, you’ll be asked to work through the Galois correspondence for an arbitrary extension F?= : F?< of ﬁnite ﬁelds, but there’s not much more to do: almost all the work is contained in the case < = 1 that we have just done.

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