Galois Theory Master's Dissertation

Victor Summers

9th May 2013 Contents

Introduction 2

1 Rings, Irreducible Polynomials and Field Extensions 5

2 Separable Polynomials and the Cyclotomic Extensions 26

3 Automorphisms, Galois Groups and the Fundamental Theorem 34

4 General Polynomials and the Discriminant 50

5 Algebraic Closures and Composite Extensions 57

6 Solvability by Radicals and Galois’ Great Theorem 63

7 Solvability of the Symmetric Groups 68

Bibliography 72

1 Introduction

Galois theory is generally considered to be the birth of modern algebra. Unlike some areas of mathematics, it is motivated by a simple clear cut question. Most secondary school children are familiar with the so-called ”quadratic formula” which expresses the two complex roots of b pb2 4ac a polynomial of degree 2; the formula = ± is instilled in us by the end of 1,2 2a high school. So why isn’t a similar looking formula for cubic polynomials also pushed into our brains as well? Does such a thing even exist? The answers to these questions are 1. yes, there does exist an analogous cubic formula and even a quartic formula, and 2. the cubic and quartic formulas are wildly complicated in comparison. Nevertheless, such formulas do exist; cubics and quartics are ”solvable by radicals”. The question arises, are there also such formulas for expressing the roots of quintics? or more generally of polynomials of degree higher than 4? The quadratic formula has been known in one form or another for thousands of years but it was as late as circa 1550 when the cubic formula was first discovered, with the quartic formula coming to light shortly after that. The methods for arriving at such general formulas seemed very much like they might generalise to polynomials of degree 5 and above, yet for almost 200 years mathematicians tried and tried but got precisely nowhere. It was only when a select few mathematicians in the 18th and 19th centuries, most notably Joseph-Louis Lagrange, Niels Henrik Abel and E´variste Galois (pictured below), began to investigate the true underlying structure of the problem in its full abstraction that significant developments in algebra were made. Lagrange introduced entirely new ways of approaching the problem. For instance, instead of trying to find clever tricks for producing explicit formulas for the roots of polynomials, he would ask what kind of structure a solution must have if it were to exist. Also, he would refer to a solution as being “known” if he could prove it to exist. This type of existential thinking is, in my opinion, the true birth of modern algebra. Galois built greatly upon the ground work put in place by Lagrange and is generally credited as the inventor of the notion of a group which, although they can be seen lurking in the background in earlier works, were not brought

2 into the light and recognised for their profound nature until then. As it turns out, there can exist no such analogous formulas for the general quintic and above! This fact is known as the Abel-Runi theorem, named after Paolo Runi and Niels Henrik Abel. Runi produced many proofs between 1799 and 1813 all of which were rejected by the mathematical community due to his use of overly awkward and unsatisfying personal notations and conventions. Abel produced a proof of the unsolvability of the general quintic and polynomials of higher degree in 1823, which in fact turned out to be slightly incomplete, but nevertheless was deemed satisfactory. The problem with the Abel-Runi theorem is that it says very little about why this is the case and gives no information about when a particular polynomial may or may not be solvable by radicals. Galois’ brilliance was in shedding light on the answer to these questions, giving a nec- essary and sucient condition under which a given polynomial is solvable by radicals! Galois’ beautiful method can be summarised as follows: ”Give me any polynomial and I will assign a group to that polynomial in such a way that the possible forms of its roots are reflected in the structure of the group. This then enables us to transform the problem to a group theoretical question which is simpler to answer”.

In this project I aim to display the foundations upon which modern Galois theory stands, illustrating the core concepts with worked examples and culminating in a rigorous proof of what has been dubbed Galois great theorem.

Throughout the project I assume a knowledge of the basic terminology and concepts from ring theory. For example, I will not provide formal definitions of a ring, Euclidean rings, prin- cipal ideal domains and polynomial rings, and I will take for granted some minor theorems and lemmas such as Euclid’s lemma and ”zero divisors if and only if the cancellation law holds”. However, in the construction of objects which are of crucial importance to the end goal I will give full formal definitions, theorems and proofs. The only rings considered in this project are assumed to be commutative rings with a multiplicative identity.

3 Chapter 1

Rings, Irreducible Polynomials and Field Extensions

Our first discussion is motivated by the following question. The polynomial x2 +1 Q[x] has 2 no roots in Q, so is there some larger field E containing Q in which it does have a root? More generally we may ask, given an arbitrary polynomial f(x) F [x] which has no roots in the 2 field F , is there some field E containing F in which f(x) has a root? The answer is yes and we begin here by drawing upon some tools from ring theory which enable us to construct such objects, as well as other tools which will be of great use to us throughout.

Let R be a ring and I R be an ideal. Define a relation on R by ⇢

a b a b I. ⇠ () 2 We see that is an equivalence relation. Indeed, ⇠ i) Reflexivity: a a =0 I for each a R so a a for each a R; 2 2 ⇠ 2 ii) Symmetry: a b a b I b a = (a b) I b a; ⇠ () 2 () 2 () ⇠ iii) Transitivity: a b and b c a b, b c I = a c =(a b)+(b c) I ⇠ ⇠ () 2 ) 2 () a c. ⇠ Let us denote the equivalence class of a R by [a]; note that b [a] b = a + i for some 2 2 () i I so that [a]= a + i i I .LetR/I denote the set of all equivalence classes of , i.e 2 { | 2 } ⇠ R/I = [a] a R , equipped with the operations of addition, multiplication and negation { | 2 } defined for each a, b R by 2 [a]+[b]=[a + b]

5 [a][b]=[ab] [a]=[ a]. We call R/I the quotient ring of I in R. Of course we need to validate the use of the word ”ring” in this definition: Since the operations are defined on equivalence classes, in order for them to be well-defined we require the definitions to be independent of representatives. Suppose a0 a and b0 b, then ⇠ ⇠

( ) a0 = a + i and b0 = b + j for some i, j I. Therefore, a0 + b0 = a + b +(i + j) and i + j I 2 2 so that a0 + b0 a + b = [a0]+[b0]=[a + b]. ⇠ )

( ) a0b0 =(a + i)(b + j)=ab +(aj + ib + ij) and aj + ib + ij I. Therefore, a0b0 ab = 2 ⇠ ) [a0][b0]=[ab].

( ) a0 = (a + i)= a i and i I. Therefore, a0 a = [a0]=[ a]. 2 ⇠ ) All the axioms of a ring follow immediately from the corresponding properties of the ring R itself. In particular, observe that the zero element is 0 := [0] = 0+i i I = I. { | 2 } Important pieces of machinery are those maps between rings which preserve certain algebraic relations.

Definition 1.1. Let R and S be rings. A function : S is called a ring homomorphism ! (or ring map for short) if it preserves the ring operations. That is, for each a, b R 2

i) (1R)=1S

ii) (ab)=(a)(b)

iii) (a + b)=(a)+(b), where 1R and 1S denote the multiplicative identities of R and S respectively.

In general, a ring map needn’t be injective or surjective. In the case where a ring map is both injective and surjective, we call it an isomorphism and say that the rings R and S are isomorphic; write R ⇠= S. In this sense we identify rings which are isomorphic to one another; although the two rings may be very di↵erent in appearance and their respective operations may even appear dissimilar, they share all ring theoretic properites and are thus the same up to relabelling of elements. We will use this mode of identification throughout this project. These isomorphisms also provide a way of assigning the afore mentioned group to a polynomial!

6 Theorem 1.2 (The first ring isomorphism theorem). Let R and S be rings amd let : R S ! be a ring map. Then, R ker() ⇠= Im() . and an isomorphism is given by the map h : R ker() Im() with ! . h([a]) := (a).

Proof. We first need to show that the definition of h is independent of representatives as it is defined on equivalence classes. To this end, suppose a0 a so that a0 = a + i for some ⇠ i ker()=: I. 2 Then,

h([a0]) = (a0) = (a + i) = (a)+(i) = (a)(asi ker()) 2 = h([a]) as required. h is a ring map:

h([a]+[b]) = h([a + b]) = (a + b)=(a)+(b)=h([a]) + h([b]) • h([a][b]) = h([ab]) = (ab)=(a)(b)=h([a])h([b]) • h(1 = [1]) = (1) = 1. • To show injectivity it suces to show that ker(h)= I , as the zero element of the quotient { } ring is I. Indeed,

h([a]) = 0 (a)=0 a I so that [a]=I. () () 2 Lastly, every member a of R belongs to some equivalence class, namely [a], and by definition b Im() a R with (a)=b. Thus b = (a)=h([a]) so h is surjective. 2 () 9 2

7 Recall the following definitions:

Definition 1.3.

A ring F is a field if every non-zero a F is a unit, that is, a has a multiplicative inverse • 2 in F .

A non-zero non-unit element r of a ring R is said to be irreducible if whenever we write • r = ab for some a, b R then either a or b is a unit; in the ring F [x] where F is a field, 2 this definition becomes:

A polynomial f(x) R[x] is irreducible over F if @(f) 1 and there is no fac- 2 torisation f(x)=g(x)h(x) where g, h F [x] are such that @(g),@(h) <@(f). Here, @(f) 2 denotes the degree of f(x).

Irreducibility of a polynomial depends of course on the field F over which it is being considered. For example, the polynomial x2 + 1 lies in both R[x] and C[x] and is irreducible over R but factorises over C as x2 +1=(x + i)(x i) An ideal P in a ring R is a prime ideal if it is a proper ideal and ab P = a P or • 2 ) 2 b P . 2 An ideal M in a ring R is a maximal ideal if it is a proper ideal, and if I is an ideal • satisfying M ( I R then I = R. ⇢ Building towards the goal of this section, we now prove some small yet very powerful lemmas which will be of great use for our purposes. Lemma 1.4. Let F be a field and suppose p(x) F [x] is of degree 1. The principle ideal 2 P =(p(x)) is a prime ideal if and only if p(x) is irreducible over F . Proof. Suppose p(x) is irreducible and that a(x)b(x) P . Then, a fortiori p(x) a(x)b(x)= 2 | ) p(x) a(x)orp(x) b(x)asp(x) is irreducible (this is a consequence of Euclid’s lemma in F [x]). | | It remains to show that P = F [x]. If P = F [x] then, in particular, 1 P so that 1 = q(x)p(x) 6 2 for some q(x) F [x]. Therefore we have the following contradiction: 2 0=@(1) = @(qp) = @(p)+@(q) @(p) 1.

8 Hence P = F [x] and P is a prime ideal. 6 Now suppose that p(x) is reducible so that p(x)=g(x)h(x) for some g(x),h(x) F [x] with 2 @(g),@(h) <@(p). Since @(pq)=@(p)+@(q) @(p)foranyq(x) F [x], we have that every 2 polynomial in P is of degree at least that of p(x). Therefore, neither g(x)norh(x) lie in P and P is not a prime ideal.

Lemma 1.5. Every non-zero prime ideal in a principle ideal domain is a maximal ideal.

Proof. Let P be a non-zero prime ideal in the principle ideal domain R. Suppose I R is an ⇢ ideal satisfying P ( I R. We show that I = R. ⇢ Since R is a PID we have that P =(a) and I =(b) for some a, b R. Now, a (a) (b)= 2 2 ⇢ ) a = rb for some r R so that rb (a). But (a)=P is prime so either r (a)orb (a). If 2 2 2 2 b (a) then (a)=(b) which is a contradiction. Therefore, r (a) and we have that a = asb 2 2 for some s R so that 1 = sb and b is a unit. Thus, I = R and P is a maximal ideal. 2 Lemma 1.6. Let R be a ring and M R be an ideal. The quotient ring R/M is a field if and ⇢ only if M is a maximal ideal.

Proof. Assume M is a maximal ideal. We want to show that every non-zero element of R/M is a unit. To this end, suppose [x] =0=M in R/M. Then, x M. Consider I =(M,x), the 6 62 ideal generated by M and x. Since x M we have that M ( I R and from the maximailty 62 ⇢ of M it follows that I = R. In particular, 1 R so there exist a, b R with 1 = am + bx. 2 2 In the quotient ring this becomes 1 = [1] = [am + bx]=[am]+[bx]=[a][m] +[b][x]=[b][x].

=0 Therefore, [x] is a unit and R/M is a field. |{z}

Now assume R/M is a field and I is an ideal satisfying M ( I R. Pick x I M. ⇢ 2 \ Then, [x] = 0 and so [x] is a unit by assumption. Therefore, [x][y] = [1] for some [y] R/M. 6 2 But this gives [x][y] [1] = 0 = M so that xy 1=m for some m M which implies that 2 1=xy m I. Hence I = R and M is a maximal ideal. 2 With these theorems at hand we are now well equipped to conclude a result which is fundamental to our exposition of Galois theory.

Lemma 1.7. Let F be a field and p(x) F [x] be a polynomial irreducible over F . Then 2 E := F [x] (p(x)) is a field containing (an isomorphic copy of) F and a root of p(x). . Proof. Since p(x) is irreducible it is non-zero. By lemma (1.4), (p(x)) is a non-zero prime ideal. Furthermore, since F [x] is a principle ideal domain, by lemma (1.5) we have that (p(x)) is in fact a maximal ideal. From lemma (1.6) we conclude that E is indeed a field.

9 Set Fˆ = [a] a F E. The map : F Fˆ, (a)=[a] is an isomorphism; Fˆ = F . { | 2 }⇢ ! ⇠ Indeed,

defines a ring map due to the very definition of the operations in E; • is plainly a surjection; • (a)=0 a (p(x)) = a = 0 (otherwise a F [x] F ), so is an injection. • () 2 ) 2 \ Finally, identifying F and Fˆ, we show that ↵ :=[x] is a root of p(x) regarded as a polynomial in E[x]: If p(x)=a + a x + a x2 + + a xn, then 0 1 2 ··· n 2 n p(↵)=[a0]+[a1]↵ +[a2]↵ ++[an]↵ =[a ]+[a ][x]+[a ][x]2 + +[a ][x]n 0 1 2 ··· n =[a ]+[a x]+[a x2]+ +[a xn] 0 1 2 ··· n =[a + a x + a x2 + + a xn]=[p(x)] = 0. 0 1 2 ··· n This completes the proof.

We end this section with two simple examples of this procedure known as ”adjoining” a root of an irreducible polynomial to its field of coecients.

Example 1.8. The polynomial p(x)=x2 3 is irreducible over Q (as will be shown in the next chapter). Therefore, the quotient ring E := Q[x] (x2 3) is a field containing Q and an element ↵ satisfying ↵2 = 3, i.e E contains a square root. of 3 which is known to be an irrational number.

Note that we can adjoin a root of any polynomial f(x) F [x] to its base field F because f(x) 2 can always be factorised into a product of irreducibles and a root of any of those factors is a root of f(x) itself!

Example 1.9. We can adjoin an nth root of 1 to Q by forming the quotient ring Q[x]/(p(x)) where p(x) is an irreducible factor of the polynomial xn 1 Q[x]. The details of this particular 2 type of field extension are fascinating and we dedicate an entire chapter to their study later on.

This spurs an investigation into irreducible polynomials. Sadly there is no general device for determining whether a given polynomial is irreducible or not. However, we can consider certain classes of polynomials and give criteria which apply to those classes. For our purposes it suces to consider almost exclusively polynomials over the field

10 of rational numbers and its extensions. We hence go about providing some irreducility criteria for polynomials f(x) Q[x] and in particular polynomials over Q with integer coecients. We 2 begin with the following elementary result.

Lemma 1.10. Let F be a field. f(x) F [x] has ↵ F as a root if and only if f(x)=(x ↵)g(x) 2 2 for some g(x) F [x]. 2 Proof. Suppose f(x)=(x ↵)g(x), then f(↵)=(↵ ↵)g(↵)=0g(↵)=0. Now suppose ↵ is a root of f(x). Set h(x) := f(x + ↵) F [x]. 2 If h(x)=a + a x + + a xm then 0 = f(↵)=h(0) = a . Therefore, o 1 ··· m 0 h(x)=a x + + a xm 1 ··· m m 1 = x(a + a x + + a x ) 1 2 ··· m so that f(x)=f((x ↵)+↵)=h(x ↵)=(x ↵)g(x). Corollary 1.11. If ↵ ,↵ ,...,↵ are distinct elements of F then the monomial (x ↵ ) 1 2 n+1 i+1 is coprime to the product (x ↵ )(x ↵ ) ...(x ↵ )foreach1 i n 1 2 i  

Proof. By the lemma, (x ↵j) - (x ↵k) if and only if ↵j is not a root of (x ↵k). Since the ↵j are distinct we hence have that (x ↵i+1) - (x ↵j)for1 j i. The claim now follows   immediately from Euclid’s lemma for a polynomial ring F [x].

Corollary 1.12. If f(x) F [x] is of degree n, then f(x) has at most n roots in F . 2 Proof. If @(f) = 1 then f(x) has 0 1 roots. Suppose @(f)=n>1 and f(x) has m>n  distinct roots in F ,say↵ ,↵ ,...,↵ . Then, f(x)=(x ↵ )g (x) for some g (x) F [x]. Now, 1 2 m 1 1 1 2 (x ↵2) - (x ↵1)so(x ↵2) g1(x) by Euclid’s lemma. Hence f(x)=(x ↵1)(x ↵2)g2(x) | for some g (x) F [x]. Continuing in this fashion we find that f(x)=(x ↵ )(x ↵ ) ...(x 2 2 1 2 ↵ )h(x) for some h(x) F [x]. This is a contradiction because the degree of the RHS is n+1 2 strictly greater than the degree of the LHS.

Theorem (1.10) leads to our first and simplest criterion:

Theorem 1.13. f(x) F [x] with @(f)=2or 3 is irreducible if and only if f(x) has no roots 2 in F .

Proof. Since f(x) is of degree 2 or 3, it is irreducible if and only if it has no linear factors. By the above, this is the case if and only if f(x) has no roots in F .

11 Prior to some more advanced criteria, we want to prove a famous lemma due to Carl Friedrich Gauss which gives a sucient condition for a polynomial having integer coecients to be irreducible as a polynomial in Q[x]. First recall that a polynomial f(x) Z[x] is called primitive if the highest common factor 2 of its coecients is 1, and that the product of two primitive polynomials is again primitive. Also recall that for any f(x) Q there is a primitive polynomial p(x) and positive constant 2 c Q such that f(x)=cp(x) and furthermore that this representation is unique and if f(x) 2 has integer coecients then c Z. 2 Lemma 1.14 (Gauss’ Lemma). Suppose f(x) Z[x] has no factorisation f(x)=g(x)h(x) in 2 Z[x] with @(g),@(h) <@(f). Then, f(x) is irreducible over Q.

Proof. We demonstrate the contrapositive. Suppose f(x) Z is reducible over Q so that 2 f(x)=f1(x)f2(x) in Q[x] with @(f1),@(f2) <@(f). Then,

fi(x)=ckik(x)fork =1, 2, where ck and ik(x) are as in the previous theorem.

Therefore, f(x)=f1(x)f2(x)

=(c1i1(x))(c2i2(x))

=(c1c2)(i1(x)i2(x)) =: cp(x) where p(x) is primitive and c Q is a positive constant. But f(x) Z[x]soc Z and we have that 2 2 2

f(x)=(ci1(x))(i2(x)) is a factorisation in Z[x] with @(ik) <@(f).

We now utilise Gauss’ lemma to provide our second irreducibility criterion.

Theorem 1.15. Let R be an integral domain, F be a field and : R F be a ring map and ! let ⇤ : R[x] F [x] be the corresponding ring map !

i i ⇤ aix = (ai)x . (1.1) ⇣X ⌘ X If f(x) R[x] satisfies @(⇤(f)) = @(f) and ⇤(f) is irreducible over F , then f(x) has no 2 factorisation f(x)=g(x)h(x) in R[x] with @(g),@(h) <@(f). In particular, if f(x) Z[x] and ⇤(f(x)) is irreducible over F then f(x) is irreducible over Q 2 by Gauss’ lemma.

12 Proof. We give a proof by contradiction. Assume ⇤(f) is ireducible over F and suppose f(x)= g(x)h(x) in R[x] with @(g),@(h) <@(f). Since ⇤ is a ring map, ⇤(f)=⇤(gh)=⇤(g)⇤(h) in F [x]. Now, ⇤(f) is irreducible over F so either ⇤(g)or⇤(h) is a constant, that is, either

@(⇤(g)) = 0 or @(⇤(h)) = 0. Without loss of generality assume @(⇤(g)) = 0.Then,

@(f)=@(⇤(f))

= @(⇤(g)⇤(h))

= @(⇤(g)) + @(⇤(h))

= @(⇤(h)) @(h)  <@(f). This is a clear contradiction so we conclude that no such factorisation exists and this completes the proof.

2 Example 1.16. f(x)=3x +5x 1 is irreducible over Q; taking F = Z2 in theorem (1.15) 2 we have ⇤(f(x)) = x + x + 1 which is irreducible over Z2 by theorem (1.13) as it has no roots in Z2.

Beware that there do exist irreducibles in Z which are reducible over Zp for every prime p! A nice property possessed by F [x] is that of being a unique factorisation domain: Theorem 1.17. Let f(x) F [x] where F is a field. Then, f(x) has a unique factorisation 2

f(x)=ci1(x)i2(x) ...in(x) in F [x] where each i (x) is a monic irreducible and c F . k 2 Proof. If f(x) is irreducible we can take c to be the leading coecient of f(x) and set i1(x)= 1 c f(x). If f(x) is reducible then we can write f(x)=f1(x)f2(x) for some f1(x),f2(x) of lower degree. Now, each fk(x) is either irreducible or reducible so we can continue in this manner until we have a factorisation f(x)=q1(x)q2(x) ...qn(x) where each qk(x) is irreducible. If ck is 1 the leading coecient of qk(x) then setting ik(x)=ck qk(x) and c = c1c2 ...cn we have that

f(x)=ci1(x)i2(x) ...in(x) is one such factorisation. Now suppose f(x)=bj1(x)j2(x) ...jm(x) is another such factorisa- tion. Since i (x) is irreducible and i (x) f(x), by Euclid’s lemma we have that i (x) j (x) 1 1 | 1 | 1 after reordering. As i1(x) and j1(x) are both monic, it follows that i1(x)=j1(x). Continuing in this manner we conclude that ik(x)=jk(x)foreachk, b = c and m = n else the degrees do not match.

13 A very well known and easy to apply criterion is called Eisenstein’s criterion: n i Theorem 1.18 (Eisenstein). Let f(x)= aix Z[x] be such that there is a prime p Z 2 2 i=0 satisfying: X

(i) p ao,a1,...,an 1 |

(ii) p - an

2 (iii) p - a0.

Then, f(x) is irreducible over Q.

The proof given here was produced by Peter Cameron and is rather elegant in its use of unique factorisation.

Proof. Assume f(x) satisfies the hypotheses of the theorem for the prime p Z. By Gauss’ 2 lemma it suces to show that there is no factorisation f(x)=g(x)h(x) in Z[x] with @(g),@(h) < @(f). Suppose there exists such a factorisation. Taking R = Z and F = Zp as in theorem (1.15) n we have that ⇤(f(x)) = anx by (i) and (ii). This is a product of a constant and n monic irreducibles, namely x. Since Zp is a field, this representation is unique so that ⇤(g(x)) and s i t i ⇤(h(x)) have the same form. If g(x)= i=0 gix and h(x)= i=0 hix then it follows that s t 2 ⇤(g(x)) = (g )x and ⇤(h(x)) = (h )x . Therefore we must have p g ,h = p g h = s t P P | 0 0 ) | 0 0 a0 which is a contradiction. The result follows.

Let us make some simple applications of these criteria.

f(x)=x10 +4x 4 is irreducible over Q by Eisentein’s criterion with p =2. • There is no prime p for which Eisenstein’s criterion can be applied to the polynomial • g(x)=3x3 +7x 5 but it is nonetheless irreducible over Q. Indeed, taking R = Z and 3 F = Z2 in theorem (1.15) we have that ⇤(g(x)) = x + x + 1 is irreducible over Z2 as it is of degree 3 with no roots in Z2.

The only prime dividing all but the leading coecient of the polynomial h(x)=x3 + • 27x2 + 9 is p = 3 but then p2 =9 9 so Eisenstein cannot be applied. However, reducing | coiecients (mod 2) as above we find that h(x) is in fact irreducible.

Creating larger fields by adjoining roots of polynomials to their fields of coecients is a partic- ular type of an algebraic construction called a field extension. In the next section we consider field extensions in a more general setting.

14 We often speak of subobjects of a given mathematical object such as subsets of a set, subspaces of a vector space and subfields of a field. Here, however, we reverse our perspective and view a field as an extension of its subfields.

Definition 1.19. Let F be a subfield of a field E. We say that E is a field extension of F and write E/F when we wish to emphasise this relationship. “E/F” reads “E over F ”.

For example, recall that the characteristic, Char(R), of a ring R is the least positive integer n such that n1=1+1+ +1= 0. The characteristic of a field is either 0 or a prime p, and ··· ntimes the prime field is the smallest subfield of F (which is isomorphic to Q when Char(F ) = 0 and | {z } is isomorphic to Zp when Char(F )=p). Therefore, we always have that a field is an extension of its prime field.

Suppose E/F is a field extension. If we momentarily forget about the multiplication in E and consider just the additive structure of E, we can view E as an F -vector space. That is, E is our set of vectors and F is the field of scalars. Indeed, we can simply define addition of vectors via the addition of elements in E and define scalar multiplication for ↵ F and v E 2 2 via the multiplication of those elements as members of E. As such, the axioms for E being an F -vector space coincide with those met by E as a field with subfield F . This connection to linear algebra brings to the table a plethora of tools from that area.

Definition 1.20. The dimension of a field E over a subfield F is called the degree of the extension E/F and is written [E : F ]. If [E : F ] is finite we say that E/F is a finite extension, else it is an infinite extension.

The word “degree” is used to describe this quantity for a reason which will be made apparent in due course.

Definition 1.21. An element ↵ of a field extension E/F is called algebraic over F if ↵ is a root of some polynomial f(x) F [x]. Otherwise, ↵ is called transcendental over F . 2 If every element of the field extension E/F is algebraic, we say that E/F is an algebraic extension.

For example, it has been shown that the real number ⇡ is transcendental over Q. On the other hand, the imaginary unit i C is algebraic over Q as it is the root of the polynomial 2 x2 +1 Q[x]. Indeed, any primitive nth root of unity is algebraic over the rational numbers as 2 we see in the next chapter.

Proposition 1.22. Let E/F be a field extension. If ↵ E is algebraic over F then, 2 15 (i) there exists a monic irreducible p(x) F [x] having ↵ as a root; 2 (ii) this polynomial is unique;

(iii) if f(x) F [x] is any polynomial having ↵ as a root then p(x) f(x). 2 | Proof. (i) Let p(x) be a polynomial of lowest degree in F [x] which has ↵ as a root; such an object exists by the well-ordering of the natural numbers. Since F is a field we can assume after multiplying by a constant that p(x) is monic. To show that p(x) is irreducible over F , assume for a contradiction that p(x)=g(x)h(x) for some g(x),h(x) F [x] with @(g),@(h) <@(p). 2 Then, 0 = p(↵)=g(↵)h(↵) in E so that g(↵)=0orh(↵) = 0 as fields have no zero divisors. In either case, the minimality of @(p) is violated. We conclude that p(x) is irreducible over F . (iii) Suppose f(x) F [x] has ↵ as a root. Since p(x) is non-zero, by the Euclidean algorithm 2 in F [x] there exist q(x),r(x) F [x] such that 2 f(x)=q(x)p(x)+r(x) where either @(r) <@(p)orr(x)=0.

If r(x) is non-zero, then since ↵ is a root of both f(x) and p(x) we have that

0=f(↵)=q(↵)p(↵)+r(↵)=r(↵) i.e ↵ is a root of r(x).

But this contradicts the minimality of @(p). Thus r(x) = 0 and f(x)=q(x)p(x) so that p(x) f(x). | (ii) Ifp ˆ(x) is any monic irreducible having ↵ as a root, then an application of (iii) gives p(x) pˆ(x). Since p(x) is irreducible and monic it follows thatp ˆ(x)=p(x) | This uniqueness of p(x) enables us to make the following definition.

Definition 1.23. Let ↵ E/F be algebraic over F . The monic irreducible polynomial of 2 lowest degree in F [x] having ↵ as a root is called the minimal polymial of ↵ over F and is denoted by p↵,F (x),p↵(x) or simply p(x) depending on whether ↵ and F are clear from the context.

As promised, we now justify the use of the term ”degree”. This itself is a rather beautiful example of a surprisingly concrete connection between two di↵erent areas/concepts (in this case linear algebra and quotient rings).

Theorem 1.24. If p(x) F [x] is an irreducble polynomial of degree n then E := F [x] (p(x)) 2 is a field extension of degree n over F . .

16 Proof. We have previously shown that the quotient ring E forms a field so it remains only to show that [E : F ]=n. Setting ↵ =[x]inE (and identifying a F with [a] E), it suces to 2 n 1 2 2 show that the set B = 1,↵,↵ ,...,↵ forms an F -basis for E. Note that we can assume { } p(x) to be monic so that it is the minimal polynomial of ↵.

Suppose B is linearly dependent, then there exist a0,a1,...,an 1 F not all zero such that 2 n 1 a0 + a1↵ + + an 1↵ =0. ··· n 1 But then ↵ is a root of the polynomial f(x)=a0 + a1x + + an 1x F [x] which is of ··· 2 degree less than p(x)=p↵(x), thus contradicting the minimality of @(p↵). Finally, we show that B spans E.Foranyf(x) F [x], by the Euclidean algorithm there exist q(x),r(x) F [x] 2 2 such that f(x)=q(x)p(x)+r(x) where @(r)

f(↵)=q(↵)p(↵)+r(↵) = r(↵)(asp(↵)=0).

Therefore, every element of E has a polynomial in ↵ of degree n 1 as a representative, i.e.  B spans E. This completes the proof.

At this point let us introduce some notation. Let ↵ be an indeterminate, then F (↵) denotes the smallest field containing F and ↵. The general form of this field is F (↵)= f(↵)/g(↵) f(x),g(x) { | 2 F [x],g(x) =0 . For indeterminates ↵ ,↵ ,...,↵ , the smallest field containing F and each 6 } 1 2 n ↵i is defined analogously. If E = F (↵) we say that E is a simple extension of F , and if

E = F (↵1,↵2,...,↵n)we say that E is finitely generated over F by the ↵i. These fields of fractions of polynomials are rather cumbersome objects, but, fortunately, in the case where ↵ is an algebraic element of a field extension E/F, it turns out that F (↵) is of a much simpler form as we now show.

Theorem 1.25. Let E/F be a field extension and suppose ↵ E is algebraic over F with 2 minimal polynomial p(x) F [x]. Then, 2 F [x] (p(x)) ⇠= F (↵). . Proof. Let : F [x] E be the evaluation map at ↵. Since p(x) is the minimal polynomial ! of ↵ we have that ker()=(p(x)). Therefore, the first isomorphism theorem tells us that the map h : F [x] (p(x)) Im() ! . 17 given by h([f(x)]) := f(↵) is an isomorphism. From theorem (1.24) we know that F [x] (p(x)) is a field containing F and ↵. Therefore, via the isomorphism h we have that Im() is. a subfield of E containing F and ↵. By the minimality of F (↵) this gives the inclusion F (↵) Im(). For the reverse inclusion note that any subfield ⇢ of E containing F and ↵ necessarily contains the set f(↵) f(x) F [x] =: Im(). Thus { | 2 } Im()=F (↵) and the conclusion follows. Definition 1.26. An algebraic element ↵ in a field extension E/F of degree n if its minimal polynomial over F is of degree n. Combining theorem (1.25) with the latter part of theorem (1.24) shows that if ↵ is an algebraic element of degree n of the field extension E/F, then F (↵)= f(↵) f(x) F [x] and @(f)

Let " ," ,...," be an E-basis for K and let µ ,µ ,...,µ be an F -basis for E. Any { 1 2 n} { 1 2 m} element ↵ K can be written 2 ↵ = a " + a " + + a " for some ↵ E, 1 1 2 2 ··· n n i 2 and each of these ai can be written as a = b µ + b µ + + b µ for some b F. i i1 1 i2 2 ··· im m ij 2 Therefore, any ↵ K has the form 2 ↵ = bijµj"i and we observe that the set i=1,2...,n j=1X,2,...,m B = µ " 1 i n, 1 j m is a spanning set for K over F . Since B = mn, it hence { j i |     } | | suces to show that B is linearly independent over F . If for some c F we have ij 2 E 2 n m c µ " = c µ " =0, ij j i z }|ij j { i i=1,2...,n i=1 j=1 ! j=1X,2,...,m X X

18 m then the linear independence of the "i over E gives j=1 cijµj =0foreachi, and the linear independence of the µ over F gives c =0foreachj and each i. Thus B is linearly independent j ij P and forms an F -basis for K of size mn.

In practise, this multiplicativity of extension degrees is most useful when we want to determine relative extension degrees as the corollary, discussion and examples to follow demonstrate.

Corollary 1.28. If K/F is a finite extension and E is a field which satisfies F E K, then ⇢ ⇢ [E : F ] [K : F ]. This is an immediate consequence of the theorem. | Example 1.29. Suppose we adjoin the field Q with a root, say ↵, of the polynomial f(x)= x3 +2x + 2 and we wish to know whether or not Q(↵) contains the imaginary unit i. Since f is irreducible over Q (by Eisenstein with p = 2), by theorems (1.24) and (1.25) we have that [Q(↵):Q]=3. Therefore, if i Q(↵), it would follow that 2 Q Q(i) Q(↵) ⇢ ⇢ Noting that the minimal polynomial of i is x2 + 1, the degree formula gives 3 = [Q(↵):Q]= [Q(↵):Q(i)][Q(i):Q]=2[Q(↵):Q(i)]. Hence [Q(↵):Q(i)] = 3/2 which is impossible as extension degrees are bound to be integer valued. We conclude that i Q(↵). 62 Example 1.30. x2n 3,xn 3 Q[x] are irreducible for any n N by Eisenstein with k 2 2 n 2n p = 3. Letting p3 denote the positive k-th root of 3, we have that Q Q(p3) Q( p3) as ⇢ ⇢ ( 2pn 3)2 = pn 3. By the degree formula,

2n 2n n n 2n n 2n =[Q( p3) : Q]=[Q( p3) : Q(p3)][Q(p3) : Q]=n[Q( p3) : Q(p3)].

2n n 2n n Therefore, [Q( p3) : Q(p3)] = 2 and the minimal polynomial of p3overQ(p3) is of degree two and hence must be p(x)=x2 pn 3. In particular, this tells us that the polynomial x2 pn 3 n is irreducible over Q(p3) which is not so easy to verify directly!

Example 1.31 (Quadratic Extensions). Let F be any field of characteristic 0 or a prime p =2 6 and suppose E/F is an extension of degree 2. For an element in E which is not in F consider the field extension F (). Since /F , we have [F ():F ] 2 but by the degree 2 formula we also have that 2 = [E : F ]=[E : F ()][F ():F ]= [F ():F ] 2. It )  follows that [F ():F ] = 2 so that [E : F ()] = 1. Therefore, E = F (). Now, the minimal polynomial of over F must be a monic quadratic: p (x)=x2 + bx + c for some b, c F . ,F 2

19 The usual quadratic formula for the roots of a quadratic is valid in any field of chacteristic b pb2 4c p =2: x = ± . For the sake of completeness let us give a proof: 6 2 b2 b2 b b2 x2 + bx + c = x2 + bx + c + =(x + )2 + c =0 4 4 2 4 b b2 (x + )2 = c () 2 4 b b2 x + = c () 2 ±r 4 b2 b x = c () ±r 4 2 pb2 4c b x = ± as claimed. () 2 pb2 4c b So = ± .Ifb2 4c is a square in F then F . But /F so b2 4c is not 2 2 2 a square in F , that is, pb2 4c does not lie in F . Since F (pb2 4c) is a field containing F and we have the inclusion F () F (pb2 4c). But any extension of F containing also ✓ contains pb2 4c so that F (pb2 4c) F (). Thus F ()=F (pb2 4c). Therefore, any ✓ field extension E/F of degree 2 is of the form E = F (pA) where A is some element in E which is not a square in F . Also, any field extension of the form E = F (pA) where A is not a square in F is of degree 2 which completely characterises extensions of degree 2. We hence called such extensions quadratic extensions.

Remark 1.32. The reason we must exclude fields of characteristic 2 is simply that we divided by 2 which of course is zero in fields of characteristic 2. The symbol “pb2 4c” is used to denote any abstract root of the polynomial x2 (b2 4c); of course there are two such roots, however, unlike square roots over the field of real numbers where we can choose pa to be either positive or negative, here there is no similar way of distinguishing between the two roots. We say the roots are algebraically indistinguishable.

Let us describe briefly the structure of the fields obtained by adjoining to a field F one or more algebraic elements of a field extension E/F.

Lemma 1.33. Let ↵, be elements of some field extension E/F. Then, F (↵,)=(F (↵))().

Proof. The field F (↵,) contains F and ↵ so by the minimality of F (↵) we have F (↵) ⇢ F (↵,). But F (↵,) also contains so that by minimality of (F (↵))() we conclude (F (↵))() ⇢ F (↵,). Finally, by the minimality of F (↵,) we have F (↵,) (F (↵))(). Thus F (↵,)= ⇢ (F (↵))().

20 Remark 1.34. Let ↵1,↵2,...,↵n belong to a field extension E/F. If we define F0 = F and

Fi = Fi 1(↵i)for1 i n then by repeated applications of the lemma and of the degree   formula we obtain

[F (↵1,↵2,...,↵n):F ]=[Fn : F ]=[Fn : Fn 1][Fn 1 : Fn 2] ...[F1 : F ].

The lemma enables us to give a general description of elements of the field F (↵1,↵2,...↵n) when the ↵i are algebraic over F . To illustrate the description, consider the simplest case F (↵,):

Suppose [F (↵,)=(F (↵))():F (↵)] = n and [F (↵):F ]=m, then every r F (↵,) is of 2 the form 2 n 1 r = a0 + a1 + a2 + + an 1 for some ai F (↵). ··· 2 In turn, each ai is of the form

2 m 1 ai = c0i + c1i↵ + c2i↵ + + c(m 1)i↵ for some cji F. ··· 2 Thus, each element of F (↵,) is of the form

i j cji↵ . j i X X It now follows that the ↵ij form a basis for F (↵,), as [F (↵,):F ]=mn by the degree formula. The problem with this description is that we don’t necessarily know the value of [F (↵,):F (↵)]. All we know is that its value is less than or equal to the degree of the minimal polynomial p,F (x)for over F , where equality holds if and only if p,F (x) is irreducible over

F (↵). Iterating this idea also gives a similar picture of F (↵1,↵2,...,↵n) when the ↵i are algebraic over F , but again the same diculties arise.

We have displayed a method for adjoining a single root of a polynomial to its field of coecients and this puts us in a position to construct a field which contains all the roots of a given polynomial. This leads to the notion of “splitting fields”.

Definition 1.35. A polynomial f(x) F [x] is said to split over F if f(x) can be written as 2 a product of linear factors in F [x]. Equivalently (by theorem (1.10)), f(x) splits over F if F contains all the roots of f(x).

The existence of a field extension E/F over which an arbitrary polynomial f(x) F [x] splits is a 2 highly non-trivial matter. For a very long time, mathematicians were reluctant to acknowledge

21 the concept of negative numbers let alone complex numbers! It was the question ”Is there a field of numbers containing R and a root of x2 +1?” which eventually led Gauss to give a proper construction of the complex numbers with which we are so familiar today.

Theorem 1.36 (Kronecker). Given a polynomial f(x) F [x] where F is a field, there exists 2 a field E containing F over which f(x) splits.

Proof. We proceed by induction on @(f). Suppose the statement holds for all polynomials of degree

Definition 1.37. Let f(x) F [x]. A field extension E/F is called a splitting field of f(x) 2 if f(x) splits over E but does not split over any proper subfield of E. In this sense, a splitting field of a polynomial f(x) F [x] is the smallest field containing F and all the roots of f(x). 2 If ↵ ,↵ ,...,↵ are the distinct roots of a polynomial f(x) F [x], then a splitting field of f(x) 1 2 n 2 is quite clearly F (↵1,↵2,...,↵n). Indeed, by the minimality of F (↵1,↵2,...,↵n), any proper subfield must omit one of the ↵i. At this point we are making no claims that splitting fields of polynomials are unique, simply that they exist.

Example 1.38. The polynomial x2 +1 Q[x] is irreducible over Q and splits over C: x2 +1 = 2 (x + i)(x i). However, C is not the splitting field of x2 + 1. Indeed, x2 + 1 splits over Q(i) which is a proper subfield of C; for example Q(i) contains no irrationals.

We now prove two theorems concerning the existence of certain extensions of isomorphisms between fields, both of which are extremely useful for studying properties of the Galois group of a polynomial. Specifically, they help us keep tabs on the number of elements of the Galois group of a polynomial! They also demonstrate uniqueness among splitting fields of a given polynomial, thus allowing one to legitimately speak of the splitting field of a polynomial.

Theorem 1.39 (First Isomorphism Extension). Let : F Fˆ be an isomorphism of the ! fields F and Fˆ. Let ⇤ : F [x] Fˆ[x] be the corresponding isomorphism of rings as described !

22 in theorem (1.15). If ↵ is a root of the irreducible polynomial f(x) F [x] and is a root of 2 ⇤(f(x)) Fˆ[x], then has an extension to an isomorphism 2 : F (↵) Fˆ() with (↵)=. ! Moreover, this extension is unique.

Before we present a proof, we need a couple of preliminary results.

Lemma 1.40. Let I1 and I2 be ideals of the rings R1 and R2, respectively, and suppose : R R is a ring isomorphism carrying I onto I , i.e (I )=I . Then, 1 ! 2 1 2 1 2

R1 R2 I1 ⇠= I2 and an isomorphism is given by the map

h : R1 I R2 I ,h([a]) = [(a)]. 1 ! 2 Proof. Firstly, h is a well-defined function since if a0 a, then a0 = a + i for some i I and ⇠ 2 1 we have that

h([a0]) = [(a0)] = [(a + i)] =[(a)+(i)] (as is a ring map) =[(a)] + [(i)]

=[(a)] + I2 (as (I1)=I2)

=[(a)] (as I2 is the zero element) = h([a]) h is a homomorphism:

h([a]+[b]) = h([a + b]) = [(a + b)] = [(a)+(b)] = [(a)] + [(b)] = h([a]) + h([b]) • h([a][b]) = h([ab]) = [(ab)] = [(a)(b)] = [(a)][(b)] = h([a])h([b]) • h(1) := h([1]) = [(1)] = [1] = 1. • h is injective:

h([a]) = I [(a)] = I 2 () 2 (a) I () 2 2 a I (as is injection with (I )=I ) () 2 1 1 2 [a]=I . () 1

23 Finally, since is surjective, for any given b R there is an a R with b = (a). For 2 2 2 1 the corresponding [b] R2 I we have h([a]) = [(a)] = [b]. Hence, h is surjective and the 2 2 conclusion follows. Lemma 1.41. If F is a field and : F (↵ ,↵ ,...,↵ ) F (↵ ,↵ ,...,↵ ) is an isomor- 1 2 n ! 1 2 n phism restricting to the identity map on F and sending each ↵ to itself, then F is the identity isomorphism.

Proof. As a vector space over F , a basis for F (↵1,↵2,...,↵n) is given by the collection B = i1 i2 in ↵1 ↵2 ,...,↵n 0 ik nk where nk =[Fk : Fk 1] and Fk is as defined in the remark { |   } following lemma (1.33). Considering as a linear map between vector spaces, we see that fixes a basis and hence must fix the whole space, i.e is the identity map.

We now render a proof of the first isomorphism extension theorem.

Proof. By theorem (1.25) there are isomorphisms

ˆ ˆ h1 : F [x] (f(x)) F (↵) and h2 : F [x] ( (f(x))) F () ! ⇤ ! . . satisfying h1([x]) = ↵ and h2([x]) = . 1 The composition := h h h : F (↵) F () is the required extension of . 2 1 !

Indeed, 1 h1 h h2 : c [c] [⇤(c)] = [(c)] (c) 7! 7! 7! showing that extends . We also have that

1 h1 h h2 : ↵ [x] [⇤(x)] = [x] as required. 7! 7! 7! To show uniqueness, assume ˆ is another such extension of and consider the map := 1 ˆ : F (↵) F (↵). Since fixes F and : ↵ ↵, from lemma (1.41) we have that ! 7! 7! is the identity isomorphism giving ˆ = .

Theorem 1.42 (Second Isomorphism Extension). Let : F Fˆ be an isomorphism of fields. ! Let f(x) F [x] be any polynomial and set g(x) := ⇤(f(x)) Fˆ[x]. 2 2 If E is any splitting field of f(x) over F and Eˆ is any splitting field of g(x) over Fˆ, then extends to an isomorphism ˆ : E E.ˆ !

24 Proof. We proceed by induction on [E : F ], using theorem (1.39) in the inductive step. If [E : F ] = 1, then E = F , f(x) splits over F and consequently g(x) splits over Fˆ. Indeed, under , any factorisation of f(x) into a product of linear terms in F [x] maps to a product of linear terms in Fˆ[x] equal to g(x). Hence Eˆ = Fˆ and ˆ := is a required extension. So assume now that [E : F ] 2 so that f(x) has at least one irreducible factor of degree 2, say p(x). Let q(x)=⇤(p(x)) and let ↵ and be roots of p(x) and q(x) respectively. By the first isomorphism extension theorem, there is an isomorphism : F (↵) F () extending and ! sending ↵ to . Since f(x) (F (↵))[x] and g(x) (Fˆ())[x], we have that E is a splitting field 2 2 of f(x)overF (↵) and Eˆ is a splitting field of g(x)overF (). Now, [F (↵):F ]=@(p) 2so the degree formula gives [E : F (↵)] = [E : F ]/[F (↵):F ] 1 [E : F ] < [E : F ]. By induction  2 there is an isomorphism ˆ : E Eˆ which restricts to on F (↵). But restricts to on F , ! so ˆ extends as required.

Note that we are most definitely not claiming uniqueness of the extension in this second second theorem!

Corollary 1.43. Any two splitting fields of a polynomial f(x) F [x] are isomorphic. 2 Proof. Let E and Eˆ be any two splitting fields of f(x). Take Fˆ = F and to be the identity isomorphism then, since fixes F , we have ⇤(f(x)) = f(x). By the theorem, there is an isomorphism ˆ : E Eˆ. !

25 Chapter 2

Separable Polynomials and the Cyclotomic Extensions

In this chapter we consider those polynomials whose roots are all distinct and we discuss some properties of the nth roots of unity in terms of their group structure and how we might go about adjoining them to an (almost) arbitrary field. These field extensions, called cyclotomic extensions, have a rich and fascinating structure all to themselves and have many far reaching applications to other areas of study such as Number Theory. Viewed as complex numbers, the nth roots of unity lie at n equally spaced points on the unit circle in the complex plane which is the reason for the word cyclotomic whose literal translation is ”circle dividing”. We start by providing some ways to identify separable polynomials.

Definition 2.1. A polynomial f(x) F [x] of degree n is called separable if it has n distinct 2 roots in its splitting field. Else we say that f(x) is inseparable (or has repeated roots).

For our purposes we wish to characterise the separable polynomials over finite fields and over fields of characteristic 0 such as Q. Definition 2.2. A polynomial f(x) F [x] is said to have repeated roots if in the splitting 2 field E of f there is a factorisation

f(x)=(x ↵)mg(x) for some ↵ E, g(x) E[x] and m 2. 2 2 Lemma 2.3. The polynomial f(x)= (x ↵ ), where the ↵ lie in F , has no repeated roots i i i if and only if f(x) is coprime to its formal derivative f (x). Q 0 Proof. If ↵ is a repeated root of f(x), then

f(x)=(x ↵)h(x) for some h(x) F [x]. Then, 2 26 2 f 0(x)=2(x ↵)h(x)+(x ↵) h0(x) =(x ↵)(2h(x)+(x ↵)h0(x)) = (x ↵) f,f0 = (f,f0) =1. ) | ) 6

Now suppose f(x) has no repeated roots so that the ↵i are mutually distinct. Then,

n f 0(x)= (x ↵ ) and we see that each monomial (x ↵ ) i i j=1 i=j X Y6 th divides all but the i summand. Hence, (x ↵i) - f 0(x)foreachi and it follows that (f,f0)= 1.

n n 1 Consider a polynomial f(x)=anx + an 1x + + a1x + a0 F [x] of degree n where F is a ··· 2 n 1 n 1 field of characteristic 0 or a prime p not dividing n. Then, f 0(x)=nanx +(n 1)an 1x + +a is of degree n 1. In particular, if f(x) is irreducible then (f,f0) = 1 so by the discussion ··· 1 above we conclude that f has no repeated roots and is thus separable. If F is of characteristic 0 then f(x) F [x] is separable if and only if it is a product of distinct irreducibles; this follows 2 from the above comments and the fact that distinct irreducibles cannot have roots in common. In fact, this holds true for any finite field as well, as we show shortly, and which completely characterises the separable polynomials over finite fields and fields of characteristic 0.

All we require for an irreducible polynomial over a field F to be separable is that (f,f0)=1.

Now, irreducibility gives (f,f0) = 1 as long as f 0(x) = 0. This plainly holds true if F has 6 characteristic 0, but it can fail if F is of characteristic p = 0. So when can it be that f 0(x)=0 6 in a field F of characteristic p =0? 6 n 1 n 1 f 0(x)=0 nanx +(n 1)an 1x + + a1 = 0 and this is the case if and () ··· mp mp 1 only if each power of x whose coecient is non-zero is some multiple of p;(x )0 = pmx =0. That is, there is a polynomial q(x) F [x] with p(x)=q(xp). 2 Lemma 2.4. In a finite field of characteristic p, every element is a pth power in F .

Proof. Consider the map : F F , (a) := ap. The standard binomial formula holds true in ! any commutative ring, so for a, b F we have 2 p p p i p i (a + b)=(a + b) = a b i i=0 ✓ ◆ X p 1 p p i p i p = a + a b + b i i=1 X ✓ ◆ = ap + bp,

27 as p p for each 1 i p 1. Also, (ab)=(ab)p = apbp = (a)(b) and (a)=0 | i   () ap =0 a = 0. We conclude that is an injective ring map. Since F is finite and is () injective, is also surjective and thus is an isomorphism. In particular, for each a F there 2 is a b F with a = bp as conjectured. 2 This puts us in a position to prove our claim:

Theorem 2.5. Let F be a finite field of characteristic p. Then, f(x) F [x] is separable if and 2 only if it is a product of distinct irreducibles.

Proof. As before, it suces to show that every irreducible polynomial over F is separable. Let f(x) F [x] be irreducible and suppose f(x) is inseparable. Then, as shown above, f(x)=q(xp) 2 m m 1 for some q(x) F [x]. Suppose q(x)=amx + am 1x + + a1x + a0. Since F is finite of 2 ··· characteristic p we have that a = bp for some b F for each i. Therefore, i i i 2 p p m p m 1 p f(x)=q(x )=am(x ) + am 1(x ) + + a1x + a0 ··· p m p p m 1 p p p p = bm(x ) + bm 1(x ) + + b1x + b0 ··· m p m 1 p p p =(bmx ) +(bm 1x ) + +(b1x) + b0 ··· m m 1 p =(bmx + bm 1x + + b1x + b0) ··· where in the last equality we have made a repeated use of the binomial formula. This factori- sation contradicts the irreducibility of f(x) and the proof is thus complete.

We now probe a particular polynomial, namely xn 1 F [x] which is separable provided the 2 characteristic of F is either 0 or a prime p not dividing n.

Definition 2.6. Consider the polynomial f(x)=xn 1 Q[x] where n N. The roots of this 2 2 polynomial (in a field extension containing the splitting field of f(x)) are called the nth roots of unity.

We now aim to show that the nth cyclotomic extension of Q obtained by adjoining the nth roots of unity is an extension of degree '(n) where ' is Euler’s totient function:

'(n)= 1 a

Theorem 2.7. Let G =(g) be a cyclic group of order n with generator g. For each divisor d of n, G has a unique subgroup of order d.

Proof. Suppose d n so that n = md for some integer m. We claim that the subgroup H | generated by gm is of order d. Since (gm)d = gmd = gn =1,gm is at most order d.If (gm)l = gml = 1 then n ml so that ml = rn for some integer r. Therefore, ml = rn = | m rmd = l = rd d and so g and hence H is of order precisely d. Now suppose H0 is any ) sugbroup of order d. Since every subgroup of a cyclic group is cyclic, we have that H0 =(h)for some h G, and h = gi for some integer i. Now, 1 = hd =(gi)d = gid = id = sn for some 2 i (n/d)s n/d s m s ) integer s. Therefore, h = g = g =(g ) =(g ) which gives the inclusion H0 H, but ⇢ H and H0 are of the same finite order so in fact H0 = H.

Lemma 2.8. If G is a cyclic group of order n generated by g, then the other generators of G are the elements gi where (i, n)=1.

Proof. Suppose i and n are coprime. Then there exist integers r and s with 1 = ri + sn.We have that g = g1 = gri+sn = grigsn =(gi)r(gn)s =(gi)r1=(gi)r (gi), from which it follows 2 that G (gi). The other inclusion holds trivially, hence G =(gi). Now suppose gi generates ⇢ i i s is is 1 G. Then, in particular g (g ) so that g =(g ) = g g = 1 for some integer s. 2 () But then n is 1= is 1=rn 1=is rn for some integer r. Thus i and n are | ) () co-prime and this completes the proof.

Letting Gen(G) denote the collection of generators for the cyclic group G of order n. By the lemma we have that Gen(G) = '(n). | | Theorem 2.9. '(d)=n for any n N. 2 d n X| Proof. Let G be any cyclic group of order n and define a relation on G by x y (x)=(y). ⇠ () is easily seen to be an equivalence relation. If X =(x) let Gen(X) denote the equivalence ⇠ class of x. Then, G can be written as the disjoint union of equivalence classes of , i.e ⇠ G = Gen(X) Xcyclic[

29 Since this union is disjoint, we have that n = G = Gen(X) = '(d) by the lemma. | | | | Since G is cyclic, there is precisely one subgroup of order d for each divisor d of n (by theorem P P (2.7)). Hence, the above equation reads

n = Gen(X) = '(d). | | d n X X|

This enables us to prove the following.

Theorem 2.10. Let G be a group of order n. Then, G is cyclic if and only if G has at most one subgroup of order d for each divisor d of n.

Proof. Theorem (2.7) proves suciency. Suppose now that G has at most one subgroup of order d for each divisor d of n. Writing G as a disjoint union of its cyclic groups as in the previous theorem, we have that

n = Gen(X) '(d)=n. | | Xcyclic d n X X| Therefore, G has precisely one subgroup of order d for each divisor d of n. Finally, n n so G | has a cyclic subgroup of order n which of course is the entire group G so G is cyclic.

Here is the result we have been aiming for:

Theorem 2.11. The group of nth roots of unity in a field F (of characteristic 0 or a prime p not dividing n) form a cyclic multiplicative group. In fact, every finite subgroup of the multiplicative group of a field F is cyclic.

Proof. Since ⌥n is a finite multiplicative subgroup of the multiplicative group of F , it suces to prove the second statement.

To this end, let G be a subgroup of order n in F ⇤ and suppose H is some subgroup of G of order d where d n.Foreachh H we have hd = 1 by an application of Lagrange’s theorem. | 2 Therefore, each member of H is a root of the polynomial xd 1. Assume there exists another subgroup of G of order d, then there exist strictly more than d roots of xd 1 in G and hence F . This is a contradiction. Thus G has at most one subgroup of order d for each divisor d of n. G is hence cyclic by theorem (2.10).

This justifies the following definition.

th Definition 2.12. Let ⇠n generate the cyclic group of n roots of unity. Then ⇠n is called a primitive nth root of unity.

30 Definition 2.13. The nth cyclotomic polynomial is the monic polynomial whose roots are precicely the primitive nth roots of unity. That is, (x) := (x ⇠). n ⇠primitiveY By lemma (2.8) there are precisely '(n) primitive nth roots of unity, and these are given by i th (⇠n) where ⇠n is any primitive n root of unity and (i, n) = 1. Therefore n(x)isofdegree '(n). Consider once again the polynomial xn 1 whose roots are all the nth roots of unity. Grouping together the primitive dth roots of unity for each divisor d of n, by theorem (2.7) we have that xn 1= (x ⇠)= (x ⇠) = (x). d ⇠ ⌥n d n ⇠ ⌥d d n Y2 Y| ⇠primitiveY2 Y|

=d(x) th This representation of n(x) makes it possible to| find{z the n }cyclotomic polynomial iteratively for any n. Let us compute the first few: x 1=(x) 1 x2 1=(x 1)(x +1)= (x) (x)=(x 1) (x)= (x)=x +1 1 2 2 ) 2 x3 1=(x) (x)=(x 1) (x)= (x)=(x3 1)/(x 1) = x2 + x +1 1 3 3 ) 3 x4 1=(x2 +1)(x2 1) =1(x)2(x)4(x) =(x 1)(x +1) (x)=(x2 1) (x)= (x)=x2 +1 4 4 ) 4 x5 1=(x) (x)= (x)=(x5 1)/(x 1) = x4 + x3 + x2 + x +1 1 5 ) 5 x6 1=(x) (x) (x) (x) 1 2 3 6 =(x2 1)(x2 + x +1) (x) 6 =(x4 + x3 x 1) (x)= (x)=x2 x +1 6 ) 6 x7 1=(x) (x)= (x)=x6 + x5 + x4 + x3 + x2 + x +1. 1 7 ) 7 Notice that when p is a prime, the pth cyclotomic polynomial is given by the formula p 1 p p 1 p 2 n (x)=(x 1)/(x 1) = x + x + + x +1 = x . p ··· n=0 X Using the binomial formula, Eisenstein’s criterion and the fact that for any c Q, a polynomial 2 f(x) Q[x] is irreducible if and only if f(x + c) is irreducible, one can relatively easily show 2 that p(x) is irreducible for each prime p. However, with a bit of hard work we can go one better than this and show that n(x) is irreducible over Q for every n N! 2 31 th Theorem 2.14. For each n N, the n cyclotomic polynomial n(x) Q[x] is a monic 2 2 irreducible polynomial of degree '(n) whose coecients lie in Z.

Proof. n(x) is monic and of degree '(n) by definition so let us first show that n(x) has integer coecients; we proceed by induction on n.1(x)=x 1 Z[x] so the statement 2 holds for the case n = 1. Suppose now that n>1 and assume the statement holds for each 1 i

To show that n(x) is irreducible over Q it suces to show there is no factorisation

n(x)=p(x)q(x) with p(x),q(x) Z[x] monic, and @(p),@(q) <@(n). 2 Assume for a contradiction that such a factorisation exists. We can assume p(x)tobean th irreducible factor. If µn is a primitive n root of unity which is a root of p(x) then p(x) is p th the minimal polynomial for µn. For any prime p not dividing n, µn is also a primitive n root p of unity and is hence either a root of p(x)orq(x). Suppose µn is a root of q(x), then µn is a root of the polynomial q(xp). By proposition (1.22), p(x) q(xp)soq(xp)=h(x)p(x)for | some h(x) Z[x]. Now, with the choices R = Z and F = Zp in theorem (1.15) and setting 2 p p p f ⇤(x) := ⇤(f(x)) we have that q⇤(x )=h⇤(x)p⇤(x) in Zp[x]. By lemma (2.4), q⇤(x )=(q⇤(x)) p so (q⇤(x)) = h⇤(x)p⇤(x). Zp[x] is a unique factorisation domain since Zp is a field, so q⇤(x) is afactorofbothh(x) and p(x). In particular, p⇤(x) and q⇤(x) share an irreducible factor, say 2 n n k⇤(x). But then (k⇤(x)) divides n(x) and hence divides x 1 in Zp[x]. This implies x 1 n has a multiple root in Zp[x], but this is impossible since x 1 is separable over any field of p characteristic p not dividing n. Thus µn is not a root of q(x) and so must be a root of p(x). For any integer k coprime to n we can write k as a product of primes, each of which does not

p1 th divide n:sayk = p1p2 ...pm. By the argument above, µn is a primitive n root of unity which p1 p2 th is a root of p(x). By the same argument, (µn ) is also a primitive n root of unity which is arootofp(x). Continuing in this fashion we find that µk = µp1p2...pm is a primitive nth root of unity which is a root of p(x). Hence, every primitive nth root of unity is a root of p(x). That is,

n(x)=p(x)son(x) is irreducible.

32 Corollary 2.15. The extension of Q obtained by adjoining the nth roots of unity is of degree '(n); [Q(µn)/Q : Q]='(n).

Proof. n(x) Q[x] is a monic irreducible of degree '(n) having µn as a root and is hence 2 the minimal polynomial for µn over Q. The conclusion follows immediately from theorem (1.25).

33 Chapter 3

Automorphisms, Galois Groups and the Fundamental Theorem

For a given polynomial with coecients lying in a field F we have shown the existence of a field extension of F which contains all the roots of that polynomial along with F itself. It was the ingenious idea of 19th century French mathematician Evariste´ Galois to compare the group of permutations of the roots of a polynomial with the structure of its splitting field. More specifically this leads to the notion of the Galois group of a polynomial which encodes information about the splitting field of the polynomial and all its subfields. Conversely, the collection of subfields of the splitting field of a polynomial holds information about its Galois group. The Fundamental Theorem of Galois Theory tells us precisely what this information is and it is this which we aim to illucidate now. Let us describe the basic machinery.

Definition 3.1. An isomorphism of a field E with itself is called an automorphism of E. Aut(E) denotes the collection of all automorphisms of E.

Definition 3.2. Aut(E) is said to fix c E if (c)=c.IfF is a subset (e.g a subfield) 2 2 of E such that (c)=c for each c F then we say that fixes F . 2 Theorem 3.3. Aut(E) forms a group under the composition of functions.

Proof.

1. Function composition is always associative

2. If ,⌫ Aut(E), then ⌫(a + b)=(⌫(a)+⌫(b)) = (⌫(a)) + (⌫(b)) = ⌫(a)+⌫(b), 2 ⌫(ab)=(⌫(a)⌫(b)) = (⌫(a))(⌫(b)) = ⌫(a)⌫(b) and ⌫(1) = (⌫(1)) = (1) = 1 so ⌫ is a ring map. Also, ⌫(c)=0 ⌫(c)=0 c =0so⌫ is injective. () ()

34 Finally, for each c E there exists b E with (b)=c and a E with ⌫(a)=b. Hence 2 2 2 ⌫(a)=(⌫(a)) = (b)=c and ⌫ is surjective. Thus Aut(E) is closed.

3. The identity automorphism serves as the indentity element.

1 4. As is a bijection, is well-defined (and is a bijection) for each Aut(E). For 2 1 any a, b E there exist c, d E with (c)=a and (d)=b. Therefore, (a + b)= 1 2 1 2 1 1 1 1 ((c)+(d)) = ((c+d)) = c+d = (a)+ (b) and (ab)= ((c)(d)) = 1 1 1 1 ((cd)) = cd = (a) (b). Hence Aut(E)soAut(E) contains all necessary 2 inverses.

Definition 3.4. Let E/F be a field extension. Aut(E/F) denotes the collection of members of Aut(E) which fix F .

1 1 If ,⌫ Aut(E/F), then ⌫(c)= (c)=c for each c F . Thus Aut(E/F) forms a 2 2 subgroup of Aut(E) and so forms a group in its own right.

A very simple to prove yet very powerful lemma which enables one to calculate the members of Aut(E/F) in the case where the extension is algebraic is the following.

Lemma 3.5. Let E/F be a field extension and let Aut(E/F).If↵ is a root of f(x) F [x], 2 2 then (↵) is also a root of f(x). In particular, if ↵ is algebraic with with minimal polynomial p(x) F [x], then (↵) is also a root of p(x), i.e members of Aut(E/F) permute the roots of 2 irreducible polynomials.

Proof. Suppose f(x)=a + a x + + a xn F [x] has ↵ as a root. Then, 0 1 ··· n 2 0=(0) = (f(↵)) = (a + a ↵ + + a ↵n) 0 1 ··· n = (a )+(a ↵)+(a ↵2) (a ↵n) (3.1) 0 1 2 ··· n = (a )+(a )(↵)+(a )(↵2)+ + (a )(↵n) (3.2) 0 1 2 ··· n = a + a (↵)+a (↵2)+ + a (↵n) (3.3) 0 1 2 ··· n = a + a (↵)+a (↵)2 + + a (↵)n = f((↵)), (3.4) 0 1 2 ··· n where (5.1), (5.2) and (5.4) hold because is a ring map and (5.3) holds because fixes F .

35 Example 3.6. Consider Aut(Q(p3)/Q), where p3 denotes the positive real square root of 3, and note that Q(p3) = a + bp3 a, b Q . The minimal polynomial for p3overQ is { | 2 } x2 3. Since the roots of this polynomial are p3 we have that (p3) = p3forany ± ± 2 Aut(Q(p3)/Q). Now, fixes Q so the elements of Aut(Q(p3)/Q)are

(a + bp3) = a bp3 and ⌫(a + bp3) = a + bp3 (the identity automorphism).

Thus Aut(Q(p3)/Q) is the cyclic group of order 2 generated by ⌫.

n n Example 3.7. Consider the group Aut(Q(p3)/Q). p3 has minimal polynomial xn 3overQ. n 2 n n 1 n th The other roots of this polynomial are µp3,µ p3,...,µ p3 where µ is a primitive n root n n of unity,. However, none of these elements belongs to Q(p3) as Q(p3) R, so any element of n n n ⇢ Aut(Q(p3)/Q) must send p3 to itself. Therefore, Aut(Q(p3)/Q) consists of just the identity automorphism!

Given a field extension E/F we now have a way of attributing to that extension a subgroup of Aut(E), namely Aut(E/F). We can also go in the other direction and associate a field extension to a subgroup of Aut(E) as we now show.

Theorem 3.8. Let G Aut(E). The collection of elements of E fixed by every member of G ✓ forms a subfield of E.

Proof. Let a, b F = c E (c)=c, g G . Then (a + b)=(a)+(b)=a + b, 2 { 2 |1 18 2 1 } (ab)=(a)(b)=ab and (a )=(a) = a so F E is closed and is hence a subfield ✓ of E.

Definition 3.9. If G is a subgroup of Aut(E) then the collection c E (c)=c, g G { 2 | 8 2 } is called the fixed field of G and is denoted by fix(G).

The correspondences “subgroups of Aut(E/F) fixed fields” and “subfields of E subgroups 7! 7! of Aut(E/F)” are inclusion reversing in the following sense: If G G Aut(E/F), then 1 ✓ 2  fix(G ) fix(G ) and if F F are subfields of E, then Aut(E/F ) Aut (E/F ). Indeed, 2 ✓ 1 1 ✓ 2 2  1 since an element fixed by every member of G2 is also fixed by every member of G1 we have that fix(G ) fix(G ), and any automorphism fixing every element of F also fixes every element 2 ✓ 1 2 of F1. If E/F is a field extension, then in general we have fix(Aut(E/F)) F . In example (3.7), n n ◆ starting with the subfield Q of Q(p3) we find that Aut(Q(p3)/Q) comprises just the identity n automorphism but, going in the other direction, the subfield of Q(p3) fixed by the identity

36 n automorphism is the entire field Q(p3). Sometimes, however, we find that these processes are the reverse of each other, as is the case in example (3.6). Indeed, Aut(Q(p3)/Q)= 1,⌫ { } (the entire group), while fix(Aut(Q(p3)/Q)) = Q. The reason for these processes not always being the reverse of each other is that sometimes there are simply not enough automorphisms of the field extension E to ensure that the fixed field of Aut(E/F) is just the base field F .In n example (3.7) it seems that Q(p3) having too few automorphisms is caused by the fact that any automorphism must permute the roots of the minimal polynomial for pn 3 and some of these roots are missing from the field extension. How many automorphisms is “enough”? We begin to address this question now. As one might have guessed, the answer is connected to splitting fields.

Definition 3.10. A finite extension E/F is said to be a Galois extension over F if Aut(E/F) = | | [E : F ]. If E/F is Galois over F we write Gal(E/F)=Aut(E/F). Here, Gal(E/F) is called the Galois group of E/F.

Theorem 3.11. If E is the splitting field of a polynomial f(x) F [x], then 2 (i) Aut(E/F) [E : F ]; | | (ii) if f(x) is separable, then Aut(E/F) =[E : F ]. | | Proof. Consider again the second isomorphism extension theorem which states that if

: F Fˆ is an isomorphism and Eˆ is the splitting field of f ⇤(x) := ⇤(f(x)) Fˆ[x], then ! 2 extends to an isomorphism ˆ : E Eˆ.If[E : F ] = 1 then E = F and Eˆ = E so the only ! extension of is itself and the number of extensions is 1. Now suppose [E : F ]=n 2, then f(x) has an irreducible factor of degree 2, say p(x). p⇤(x) is of the same degree as p(x) and is also irreducible as any factorisation of p⇤(x) in Fˆ[x] would immediately lead to a factorisation of p(x) in F [x] via the isomorphism ⇤.Let↵ be some fixed root of p(x). If is the restriction to F (↵) of any extension ˆ : E Eˆ, then is an isomorphism with some subfield of Eˆ. Now, ! is determined completely by its value at ↵ and, by lemma (3.5), (↵) is some root of p⇤(x). That is, : F (↵) Fˆ() is an isomorphism. By the first isomorphism theorem, to each root ! of p⇤(x) there exists such an isomorphism , and by the second isomorphism extension theorem each of those ’s extends to an isomorphism ˆ : E Eˆ. We hence have the diagram !

37 ˆ E - Eˆ

F (↵) - Fˆ()

- F Fˆ

Therefore, the number of extensions ˆ is equal to the number of distinct roots of p⇤(x). Now,

@(p⇤)=@(p)=[F (↵):F ] by theorem (1.24), so p⇤(x) has [F (↵):F ] distinct roots with  equality if p⇤(x), and hence p(x), is separable. E is the splitting field of f(x)overF and Eˆ is the splitting field of f ⇤(x)overFˆ, so by the degree formula [E : F (↵)] = [E : F ]/[F (↵):F ]

Corollary 3.12. If E is the splitting field of a polynomial f(x) F [x], then E/F is a Galois 2 extension.

Proof. The splitting field of a polynomial is equal to the splitting field of the product of its distinct irreducible factors which is separable since distinct irreducibles cannot share roots.

We are now able to assign a group to a polynomial!

Definition 3.13. Let f(x) F [x] be a separable polynomial with splitting field E. The Galois 2 group of f(x) is Gal(f) := Gal(E/F)

Example 3.14. As we have seen, if F is a field of characteristic p = 2, then any extension 6 E/F of degree 2 is of the form E = F (pA) for some A E which is not a square in F . 2 F (pA) is evidently the splitting field of the polynomial x2 A F [x] since the roots of 2 x2 A are pA, both of which lie in F (pA). Therefore, E/F is a Galois extension and ± Gal(E/F) =[E : F ] = 2. By lemma (3.5), the two possible images of pA under an element | |

38 in Gal(F (pA)/F )are pA. Since Gal(E/F) = 2, these must both define automorphisms. ± | | These are given explicitly by

: a + bpA a bpA and : a + bpA a + bpA. 1 7! 2 7! We see that 2 = = 1 so that Gal(F (pA)/F )= 1, is cyclic of order 2. 1 2 { 1} n n Example 3.15. The extension Q(p3)/Q is of degree n as p3 has minimal polynomial xn n 3overQ. We have seen that Aut(Q(p3)/Q) contains just a single element, the identity n automorphism. Therefore, Q(p3)/Q is not a Galois extension.

Example 3.16. The polynomial x4 2 is irreducible over Q by Eisenstein’s criterion with 4 p =2,so[Q(p2) : Q] = 4 where p2 denotes, say, the positive real root of 2. Since the other 4 4 4 4 roots of x4 2areµp2,µ2p2 and µ3p2 where µ is a primitive 4th root of unity, Q(p2) is 4 not the splitting field of x4 2overQ as Q(p2) contains no primitive 4th roots of unity. The 4 4 4 splitting field of x4 2overQ is Q(p2,µp2) = Q(p2,µ) which in light of the degree formula is an extension of degree 4 '(4) = 4 2=8overQ because 4(x) remains irreducible over 4 · · 4 Q(p2). The extension is Galois by corollary (3.12), so Gal(Q(p2,µ)/Q) = 8. The mini- 4 | 2 | mal polynomial of µ over Q(p2) is the cyclotomic polynomial 4(x)=x +1,soµ satisfies µ2 +1=0 µ2 = 1 and the possible images of µ under an automorphism are µ and () µ. The minimal polynomial of p4 2 is of course x4 2 so the possible images of p4 2 under an automorphism are p4 2,µp4 2,µ2p4 2 and µ3p4 2. This gives rise to 8 possible maps all of which 4 must define members of Gal(Q(p2,µ)/Q):

p4 2 p4 2 p4 2 µp4 2 p4 2 p4 2 p4 2 µp4 2 1: ! 1 : ! 2 : ! 3 : ! µ µ µ µ µ µ µ µ ( ! ( ! ( ! ( ! p4 2 p4 2 p4 2 µp4 2 p4 2 p4 2 p4 2 µp4 2 4 : ! 5 : ! 6 : ! 7 : ! µ µ µ µ µ µ µ µ ( ! ( ! ( ! ( !

We will see in a later section that this Galois group is isomorphic to D8, the dihedral group of order 8. Suppose we had stuck with the original generators p4 2 and µp4 2. This would have given rise to 4 4 16 possible maps, only 8 of which define members of Gal(Q(p2,µp2)/Q). For instance, the map defined by

p4 2 p4 2 ! µp4 2 p4 2 ( ! 39 cannot define an automorphism as it is not injective. The reason that not all of the possible images of p4 2 and µp4 2 define an automorphism is that an automorphism must respect algebraic relations that exist between the generators of the extension. For this particular mapping we see that µ is mapped to 1: µp4 2 p4 2 µ = =1 p4 2 7! p4 2 but 1 is not a root of the minimal polynomial of µ. That is, the map does not respect the relation which is enforced by lemma (3.5). For this reason, this map cannot define a member 4 4 of Gal(Q(p2,µp2)/Q).

Example 3.17. The extension Q(p2, p3)/Q is Galois because it is the splitting field of the polynomial f(x)=(x2 2)(x2 3) over Q. The possible images of p2 and p3 under an automorphism are p2 and p3, respectively. Therefore, there are 4 possible maps all of ± ± which must define automorphisms as [Q(p2, p3) : Q]=4:

p2 p2 p2 p2 p2 p2 p2 p2 1: ! : ! ⌫ : ! ⇢ : ! . p3 p3 p3 p3 p3 p3 p3 p3 ( ! ( ! ( ! ( !

Since Q(p2, p3) = a + bp2+cp3+dp6 a, b, c, d Q , these maps are given explicitly by { | 2 }

1:a + bp2+cp3+dp6 a + bp2+cp3+dp6 7! : a + bp2+cp3+dp6 a bp2+cp3 dp6 7! ⌫ : a + bp2+cp3+dp6 a + bp2 cp3 dp6 7! ⇢ : a + bp2+cp3+dp6 a bp2 cp3+dp6 7!

Noting that ⇢ = ⌫ and 2 = ⌫2 =(⌫)2 = 1, we have that

Gal(Q(p2, p3)/Q)= 1,,⌫,⌫ { } is isomorphic to the Klein-4 group. Let us compute the fixed fields of the subgroups of Gal(Q(p2, p3)/Q). The subgroups are given by

H = 1,,⌫,⌫ ,H= 1, ,H= 1,⌫ ,H= 1,⌫ and H = 1 . 1 { } 2 { } 3 { } 4 { } 5 { }

(a + bp2+cp3+dp6) = a bp2+cp3 dp6 = a + bp2+cp3+dp6 b = d =0 fix(H2)=Q(p3) () )

40 ⌫(a + bp2+cp3+dp6) = a + bp2 cp3 dp6 = a + bp2+cp3+dp6 c = d =0 fix(H3)=Q(p2) () )

⌫(a + bp2+cp3+dp6) = a bp2 cp3+dp6 = a + bp2+cp3+dp6 b = c =0 fix(H4)=Q(p6) () )

Since Q fix(H1) Q(p2) Q(p3) = Q, we have that fix(H1)=Q. Finally, fix(H5) is the ✓ ✓ \ entire field Q(p2, p3) and we summarise these correspondences with the following table:

Subgroup Fixed Field

1,,⌫,⌫ Q { } 1, Q(p3) { } 1,⌫ Q(p6) { } 1,⌫ Q(p2) { } 1 Q(p2, p3) { }

Observe the afore mentioned inclusion reversing nature of the correspondence. In this example, every subgroup is normal and every subfield is Galois over Q, however, this is not always the case. Relations such as this are the essence of the fundamental theorem of Galois theory. Before presenting the fundamental theorem, we first discuss some properties of characters of a group and provide four equivalent characterisations of Galois extension.

Definition 3.18. 1. Let G be a group and E be a field. A group character of G in E is a group homomorphism : G E⇤ where E⇤ is the multiplicative group of E. We say that ! the characters , ,..., are mutually distinct if for each i = j there is an x G with 1 2 n 6 2 (x) = (x). i 6 j

2. A collection , ,..., of characters of a group G in E is said to be linearly in- { 1 2 n} dependent if a (x)+a (x)+ + a (x)=0 1 1 2 2 ··· n n for some a E and each x G implies a = a = = a =0. i 2 2 1 2 ··· n Theorem 3.19. If , ,..., are distinct characters, then , ,..., is linearly in- 1 2 n { 1 2 n} dependent.

41 Proof. We proceed by induction on n.Forn =1,a(x)=0 = a =0as(x) = 0 and ) 6 E is an integral domain. Now suppose n 2 and assume the theorem holds true for all of collections of characters of size

As the i are mutually distinct there is a y G with 1(y) = n(y). 2 1 6 Replacing x with xy G and multiplying by (y) yields 2 n 1 a (x) (y) (y) + + (x)=0 foreachx G (3.6) 1 1 1 n ··· n 2 Equations (3.5) and (3.6) now give

1 1 a1(1 1(y)n(y) )1(x)+ + an 1(1 n 1(y)n(y) )n 1(x)=0 foreachx G. ··· 2

Since 1,2,...,n 1 is a collection of size

But each ai is non-zero so this gives

1 1 (y) (y) =0 (y)= (y)foreachi =1, 2,...,n 1. i n () i n This is a contradiction as (y) = (y) and the proof is complete. 1 6 n Now, an injective homomorphism of a field F into a field E (called an embedding of F into E) restricted to the multiplicative group F ⇤ defines a group character of F ⇤ in E.By injectivity of such a map, the only element which can be, and is, mapped to 0 E is 0 F , 2 2 so all the information we want to know about as a function on F into E is given by the character . In the light of this theorem we hence have the easy but important corollary:

Corollary 3.20. Distinct automorphisms of a field E are linearly independent as functions on E.

This follows by observing that distinct automorphisms of a field E are distinct embeddings which are linearly independent as functions on E by the above. This corollary puts us in a position to display the fundamental relationship between the size of a subgroup H of automorphisms of a field E and the degree of the extension E/F, in the situation where F is the fixed field of H.

42 Theorem 3.21. Let E be a field. If H = , ,..., is a subgroup of Aut(E) and F = { 1 2 n} fix(H), then [E : F ]= H = n. | | Proof. The are two stages to the proof. We first assume [E : F ] nand use properties obeyed by H as a group in order to reach a contradiction and conclude that [E : F ] n. This of course will immediately lead to the result.  Assume [E : F ]=m

(↵ )x + (↵ )x + + (↵ )x = 0 (3.7) 1 1 1 2 1 2 ··· n 1 n (↵ )x + (↵ )x + + (↵ )x =0 1 2 1 2 2 2 ··· n 2 n . . (↵ )x + (↵ )x + + (↵ )x =0. 1 m 1 2 m 2 ··· n m n

Let y1,y2,...,yn E be such a solution. Now, any collection of elements c1,c2,...,cm of F 2 th are by definition fixed by each i, so multiplying the i equation in the system by ci yields the system

(c ↵ )y + (c ↵ )y + + (c ↵ )y = 0 (3.8) 1 1 1 1 2 1 1 2 ··· n 1 1 n (c ↵ )y + (c ↵ )y + + (c ↵ )y =0 1 2 2 1 2 2 2 2 ··· n 2 2 n . . (c ↵ )y + (c ↵ )y + + (c ↵ )y =0. 1 m m 1 2 m m 2 ··· n m m n Adding all of these equations gives

(c ↵ + c ↵ + + c ↵ )y + + (c ↵ + c ↵ + + c ↵ )y = 0 (3.9) 1 1 1 2 2 ··· m m 1 ··· n 1 1 2 2 ··· m m n However, every element E is of the form = c ↵ + c ↵ + ...+ c ↵ for some c F as 2 1 1 2 2 m m i 2 the ↵i form an F -basis for E. Therefore, equation (3.9) reads

()y + ()y + + ()y =0 for all E 1 1 2 2 ··· n n 2 for some y1,y2,...,yn not all zero. That is, the distinct automorphisms 1,2,...,n are not linearly independent. This contradicts corollary (3.20). Hence [E : F ] n. Now assume

43 [E : F ] >nand let ↵1,↵2,...,↵n,↵n+1 be n + 1 elements of E that are linearly independent over F . The following system of n equations in n + 1 unknowns has a non-trivial solution.

(↵ )x + (↵ )x + + (↵ )x = 0 (3.10) 1 1 1 1 2 2 ··· 1 n+1 n+1 (↵ )x + (↵ )x + + (↵ )x =0 2 1 1 2 2 2 ··· 2 n+1 n+1 . . (↵ )x + (↵ )x + + (↵ )x =0. n 1 1 n 2 2 ··· n n+1 n+1

If y1,y2,...,yn+1 is a non-trivial solution of this system with each yi lying in F , then, since each i fixes F we have that

(↵ )y + (↵ )y + (↵ )y =0 1 1 1 1 2 2 ··· 1 n+1 n+1 (↵ y )+ (↵ y )+ (↵ y )=0 () 1 1 1 1 2 2 ··· 1 n+1 n+1 (↵ y + ↵ y + + ↵ y )=0 () 1 1 1 2 2 ··· n+1 n+1 ↵ y + ↵ y + + ↵ y =0. () 1 1 2 2 ··· n+1 n+1

But this contradicts the linear independence of the ↵i over F . Therefore, at least one of the yi lies in E not F .Letz1,z2,...,zn+1 be a solution with the fewest zi being non-zero, say 1 z ,z ,...,z = 0 after renumbering. Without loss of generality (or after multiplying by z ) 1 2 s 6 s we may assume zs = 1. We hence have the system

i(↵1)z1 + i(↵2)z2 + + i(↵s 1)zs 1 + i(↵s)=0 fori =1, 2,...,n. (3.11) ··· As we have established, not all of the z’s are fixed by every , so suppose z / F . Then, i 1 2 there is some k 1, 2,...,n with (z ) = z . Now, since the form a group, the elements 2{ } k 1 6 1 i k1,k2,...,kn are the same as the elements 1,2,...,n in some order. Applying k to the system (3.11) thus gives

i(↵1)k(z1)+i(↵2)k(z2)+ + i(↵s 1)k(zs 1)+i(↵s) = 0 (3.12) ··· for i =1, 2,...,n. Finally, taking the di↵erence of systems (3.11) and (3.12) we find that

i(↵1)(k(z1) z1) +i(↵2)(k(z2) z2)+ + i(↵s 1)(k(zs 1 zs 1)=0 ··· =0 6 for i =1, 2,...,n| gives{z a non-trivial} solution to the system (3.10) with

44 Corollary 3.22. For any finite extension E/F, we have the inequality Aut(E/F) [E : F ] | | where equality holds if and only if F = fix(Aut(E/F)). Equivalently, E/F is a Galois extension if and only if F is precisely the fixed field of Aut(E/F). Recall that in general we are only guarenteed that F fix(Aut(E/F))! Indeed, if B = fix(Aut(E/F)), then F B E so by ✓ ✓ ✓ theorem (3.21) and the degree formula we have

[E : F ] [E : F ]=[E : B][B : F ]= Aut(E/F) [B : F ] Aut(E/F) = [E : F ] | | ()| | [B : F ]  where equality holds if and only if [B : F ] = 1 if and only if B = F .

Corollary 3.23. If H is a finite subgroup of Aut(E) and F = fix(H), then H contains every automorphism of E which fixes F , that is, H = Aut(E/F) and E/F is Galois with Gal(E/F)=H.

Proof. By theorem (3.21) we have that [E : F ]= H and by the corollary we have Aut(E/F) | | | | [E : F ]. Of course, H Aut(E/F) so that H Aut(E/F) . Putting this all together yields  | || | [E : F ]= H Aut(E/F) [E : F ] | || | so each inequality is in fact an equality. Thus, H = Aut(E/F) and both groups are finite so | | | | H = Aut(E/F).

Corollary 3.24. If H and H are distinct subgroups of Aut(E), then fix(H ) = fix(H ). 1 2 1 6 2

Proof. We demonstrate the contrapositive. Let B1 = fix(H1) and B2 = fix(H2). If B1 = B2, then B2 = fix(H1) and every automorphism fixing B2 lies in H1 by the previous corollary, hence H H . Similarly, every automorphism fixing B lies in H so that H H . Thus 2 ✓ 1 1 2 1 ✓ 2 H1 = H2.

As a corollary to corollary (3.22) we have

Theorem 3.25. If E/F is a Galois extension, then any irreducible polynomial p(x) F [x] 2 having a root in E has all its roots in E, i.e p(x) splits over E.

Proof. Let p(x) F [x] be an irreducible having ↵ E as a root and let Gal(E/F)= 2 2 1, ,..., . Then, ↵, (↵),..., (↵) are also (not necessarily distinct) roots of p(x). Sup- { 2 n} 2 n pose ↵,↵ ,...,↵ are the distinct elements on this list. Since Gal(E/F) is a group, ⇢ 2 s 2 Gal(E/F) permutes the elements of Gal(E/F), that is, ⇢ ,⇢ ,...,⇢ = , ,..., . { 1 2 n} { 1 2 n} Therefore, ⇢(↵),⇢(↵2),...,⇢(↵s) is the same list as ↵,↵2,...,↵s in some order. Consider the polynomial f(x)=(x ↵)(x ↵ ) ...(x ↵ ) E[x]. By the discussion above, ⇢ fixes all 2 s 2 45 coecients of f(x) so by corollary (3.22) it follows that f(x) lies in F [x]. Since p(x) is ir- reducible, after multiplying by a constant we can assume p(x) to be monic so that it is the minimal polynomial for ↵.Therefore,p(x) f(x) in F [x] (and hence in E[x]) by theorem (1.22 | (iii)). But clearly f(x) p(x) in E[x] so that p(x)=f(x). Thus, p(x) splits over E. | Notice that p(x) is actually separable as the ↵’s are distinct. In particular, this means that the minimal polynomial of any element of a Galois extension E/F is separable and has all its roots in E. Our last corollary prior to the fundamental theorem is the converse to theorem (3.11 (ii)) which states that the splitting field of a separable polynomial is a Galois extension.

Corollary 3.26. If E/F is a Galois extension, then it is the splitting field of some separable polynomial q(x) F [x]. 2 Proof. Since E/F is Galois it is finite and hence generated over F by finitely many algebraic elements 1,2,...,m.Letqi(x) be the minimal poynomial for i over F for each i, then each qi(x) is separable with all its roots lying in E by the previous theorem. Now, E =

F (1,2,...,m) is evidently the splitting field over F of the product q1(x)q2(x) ...qm(x), so the polynomial q(x) F [x] obtained by omitting any repeated factors of this product is 2 separable and has the same splitting field, namely E.

We now have four equivalent characterisations of a Galois extension E/F:

1. Aut(E/F) =[E : F ] | | 2. the fixed field of Aut(E/F) is precisely F

3. E/F is a finite, separable extension in which the minimal polynomial of every element of E splits over E

4. E is the splitting field of some separable polynomial.

So why are Galois extensions of such particular interest to us? Galois extensions are the extensions which we intuitively described before as having “enough” automorphisms! It turns out that these extensions along with their Galois groups have some quite remarkable properties, and these are the content of the Fundamental Theorem of Galois Theory:

Theorem 3.27 (The Fundamental Theorem of Galois Theory). Let E/F be a Galois extension with Galois group G = Gal(E/F). Then, there is a bijection between subfields B of E containing

46 F and subgroups H of G given by

B all elements of G which fix B 7! H fix(H) and these correspondences are inverse[ to one another. Moreover, this correspondence has the following properties:

1. If B H and B H , then B B H H . 1 7! 1 2 7! 2 1 ✓ 2 () 2  1 2. E/B is always a Galois extension and has Galois group Gal(E/B)=H. G 3. [E : B]= H and [B : F ]= G : H = | | , where G : H denotes the index of H in G. | | | | H | | | | 4. B/F is a Galois extension if and only if H is a normal subgroup of G. In this case we have that Gal(E/F) G Gal(B/F) ⇠= Gal(E/B) = /H . 5. If B H and B H , then B B (H ,H ) and for the composite field B B 1 7! 1 2 7! 2 1 \ 2 7! 1 2 1 2 we have B B H H . 1 2 7! 1 \ 2 Proof. By corollary (3.24), for each subgroup H of G there is a unique subfield B of E so the correspondence is injective from right to left. Since E/F is Galois, E is the splitting field of some separable polynomial f(x) F [x]. Viewing f(x) as a polynomial over a subfield B of E 2 containing F , we have that E is the splitting field of f(x)overB,soE/B is a Galois extension. Since H := Gal(E/B) G has fixed field B, it follows that each subfield appears as the fixed  field of some subgroup of G. That is, the correspondence is surjective from right to left and is hence a bijection. The fact that the correspondences are inverse to one another follows directly from the group of automorphisms of a Galois extension E/B having precisely B as its fixed field.

1. The inclusion reversing nature of the correspondence was shown previously.

2. An application of theorem (3.21) gives [E : F ]= G and [E : B]= H , so the degree | | | | formula yields [E : F ] G [B : F ]= = | | = G : H [E : B] H | | | | 3. As above, E/B is the splitting field of some f(x) B[x] and so is a Galois extension. 2 Since Gal(E/B) H and Gal(E/B) has fixed field B, we have that Gal(E/B)=H as  a direct consequence of corollary (3.23).

47 4. Let B = fix(H). Each G gives rise to an embedding of B into the subfield (B) E 2 ✓ via its restriction to B. Conversely, suppose F is some fixed algebraic closure of F containing E and let : B (B) F be any embedding of B fixing F .If↵ B has ! ✓ 2 minimal polynomial p(x), then (↵) is another a root of p(x). But E contains all the roots of p(x) by corollary (3.26) so that (↵) E for each ↵ E, i.e (B) E.LetE be the 2 2 ✓ splitting field of f(x) B[x] as in the proof of (3), then E is also the splitting field field of 2 (f(x)) = f(x)over(B). Now, by the second isomorphism extension theorem there is an isomorphism : E E extending . Therefore, conversely, each embedding of B fixing ! F is of the form B for some automorphism G. It is clear that two automorphisms 2 1 , G restrict to the same embedding if and only if ( ) = 1. This is equivalent 1 2 1 2 B 2 1 to the statement that H if and only if H. Therefore, = 1 2 2 1 1 B 2 B 2 2 () 1H = 2H, that is, the distinct embeddings of B are in a bijection with the cosets of H in G. Letting emb(B/F) denote the collection embeddings of B fixing F , by the arguments above and (2) we hence have the equality emb(B/F) = G : H =[B : F ]. | | | | Now, B/F is a Galois extension if and only if Aut(B/F) =[B : F ]= emb(B/F , which, | | | | together with the inclusion Aut(B/F) emb(B/F), implies that B/F is Galois if and  only if Aut(B/F)=emb(B/F). Thus, (B)=B for every G. Observe that for any 1 2 G and ↵ B we have that (h )((↵)) = (h(↵)) = (↵)foreachh H (as 2 2 1 1 2 h(↵)=↵ for each ↵ B). Therefore, H fixes (B), giving H H. Since the 2  order of the group fixing (B) is equal to the degree of the extension E/(B), which in 1 turn is equal to the degree of the extension E/B (as Bcong(B)), it follows that H 1 and H are of the same finite order, so H = H. As we have seen, two subfields of E have the same fixing subgroup if and only if those subfields are equal, which is to say: 1 (B)=B for all G if and only if H = H for all G. We hence conclude that 2 2 B/F is a Galois extension if and only if H is a normal subgroup of G. Finally, having identified emb(B/F) with the collection of cosets of H in G, and having deduced that emb(B/F)=Aut(B/F) when H is normal in G, it follows that when B/F is Galois the group of automorphisms of B/F is isomorphic to the group of cosets of H in G. That is G Gal(B/F) ⇠= /H .

5. If B1 and B2 are the fixed fields of the subgroups H1 and H2, respectively, then any member of H H fixes every element of B , B and hence fixes the composite field 1 \ 2 1 2 B B . On the other hand, every element G fixing the composite B B fixes B and 1 2 2 1 2 1 B2 and so lies in H1 H2. Therefore, B1B2 corresponds to the subgroup H1 H2. Similarly, any member of B B is fixed by H , H and is hence fixed by the group they T 1 2 1 2 T generate; (H ,H ). Conversely, the elements fixed by (H ,H ) are those elements fixed by 1 2 T 1 2

48 both H1 and H2 and these are the elements of B1 B2. Therefore, B1 B2 corresponds to the subgroup (H ,H ). 1 2 T T

Example 3.28. Consider the Galois extension Q(p2, p3) with Galois group G = Gal(Q(p2, p3)/Q).

p2 p2 p2 p2 If : ! and ⇢ : ! , then we have the following table of p3 p3 p3 p3 ( ! ( ! subfields and subgroups:

Subgroup Fixed Field

1,,⇢,⇢ Q { } 1, Q(p3) { } 1,⇢ Q(p6) { } 1,⇢ Q(p2) { } 1 Q(p2, p3) { }

In this case, we see that all subfields are Galois over Q which corresponds to each subgroup being normal in G.

3 Example 3.29. Consider the Galois extension Q(!, p2) where ! is a cube root of unity and p3 2 is the real positive cube root of 2.

p3 2 !p3 2 p3 2 p3 2 If : ! and ⇢ : ! , then we have the table ! ! ! !2 ( ! ( !

Subgroup Fixed Field

G =(,⇢) Q () Q(!) 3 (⇢2) Q(!2p2) 3 (⇢) Q(!p2) 3 (⇢) Q(p2) 3 1 Q(!, p2) { }

In this case, the only normal subgroup is H =() which is in accordance with Q(!)/Q being the only Galois extension on the list!

49 Chapter 4

General Polynomials and the Discriminant

Consider a separable polynomial f(x) F [x]. We defined the Galois group of such an f to 2 be the Galois group of its splitting field. We also showed that an arbitrary Galois extension

E of a field F is the splitting field of some separable polynomial. Suppose ↵1,↵2,...,↵n are the distinct roots of f(x) so that E = F (↵1,↵2,...,↵n). As we know, any automorphism of Gal(E/F) maps a root of an irreducible factor of f(x) to another root of that irreducible factor, and since E is generated over F by the ↵i, is completely determined by its values at these generators. Thus, defines a permutation of the letters ↵1,↵2,...,↵n which is unique to each member of Gal(E/F) by lemma (1.41). If we fix a particular ordering of ↵1,↵2,...,↵n, then each Gal(E/F) defines a unique permutation of the set of indices 1, 2,...,n .We 2 { } state this observation formally as

Theorem 4.1. If f(x) F [x] has n distinct roots in its splitting field E, then Aut(E/F) is 2 isomorphic to a subgroup of the symmetric group Sn. Proof. Let R = ↵ ,↵ ,...↵ be the collection of roots of f(x). If Aut(E/F), then { 1 2 n} 2 (R)=R. Since is a ring map, the map

: Aut(E/F) S ,() := ! R R restricting each Aut(E/F)toR is a group homomorphism into the group of permutations 2 of the letters ↵1,↵2,...,↵n. Since the splitting field of f(x), E = F (↵1,↵2,...,↵n), is generated over F by the ↵j, is determined completely by its values at the ↵j. Thus, if (1)=(2) then 1 and 2 have the same values at each ↵j and hence are equal by theorem (1.41), so is injective. Therefore, Aut(E/F) is isomorphic to a subgroup of SR. Finally, SR ⇠= Sn so composing isomorphisms gives Aut(E/F) ⇠= Sn and this completes the proof.

50 Corollary 4.2. If f(x) F [x] is a separable polynomial of degree n with splitting field E, 2 then Aut(E/F) is isomorphic to a subgroup of the symmetric group Sn.

Note that in an earlier chapter we inadvertently proved that the splitting field E/F of a polynomial f(x) F [x] was an extension of degree n!. The fundamental theorem tells us 2  that the degree of E/F is equal to the order of its Galois group. Since this Galois group is a subgroup of S and S = n!, this provides a group-theoretic reasoning for such a bound on n | n| the degree of E/F.

Remark 4.3. If we were to fix di↵erent labellings of the roots of f(x) this would give rise to an isomorphism of Gal(E/F) with a di↵erent subgroup of Sn. Of course these subgroups will still be isomorphic to one another.

4 Example 4.4. Let us revisit the field extension Q(p2,µ)/Q, the splitting field of the separable 4 4 polynomial x 2. By the above, Gal(Q(p2,µ)/Q) is isomorphic to a subgroup of S4.If we fix a labelling of the four roots: ↵ = p4 2,↵ = µp4 2,↵ = p4 2 and ↵ = µp4 2, 1 2 3 4 then : ↵ ↵ ↵ ↵ , so we identify with the element (1 2 3 4) in S . Also, 1 1 7! 2 7! 3 7! 4 1 4 : ↵ ↵ ↵ , so we identify with the element (1 3) in S . This gives the explicit 6 1 7! 3 7! 1 6 4 isomorphism:

4 2 3 2 3 Gal(Q(p2,µ)/Q)= 1,1,1,1,6,16,16,16 { } = 1, (1 2 3 4), (1 3)(2 4), (1 4 3 2), (1 3), (1 2)(3 4), (2 4), (1 4)(2 3) ⇠ { } ⇠= D8. Galois’ great theorem relates strongly the notion of a “solvable group” to the “solvability by radicals” of a polynomial. Since the question of solvability of the symmetric groups Sn is relatively simple to answer, it would be of tremendous advantage to us if a general polynomial of degree n were to have the full group Sn as its Galois group. Happily this is indeed the case! In order to see this, we need to make precise what we mean by a general polynomial of degree n.

th Definition 4.5. Let x1,x2,...,xn be indeterminates. The i elementary symmetric func- tion is the sum of all products of the xj taken i at a time. That is,

s = x + x + + x 1 1 2 ··· n s2 = x1x2 + x1x3 + + x1xn + x2x3 + x2x4 + + x2xn + + xn 1xn ··· ··· ··· . .

sn = x1x2 ...xn.

51 Definition 4.6. The general polynomial of degree n over a field F is the monic polyno- mial whose collection of roots comprises the indeterminates x1,x2,...,xn:

g(x)=(x x )(x x ) (x x ), 1 2 ··· n considered as a polynomial over the field of rational functions in the si F (s1,s2,...,sn). Expanding this expression, it is clear that the coecients of the general polynomial of degree n are the elementary symmetric functions in the roots xj:

n n 1 2 n 2 n g(x)=x +( 1)s x +( 1) s x + +( 1) s 1 2 ··· n n n 1 n 2 n = x s x + s x + +( 1) s . 1 2 ··· n

Consider the field F (s1,s2,...,sn) of rational functions in the elementary symmetric poly- nomials. The splitting field of the general polynomial of degree n over F (s1,s2,...,sn) is

F (x1,x2,...,xn). Therefore, F (x1,x2,...,xn)/F (s1,s2,...,sn) is a Galois extension.

As above, any member of the symmetric group Sn defines a unique permutation of the set 1, 2,...,n and hence of the set of roots x ,x ,...x . Such a permutation acts on rational { } { 1 2 n} functions in the xj by simply permuting the indeterminates. It is easily seen, therefore, that each member of Sn defines a unique automorphism of F (x1,x2,...,xn). This means we can identify Sn as a subgroup of Aut(F (x1,x2,...,xn)). What is the associated fixed field of Sn?

First observe that every elementary symmetric polynomial si remains unchanged by elements of Sn, therefore, so does any rational function in the si. That is, F (s1,s2,...,sn) is a subfield of fix(Sn). Now, since F (x1,x2,...,xn) is the splitting field of the general polynomial of degree n we have that [F (x ,x ,...,x ):F (s ,s ,...,s )] n!, but by the fundamental theorem of 1 2 n 1 2 n  Galois theory we also have that [F (x ,x ,...,x ):fix(S )] = S = n!. Therefore, the degree 1 2 n n | n| formula yields

n! [F ( ,x ,...,x ):F (s ,s ,...,s )] 1 2 n 1 2 n =[F (x ,x ,...,x ):fix(S )][fix(S ):F (s ,s ,...,s )] n! 1 2 n n n 1 2 n =n! [fix(S ):F (s ,s ,...,s )] = 1 ()| n {z 1 2 } n fix(S )=F (s ,s ,...,s ). () n 1 2 n

We conclude that in fact F (s1,s2,...,sn) constitutes the entire fixed field of Sn; this is the content of the fundamental theorem of symmetric functions.

Definition 4.7. A member f(x1,x2,...,xn) of the field F (x1,x2,...,xn) of rational func- tions in the indeterminates x1,x2,...,xn is a symmetric function if the function remains unchanged under a permutation of the xi.

52 Theorem 4.8 (Fundamental Theorem of Symmetric Functions). Every symmetric function in the indeterminates x1,x2,...,xn is a rational function of the elementary symmetric functions in the xi.

In truth, a stronger statement holds true: If the symmetric function f in the above happens to be a polynomial function of the xi, then f is a polynomial function of the elementary symmetric polynomials! Let us now reverse our perspective and assume we are given a general polynomial

n n 1 n 2 n g(x)=x s x + s x + +( 1) s 1 2 ··· n of degree n over the field of rational functions F (s1,s2,...,sn) and we view the si as indeter- minates. By this we mean that there exists no non-trivial polynomial relation between the si. That is, p(s ,s ,...,s )=0 p 0. 1 2 n () ⌘

Now, defining the roots of g(x)tobex1,x2,...,xn leads to the coecients si being (up to sign) the elementary symmetric polynomials in x1,x2,...,xn. We claim that the xi are also indeterminates in the same sense as the s . Indeed, if p(y ,y ...,y ) F [y ,y ,...,y ] is a i 1 2 n 2 1 2 n non-trivial polynomial satisfying p(x1,x2,...,xn) = 0, then, if 1,2,...,n! are the members of Sn, we have that

q(y ,y ,...,y ) := p(y ,...,y )p(y ,...,y ) p(y ,...,y )=0 1 2 n 1(1) 1(n) 2(1) 2(n) ··· n!(1) n!(n) is a non-trivial symmetric polynomial in F [y1,y2,...,yn] which also satisfies q(x1,x2,...,xn)= 0. But this is a contradiction because, by the fundamental theorem of symmetric functions, it produces a non-trivial polynomial relation between the si where none exists. It can also be shown that the coecients of a polynomial whose roots are indeterminates are themselves inde- terminates. In conclusion to this discussion, we may equivalently define the general polynomial over F as having either indeterminate coecients or indeterminate roots. In this sense we are fully justified in our use of the term general polynomial.

Through this discussion we are now able to conclude our original suggestion:

Theorem 4.9. The general polynomial of degree n over a field F is separable with Galois group isomorphic to the full symmetric group Sn.

Certain properties of the Galois group of a polynomial can be deduced and restated in terms of a quanity known as the discriminant.

53 Definition 4.10. The discriminant of x1,x2,...,xn is the quantity

D = (x x )2. i j i

Observe that the discriminant is a symmetric function of the xi,soD lies in the fixed field

F (s1,s2,...,sn)ofSn.

Let = pD = (x x ) and note that / fix(S ) (with the exception of the case n =1. i j 2 n il. Specifically, if there i j k l are an even number of such occurrences then ()=and if there are an odd number then () = . For this reason we say that S is even if ()=and is odd if () = . 2 n Naturally, the even permutations form a subgroup whereas the odd permutations do not. The subgroup of even permutations is denoted An. In light of our discussion on quadratic extensions, from the fundamental theorem of Galois theory it follows that if F is a field of characteristic p = 2, then fix(A )=F () and generates a quadratic extension of F (s ,s ,...,s )as 6 n 1 2 n / F (s ,s ,...,s ) and the only (normal) subgroup of index 2 in S is A . We conclude that 2 1 2 n n n if Char(F ) = 2, then A if and only if ()=. 6 2 n Let us now consider separable polynomials over a field F of characteristic 0 and their Galois groups.

Suppose f(x) has roots ↵1,↵2,...,↵n. By definition, the discriminant of f(x) is

D = (↵ ↵ )2. i j i

D =(↵ ↵ )2 =(↵ + ↵ )2 4↵ ↵ =( b)2 4c = b2 4c, 1 2 1 2 1 2 the discriminant familiar from elementary algebra. f(x) has two distinct roots if and only if b2 4c =0. 6 Now, since Gal(f) is a subgroup of S , there are two possible Galois groups, namely A = 1 2 2 { } and S itself. Gal(f)=A if and only if pb2 4c lies in F so that Gal(f)=S if and 2 2 2 only if pb2 4c/F . This reflects our previous calculations of the possible Galois groups of 2 quadratics: If pD/F then f(x) does not split over F and so the splitting field of f(x) is the 2 quadratic extension F (pD)/F and Gal(f) = S , whereas if pD F then F is the splitting ⇠ 2 2 field of F and Gal(f) is trivial. Now consider the cubic polynomial

f(x)=x3 + bx2 + cx + d.

First simplify the problem by performing what is known as a Tschirnhaus transformation: b Setting y = x + 3 and substituting into f gives 1 1 g(y) := f(x(y)) = y3 + py + q where p = (3c b2) and q = (27d +2b2 9bc). 3 27 Since x and y di↵er by an element of F , the splitting fields of f and g coincide. Also, if f has b b b roots ↵1,↵2 and ↵3, then the roots of g are ↵1 + 3 ,↵2 + 3 and ↵3 + 3 , so the discriminant of g b b [(↵ + ) (↵ + )] = (↵ ↵ ) i 3 j 3 i j i

g0(y)=(y )(y )+(y )(y )+(y )(y ) 2 3 1 3 1 2 so that

g0( )=( )( ) 1 1 2 1 3 g0( )=( )( ) 1 2 1 2 3 g0( )=( )( ). 1 3 1 3 2

55 and

D =( )2( )2( )2 1 2 1 3 2 3 = [( )( )][( )( )][( )( )] 1 2 1 3 1 2 2 3 1 3 2 3 = g0( )( g0( ))g0( )= g0( )g0( )g0( ). 1 2 3 1 2 3 2 Also, from the original form of g we have that g0(y)=3y + p which gives

D = (32 + p)(32 + p)(32 + p) 1 2 3 = [27 (222) +9p (22 + 22 + 22) +3p2 (2 + 2 + 2) +p3] 1 2 3 1 2 1 3 2 3 1 2 3 2 2 2 2 2 =s =q s 2s1s2=p s 2s2= 2p 3 2 1 = 27q|2 {z4p2 } | {z } | {z } = b2c2 4c3 4b3d 27d2 +18bcd. As with the quadratic, we can calculate the possible Galois groups of the cubic f(x)=x3 + bx2 + cx + d F [x] and classify them in terms of the discriminant. If f(x) is reducible, then 2 either f(x) splits over F and the Galois group of f is trivial, or f(x) has an irreducible quadratic factor and the Galois group is either trivial or cyclic of order 2 as above. If f(x) is irreducible, let ↵ denote any root of f(x). Then, [F (↵):F ] = 3 so that 3 divides the degree of the splitting field of f(x). But Gal(f) S and the only subgroups of S divisible by 3 are A and S itself.  3 3 3 3 Therefore, there are two possible Galois groups: if pD F , then Gal(f)=A and if pD/F , 2 3 2 then Gal(f)=S3. To see how this works directly, recall that pD always lies in the splitting field of f(x). If pD F then the splitting field of F is F (↵) where ↵ denotes any root of f(x). 2 F (↵)/F is Galois of degree 3 so Gal(f) is of order 3 and is hence A .IfpD/F then we must 3 2 also adjoin pD to F (↵). F (↵, pD)/F is Galois of degree 6, so in this case Gal(f) is of order

6 and is hence S3. Note that in either case, the splitting field f could be obtained by adjoining pD and any root of f(x)toF .

56 Chapter 5

Algebraic Closures and Composite Extensions

We have solved the problem of constructing field extensions containing all the roots of a single polynomial, so we might ask is it possible to construct a field containing all the roots of all polynomials with coecients lying in the base field?

Definition 5.1. A field F is called an algebraic closure of the field F if F/F is an algebraic extension and every polynomial f(x) F [x] splits over F . 2 Definition 5.2. A field E is called algebraically closed if every polynomial f(x) E[x] has 2 a root in E.

Of course, if E is an algebraic closure of itself then E is algebraically closed. The converse is also true. Indeed, if E is algebraically closed and f(x) E[x] is any polynomial then f(x)as 2 aroot↵ E, so has a factorisation (x ↵)g(x) in E[x]. Now, g(x) E[x] has a root E, 2 2 2 so f(x)=(x ↵)(x )h(x) in E[x]. Continuing in this manner we find that f(x) splits over E so that E is an algebraic closure of itself. This tells us that taking the algebraic closure of an algebraic closure adds no new elements. At this point we have no reason to believe that an algebraic closure of an arbitrary field F exists or that algebraically closed fields exist. If we want to adjoin to F all the roots of a finite collection of polynomials over F , then the splitting field of each member of that collection is contained in the splitting field E of the product of all the members. Therefore, the field generated by the individual splitting fields is the smallest subfield of E containing them. Now, in order to construct an algebraic closure of a field F we would like to construct a field containing the roots of infinitely many polynomials (in general), so we no longer have a clear idea of into what might we generate an algebraic closure.

57 Theorem 5.3. Given a field F , there exists a field E containing F which is algebraically closed.

Proof. Let A = f(x) F [x] f is monic,@(f) 1 F [x]. For each f(x) A let x be { 2 | }✓ 2 f an indeterminate and consider the ring of polynomials over F in the xf ’s; F [...,xf ,...]. Let I F [...,x ,...] be the ideal generated by f(x ). We claim that I is a proper ideal. If not, ✓ f f then 1 I and there exist h ,h ,...,h F [...,x ,...] such that 2 1 2 k 2 f h f (x )+h f (x )+ + h f (x )=1. (5.1) 1 1 f1 2 2 f2 ··· k k fk

For the sake of clarity set xi := xfi for i =1, 2,...,k and suppose the indeterminates occurring in the hi and not the fi are xk+1,...,xn (there are only finitely many as the hi are polynomials by definition). We then have the equation

h (x ,x ,...,x )f (x )+ + h (x ,x ,...,x )f (x )=1. (5.2) 1 1 2 n 1 1 ··· k 1 2 n k k

If ↵i is a root of fi(x) in a splitting field of fi, consider the extension F (↵1,↵2,...,↵k)/F .

In this extension, setting xi = ↵i for i =1, 2,...,n in the above equation yields 0 = 1 in

F (↵1,↵2,...,↵k)/F ; a contradiction. Hence, I is a proper ieal. Next we claim that I is contained in a maximal ideal and to show this we invoke Zorn’s lemma.

To this end let S denote the collection of all proper ideals of F [...,xf ,...] containing I. S is certainly non-empty as it contains I itself. Let C = I be a chain in S where is { | 2A} A some index set and set U = I.Foranya, b U we have that a I1 and b I2 for 2A 2 2 2 some , , but C is totally ordered so without loss of generality we may assume I I 1 2 2A S 1 ✓ 2 so that a, b I . Now, I is an ideal, so a b lies in I and hence in U. Also, for any 2 2 2 2 r F [...,x ,...] and a U, a I for some so that ra lies in I and hence in U. 2 f 2 2 2A Therefore, U is an ideal. If U is not a proper ideal then 1 U so that 1 I for some , 2 2 2A but this is impossible since C contains only proper ideals. Thus, U is a proper ideal containing I which is evidently an upper bound for C. By Zorn’s lemma, S contains a maximal element, M. This element is plainly the maximal ideal we were looking for. Consider now the quotient ring

F1 := F [...,xf ,...] /M .

Since M is maximal, F1 is a field. F1 contains F (an isomorphic copy thereof) and, letting

↵f denote the equivalence class of xf in the quotient, F1 contains a root ↵f of f(x)foreach f(x) F [x]asf(x ) I M(= 0 in F ). By the same procedure we can construct a field F 2 f 2 ✓ 1 2 containing F1 in which every polynomial in F1[x] has a root. Continuing this process we obtain a sequence of fields F = F F F F F 0 ✓ 1 ✓ 2 ✓···✓ n ✓ n+1 ✓··· 58 with the property that for each n N0, every polynomial in Fn[x] has a root in Fn+1. Now, 2 setting E = F we have that E is a field containing F . The coecients of a polynomial n 0 n 2N p(x) E[x], say a ,a ,...,a lie in some F ,F ,...,F , respectively. But F F for each 2 S 0 1 j n0 n1 nj ni ✓ N i =0, 1,...,j where N := max n ,n ,...,n . Therefore, p(x) lies in F [x], so has a root in { 0 1 j} N FN+1 and hence in E. Thus, E is an algebraically closed field containing F .

We now use the existence of an algebraically closed field containing F to give an internal construction of an algebraic closure of F .

Theorem 5.4. The collection F of elements of an algebraically closed field E containing F which are algebraic over F forms an algebraic closure of F .

Proof. F is algebraic over F by construction so it remains to show that every polynomial in F [x] splits over F . By definition, each f(x) F [x] splits over E, so has all its roots in E. 2 Therefore, each of these roots is algebraic over F and hence lies in F . Thus, f(x) has all its roots in F and so splits over F .

Definition 5.5. Let E be an extension of the fields E1 and E2. The compostie field of E1 and E2 is the intersection of all subfields of E containing both E1 and E2, denoted E1E2.In this sense, E1E2 is the smallest subfield of E containing E1 and E2. Similarly, we define the composite of any number of subfields of E to be the smallest subfield containing all those fields.

Theorem 5.6. If E1/F and E2/F are finite extensions where E1 and E2 are contained in a field E, then [E E : F ] [E : F ][E : F ] 1 2  1 2 where equality holds when and only when a basis for E1/F remains linearly independent over

E2, or vice versa.

Proof. Let ↵1,↵2,...,↵n and 1,2,...,m be bases for E1 and E2 over F , respectively. Then,

E1E2 = F (↵1,↵2,...,↵n,1,2,...,m). Now, the set of all F -linear combinations of the form

n m

cij↵ij i=1 j=1 X X is closed as any power of the ↵’s and ’s is an F -linear combination of ↵1,...,↵n and 1,...,m), respectively. It follows that the collection B := ↵ 1 i n, 1 j m is an F - { i j |     } spanning set for E E . This gives [E E : F ] nm =[E : F ][E : F ]. Finally, since E E = 1 2 1 2  1 2 1 2 E ( , ,..., ) we have that [E E : E ] [E : F ] with equality when and only when the 1 1 2 m 1 2 1  2 j remain linearly independent over E1. The degree formula gives [E1E2 : F ]=[E1E2 : E1][E1 : F ] from which we conclude the result.

59 Theorem 5.7. Let E/F be a Galois extension and let Fˆ be any extension of F . Then, EF/ˆ Fˆ is a Galois extension. Moreover, Gal(EF/ˆ Fˆ) is isomorphic to the subgroup Gal(E/E Fˆ) of \ Gal(E/F): Gal(EF/ˆ Fˆ) = Gal(E/E Fˆ). ⇠ \ Proof. Since E/F is Galois, E is the splitting field of some f(x) F [x]. By the minimality of 2 EFˆ as a field containing Fˆ and all the roots of f(x), EF/ˆ Fˆ is the splitting field of f(x) viewed as a polynomial in Fˆ[x]. Hence, EF/ˆ Fˆ is also Galois. Consider the map

:Gal(EF/ˆ Fˆ) Gal(E/F) ! ()= E. By the considerations in the proof of the fundamental theorem, every embedding of E fixing F defines an automorphism of E. Therefore, is well-defined and defines a group homomorphism. Any Gal(EF/ˆ Fˆ) fixes Fˆ by definition, so if restricts to the identity map on E, then it 2 restricts to identity on both E and Fˆ and hence fixes the composite field EFˆ. Therefore, is an injection. Now, let H = Im( ) Gal(E/F) and set B = fix(H). Every element of Fˆ is  fixed by each member of H,soE Fˆ B. Also, every member of Gal(EF/ˆ Fˆ) restricted to \ ✓ B fixes B, so every member of Gal(EF/ˆ Fˆ) fixes both B and Fˆ and hence fixes the composite field BFˆ. But by the fundamental theorem we have that BFˆ = Fˆ which implies B Fˆ, and ✓ together with the trivial inclusion B E this yields B E Fˆ. Thus, B = E Fˆ. Finally, ✓ ✓ \ \ the fundamental theorem and the first isomorphism theorem for groups give

Gal(EF/ˆ Fˆ) = Im( ) = H = Gal(E/B)=Gal(E/E Fˆ). ⇠ \

[E : F ][Fˆ : F ] Corollary 5.8. If E/F is Galois and F/Fˆ is finite, then [EFˆ : F ]= . [E Fˆ : F ] \ Proof. From the previous theorem and the fundamental theorem we have that

[EFˆ : Fˆ]= Gal(EF/ˆ Fˆ) = Gal(E/E Fˆ) =[E : E Fˆ]. | | | \ | \ [E : F ] The degree formula gives [E : E Fˆ]= . \ [E Fˆ : F ] Thus, \ [E : F ][Fˆ : F ] [EFˆ : F ]=[EFˆ : Fˆ][Fˆ : F ]=[E : E Fˆ][Fˆ : F ]= . \ [E Fˆ : F ] \

60 Theorem 5.9. If E1/F and E2/F are Galois extensions, then the following hold:

1. E E /F and E E /F are Galois extensions 1 \ 2 1 2 2. Gal(E E /F ) = S := ( , ) = on E E Gal(E /F ) Gal(E /F ). 1 2 ⇠ { 1 2 | 1 2 1 \ 2} 1 ⌦ 2 [E1 : F ] Proof. (i) Since E1/F is Galois it is finite, and [E1 E2 : F ]= [E1 : F ] < \ [E1 : E1 E2]  1 so E E /F is also finite. Also, any irreducible polynomial p(x) F [x]\ having ↵ E E 1 \ 2 2 2 1 \ 2 ✓ E ,E as a root is separable with all its roots lying in E and E and hence E E by theorem 1 2 1 2 1 \ 2 (3.25). Therefore, E E /F is a finite, separable extension over which the minimal polynomial 1 \ 2 of every element of E E /F splits. Thus, E E /F is Galois. 1 \ 2 1 \ 2

Suppose E and E are splitting fields of the separable polynomials p (x) F [x] and p (x) 1 2 1 2 2 2 F [x], respectively. The, E E is the splitting field of the separable polynomial p(x) F [x] ob- 1 2 2 tained by removing any repeated factors in the product p1(x)p2(x). Hence, E1E2/F is Galois. (ii) Consider the map

:Gal(E E /F ) Gal(E /F ) Gal(E /F ) 1 2 ! 1 ⌦ 2 () :=( , ). E1 E2 If ↵1,↵2,...,↵n are the roots of p1(x), then E1= F (↵1,↵2,...,↵n) so that

(E1)=(F (↵1,↵2,...,↵n))

= F ((↵1),(↵2),...,(↵n)=F (↵1,↵2,...,↵n)=E1.

Therefore, does indeed define an element of Gal(E1/F ). Similary, Gal(E2/F )so E1 E2 2 is well-defined and is easily seen to be a group homomorphism. Now, any Gal(E E /F ) 2 1 2 fixing E1 and E2 also fixes the composite field E1E2, so the kernel of is trivial and is an injection. Since ( ) = =( ) , it follows that Im( ) S. In order to count the E1 E1 E2 E1 E2 E2 E1 E2 \ \ \  elements of S, note that for any fixed1 Gal(E1/F ) there are precisely Gal(E2/E1 E2) elements 2 | \ | 2 Gal(E2/F ) satisfying 1 = 2 . Therefore, E1 E2 E1 E2 2 \ \ S = Gal(E /E E ) Gal(E /F )) | | | 2 1 \ 2 |⇥| 1 | =[E : E E ] [E : F ] 2 1 \ 2 ⇥ 1 [E : F ][E : F ] = 2 1 [E E : F ] 1 \ 2 =[E1E2 : F ] (by the previous corollary) = Gal(E E /F ) . | 1 2 |

61 That is, S and Gal(E1E2/F ) are of the same finite order and so are S and Im( ). Thus, S = Im( ) and this completes the proof.

Corollary 5.10. If E /F and E /F are Galois extensions and F = E E , then Gal(E E /F ) = 1 2 1\ 2 1 2 ⇠ Gal(E /F ) Gal(E /F ). 1 ⌦ 2 Proof. Any pair , Gal(E E /F )agreeonE E so that ( , ) S and is a 1 2 2 1 2 1 \ 2 1 2 2 surjection.

Corollary 5.11. If E/F is a finite separable extension, then there is a Galois extension E/Fˆ with E Eˆ where Eˆ is minimal with respect to inclusion (in some fixed algebraic closure of ✓ Eˆ).

Proof. Since E/F is finite, E is finitely generated over F by some ↵ ,↵ ,...,↵ E.LetE be 1 2 n 2 i the minimal polynomial of ↵ . Then, the composite field E E E is Galois by induction on i 1 2 ··· n n using theorem (5.9(i)). Now take the intersection over all Galois extensions of F containing E.

62 Chapter 6

Solvability by Radicals and Galois’ Great Theorem

In this section we show how properties of the Galois group of a polynomial determine whether or not its roots can be expressed in terms of the field operations and taking nth roots for various n N. When this can be done we refer to the polynomial as being “solvable by radicals”; 2 we will give a concrete definition of this property in due course. The quadratic, cubic and quartic formulas provide a way to solve for the roots by radicals for entire classes of polynomial equations. We will see that no such analogous formula for solving polynomial equations of higher degree by radicals can exist. The reason for this is simple: a polynomial f(x) of degree n is solvable by radicals if and only if Gal(f) is a solvable group, but for the general polynomial of degree n over a field F we have Gal(f) = S which is not solvable for n 5 as we will show. ⇠ n Perhaps the simplest extensions of a field F are those obtained by adjoining an nth root of an element c F , that is, adjoining some root of the polynomial f(x)=xn c. Of course if 2 pn c denotes any root of xn c, then F (pn c)/F is a Galois extension when and only when F contains the nth roots of unity. For this reason, these extensions are most well-behaved when F contains the necessary roots of unity.

Definition 6.1. The field extension E/F is called pure of type n if E = F (↵) where ↵n F 2 for some n N. That is, E is the extension of F obtained by adjoining the nth root of some 2 element of F .

Definition 6.2. Let ↵ be algebraic over F . We say that ↵ is expressible by radicals if there is a finite sequence of field extensions

F = E E E E E 0 ✓ 1 ✓ 2 ✓···✓ r ✓

63 with ↵ E and where each Ek/Ek 1 is pure of type nk for some nk N. 2 2 Here, E is called a root extension of F .

Definition 6.3. A polynomial f(x) F [x] is said to be solvable by radicals if each of its 2 roots are expressible by radicals.

Lemma 6.4. Let E = F (!) where ! is a primitive nth root of unity. Then, E/F is Galois with Galois group isomorphic to a subgroup of the multiplicative group of units Zn⇤ of Zn.

Proof. E/F is Galois as it is the splitting field of the cyclotomic polynomial n(x). Any Gal(E/F) maps ! to another root of (x), i.e (!)=!i for some 0

:Gal(E/F) (Zn, ) ! ⇥ [i] i 7! is well-defined. is a homomorphism as follows:

j j i j ij ij(!)=i(! )=(i(!)) =(! ) = !

That is, (ij)= (i) (j). If (!)=⌫(!), then = ⌫ by lemma (1.41), so is an injection and the result follows.

Corollary 6.5. Let ! be a primitive pth root of unity over a field F , p a prime. Then, Gal(F (!)/F ) is a cyclic group.

Proof. Gal(F (!)/F ) is isomorphic to a subgroup of Zp⇤ which is cyclic by theorem (2.11). Definition 6.6. Let E/F be a Galois extension. E/F is called a cyclic extension if Gal(E/F) is a cyclic group.

Theorem 6.7. If F is a field of characteristic p - n containing the nth roots of unity and c belongs to F , then the pure extension F (pn c)/F of type n is a cyclic extension of degree dividing n.

Proof. As above, F (pn c)/F is Galois as it is the splitting field of the polynomial xn c F [x]. 2 If Gal(F (pn c)/F ), then (pn c)=µpn c for some µ ⌥ . Since is completely determined 2 2 n n n n by its value at pc, write µ = µ in case (pc)=µpc. This gives a well-defined mapping

' : Gal(F (pn c)/F ) ⌥ ! n µ 7!

64 ' is a group homormorphism. Indeed, since ⌥ F , every member of the Galois group fixes n ✓ ⌥ . Therefore, for , Gal(F (pn c)/F ) n 1 2 2 : pn c (pn c)=µ pn c 2 1 7! 1 1 (µ pn c)= (µ ) (pn c) 7! 2 1 2 1 2 n = µ1 µ2 pc

n = µ2 µ1 pc. If ker('), then fixes pn c and so fixes the full field F (pn c) and is the identity map. 2 Therefore, ' is an injection and the result follows from the first isomorphism for groups and the fact that every subgroup of a cyclic group is cyclic.

In fact, the converse also holds true, so that when a field contains the necessary nth roots of unity, cyclic extensions are pure extensions and vice versa. Theorem 6.8. Let E/F be a Galois extension. If E/F is cyclic of degree n not dividing the characteristic of F and F contains the nth roots of unity, then E is a pure extension of type n. That is, E = F (pn c) for some c F . 2 Proof. Let generate the cyclic group Gal(E/F) and µ be a primitive nth root of unity. Consider the element

2 2 n 1 n 1 (c, µ) := c + µ(c)+µ (c)+ + µ (c) E, ··· 2 Since n = 1 and µ lies in F , fixes µ and we have that

2 2 3 n 2 n 1 n 1 ((c, µ)) = (c)+µ (c)+µ (c)+ + µ (c)+µ c ··· 1 2 2 n 1 n 1 = µ (c + µ(c)+µ (c)+ + µ (c)) ··· 1 = µ (c, µ).

n 1 n 1 n n n Therefore, ((c, µ)) = (c, µ) =[µ (c, µ)] =(µ ) (c, µ) =(c, µ) and we see that fixes (c, µ)n. But generates Gal(E/F)so(c, µ)n lies in the fixed field of Gal(E/F). Thus, (c, µ) F . 2 Observe now that the linear independence of the auotmorphisms of the Galois group implies k 1 k 1 1 k the existence of a c E with (c, µ) = 0. Now, ((c, µ)) = µ ((c, µ)) = =(µ ) (c, µ) 2 6 1 k ··· for any natural number k and the elements (µ ) are distinct for 0 1 ✓ and F ((c, µ)) is the fixed field of a subgroup of Gal(E/F) of order > 1 by the fundamental theorem. But we have shown this not to be the case, so E = F ((c, µ)) = F (pn b) where b =(c, µ)n F . This proves the assertion. 2 65 Theorem 6.9. Suppose ↵ is algebraic over F and that E/F is a root extension containing ↵, as in (6.2). Then, there exists a Galois root extension E0 in which each intermediate extension is cyclic.

Proof. Let L be the Galois extension of F as constructed in corollary (5.11). If Gal(L/F ), 2 then the sequence of subfields

F = (E ) (E ) (E ) (E )=(E) 0 ✓ 1 ✓ 2 ✓···✓ r is such that each (Ek)/(Ek 1) is again a pure extension of type nk. Indeed, (Ek) is generated nk over (Ek 1) by a root of the polynomial x (ck). Now, the composite of two root extensions of F is again a root extension of F as follows: if

F = B B = B 0 ✓···✓ l

mk is another root extension where Bk = Bk 1( pbk) for some bk Bk 1, then taking the com- 2 posite of each of these subfields with E1 we see that

m1 m2 ml E1B0 = E1 E1B1 = E1B0( b1) E1B2 = E1B1( b2) E1Bl = E1Bl 1( bl) ✓ ✓ ✓···✓ p p p so that E1B is a root extension of E1. If we now take the composite of each of these subfields with E2 we obtain a root extension of E2 in the same manner. Continuing this process provides a root extension EB of E, but E is a root extension of F so EB is a root extension of F . Now, by induction, the composite of any finite collection of root extensions is again a root extension. In particular, if Gal(L/F )= , ,..., , then the composite field (E) (E) ... (E) is { 1 2 n} 1 2 n a root extension of F . But 1(E)2(E) ...n(E) is the entire field L,so↵ is contained in the Galois root extension L.

Let m be the least common multiple of n1,n2,...,nr and m1,m2,...,ml and write m = p p p as a product of primes. We can adjoin all the nth and mth roots of unit to the 1 2 ··· s i i field F by adjoining a primitive mth root of unity !; F (!)/F . We can refine this extension to a

p1 pi 1pi+1 ps th sequence of cyclic extensions as follows: ! ··· ··· is a primitive pi root of unity so the ex- p2 ps p1 pi 1pi+1 ps tension F1 = F (! ··· )/F is a cyclic extension. We then simply define Fi = F (! ··· ··· ) and it is clear that F F F F = F (!) is the required sequence of cyclic exten- ✓ 1 ✓ 2 ✓···✓ s sions. Now, F (!)L is a Galois root extension, being the composite of two Galois root extensions, in which each intermediate extension is cyclic as F (!) contains the necessary roots of unity. The required extension is thus E0 = F (!)L.

Definition 6.10. Let G be a group.

66 1. A normal series is a sequence of normal subgroups G = H H H = 1 0 1 ··· n { } of G, where each factor group Hk/Hk 1 is cyclic. 2. G is said to be a solvable group if it has a normal series. We finally arrive at Galois Great Theorem! Theorem 6.11. A polynomial f(x) F [x] is solvable by radicals if and only if Gal(f) is a 2 solvable group.

Proof. Suppose f(x) is solvable by radicals. If ↵1,↵2,...,↵n are the roots of f(x), then each

↵i is expressible by radicals and so is contained in a Galois root extension Ei0 by the previous theorem. The composite field C := E0 E0 E0 is also a Galois root extension of the same 1 2 ··· n type. If C has associated sequence of subfields F = C C C C = C, 0 ✓ 1 ✓ 2 ✓···✓ t then

1 Gal(C/Ct 1) Gal(C/Ct 2) Gal(C/C1) Gal(C/F) { }  ···  is a normal series for Gal(C/F). Indeed, each C/C is Galois and Gal(C/Ci 1) = i Gal(C/Ci) ⇠ Gal(Ci/Ci 1) by the fundamental theorem, and each Gal(Ci/Ci 1) is cyclic by. assumption. Therefore, Gal(C/F) is a solvable group. Lastly, C contains a splitting field of f(x)overF , Gal(C/F) say S, so that Gal(f)=Gal(S/F ) ⇠= Gal(C/S), and every quotient group of a solvable group is solvable. Thus, Gal(f) is a solvable. group. Suppose now that G = Gal(f)=Gal(E/F) is a solvable group with normal series G = H H H H = 1 . 0 1 2 ··· l { } Setting Ei = fix(Hi) gives the sequence of subfields F = E E E E = E 0 ✓ 1 ✓ 2 ✓···✓ l where each Ei/Ei 1 is cyclic by the fundamental theorem, and of degree say ni for i =1, 2,...,l. As in the previous theorem, let m be the least commnon multiple of the ni and let ! be a primitive mth root of unity. Then, F (!)/F is a root extension of F with a decomposition into a sequence of cyclic pure extensions. Now, forming the composite fields F (!)Ei gives a sequence F F (!)=F (!)E F (!)E F (!)E = F (!)E ✓ 0 ✓ 1 ✓···✓ l in which each F (!)Ei/F (!)Ei 1 is a cyclic extension. But the necessary roots of unity are now present so that each F (!)Ei/F (!)Ei 1 is in fact a pure extension. Thus, each root of f(x) is contained in the roots extension F (!)E and f(x) is solvable by radicals.

67 Chapter 7

Solvability of the Symmetric Groups

1 1 Definition 7.1. The commutator of elements a and b of a group G is the element aba b .

The commutator subgroup of G, denoted G0, is the subgroup generated by the collection 1 1 of all commutators in G. That is, G0 =(aba b a, b G). | 2 Since G0 forms a group in its own right, we can consider its commutator subgroup, (G0)0. Similarly, we define the higher commutator subgroups of G inductively via G(0) = G and G(k) := (k 1) (G )0 for each k N. 2 Lemma 7.2. The commutator subgroup of a group G is a normal subgroup of G. Futhermore, if N is any normal subgroup of G then N 0 is also a normal subgroup of G.

Proof. For any c G0 and g G we have 2 2 1 1 1 1 1 gcg = gcg (c c)=(gcg c ) c G0, so G0 is normal in G. 2 G 2 0 1 1 Now suppose N is normal in G and let| n{z= aba} b N 0 for some a, b N. Then, for any 2 2 g G we have 2 1 1 1 1 1 1 1 1 1 1 gng = gaba b g =(gag )(gbg )(ga g )(gb g ) 1 1 1 1 1 1 =(gag )(gbg )(gag ) (gbg )

N 0 by the normality of N in G. 2

Thus, N 0 is normal in G.

(k) (k 1) Corollary 7.3. G is a normal of both G and G.

(k) (k 1) (k 1) Proof. By the first statement of the lemma, G =(G )0 is normal in G . Since G0 is (2) (1) normal in G, by the second statement we have that G =(G )0 is also normal in G so by (k) (k 1) induction G =(G )0 is normal in G for each k N. 2 68 Lemma 7.4. The quotient group G/G0 is abelian and if N is any normal subgroup of G such that the quotient group G/N is abelian, then G0 N.  Proof. For any a, b G we have 2 1 1 (aG0)(bG0) := abG0 =(baa b )abG0 1 1 = ba(a b ab)G0 1 1 = baG0 (as a b ab G0) 2 =: (bG0)(aG0)soG/G0 is abelian.

If N is normal in G with G/N abelian, then for any a, b G we have 2 1 1 1 1 (aN)(bN)=(bN)(aN) abN = baN = a b abN = N = a b ab N. () ) ) 2 Therefore, every commutator of G lies in N and so too must the subgroup generated by them, i.e G0 N.  We are now ready to give a characterisation of solvable groups in terms of commutator sub- groups.

Theorem 7.5. A group G is solvable if and only if G(n) = 1 for some n N. { } 2 Proof. Suppose G is solvable. Then, there is a finite sequence of subgroups

G = H0 H1 H2 Hn = 1 for some n N ◆ ◆ ◆······◆ { } 2 where each Hk is a normal subgroup of Hk 1 and each quotient group Hk 1/Hk is abelian. By the previous lemma,

H H0 = G0 1 ◆ 0 (2) H H H0 (G0)0 = G 2 ◆ 1 ◆ 1 ◆ (2) (3) H H H0 (G )0 = G 3 ◆ 2 ◆ 2 ◆ . . (n 1) (n) 1 = Hn Hn 1 Hn0 1 (G )0 = G . { } ◆ ◆ ◆ Thus, G(n) = 1 as claimed. { (}n) (k) Now suppose G = 1 for some n N and define H0 = G and Hk = G for k =1, 2,...,n. { } 2 Then, G = H H ...... H = 1 0 ◆ 1 ◆ ◆ n { } (k 1) (k 1) is a finite sequence of subgroups with each Hk normal in Hk 1 and each Hk 1/Hk = G /(G )0 abelian. Hence, G is a solvable group.

69 Theorem 7.6. The symmetric group S is not a solvable group for n 5. n

Proof. Since G = Sn is a normal subgroup of itself, it suces, by the previous theorem, to show that N 0 contains all 3-cycles whenever N contains all 3-cycles. Indeed, this would lead us to conclude G(n) = 1 for all n N. To this end, let N be any normal subgroup of G containing 6 { } 2 all 3-cycles. Since n 5, N contains the elements a = (1 2 3) and b =(345).Therefore, 1 1 (2 3 5) = (1 2 3)(3 4 5)(3 2 1)(5 4 3) = aba b N 0. i.e N 0 contains a 3-cycle. 2 1 Now, N 0 is normal in G so g(2 3 5)g N 0 for any g G.If(ijk) is any 3-cycle in G, then 2 2 picking g G such that g(i)=2,g(j) = 3 and g(k) = 5 gives 2 1 (ijk)=g(2 3 5)g N 0 so N 0 contains all 3-cycles. 2 This completes the proof.

Corollary 7.7. The general polynomial of degree n over a field F is not solvable by radicals!

Proof. The general polynomial of degree n has Galois group isomorphic to the full symmetric group S which is not solvable for n 5. n

This is to say, if F is a field and a0,a1,...,an are indeterminates then the particular polynomial

f(x)=a + a x + + a xn is not solvable by radicals 0 1 ··· n over the field F (a ,a ,...,a ) of rational functions of the a with coecients in F ,forn 5. 0 1 n i This level of abstractness is a little unsatisfying and it would be nice to exhibit a specific example of a polynomial over a more familiar field which is not solvable by radicals. With the aid of the lext couple of lemmas we will be able to do just this!

Lemma 7.8. Let p be a prime. The symmetric group Sp is generated by any transposition and any p-cycle.

Proof. Let ↵ =(x x x )beanyp-cycle and let =(y y ) be any transposition in S . 1 2 ··· p 1 2 p Any relabelling of the xi evidently defines an automorphism of Sp, so we may assume y1 =1 so that ↵ =(x x 1 x )=(1 x x x ) and =(1y ) 1 2 ··· ··· p ··· p 1 2 ··· 2 Since p is a prime, ↵k is a p-cycle for any k 12 ... p 1 so, in particular, ↵k =(1y ) 2{ } 2 ··· for some k 1, 2,...,p 1 . Therefore, we can assume that our p-cycle and transposition are 2{ } of the form ↵ =(1a ) and =(1a). Continuing in a similar vein, we may assume ··· ↵ =(12 (p 1) p) and =(12). ··· 70 Consider the elements

1 ↵↵ =(1p) 1 ↵(1 p)↵ =((p 1) p) 1 ↵((p 1) p)↵ =((p 1) (p 2)) . . 1 ↵(3 4)↵ =(23).

We see that ↵ and generate the set (1 2), (2 3),...,((p 2) (p 1)), ((p 1) p) , and it is { } a simple exercise to show that this set generates all transpositions in Sp and thus generates Sp in its entirety. This completes the proof.

Lemma 7.9. Let p(x) Q[x] be irreducible and of degree p, p a prime. Moreover, suppose p(x) 2 has precisely two non-real roots. Then, the Galois group of p(x) is the full symmetric group Sp.

Proof. Let E/Q be the splitting field of p(x). If ↵ is any root of p(x), then Q Q(↵) E ✓ ✓ and Q(↵)/Q is an extension of degree p. The degree formula yields [E : Q]=[E : Q(↵)][Q(↵): Q]=p[E : Q(↵)], so p divides [E : Q]= Gal(p) . Therefore, Gal(p) has a subgroup of order | | p by Cauchy’s and so contains an element of order p. Now, let 1 and 2 be the two non-real roots of p(x). Since the non-real roots of a polynomial in Q[x] appear in complex conjugate pairs, we have that 1 = 2 and 2 = 1.If3,4,...,p are the other roots of p(x), then

i = i for i =1, 2,...,p. That is, the map of complex conjugation defines an element of

Gal(p) which fixes 3,4,...,p and interchanges 1 and 2,so is the transposition (1 2) when identifying Gal(p) as a subgroup of Sp. Since the only elements of order p in Sp are the p-cycles, it follows that Gal(p) conatins a transposition and a p-cycle. By the previous lemma it follows that Gal(p) ⇠= Sp. We finish with an example of a polynomial over Q of degree 5 which is not solvable by radicals. Consider f(x)=2x5 10x+5 Q[x]. Since f( 2) = 39 < 0, f( 1) = 13 > 0, f(1) = 3 < 0 2 and f(2) = 49 > 0, it follows from the intermediate value theorem that f(x) has at least one real root in each of the intervals ( 2, 1), ( 1, 1) and (1, 2). If f(x) were to have at least 4 real 4 roots, then f 0(x)=10(x 1) would have at least three real roots which it does not. Therefore, f(x) has precisely 2 non-real roots and Gal(f(x)) ⇠= S5 by the lemma.

Thus, 2x5 10x + 5 is not solvable by radicals by Galois’ great theorem!

71 Bibliography

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