University of Cambridge

Mathematics Tripos

Part II Galois Theory

Michaelmas, 2017

Lectures by C. Brookes

Notes by Qiangru Kuang Contents

Contents

0 History 2

1 Field Extensions 3 1.1 Field Extensions ...... 3 1.2 Digression on (non-)constructability ...... 7

2 Separable, Normal and Galois Extensions 15 2.1 Separable Extension ...... 15 2.2 Trace & Norm ...... 19 2.3 Normal Extensions ...... 22

3 Fundamental Theorem of Galois Theory 25 3.1 Galois Group of Polynomials ...... 28 3.2 Galois Theory of Finite Fields ...... 31

4 Cyclotomic and Kummer Extensions, Cubics and Quartics, So- lution by Radicals 33 4.1 Cyclotomic Extensions ...... 33 4.2 Kummer Theory ...... 36 4.3 Cubics ...... 40 4.4 Quartics ...... 42 4.5 Solubility by Radicals ...... 44

5 Final Thoughts 50 5.1 Algebraic Closure ...... 50 5.2 Symmetric Polynomials & Invariant Theory ...... 53

1 0 History

0 History

The primary motivation of this course is to study polynomial equations in one variable and to consider whether there is a formula involving roots, i.e. solution by radicals. Quadratics have been well understood since long long time ago and we have already studied them at school. For cubics and quartics, it took long time before people discovered how to solve them by radicals. In 1770 Lagrange studied why it worked. However, In 1799 Ruffini claimed that there were some quintics that can’t be solved by radicals, i.e. there is no general formula, although his proof had gaps. In 1824, Abel (1802 – 1829) first accepted proof of insolubility using existing ideas about permutations of roots. In 1831, Galois (1811 – 1832) first explained why some polynomials are soluble by radicals and others are not. He made use of a group of permutations of the roots and he realised in particular the importance of normal subgroups. Galois’ work was not known in his lifetime — it was only published by Liouville in 1846 who realised that it fit in well with the work of Cauchy on permutations. Galois had submitted his work for various competitions and for entry into 1 the École Polytechnique. He died in a duel, leaving a 6 2 page letter indicating his thoughts about future development. Most of this course is Galois Theory but it is presented in a slightly more modern way in terms of field extensions. Recall from IB Groups, Rings and Modules that if f(t) is an irreducible polynomial in K[t] for some field K then K[t]/(f(t)) is a field where f(t) is the ideal generated by f(t). This is the starting point of this course. This course requires quite a lot of IB Groups, Rings and Modules but no content about module is required except in one place where it is useful to know the Structural Theorem of Finitely Generated Abelian Groups.

2 1 Field Extensions

1 Field Extensions

1.1 Field Extensions

Definition (field extension). A field extension K ≤ L is the inclusion of a field K into another field L, with the same 0 and 1 and the restriction of + and · in L to K gives + and · in K. √ √ Example. Q ≤ R, R ≤ C, Q ≤ Q( 2) = {λ + µ 2 : λ, µ ∈ Q} and Q(i) = {λ + µi : λ, µ ∈ Q} ≤ C are all field extensions. Suppose K ≤ L is a field extension. Then L is a K-vector space with addition given by the field and scalar multiplication given by the multiplication in the field L.

Definition (degree of extension). The degree of L over K is dimk L, the dimension of the K-vector space L. It is denoted by |L : K|. It may or may not be finite.

Definition (finite extension). If |L : K| < ∞ the extension is finite. Oth- erwise it is infinite.

Example.

1. |C : R| = 2 since {1, i} is a basis. 2. Similarly |Q(i): Q| = 2. 3. Q ≤ R is an infinite extension.

Theorem 1.1 (Tower Law). Suppose K ≤ L ≤ M are field extensions. Then |M : K| = |M : L||L : K|.

b Proof. Assume |M : L| < ∞, |L : K| < ∞. Then we take an L-basis {fi}i=1 a and a K-basis {ej}j=1. Pb Now take m ∈ M. m = i=1 µifi for some µi ∈ L. For each µi, we can P∞ write µi = j=1 λijej for some λij ∈ K. Thus

b a X X m = λijfiej i=1 j=1 so {fiej} span M. To show linear independence, it suffices to show that if m = 0 then each of the λij is zero. When m = 0, the linear independence of fi forces each µi to be zero. Then the linear independence of ei forces λij to be zero as required. The proof for infinite extensions is omitted. Observe (not very rigorously) that if M is an infinite extension of L then it is an infinite extension of K, and if L is an infinite extension of K then the larger field M must also be an infinite extension of K.

3 1 Field Extensions

Example. Consider √ √ Q ≤ Q( 2) ≤ Q( 2, i). √ √ √ √ Q(√2) has {1, 2} as√ a Q-basis.√ Q( 2, i) has {1, i} as a Q( 2)-basis. Now Q( 2, i) has basis {1, 2, i, i 2} over Q. Thus √ √ √ √ |Q( 2, i): Q| = 4 = 2 · 2 = |Q( 2, i): Q( 2)||Q( 2) : Q|. √ What√ are the intermediate√ fields between Q and Q(i, 2)? Obviously we have Q( 2), Q(i) and Q(i 2). Are there any more? To answer, this, the Galois correspondence arising in the Fundamental The- orem of Galois Theory gives an order-reversing bijection between the lattice of intermediate subfields √and the subgroups of a group of ring automorphisms of the extension field (Q(i, 2) here)√ that fix the smaller field element-wise.√ √ Ring automorphisms of Q(i, 2) that√ fixes√Q include the identity e : 2 7→ 2, i 7→ i and complex conjugation g√: 2 7→ 2, i 7→ −i. Notice that i and −i play the same role in the field Q(i, 2) — they are both roots of t2 + 1 = 0. There is another automorphism √ √ h : 2 7→ − 2 i 7→ i which switches the roots of t2 − 2 = 0. Notice that the composition √ √ gh : 2 7→ − 2 i 7→ −i √ These form a group of order 4, which equals to |Q(i, 2) : Q|.

√ Q(i, 2) {e} 2 2 2 2 2 2 √ √ Q( 2) Q(i) Q(i 2) {e, g} {e, h} {e, gh} 2 2 2 2 2 2 Q {e, g, h, gh}

The recipe for producing√ an intermediate subfield from a subgroup isto take the elements√ of Q(i, 2) which are fixed by all elements of the subgroup. For example, Q(i 2) is the field of elements fixed by both e and gh. This correspondence doesn’t always work for all finite field extensions: it only works for Galois extensions. In the correspondence, normal extensions correspond to normal subgroups. In this example all subgroups are normal and hence the extensions are normal. We will also prove Primitive Element Theorem, which in the context of finite extensions of Q tells√ us that they√ are necessarily of the form Q(α) for some α. For example, Q(i, 2) = Q(i + 2). We now do a review of IB Groups, Rings and Modules. Suppose K ≤ L is a field extension. Take α ∈ L. Define

Iα = {f ∈ K[t]: f(α) = 0}.

4 1 Field Extensions

Definition (algebraic, transcendental). α is algebraic over K if Iα 6= {0}. Otherwise α is transcendental over K.

Definition (algebraic extension). L is algebraic over K if for all α ∈ L, α is algebraic over K.

Remark. Iα is the kernel of the ring homomorphism K[t] → L f 7→ f(α) evaluation at α. It follows that Iα is an ideal of K[t]. Example. √ 1. 2 is algebraic over Q since it is a root of t2 − 2. 2. π is transendental over Q.

Lemma 1.2. Let K ≤ L be a finite field extension. Then L is algebraic over K. Proof. Let |L : K| = n. Take α ∈ L. Consider 1, α, α2, . . . , αn. These must be linearly dependent in the n-dimensional K-vector space L so

n X i λiα = 0 i=0

Pn i for some λi ∈ K not all zero. Then α is a root of f(t) = i=0 λit . Thus α is algebraic over K.

The non-zero ideal Iα (where α is algebraic over K) is principal since K[t] is a PID. Then

Definition (minimal polynomial). Iα = (fα(t)) and fα(t) can be assumed to be monic. Such a monic fα(t) is the minimal polynomial of α over K.

Remark. Multiplication by α within the field L gives a K-linear map L → L, an automorphism of L if α 6= 0. In IB Groups, Rings and Modules, we proved that the minimal polynomial of a linear map is unique. Example. √ 1. The minimal polynomial of 2 over Q is t2 − 2. √ √ 2. The minimal polynomial of 2 over R is t − 2.

Lemma 1.3. Suppose K ≤ L is a field extension, α ∈ L and α is algebraic over K. Then the minimal polynomial fα(t) of α over K is irreducible in K[t] and so Iα is a prime ideal.

5 1 Field Extensions

Proof. Suppose not and fα(t) = p(t)q(t). We must show that either p(t) or q(t) is a unit in K[t]. Note that 0 = f(α) = p(α)q(α) and so p(α) = 0 or q(α) = 0. Assume wlog p(α) = 0. Thus p(t) ∈ Iα and so p(t) = fα(t)r(t) since Iα = (fα(t)). Thus fα(t) = fα(t)r(t)q(t) and so r(t)q(t) = 1 in K[t]. Thus q(t) is a unit as required. Recall that irreducible elements in an integral domain are prime and generate prime ideals of K[t]. Thus Iα is a prime ideal.

Definition (generated field). Suppose K ≤ L is a field extension and α ∈ L. K(α) is defined to be the smallest subfield of L containing K and α. It is called the field generated by K and α.

Definition (simple extension). L is a simple extension if L = K(β) for some β ∈ L.

Given α1, . . . , αn ∈ L, K ≤ L, K(α1, . . . , αn) is the smallest intermediate field containing α1, . . . , αn. It is the field generated by K and α1, . . . , αn. On the other hand, K[α] = im(K[t] → L, f 7→ f(α)) is the ring generated by K and α.

Theorem 1.4. Suppose K ≤ L is a field extension and α ∈ L is algebraic. Then

1. K(α) = K[α] = im(f 7→ f(α)).

2. |K(α): K| = deg fα(f) where fα(t) is the minimal polynomial of α over K. Proof. 1. Clearly K[α] ≤ K(α). We need to show that if 0 6= β ∈ K[α] then it is a unit in K[α] and so K[α] is a field. Let β = g(α) for some g(t) ∈ K[t]. Since β = g(α) 6= 0, g(t) ∈/ Iα = (fα(t)). Therefore fα(t) - g(t). Since fα(t) is irreducible and K[t] is a PID, there exists r(t), s(t) ∈ K[t] with

r(t)fα(t) + s(t)g(t) = 1 ∈ K[t].

Thus s(α)g(α) = 1 ∈ K[α] and thus β = g(α) is a unit as required.

n−1 2. Let n = deg fα(t). We will show that {1, α, . . . , α } is a K-vector space basis of K[α]. • spanning: suppose

n n−1 fα(t) = t + an−1t + ··· + a0

with ai ∈ K. Then

n n−1 α = −an−1α − · · · − a0.

n i n−1 This implies that α is a linear combination of {α }0 . An easy induction shows that am for m ≥ n is likewise a linear combination i n−1 of {α }0 . Spanning thus follows.

6 1 Field Extensions

Pn−1 i • linear independence: suppose there is a relation i=0 λiα = 0. Let Pn−1 i g(t) = i=0 λit . Since g(α) = 0, we have g(t) ∈ Iα = (fα(t)). Thus g(t) = 0 or fα(t) | g(t). The latter is not possible by degree consideration. Thus g(t) = 0 ∈ K[α] and all the λi’s are zero.

Corollary 1.5. If K ≤ L is a field extension and α ∈ L then α is algebraic over K if and only if K ≤ K(α) is finite.

Proof. Straightforward by combining previous results.

Corollary 1.6. Let K ≤ L be a field extension with |L : K| = n and α ∈ L. Then deg fα(t) | n.

Proof. By Tower Law on K ≤ K(α) ≤ L, we deduce that |K(α): K| divides |L : K|. The result follows from the corollary above.

1.2 Digression on (non-)constructability In schedules it mentions “other classical problems” and we are in a position to tackle some of these using Corollary 1.6. A classical problem from Greek geometry concerns the existence or otherwise of constructions using ruler and compasses. Here “ruler” means single unmarked straightedge. If you’re an expert you can divide a line between two points into arbitrarily many equal segments, bisect an angle and produce parallel lines. In addition, given a polygon, you can produce a square of the same area or double the area. However, you cannot 1. duplicate the cube (i.e. given a cube you cannot produce a cube of double the volume); 2. trisect the angle π/3; 3. square a circle (i.e. given a circle you cannot construct a square of the same area).

2 Assume we are given a set P0 of points in R ,

• ruler operation draws a straight line through any points in P0,

• compass operation draws a circle with centre being a point in P0 and radius being distance between a pair of points in P0.

Definition (constructible number). The points of intersection of any two distinct lines or circles drawn using these operations are constructible in one step from P0. 2 A point r ∈ R is constructible from P0 if there is a finite sequence r1, r2, . . . , rn = r such that ri is constructible in one step from P0∪{r1, . . . , ri−1}.

Exercise. Construct the midpoint of a line between two points.

7 1 Field Extensions

Let K0 be the subfield of R generated by Q and the coordinates of points in P0. Let r0 = (xi, yi). Set Ki = Ki−1(xi, yi). Then

K0 ≤ K1 ≤ · · · ≤ Kn ≤ R.

Lemma 1.7. xi, yi are both roots in Ki of quadratic polynomials in Ki−1[t]. Proof. There are three cases: • line intersects line. • circle intersects circle, • circle intersects circle. For the line intersecting circle case, the line passes through A = (p, q) and B = (r, s) and the circle centres at C = (t, u) and has radius w. Then the equation of line is x − p y − q = r − p s − q and the equation of circle is (x − t)2 + (y − u)2 = w2. Solving gives coordinates of the intersection satisfying quadratic polynomials over Ki−1. The other cases are similar.

2 Theorem 1.8. If r = (x, y) is constuctible from a set P0 of points in R and if K0 is the subfield of R generated by Q and the coordinates of the points in P0 then the degrees |K0(x): K0| and |K0(y): K0| are powers of 2.

Proof. Use the previous notation Ki = Ki−1(xi, yi). By Tower Law,

|Ki : Ki−1| = |Ki−1(xi, yi): Ki−1(xi)||Ki−1(xi): Ki−1|.

But the previous lemma tells us that |Ki−1(xi): Ki−1| = 1 or 2, depending on whether the quadratic polynomial satisfied by the point of intersection is irreducible or not. Similarly yi satisfies a quadratic polynomial over Ki−1, and hence over Ki−1(xi) and so Ki−1(xi, yi): Ki−1(xi)| = 1 or 2. Therefore |Ki : Ki−1| = 1, 2 or 4 (but in fact 4 does not happen). Therefore by Tower Law, |Kn : K0| = |Kn : Kn−1| · · · |K1 : K0| is a power of 2. If r = (x, y) is contrucible from P0 then x, y ∈ Pn for some n and therefore

K0 ≤ K0(x) ≤ Kn

K0 ≤ K0(y) ≤ Kn and the Tower Law again gives that |K0(x): K0| and |K0(y): K0| are powers of 2. To use this for proofs about non-constructibility we need to be reasonably expert at working out minimal polynomials. Recall from IB Groups, Rings and Modules

8 1 Field Extensions

Theorem 1.9 (Gauss’ Lemma). Let f(t) be a primitive integral polynomial. Then f(t) is irreducible in Q[t] if and only if it is irreducible in Z[t].

n n−1 Theorem 1.10 (Eisenstein’s criterion). Let f(t) = ant + an−1t + ··· + a0 ∈ Z[t]. Suppose there is a prime p such that

• p - an,

• p | an−1, p | an−2, . . . , p | a0,

2 • p - a0,

then f(t) is irreducible in Z[t].

Example. Suppose p is prime. Then f(t) = tp−1 + tp−2 + ··· + 1 is irreducible over Q by considering f(t + 1) and using p as the prime in Eisen- stein. Another method to determine reducibility is to consider an integral polyno- mials f(t) (mod p). If f(t) is reducible in Z[t] then it is reducible over Z/pZ. So if we find a prime p such that f(t) (mod p) is irreducible then f(t) is irreducible in Z[t]. Example. t3 + t + 1 is irreducible mod 2: if it were reducible it would have a linear factor and so the polynomial would have a root mod 2, but 0, 1 are not roots. So t3 + t + 1 is irreducible in Z[t] and hence irreducible in Q[t]. Remark. On example sheet we will meet an irreducible polynomials in Z[t] which is reducible mod p for all primes p. Now back to construcibility.

Theorem 1.11. The cube cannot be duplicated by rulers and compasses.

Proof. The problem amounts to whether given a unit distance one can construct points distance α apart where α satisfies t3 − 2 = 0. Starting with points P0 = {(0, 0), (1, 0)}. Can we produce (α, 0)? No. If we could, |Q(α): Q| would be a power of 2. We show |Q(α): Q| = 3: since α satisfies t3 − 2, which by Eisenstein is irreducible over Z and hence over Q. Thus it is the minimal polynomials of α over Q.

Theorem 1.12. The circle cannot be squared using ruler and compasses. √ Proof. Starting with (0,√0) and (1, 0), can we construct ( π, 0) so that we have a square of side length π, and hence the area of the square equals to the are of the unit circle? √ The answer is no since π and hence π is transcendental over Q. Now return to further theory develpment.

9 1 Field Extensions

Lemma 1.13. Let K ≤ L be a field extension. Then

1. α1, . . . , αn ∈ L are algebraic over K if and only if K ≤ K(α1, . . . , αn) is a finite extension. 2. If K ≤ M ≤ L is such that K ≤ M is finite, then there exists α1, . . . , αn ∈ L such that K(α1, . . . , αn) = M.

Proof. 1. We know α is algebraic over K if and only if K ≤ K(a) is a finite extension. Suppose for each i, αi is algebraic over K, hence over K(α1, . . . , αi−1) and so |K(α1, . . . , αi): K(α1, . . . , αi−1)| < ∞. By Tower Law applied to

K ≤ K(α1) ≤ · · · ≤ K(α1, . . . αn),

we get |K(α1, . . . , αn): K| < ∞. Conversely, consider

K ≤ K(αi) ≤ K(α1, . . . , αn).

Then the Tower Law says that if |K(α1, . . . , αn): K| < ∞ then |K(αi): K| < ∞ and so αi is algebraic over K. 2. If |M : K| = n then M is an n-dimensional K-vector space so there exists a K-basis {α1, . . . , αn} of M. Then K(α1, . . . , αn) ≤ M. However, any elements of M is a K-linear combination of α1, . . . , αn and so lies in K(α1, . . . , αn) so equality.

Definition (K-homomorphism). Suppose K ≤ L, K ≤ L0 are field exten- sions. A K-homomorphism φ : L → L0 is a ring homomorphism such that φ|K = id.

Notation. Let

0 0 HomK (L, L ) = {K-homomorphisms L → L }.

A K-homomorphism φ : L → L0 is a K-isomorphism if it is a ring isomorphism.

0 Notation. AutK (L) = {K-isomorphism L → L } which has a group structure.

Lemma 1.14. Suppose K ≤ L, K ≤ L0 are field extensions. Then 1. any K-homomorphism φ : L → L0 is injective and K ≤ φ(L) is a field extension; 2. if |L : K| = |L0 : K| < ∞, then a K-homomorphism φ : L → L0 is a K-isomorphism.

Proof.

10 1 Field Extensions

1. L is a field and ker φ is an ideal of L. Note 1 7→ 1 and so ker φ 6= L, so ker φ = 0. Therefore φ(L) is a field and K ≤ φ(L) is a field extension. 2. φ is an injective K-linear map and so |φ(L): K| = |L : K|, which, by considering the dimensions of K-vector spaces, is smaller than |L0 : K| = |L : K| by assumption. Thus φ : L → L0 is a K-isomorphism. In particular if L = L0 then φ is a K-automorphism of L.

Notation. If K ≤ L is a field extension and if f(t) ∈ K[t], we denote the set of roots of f in L by Rootf (L).

Definition (splitting polynomial, splitting field). Let K ≤ L be a field extension and f(t) ∈ K[t]. We say f splits over L if

f(t) = a(t − α1)(t − α2) ··· (t − αn)

where a ∈ K, α1, . . . , αn ∈ L. We say L is a splitting field for f over K if L = K(α1, . . . , αn).

Remark. This is equivalent to saying that L is a splitting field for f over K if and only if 1. f splits over L, 2. if K ≤ M ≤ L and f splits over M then M = L. Example. √ √ √ √ 3 3 3 3 1. f(t) = t3−2 over Q: Q( 2) is not a splitting field over Q but Q( 2, ω 2, ω2 2) 2π i is where√ √ω is a√ primitive√ cubic root of unity, e.g. ω = e 3 . Note that 3 3 3 3 Q( 2, ω 2, ω2 2) = Q( 2, ω). √ √ 3 3 3 As 2 satisfies t − 2 over Q and hence over Q(ω), we have √|Q( 2, ω): 2 3 Q(ω√)| ≤ 3. ω √satisfies t + t + 1 over Q and hence over Q( 2) and so 3 3 3 2 |Q( 2,√ ω): Q( 2)| ≤ 2. Since t − 2 and t + t + 1 are irredcuible over 3 Q, |Q( 2) : Q| = 3 and√|Q(ω): Q| = 2 so by tower√ law we conclude√ that 3 3 3 equality holds, i.e. |Q( 2, ω): Q(ω)| = 3 and |Q( 2, ω): Q( 2)| = 2. √ 3 Q( 2, ω) ≤3 ≤2 √ 3 Q( 2) Q(ω) 3 2 Q

2. f(t) = (t2 − 3)(t3 − 1) over Q: the splitting field is √ √ √ √ 2 Q( 3, −3, ω, ω , 1) = Q( 3, ω) = Q( 3, i) √ −1+i 3 as ω = 2 .

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√ 3. t2 − 3 and t2 − 2t − 2 both have Q( 3) as the splitting field over Q. 2 4. f(t) = t + t + 1 over F2[t]: f(t) is irreducible over F2 since it has no 2 roots and hence no linear factors. Thus F2[t]/(t + t + 1) is a field. Set 2 2 2 α = t + (t + t + 1) ∈ F2[t]/(t + t + 1), Then F2[t]/(t + t + 1) = F2(α). 2 The elements are 0, 1, α and 1 + α. Note that α = α + 1 since F2 has characteristic 2. Now f(t) splits over F2(α) as f(t) = (t − α)(t − 1 − α)

so F2(α) is a splitting field of f over F2. We use the following construction to produce splitting field in general.

Theorem 1.15 (existence of splitting field). Let K be a field and f(t) ∈ K[t]. Then there exists a splitting field for f over K.

Proof. Induction of deg f. If deg f = 1 then K is the splitting field of f over K. Suppose deg f > 1 and pick an irreducible factor g(t) of f(t) over K. Note that K ≤ K[t]/(g(t)) is a field extension. Let α1 = t + (g(t)) ∈ K[t]/(g(t)) so K[t]/(g(t)) = K(α1) and g(α1) = 0 in K(α1). Therefore f(α1) = 0 in K(α1) and we can write f(t) = (t − α1)h(t) in K(α1)[t]. Repeat, note that deg h < deg f, and we get

f(t) = (t − α1)(t − α2) ... (t − αn)a where a is a constant which is in K (by considering the top coefficient of f). Thus we have a factorisation of f(t) in K(α1, . . . , αn)[t]. K(α1, . . . , αn) is a splitting field for f over K.

Theorem 1.16 (uniqueness of splitting field). If K is a field and f(t) ∈ K[t], then the splitting field for f over K is unique up to K-isomorphism, i.e. if there are two such splitting fields L and L0 then there is a K-isomorphism φ : L → L0.

Proof. Suppose L and L0 are splitting fields for f(t) ∈ K[t] over K. We need to show that there is a K-isomorphism L → L0. Suppose K ≤ M ≤ L and there exists K ≤ M 0 ≤ L0 and a K-isomorphism ψ : M → M 0. Clearly we can take M = K so such intermediate fields exist. Pick M so that |M : K| is maximal among such M,M 0 and ψ. Now we want to show that M = L and M 0 = L0. Note that if M = L then f(t) splits over M, say

f(t) = a(t − α1) ... (t − αm) ∈ M[t].

Apply ψ we get an induced map M[t] → M 0[t] which maps f(t) to

ψ(f)(t) = ψ(a)(t − ψ(α1) ... (t − ψ(αm)) so f(t) splits over ψ(M) = M 0. But L0 is a splitting field and L0 ≤ M 0 so we have equality.

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Now suppose M 6= L and we’ll arrive at a contradition of maximality of M. Since M 6= L, there is a root α of f(t) in L which isn’t in M. Factorise f(t) = g(t)h(t) ∈ M[t] so that g(t) ∈ M[t] is irreducible and g(α) = 0 in L. Then there exists a K-homomorphism

M[t]/(g(t)) → L t + (g(t)) 7→ α

The image of the map is M(α). The K-isomorphism M[t] → M 0[t] induced by ψ maps g(t) ∈ M[t] to some irreducible γ(t) ∈ M 0[t] so

f(t) = g(t)h(t) ∈ M[t] yields f(t) = γ(t)δ(t) ∈ M 0[t]. We have a field extension M 0 ≤ M 0[t]/(γ(t)) and there exists an M 0-homomorphism

M 0[t]/(γ(t)) → L0 t + (γ(t)) 7→ α0 by picking a root α0 of γ(t) in L0. However γ(t) | f(t) in M 0[t] and hence in L0[t] and so α0 is also a root of f(t) in L0. The M 0-homomorphism gives a K-isomorphism M 0[t]/(γ(t)) → M 0(α0) and we have a K-homomorphism M(α) → M 0(α0), contradicting the maximality of M.

Definition (normal extension). An algebraic field extension K ≤ L is nor- mal if for every α ∈ L, the minimal polynomial fα(t) of α over K splits over L.

Theorem 1.17. Let K ≤ L be a finite field extension, then K ≤ L is normal if and only if L is the splitting field for some f(t) ∈ K[t].

Proof. Later.

Example. Let F be a finite field with |F| = m. F has characteristic p for some r prime p and Fp ≤ F. Therefore m = p for some r. The nonzero elements of F form the multiplicative group or order n = m − 1 n n n Qn and they satisfy t − 1 so they are roots of t − 1. Thus t − 1 = i=1(t − αi) where the αi’s are the nonzero elements of F. Thus F is the splitting field for n t − 1 over Fp. By uniqueness of splitting field, any other field with m elements is Fp- isomorphic to F. (We will later show that there exists a field of m elements.)

Theorem 1.18. Let G be a finite subgroup of the multiplicative group ofa field K. Then G is cyclic. In particular the multiplicative group of a finite field is cyclic.

13 1 Field Extensions

Proof. Let |G| = n. By the structure theorem of finite abelian groups from IB Group, Rings and Modules, ∼ G = C m1 × C m2 × · · · × C mr q1 q2 qr with qi’s (not necessarily distinct) prime. 2 If q = qi = qj for some i 6= j then there are at least q distinct solutions of q t − 1 in K (since Cq × Cq is isomorphic to a subgroup of G). However, in a field, which is an integral domain, a polynomial of degree q has at most q roots, contradiction. Thus all the qi’s are distinct and G is cyclic and is generated by g1 . . . gr where each gi generates C mi . qi

14 2 Separable, Normal and Galois Extensions

2 Separable, Normal and Galois Extensions

2.1 Separable Extension

Definition (separable polynomial). Let K be a field and f(t) ∈ K[t]. Sup- pose f(t) is irreducible in K[t] and L is a splitting field of f(t) over K. Then f(t) is separable over K if f(t) has no repeated roots in L. For general f(t), we say f(t) is separable over K if every irreducible factor in K[t] is separable over K. All constant polynomials are deemed to be separable.

Definition (formal differentiation). If K is a field, the formal differentiation is the K-linear map

D : K[t] → K[t] tn 7→ ntn−1

Notation. Denote D(f(t)) by f 0(t).

Lemma 2.1. Let K be a field, f(t), g(t) ∈ K[t]. Then

(f(t)g(t))0 = f 0(t)g(t) + f(t)g0(t)

and if f(t) 6= 0, f(t) has a repeated root in a splitting field if and only if f(t) and f 0(t) have a common irreducible factor in K[t].

Proof. D is a K-linear map so we only need to check for f(t) = tn. It is left as an exercise. Let α be a repeated root in a splitting field L. Then f(t) = (t−α)2g(t) ∈ L[t] so f 0(t) = (t − α)2g0(t) + 2(t − α)g(t) and f 0(α) = 0. Therefore the minimal 0 polynomial fα(t) of α in K[t] divides both f(t) and f (t), so it is a common irreducible factor of f(t) and f 0(t). Conversely, let h(t) be a common irreducible factor of f(t) and f 0(t) in K[t]. Pick a root α ∈ L of h(t). Then f(α) = f 0(α) = 0. Then f(t) = (t−α)g(t) ∈ L[t] and f 0(t) = (t − α)g0(t) + g(t). Since f 0(α) = 0, we have (t − α) | f 0(t) so (t − α) | g(t). Hence (t − α)2 | f(t).

Corollary 2.2. If K is a field and f(t) ∈ K[t] is irreducible, then 1. if char K = 0 then f(t) is separable over K,

2. if char K = p > 0 then f(t) is not separable if and only if f(t) ∈ K[tp].

Proof. By Lemma 2.1, f(t) is not separable if and only if f(t) and f 0(t) have a common irreducible factor. But since f(t) is irreducible, the only possible factor is f(t) itself, i.e. f(t) | f 0(t). f 0(t) = 0 as it has a smaller degree. But if Pn i 0 Pn i−1 0 f(t) = i=0 ait then f (t) = i=1 iait so f (t) = 0 if and only if iai = 0 for all i ≥ 1. Thus

15 2 Separable, Normal and Galois Extensions

1. if char K = 0, f 0(t) 6= 0 for non-constant f(t) so f(t) is separable over K,

0 2. if char K = p > 0, then if f (t) = 0 we must have iai = 0 for all i ≥ 1, i.e. f(t) is not separable if and only if f(t) ∈ K[tp].

Definition (separable element). If K ≤ L is a field extension, we say α ∈ L is separable over K if its minimal polynomial is separable over K.

Definition (purely inseparable). If the minimal polynomial of α is fα(t) = (t − α)n = tn − αn where n is a power of char K, α is said to be purely inseparable over K.

Definition (separable extension). Given K ≤ L, L is separable over K if all elements of L are separable over K.

Example.

1. Let Q ≤ L be an algebraic extension. Then L is separable over Q.

2. Let L = Fp(X), the field of rational functions in X over Fp. It has p K = Fp(X ) as a subfield. Then K ≤ L is not separable:

Proof. Observe that if f(t) = tp − Xp ∈ K[t] then f 0(t) = 0. But tp − Xp = (t − X)p ∈ L[t]. However, f(t) ∈ K[t] is irreducible: suppose f(t) = g(t)h(t) ∈ K[t] ⊆ L[t], we get g(t) = (t − X)r for some 0 < r < p if the factorisation is non-trivial. But this would mean Xr ∈ K. However r and p are coprime so there exist a, b ∈ Z such that ar + bp = 1, so X = (Xr)a(Xp)b ∈ K. Thus we would have X = u(Xp)/v(Xp), absurd. Thus f(t) = tp − Xp is the minimal polynomial of X over K. α is purely inseparable over K and it follows that K ≤ L is not separable.

3. Let F be a finite field with |F| = m, a power of char F, and f(t) = tn − 1 where n = m − 1. We know this is separable over Fp since we saw that f(t) has distinct linear factors in F[t]. Remark. It is useful to have an alternative approach to separability of field extensions without having to check separability of minimal polynomials for all elements of the larger field. This is where we start thinking about K- homomorphisms.

Lemma 2.3. Let M = K(α) where α is algebraic over K. Let fα(t) be the minimal polynomial of α over K. For any field extension K ≤ L, the number of K-homomorphisms M → L is equal to the number of distinct roots of fα(t) in L. Thus

| HomK (M,L)| ≤ deg fα(t) = |K(α): K| = |M : K|.

16 2 Separable, Normal and Galois Extensions

Proof. We know that any K-homomorphism M → L is injective. Since K(α) =∼ K[t]/(fα(t)), for any root β of fα(t) in L, we can define a K-homomorphism

K[t]/(fα(t)) → L

t + (fα(t)) 7→ β

Conversely, for any K-homomorphism φ : M → L, the image φ(α) must satisfy fα(φ(α)) = 0. These two maps are inverses to each other. Thus there is a one-to-one correspondence

{K-homomorphism M → L} ↔ {root of fα(t) in L}.

√ √ 3 3 Example. Let K = Q,L = Q( 2). Then α = 2 has minimal polynomial√ 3 3 fα(t) = t − 2 over Q. There is only one K-homomorphism M = K( 2) → L, i.e. the identity map.

Corollary 2.4. In the previous lemma, the number of K-homomorphisms K(α) → L equals to deg fα(t) if and only if L is large enough (so that it contains a splitting field for fα(t)) and α is separable over K.

Proof. Trivial from Lemma 2.3.

Lemma 2.5. Let K ≤ M be a field extension and M1 = M(α), where α is algebraic over M. Let f(t) be the minimal polynomial of α over M and let K ≤ L. Let φ : M → L be a K-homomorphism. Then there is a one-to-one correspondence

{extension φ1 : M1 → L of φ} ↔ {root of φ(f(t)) in L}.

Remark. Lemma 2.3 is a special case where M = K and φ = idK . Proof. f(t) ∈ M[t] is irreducible so φ(f(t)) ∈ φ(M)[t] is irreducible. Any exten- sion φ1 : M1 → L of φ produces a root φ1(α) of φ(f(t)). Conversely, given a root γ of φ(f(t)) in L, ∼ ∼ ∼ M1 = M(α) = M[t]/(f(t)) = φ(M)[t]/(φ(f(t))) = φ(M)(γ) ≤ L so we get an extension φ1 of φ as required.

Corollary 2.6. Given L contains a splitting field of f(t), the number of φ1 extending φ equals to the number of distinct roots of f(t) in L. This equals to |M1 : M| if and only if α is separable over M.

Corollary 2.7. Let K ≤ M ≤ N be finite field extensions and K ≤ L. Let φ : M → L be a K-homomorphism. Then the number of extensions of φ to maps θ : N → L is less than or equal to |N : M|. Moreover, such θ exists if L is large enough.

17 2 Separable, Normal and Galois Extensions

Proof. Pick α1, . . . αn so that N = M(α1, . . . , αr) and set Mi = M(α1, . . . , αi). Thus we have got M ≤ M1 ≤ · · · ≤ Mr = N.

Use Lemma 2.5 repeatedly, there are at most |M1 : M| extensions φ1 : M1 → L of φ and |Mi+1 : M| extensions φi+1 : Mi+1 → L of φi for 1 ≤ i < r so by Tower Law the number of extensions θ : N → L of φ is less than or equal to

|Mr : Mr−1||Mr−1 : Mr−2| · · · |M1 : M| = |N : M|.

The last bit come from the proof of Lemma 2.5 since we need L to contain a root.

Remark. The proof shows that the number of extensions θ of φ is |N : M| if and only if L is large enough and αi is separable over M(α1, . . . , αi−1) for all i.

Theorem 2.8. Let K ≤ N be a field extension with |N : K| = n and N = K(α1, . . . , αi), say. Then TFAE: 1. K ≤ N is separable.

2. Each αi is separable over K(α1, . . . , αi−1). 3. If L is large enough then there are exactly n distinct K-homomorphisms N → L. Proof.

• 1 ⇒ 2: If K ≤ N is separable then each αi is separable over K. As K ≤ K(α1, . . . , αi−1), the minimal polynomial of αi over K(α1, . . . , αi−1) divides that over K. So if the latter has distinct roots in a splitting field then the former does as well. • 2 ⇒ 3: Corollary 2.7. • 3 ⇒ 1: Suppose 3 is true and 1 is false, we shall get a contradiction. Suppose there exists β ∈ N that is not separable over K. Then there are strictly less than |K(β): K| K-homomorphisms φ : K(β) → L. Each φ extends to at most |N : K(β)| extensions θ : N → L. Thus there are stricly less than |N : K(β)||K(β): K| = |N : K| = n K-homomorphisms N → L. Absurd.

Corollary 2.9. A finite extension is separable if and only if it is separably generated.

Proof. 1 ⇔ 2 above.

18 2 Separable, Normal and Galois Extensions

Lemma 2.10. If K ≤ M ≤ L are finite extensions then M ≤ L and K ≤ M are both separable if and only if K ≤ L is separable.

Proof. Example sheet.

Example. Let F be a finite field with |F| = m. Then the multiplicative group of order n = m − 1 is cyclic. Take a generator α, then F = F(α). Since αn = 1, the minimal polynomial of α divides tn − 1. Since tn − 1 has distinct roots (all the non-zero elements of F), the minimal polynomial of α is separable so F = F(α) is separable over F.

Theorem 2.11 (Primitive Element Theorem). Any finite separable exten- sion K ≤ M is simple, i.e. M = K(α) for some α, which is called the primitive element.

Proof. If K is a finite field then M is also finite. So we can take α to be generator of the multiplicative group of M, which is cyclic. Now assume K is an infinite field. Since K ≤ M is a finite extension, M = K(α1, . . . , αn) for some αi. It suffices to show that any field M = K(α, β) with β separable over K is of the form K(γ). Take f(t) and g(t) to be the minimal polynomials of α and β over K respec- tively. Let L be the splitting field for f(t)g(t) over K(α, β). The distinct zeros of f(t) in L are α1 = α, α2, . . . , αa and those of g(t) are β1 = β, β2, . . . , βb. By separability we know b = deg g(t). Choose λ ∈ K such that all αi + λβj are distinct (this is possible since K is infinite). Now we set γ = α + λβ (remember α is α1 and β is β1). Let F (t) = f(γ − λt) ∈ K(γ)[t]. We have g(β) = 0 and F (β) = f(α) = 0. Thus F (t) and g(t) have a common zero. Any other common zero would have to be βj for some j > 1. But then F (βj) = f(α + λ(β − βj)). By assumption α + γ(β − βj) is never an αi so f(βj) 6= 0. Now separability of g(t) says that the linear factors are all disinct. So t − β is a highest common factor of F (t) and g(t) in L[t]. However, the minimal polynomial h(t) of β over K(γ) divides F (t) and g(t) in K(γ)[t], and hence in L[t]. This imples h(t) = t − β, and so β ∈ K(γ). Therefore α = γ − λβ ∈ K(γ). So K(α, β) ≤ K(γ). The other direction is obvious. √ Exercise. In our√ example on page√4 we have Q ≤ Q( 2, i). We had interme- diate subfields Q( 2),√Q(i) and Q(i 2). If we follow the procedure of the proof 2 2 √of Theorem 2.11, √α = 2, β = i, f(t) = t − 2 and g(t) = t + 1. Consider some 2 + λi√where ± √2 ± λi are all distinct, for example λ = 1. The proof show that Q( 2, i) = Q( 2 + i).

2.2 Trace & Norm These will also be used in IID Number Fields.

Definition (trace, norm). Let K ≤ M be a finite field extension and α ∈ M. Multiplication by a is a K-linear map θα : M → M. The trace of α over K is TrM/K (α) = Tr θα ∈ K

19 2 Separable, Normal and Galois Extensions

and the norm of α over K is

NM/K (α) = det θα ∈ K. Note. Trace and norm depend on the ground field.

s s−1 Theorem 2.12. Suppose fα(t) = t + as−1t + ··· + a0 is the minimal polynomial of α over K. Let r = |M : K(α)|. Then the characteristic r polynomial of θα is (fα(t)) and

TrM/K (α) = −ras−1 s r NM/K (α) = ((−1) a0)

Proof. Regard M as a K(α)-vector space with basis β1 = 1, β2, . . . , βr. Now 2 s−1 i take the K-vecotr space basis 1, α, α , . . . , α of K(α). Then {α βj}i

A 0 ··· 0 0 A ··· 0    . . .   . .. .  0 0 ··· A

r whose characteristic polynomial is (fα(t)) . By inspecting the coefficients of the characteristic polynomial we get the trace and norm.

Theorem 2.13. Let K ≤ M be a finite separable field extension and |M : K| = n. Let α ∈ M and K ≤ L large enough so that there are n distinct K-homomorphisms σ1, . . . , σn : M → L. Then the characteristic polynomial of θα : M → M is n Y (t − σi(α)). i=1 Thus n X TrM/K (α) = σi(α) i=1 n Y NM/K (α) = σi(α) i=1

20 2 Separable, Normal and Galois Extensions

Proof. Let s Y s s−1 fα(t) = (t − αi) = t + as−1t + ··· + a0 i=1 be the minimal polynomial of α over K in L[t] where L is large enough that fα(t) splits in L. There are s K-homomorphisms K(α) → L, corresponding to maps sending α to αi. Each of these extends in r = |M : K(α)| ways to give K(α)-homomorphisms M → L. However, each of such extension mapping α 7→ αi still does so. So there are r maps sending α → αi for each i. Thus if the n = rs distinct K-homomorphism M → L are σ1, . . . , σn then n X σi(α) = r(α1 + ··· + αs) = −ras−1 = TrM/K (α) i=1 since the sum of roots of fα(t) is −as−1, and n Y s r σi(α) = ((−1) a0) = NM/K (α). i=1

r Recall from Theorem 2.12 that the characteristic polynomial of θα is (fα(t)) where fα(t) is the minimal polynomial of α over K and r = |M : K(α)|. Qs r The characteristic polynomial is n=1(t − αi) where αi are the roots of fα(t) in a splitting field. We also saw that those root are σ1(α), . . . , σn(α) so the characteristic polynomial is n Y (t − σi(α)). i=1

Theorem 2.14. Let K ≤ M be a finite separable extension. We define a K-bilinear form

T : M × M → K

(x, y) 7→ TrM/K (xy)

where xy is the product in M. Then this is non-degenerate. In particular, the K-linear map TrM/K : M → K is non-zero. Thus it is surjective.

Remark. If K ≤ M is a finite extension which is not separable then TrM/K is always zero. It follows that T is degnerate. See example sheet. Proof. By Primitive Element Theorem, separability implies that M = K(α) i n−1 for some α. We have a K-basis {α }i=0 of K(α) where n = |M : K|. The K-bilinear form T is represented by the matrix   TrM/K (1) TrM/K (α) ··· 2 A = TrM/K (α) TrM/K (α ) ··· ·········

Let L be a splitting field of the minimal polynomial f(t) of α over K. Then Qn fα(t) = i=1(t − αi) with α1, . . . , αn ∈ L. The entries in A are of the form ` ` ` TrM/K (α ) which is α1 + ··· + αn by Theorem 2.13.

21 2 Separable, Normal and Galois Extensions

Q Now consider ∆ = iVandermonde matrix  1 1 ··· 1   α1 α2 ··· αn   2 2 2   α1 α2 ··· αn  V =    . . .   . .. .  n−1 n−1 n−1 α1 α2 . . . αn

P i+j−2 i−1 Observe that Aij = n αn ,Vij = αj so

T (VV )ik = VijVkj i−1 k−1 = αj αj X i+k−2 = αn n

= Aik so VV T = A. Thus 0 6= D = ∆2 = |VV T | = |A|. Thus A is non-singular and therfore the bilinear form T is non-degenerate.

Remark. We will meet D again shortly. It is the discriminant of the polynomial fα(t).

2.3 Normal Extensions We met this definition before:

Definition (normal extension). An extension K ≤ M is normal if for every α ∈ M, the minimal polynomial fα(t) of α over K splits over M.

Theorem 2.15. Let K ≤ M be a finite field extension. Then K ≤ M is normal if and only if M is the splitting field for some f(t) ∈ K[t].

Proof.

• ⇒: suppose K ≤ M is normal. Pick α1, . . . , αr ∈ M such that M =

K(α1, . . . , αk). Let fαi (t) be the minimal polynomial of αi over K. Let Qk f(t) = n=1 fαi (t).

By normality, each fαi (t) splits over M and so is f(t). M is the splitting field of f(t) over K since if β1, . . . , βm are the roots of f(t) then M = K(β1, . . . , βm).

• ⇐: suppose M is a splitting field for f(t) over K. Then M = K(β1, . . . , βm) where βj are the roots of f(t) over M. Take α ∈ M. Let fα(t) be the minimal polynomial of α over K. Let M ≤ L be large enough so that fα(t) splits over L.

Now consider a K-homomorphism φ : M → L: φ(βj) is also a root of f(t) and is therefore one of the βj. Injectivity of K-homomorphisms implies

22 2 Separable, Normal and Galois Extensions

that φ permutes βj. However, M = K(β1, . . . , βm) and so φ is determined by the images of βj, thus φ(M) = M.

However, if αi is a root of fα(t) in L, then there is a K-homomorphism

K(α) → K(αi)

α 7→ αi

This extends by 2.7 to a K-homomorphism φ : M → L. But φ(M) = M so αi ∈ M. Thus M is normal over K.

Remark. As the same for separability, being normal is equivalent to being normally generated. We will show in example sheet that K ≤ L is a normal and finite extension if and only if L = K(α1, . . . , αr) with the minimal polynomial of each adjoined element splitting over L.

Definition (K-automorphism group). Let K ≤ M be a finite extension. Its K-automorphism group is

AutK (M) = HomK (M,M).

From before we know that such K-automorphisms are isomorphisms and thus have inverses (well, it name says so).

Lemma 2.16. | AutK (M)| ≤ |M : K|.

Proof. Corollary 2.7.

Theorem 2.17. Let K ≤ M be a finite field extension, then

| AutK (M)| = |M : K|

if and only if the extension is both normal and separable.

Definition (Galois extension). A finite field extension that is normal and separable is a Galois extension.

Definition (Galois group). Let K ≤ M be a Galois extension. Then the K-automorphism group of M is the Galois group of M over K, denoted by

Gal(M/K).

Remark. Some authors use the term “Galois group” as a synonym for auto- morphism group even when the extension is not Galois.

23 2 Separable, Normal and Galois Extensions

Proof of Theorem 2.17. Suppose | AutK (M)| = |M : K| = n. Let M ≤ L be large enough. The n distinct K-automorphisms φ : M → M extend to n K- homomorphisms φ : M → L and Theorem 2.8 says that M is separable over K. For normality, pick α ∈ M with minimal polynomial fα(t) over K. Take M = K(α1, . . . , αm) as in the proof of Corollary 2.7 with α = α1 and L = M. We only get |M : K| extensions of the inclusion K,→ M if each inequality in the proof is an equality. In particular, we need the number of K-homomorphisms K(α1) → M to be |K(α1): K|. But then root correspondence says we have |K(α): K| distinct roots of fα(t) in M. Thus fα(t) splits over M. Conversely, suppose K ≤ M is separable and normal. Then for K ≤ M ≤ L with L large enough, separability implies there are |M : K| K-homomorphisms φ : M → L by Theorem 2.8. However, K ≤ M is normal implies that it is the splitting field for some polynomial f(t) ∈ K[t] and thus M = K(α1, . . . , αn) Qn where f(t) = i=1(t−αi). Note that φ(αj) is also a root of φ(f(t)) = f(t), and is therefore one of the αj. Thus φ(M) = M. Thus | AutK (M)| = |M : K|.

Remark. In the previous proof we have shown that if K ≤ M ≤ L, φ ∈ HomK (M,L) and K ≤ M is normal then φ(M) = M.

Example. √ 1. Consider Q ≤ Q( 2, i), which is Galois: √ √ √ ∼ Gal(Q( 2, i)/Q) = hσ : 2 7→ − 2, τ : i 7→ −ii = C2 × C2.

All non-identity elements have order 2. √ 3 3 2. Let f(t) = t − 2. The splitting field of f(t) over √Q is Q( 2, ω) where 3 ω is a primitive√ cubic root√ of unity. Thus Q ≤ Q( 2, ω) is Galois with 3 3 | Gal(Q( 2, ω)/Q|) = |Q( 2, ω): Q| = 6. The Galois group contains √ √ σ : 3 2 7→ ω 3 2, ω 7→ ω √ √ τ : 3 2 7→ 3 2, ω 7→ ω2 (complex conjugation)

and is generated√ by these two elements. It is an exercise to show that 3 ∼ ∼ Gal(Q( 2, ω)/Q) = D6 = S3.

24 3 Fundamental Theorem of Galois Theory

3 Fundamental Theorem of Galois Theory

Definition (fixed field). Let K ≤ L be a field extension and H ≤ AutK (L). The fixed field of H is

LH = {α ∈ L : σ(α) = α for all σ ∈ H}.

The fixed field is a field and K ≤ LH ≤ L.

Theorem 3.1 (Fundamental Theorem of Galois Theory). Let K ≤ L be a finite Galois extension. Then 1. There is a one-to-one correspondence

 intermediate subfield   subgroup  ←→ K ≤ M ≤ L H ≤ Gal(L/K)

M 7→ AutM (L) LH ← H [ 2. H is a normal subgroup of Gal(L/K) if and only if K ≤ LH is normal if and only if K ≤ LH is Galois.

3. If H E Gal(L/K) then the map θ : Gal(L/K) → Gal(LH /K)

given by restriction to LH is a surjective group homomorphism with kernel H.

Remark. 1. Observe that M ≤ L is Galois and so we could have written Gal(L/M) instead of AutM (L). To see this, • separability: follows from Lemma 2.10. • normality: if α ∈ L then the minimal polynomial of α over M divides the minimal polynomial of α over K. But the latter splits over L (see example sheet). 2. If K ≤ M is normal then the remark after proof of Theorem 2.17 says if σ : L → L then σ(M) = M and so we can talk about the restriction of α to M giving an automorphism of M.

Example. √ 1. Q ≤ Q( 2, i). We saw in example√ on page 4 the lattices of intermediate ∼ subfields and subgroups Gal(Q( 2, i)/Q) = C2 × C2 which is abelian. Thus all subgroups are normal and all intermediate subfields are normal extensions of Q.

25 3 Fundamental Theorem of Galois Theory

√ 3 2. Q ≤ Q( 2, ω). √ 3 Q( 2, ω) 2 2 2 3 √ √ √ 3 3 3 Q(ω) Q( 2) Q(ω 2) Q(ω2 2)

2 3 3 3 Q

1 2 2 2 3 hσi hτi hσ2τi hστi 2 3 3 3 ∼ hσ, τi = D6

The subgroup H√of order 3 is normal but those of order 2 are not so 3 the map Gal(Q( 2, ω)/Q) → Gal(Q(ω)/Q), i.e. D6 → C2, generated by conjugation has kernel hσi.

Theorem 3.2 (Artin). Let K ≤ L be a field extension and H ≤ AutK (L) be a finite subgroup. Let M = LH . Then M ≤ L is a finite Galois extension and H = Gal(L/M).

Remark. This implies one way of Galois correspondence:

H 7→ LH 7→ Gal(L/LH ) = H.

Proof. Take α ∈ L. We first show that |M(α): M| ≤ |H|. Let {α1, . . . , αn} be the distinct images of α under H, i.e. {φ(α): φ ∈ H}. Define g(t) = Qn i=1(t − αi). Each φ induces an endomorphism on L[t] under which g(t) is H invariant since φ permutes the αi. Thus the coefficients lie in L = M and so g(t) ∈ M[t]. By definition g(α) = 0 since α is one of the αi. Hence the minimal polynomial fα(t) of α over M divides g(t). Thus

|M(α): M| = deg fα(t) ≤ deg g(t) ≤ |H|.

This step shows that α is algebraic over M and fα(t) is separable since g(t) is. Thus M ≤ L is a separable extension. Next we show that M ≤ L is a simple extension. Pick α ∈ L with |M(α): M| maximal. We will show that L = M(α) for this α. Suppose β ∈ L. Then M ≤ M(α, β) is finite and separably generated and hence is a finite separable extension. By Primitive Element Theorem M(α, β) = M(γ) for some γ. But then M ≤ M(α) ≤ M(γ) so by the maximality of |M(α): M|, M(α) = M(γ). Thus β ∈ M(γ) = M(α) so L = M(α). It follows that |L : M| ≤ |H|.

26 3 Fundamental Theorem of Galois Theory

Finally, H ≤ AutM (L) so

|L : M| = |M(α): M| ≤ |H| ≤ | AutM (L)| ≤ |L : M| so we must have equality throughtout, i.e. |L : M| = | AutM (L)| = |H| so M ≤ L is a finite Galois extension and H = Gal(L/M).

Theorem 3.3. Let K ≤ L be a finite field extension. TFAE: 1. K ≤ L is Galois,

H 2. L = K where H = AutK (L).

Remark. The theorem allows some authors to give yet another definition of a Galois extension. Proof. H 1. 1 ⇒ 2: Let M = L where H = AutK (L). By Artin M ≤ L is a Galois extension. Now we have two Galois extensions, giving the equalities |H| = | Gal(L/K)| = |L : K| = | Gal(L/M)| = |L : M|

As K ≤ LH = M, equality. 2. 2 ⇒ 1: Artin.

Proof of Fundamental Theorem of Galois Theory. 1. Composing the maps H 7→ LH M 7→ Gal(L/M)

gives H → H by Artin. Composition the other way M 7→ Gal(L/M) 7→ LH where H = Gal(L/M) gives M by Theorem 3.3.

−1 2. Take H ≤ Gal(L/K). Then for φ ∈ Gal(L/K), LφHφ = φ(LH ) so by 1 H H H H E Gal(L/K) if and only if φ(L ) = L . Let M = L . We will show that K ≤ M is normal if and only if φ(M) = M for every φ ∈ Gal(L/K): • ⇒: remark 2 after Theorem 3.1. • ⇐: suppose φ(M) = M for all φ ∈ Gal(L/K). Pick α ∈ M and let fα(t) be its minimal polynomial over K. We take β as a root for fα(t) in L (which is possible by normality). Then there is a K-homomorphism K(α) → K(β) α 7→ β

This extends to a K-homomorphism φ : L → L. Since we assume φ(M) = M, φ(α) = β ∈ M. Thus K ≤ M is normal.

27 3 Fundamental Theorem of Galois Theory

Note that K ≤ M is separable since K ≤ M ≤ L and K ≤ L is separable. Thus K ≤ M is Galois. 3. By remark 2 after Theorem 3.1, the restriction map θ : Gal(L/K) → Gal(LH /K) is well-defined. Surjectivity follows from being able to extend a K-homomorphism LH → LH ≤ L to a K-homomorphism L → L. Clearly H ≤ ker θ. However |L : K| | Gal(L/K)| = | ker θ| | ker θ| = | Gal(LH /K)| by surjectivity of θ = |LH : K| since K ≤ LH is Galois |L : K| = by Tower Law |L : LH | Then | ker θ| = |L : LH | = | Gal(L/LH )| = |H| by Artin. Thus ker θ = H.

3.1 Galois Group of Polynomials

Definition (Galois group). Let f(t) ∈ K[t] be a separable polynomial and K ≤ L with L a splitting field for f(t). Then the Galois group of f(t) over K is Gal(f) = Gal(L/K).

Since L is a splitting field for f(t), L = K(α1, . . . , αn) where α1, . . . , αn are the roots of f(t) in L. Observe that if φ ∈ Gal(L/K) it maps bijectively roots of f(t) to itself. Thus φ permutes the αi. Moreover if φ fixes each αi it also fixes all elements of L and so it is the identity map. Thus Gal(f) may be regarded as a permutation group of the n roots, so in particular admitting a permutation representation in Sn.

Lemma 3.4. Suppose separable f(t) = g1(t) ··· gs(t) with gi(t) irreducible in K[t] is a factorisation in K[t]. Then the orbits of Gal(f) acting on the roots of f(t) correspond to the factors gj(t): two roots are in the same orbit if and only if they are roots of the same gi(t). In particular if f(t) ∈ K[t] is irreducible, there is only one orbit, i.e. Gal(f) acts transitively on the roots of f(t).

Proof. Let αk and α` be in the same orbit under Gal(f). Thus there is φ ∈ Gal(f) with α` = φ(αk). But if αk is a root of gj(t) then φ(αk) is also a root of gj(t). Conversely if αk and α` are roots of gj(t), then ∼ ∼ K(αk) = K[t]/(gj(t)) = K(α`) ≤ L

Denote by φ0 the isomorphsim K(αk) → K(α`). φ0 extends to φ ∈ Gal(L/K). Thus αk and α` are in the same orbit.

28 3 Fundamental Theorem of Galois Theory

Lemma 3.5. The transitive subgroups of Sn for n ≤ 5 are

n ∼ 2 S2 = C2 ∼ 3 A3 = C3,S3 4 C4,V4,D8,A4,S4 5 C5,D10,H20,A5,S5

where H20 is generated by a 5-cycle and a 4-cycle.

Theorem 3.6. Let p be a prime and f(t) ∈ Q[t] irreducible of degree p. Suppose f(t) has exactly 2 non-real roots in C. Then ∼ Gal(f) = Sp.

Proof. Gal(f) acts on the p distinct roots of f(t) in a splitting field L of f(t) (in C). By Lemma 3.4 the irreducibility of f(t) implies that Gal(f) acts transitively on p roots so by orbit-stabiliser p | | Gal(f)| but

| Gal(f)| ≤ |Sp| = p! and so Gal(f) has a Sylow p-subgroup of order p, necessarily cyclic. Thus Gal(f) contains a p-cycle. Supposition that there are exactly 2 non-real roots gives that complex conjugation yields a transposition in Gal(f). The p-cycle and the transposition generate Sp. 5 ∼ Example. Let f(t) = t − 6t + 3 ∈ Q[t], then claim Gal(f) = S5: Proof. f(t) is irreducible by Eisenstein with p = 3. We want to show that f(t) has 3 real roots (so 2 non-real roots) and apply the previous theorem. Now we apply knowledge from analysis: f(−2) = −17, f(−1) = 8, f(1) = −2, f(2) = 23 and f 0(t) = 5t4 − 6 which has two real roots. Thus by Intermediate Value Theorem there are 3 real roots while by Rolle’s Theorem there are at most 3 real roots.

Definition (discriminant). Let f(t) ∈ K[t] have distinct roots α1, . . . , αn Q (in a splitting field) (f(t) need not be irreducible). Set ∆ = i

2 Y 2 n(n−1) Y D(f) = ∆ = (αi − αj) = (−1) 2 (αi − αj). i

Remark. We have already met this in the proof of Theorem 2.14.

Lemma 3.7. Let f(t) ∈ K[t] be separable of degree n with char K 6= 2. Then Gal(f) ≤ An ⇔ D(f) is a square in K.

29 3 Fundamental Theorem of Galois Theory

Proof. Let L be a splitting field of f(t) over K. Then D(f) 6= 0 and is fixed by all elements of G = Gal(L/K) as the latter permutes the roots. Thus D ∈ K since LG = K by Galois correspondence. If σ ∈ G then σ(∆) = (sgn σ)∆ where we are regarding G as a subgroup of Sn and sgn is the signature (this is where we need char K 6= 2). Thus if G ≤ An we got that ∆ is fixed by all σ ∈ G. Thus ∆ ∈ K = LG. On the other hand if G  An we get σ(∆) = −∆ if σ is odd and so ∆ ∈/ K = LG. Finally note that if D has square roots they must be ±∆.

Example.

2 1. n = 2: f(t) = t + bt + c = (t − α1)(t − α2) has discriminant

2 2 2 D(f) = (α1 − α2) − (α1 + α2) − 4α1α2 = b − 4c

2. n = 3: f(t) = t3 + ct + d has discriminant

D(t) = −4c3 − 27d2

Remark. Any general monic cubic g(t) can be put into this form by a suitable substitution f(t) = g(t1 + λ) for suitable λ. Note D(f) = D(g).

Example.

• f(t) = t3 − t − 1 ∈ Q[t] irreducible in Z[t] since irreducible mod 2. D(f) = ∼ −23 which is not a square in Q. Thus Gal(f) = S3. • f(t) = t3 − 3t − 1 ∈ Q[t] irredicible since irreducible mod 2. D(f) = 81 ∼ which is a square so Gal(f) = A3. Now we move on to irreducible quartics. We saw that the possible Galois groups are V4,A4,C4,D8,S4 with the first two being subgroups of A4. From looking at the discriminant one gets information as whether the group is one of V4,A4 or one of C4,D8,S4. We need further methods to pin down which group we are dealing with.

Theorem 3.8 (mod p reduction). Let f(t) ∈ Z[t] be monic of degree n with n distinct roots in a splitting field. Let p be a prime such that f(t), the reduction of f(t) mod p, also has n distinct roots in a splitting field. Let f(t) = g1(t) ... gs(t) be the factorisation into irreducible in Fp[t] with nj = deg gj(t). Then Gal(f) ,→ Gal(f)

and has an element of cycle type (n1, n2, . . . , ns).

Proof. See Remark after discussion about Galois groups over finite fields. The fact that Gal(f¯) ,→ Gal(f) is from IID Number Fields. Look at Tony Scholl’s teaching page on Galois Theory.

30 3 Fundamental Theorem of Galois Theory

Example. Given a quartic of the form f(t) = t4 + dt + e, its discriminant is D(t) = −27d4 + 256e3. Consider f(t) = t4 − t − 1, irreducible since irreducible mod 2 and D(f) = −283 which is not a square. Consider mod 7, f(t) = t4 − t − 1 = (t + 4)(t3 + 3t2 + 2t + 5) the second factor is irreducible over F7 since it has no roots in F7. By Theo- rem 3.8 Gal(f) contains an element of cycle type (1, 3), i.e. a 3-cycle. We deduce ∼ that Gal(f) = S4 as it is the only transitive subgroup of S4 that contains an odd permutation and a 3-cycle.

3.2 Galois Theory of Finite Fields Recall what we already know from ??: a finite field F is of characteristic p > 0 where p is a prime and |F| = pr for some r. The multiplicative group of F n r is cyclic. It is a splitting field for t − 1 over Fp where n = p − 1. By the uniqueness of splitting fields this is unique. Observe we could also describe F pr as the splitting field of t − t over Fp. What we haven’t shown yet is that for any pr there is a field with |F| = pr.

Definition (Frobenius automorphism). Let F be a finite field of character- istic p. The Frobenius automorphism of F is

φp : F → F α 7→ αp

Remark. (α + β)p = αp + βp since all terms in binomial expansion is divisible by p. Also Fp is fixed under this so this isan Fp-automorphism. r Since tp − t splits as a product of linear factors (t − α) in F, we have that Fp ≤ F is a Galois extension and so we consider G = Gal(F/Fp). It has order r since |F : Fp| = r.

Theorem 3.9 (Galois group of finite fields). Let F be a finite field with r |F| = p . Then Fp ≤ F is a Galois extension with ∼ Gal(F/Fp) = hφpi = Cr.

Proof. It remains to show that the order of Frobenius automorphism is r. Sup- s ps ps s pose φp = id. Then α = α for all α ∈ F. But t − t has at most p roots in r pr F. So we conclude s ≥ r. Observe that φp = id since α = α for all α ∈ F. Now apply Fundamental Theorem of Galois Theory,

{intermediate fields Fp ≤ M ≤ F} ↔ {subgroups H ≤ G} where G = Gal(F/Fp) is cyclic. But we know all about subgroups of a cyclic group with generator φp with order r. There is exactly one group of order s for each s | r generated by r/s hφr/si φp . The corresponding intermediate subfields are the fixed fields F p and hφr/si hφr/si |F : F p | = s. By Tower Law |F p : Fp| = r/s.

31 3 Fundamental Theorem of Galois Theory

Observe that all subgroups of cyclic groups are normal and therefore all r/s hφp i ∼ our intermediate fields are normal extensions of Fp. Thus Gal(F /Fp) = r/s Gal(F/Fp)/H where H = hφp i.

Corollary 3.10. Let Fp ≤ M ≤ F be finite fields. Then Gal(F/M) is cyclic, n n generated by φp where φp is the Frobenius map and |M| = p and M is the n fixed field of hφp i.

Proof. Set n = r/s.

Theorem 3.11 (existence of finite fields). Let p be a prime and n ≥ 1. Then there is a field of order pn unique up to isomorphism.

pn Proof. Consider the splitting field L of f(t) = t − t over Fp. Fp ≤ L is a finite n Galois extension. However the roots of f(t) form a field, the fixed field of φp . Set L = F and |F : Fp| = n. Remark (mod p reduction). We will discuss in IID Number Fields that Gal(f) ,→ Gal(f) if f(t) ∈ Z[t]. We factorised f(t) = g1(t) ··· gs(t) as a product of irre- ducibles (actually gs(t) lives in p-adic integers). We knew from Lemma 3.4 that the orbits of Gal(f) correspond to the factorisation. We know Gal(f) is cyclic generated by the Frobenius map, which must have cyclic type (n1, . . . , ns) where nj = deg gj(t).

32 4 Cyclotomic and Kummer Extensions, Cubics and Quartics, Solution by Radicals

4 Cyclotomic and Kummer Extensions, Cubics and Quartics, Solution by Radicals

4.1 Cyclotomic Extensions

Definition (cyclotomic extension). Suppose char K = 0 or p is prime where p - m. The mth cyclotomic extension of K is the splitting field L of tm − 1.

Remark. f(t) = tm − 1 and f 0(t) = mtm−1 have no common roots and so the roots of f(t) are distinct, which are the mth roots of unity. They form a finite × subgroup µm of L and hence by (1.28) ?? a cyclic group hζi. Thus L = K(ζ) is a simple extension.

Definition (primitive root of unity). An element ζ ∈ µm is a primitive mth root of unity if µm = hζi.

Choosing a primitive mth root of unity determines an isomorphism

µm → Z/mZ.

i Note that ζ is a generator of µm if and only if (i, m) = 1 and so the primitive roots of unity correspond to elements of (Z/mZ)×, the unit group of Z/mZ. Now consider Galois groups of cyclotomic extensions. We will see that they must be abelian. Observe f(t) = tm −1 is separable and so the extension K ≤ L is Galois. Let G = Gal(L/K). An element σ ∈ G sends a primitive mth root of unity ζ to another mth root of unity ζi where (i, m) = 1. Then ζ 7→ ζi determines a K-homomorphism

K(ζ) → K(ζ) which is an injective map.

Definition. Define

× θ : G → (Z/mZ) σ 7→ i where σ(ζ) = ζi

It is a group homomorphism since if σ(ζ) = ζi, φ(ζ) = ζj then (σφ)(ζ) = σ(ζj) = ζij.

Thus G is abelian and we regard G as a subgroup of (Z/mZ)×.

Definition (cyclotomic polynomial). The mth cyclotomic polynomial is

Y i Φm(t) = (t − ζ ), × i∈(Z/mZ)

the product of the linear factors of tm − 1 corresponding to the primitive mth roots of unity.

33 4 Cyclotomic and Kummer Extensions, Cubics and Quartics, Solution by Radicals

Remark. f(t) = tm − 1 Y = (t − ζi)

i∈Z/mZ Y = Φd(t) d|m

Example. Take K = Q, then

Φ1(t) = t − 1

Φ2(t) = t + 1 2 Φ3(t) = t + t + 1 2 Φ4(t) = t + 1 and from which we can deduce that

4 Φ8(t) = t + 1 since t8 − 1 = (t − 1)(t + 1)(t2 + 1)(t4 + 1).

Lemma 4.1.

• Φm(t) ∈ Z[t] if char K = 0 (i.e. Q ,→ K).

• Φm(t) ∈ Fp[t] if char K = p (i.e. Fp ,→ K).

Proof. By induction on m. If m = 1 then done. For m > 1, consider

m Y f(t) = t − 1 = Φm(t) Φd(t) d|m d6=m Q Note that d|m,d6=m Φd(t) is monic and is in Z[t] or Fp[t] by induction hypoth- esis. Now

• if char K = 0 we deduce Φm(t) ∈ Q[t] by division of polynomials. By Gauss’ Lemma it is in Z[t].

• if char K = p > 0 we deduce by division that Φm(t) ∈ Fp[t].

Lemma 4.2. The homomorphism θ : G → (Z/mZ)× defined above is an isomorphism if and only if Φm(t) is irreducible.

Proof. We know from Lemma 3.4 that the orbits of G = Gal(L/K) correspond to the factorisation of f(t) in K[t]. In particular the primitive mth roots of unity form one orbit if and only if Φm(t) is irreducible. Then θ is surjective if and only if Φm(t) is irreducible.

34 4 Cyclotomic and Kummer Extensions, Cubics and Quartics, Solution by Radicals

Theorem 4.3. Let L be the mth cyclotomic extension of finite fields F = Fq where q = pn. Then the Galois group G = Gal(L/F) is isomorphic to the cyclic subgroup of (Z/mZ)× generated by q.

n Proof. We know from Corollary 3.10 that G is generated by α 7→ αp so

pn n × θ(G) = θ(hα 7→ α i) = hp i = hqi ≤ (Z/mZ) .

Remark. If (Z/mZ)× is not cyclic then θ is not surjective for any finite field and Φm(t) is reducible over F.

Example. Let F = F3. Then 4 2 2 Φ8(t) = t + 1 = (t + t − 1)(t − t − 1) so t8 − 1 factorises as a product of linear and quadratic polynomials mod 3. So L = F9, the unique field of order 9, whose multiplicative group is isomorphic to C8. As | Gal(L/F3)| = 2, it is isomorphic to C2. But × ∼ (Z/8Z) = {1, 3, 5, 7} = C2 × C2 so θ is not surjective and Φ8(t) is reducible. So far we have worked exclusively with finite fields with non-zero character- istic. The following theorem is a result regarding Q:

Theorem 4.4. For all m > 0, Φm(t) is irreducible in Z[t] and hence in Q[t]. Thus θ is an isomorphism and

× Gal(Q(ζ)/Q) =∼ (Z/mZ)

where ζ is a primitive mth root of unity.

Remark. We already knew this when m = p for p prime by substitution and Eisenstein.

Proof. Gauss’ Lemma implies that irreducibility in Z[t] gives irreducibility in Q[t]. Lemma 4.2 says that irreducibility corresponds to surjectivity of θ. Thus it is left to show that Φm(t) is irreducible in Z[t]. Suppose not and Φm(t) = g(t)h(t) in Z[t], with g(t) irreducible and monic with deg g(t) < deg Φm(t). Let Q ≤ L be the mth cyclotomic extension and ζ be a primitive mth root and unity and also a root of g(t). Claim that if p - m and p is prime then ζp is also a root of g(t) in L:

Proof. Suppose not, then ζp is also a primitive mth root of unity (since p - m) p and a root of Φm(t). The supposition implies that ζ is a root of h(t). Define r(t) = h(tp), then r(ζ) = 0. But g(t) is the minimal polynomial of ζ over Q and so g(t) | r(t) in Q[t]. By Gauss’ Lemma r(t) = g(t)s(t) with s(t) ∈ Z[t]. Now reduce mod p, r(t) = g(t)s(t).

35 4 Cyclotomic and Kummer Extensions, Cubics and Quartics, Solution by Radicals

p p But r(t) − h(t ) = (h(t)) . If a(t) is any irreducible factor of g(t) in Fp[t] then p 2 a(t) | (h(t)) and so a(t) | h(t). But then (a(t)) | g(t)h(T ) = Φm(t). So Φm(t) has a repeated root and thus tm − 1 has repeated roots mod p. Absurd. So the claim is true. Now consider a root γ of h(t). It is also a primitive mth root of unity and γ = ζi for some i coprime with m. Factorise it as i = p1 ··· pk with pj prime and not necessarily distinct and pj - m. Apply the claim repeatedly, we get that γ is also a root of g(t) and so Φm(t) has a repeated root. Absurd. Thus Φm(t) is irreducible over Q.

Definition (cyclic extension). An extension K ≤ L is cyclic if the extension is Galois and Gal(L/K) is cyclic.

Definition (abelian extension). An extension K ≤ L is abelian if the ex- tension is Galois and Gal(L/K) is abelian.

Example.

1. We saw in Corollary 3.10 that for finite fields F ≤ L is cyclic. 2. Cyclotomic extensions are abelian.

3. Theorem 4.4 says Gal(Q(ζ)/Q) =∼ (Z/mZ)× and so if m = 8, Q ≤ Q(ζ) is a non-cyclic abelian extension.

4.2 Kummer Theory Rather than consider the mth root of unity, in this section we consider Galois extensions K ≤ L where L is the splitting field of a polynomial of the form tm − λ where λ ∈ K.

Theorem 4.5. Let f(t) = tm − λ ∈ K[t] and char K - m. Then the splitting field L of f(t) over K contains a primitive mth root of unity ζ and Gal(L/K(ζ)) is cyclic of order dividing m. Moreover f(t) is irreducible over K(ζ) if and only if |L : K(ζ)| = m.

Remark. We have the following Galois correspondence   L 1 cyclic | |  K(ζ) Gal(L/K(ζ)) | | K Gal(L/K) with Gal(L/K(ζ)) E Gal(L/K). By Fundamental Theorem of Galois Theory Gal(L/K)/ Gal(L/K(ζ)) =∼ Gal(K(ζ)/K) which is abelian.

36 4 Cyclotomic and Kummer Extensions, Cubics and Quartics, Solution by Radicals

Proof. Since tm − λ and mtm−1 are coprime, we know that tm − λ has distinct roots, say α1, . . . αm in the splitting field L. Thus K ≤ L is Galois. Since −1 m −1 (αiαj ) = λλ = 1, the elements

−1 −1 −1 α1α1 = 1, α2α1 , . . . , αmα1 are m distinct mth roots of unity in L and so tm − λ = (t − β)(t − ζβ)(t − ζ2β) ··· (t − ζn−1β) ∈ L[t] where β = α1 and ζ is a primitive mth root of unity. Thus L = K(ζ, β).

Let σ ∈ Gal(L/K(ζ)). It is determined by its action on β. Note that σ(β) is also a root of tm − λ so σ(β) = ζj(σ)β where 0 ≤ j(σ) < m. Also if σ, τ ∈ Gal(L/K(ζ)) then τσ(β) = τ(ζj(σ)β) = ζj(σ)τ(β) = ζj(σ)ζj(τ)β where the second inequality comes from the fact that ζ is fixed by τ. Thus there is a group homomorphism

Θ : Gal(L/K(ζ)) → Z/mZ σ 7→ j(σ)

Note that j(σ) = 0 only if σ is the identity and so Θ is injective. Thus Gal(L/K(ζ)) is isomorphic to a subgroup of Z/mZ, so a cyclic group of or- der dividing m. Finally, |L : K(ζ)| = | Gal(L/K(ζ))| ≤ m with equality precisely when the action of Gal(L/K(ζ)) is transitive on the roots and that is when tm − λ is irreducible over K(ζ) by Lemma 3.4. Example. 1. f(t) = t6 + 3 over Q. Let ζ = −ω be a primitive 6th root of unity where ω is a primitive cubic root of unity. Then 1 √ √ (ζ) = (ω) = ( (1 + −3)) = ( −3). Q Q Q 2 Q

f(t√) is irreducible over Q by Eisenstein with 3. However over Q(ζ) = Q( −3) f(t) factorises as √ √ f(t) = (t3 + −3)(t3 − −3)

so the Galois group of the splitting field L of f(t) over Q(ζ) is a proper subgroup of Z/6Z. We have the following Galois correspondence: L 1 | | 3 Q(ζ) Gal(L/Q(ζ)) | | 2 Q Gal(L/Q)

37 4 Cyclotomic and Kummer Extensions, Cubics and Quartics, Solution by Radicals

Note that

Gal(L/Q(ζ)) =∼ Z/3Z ,→ Z/6Z Gal(Q(ζ)/Q) = hi 7→ −ii =∼ Z/2Z

Let β be a root of f(t). Then the roots are

β ζβ ζ2β ζ3β ζ4β ζ5β β ω2β ωβ −ωβ −β −ω2β

where a 3-cycle is given by permuting β, ω2β, ωβ and the orbits are shown as the last two rows above. Denote the 3-cycle σ and complex conjugation τ, since τστ −1 = σ−1, we get dihedral relation so Gal(L/Q) is dihedral of order 6. 2. f(t) = t5 − 2 over Q. It is irreducible by Eisenstein. Let L be the splitting field of f(t) over Q and ζ be a primitive 5th root of unity. We have the following Galois correspondence:

L 1 | | 5 Q(ζ) Gal(L/Q(ζ)) | | 4 Q Gal(L/Q)

Let β be a root of f(t). Then Q ≤ Q(β) ≤ L so 5 | |L : Q| and thus 5 | | Gal(L/Q(ζ))|. Since we also have Gal(L/Q(ζ)) ,→ Z/5Z, it follows that Gal(L/Q(ζ)) =∼ Z/5Z. We can thus deduce that f(t) remains irreducible over Q(ζ) and | Gal(L/Q)| = 20. By irreducibility of f(t) over Q, Gal(L/Q) is a transitive subgroup of S5. By Lemma 3.5, it is isomorphic to H20, the subgroup generated by a 5-cycle and a 4-cycle. By Fundamental Theorem of Galois Theory,

× Gal(L/Q)/ Gal(L/Q(ζ)) =∼ Gal(Q(ζ)/Q) =∼ (Z/5Z)

which is cyclic and contains a 4-cycle. Thus without a priori knowledge of H20 we know our subgroup of S5 contains a 4-cycle. Theorem 4.5 tells us the property of the splitting field of tm − λ. The following theorem gives its converse:

Theorem 4.6. Suppose K ≤ L is a cyclic extension with |L : K| = m where char K - m and K contains a primitive mth root of unity. Then there exists λ ∈ K such that tm − λ is irreducible over K, and L is the splitting field of tm − λ over K. If β is a root of tm − λ in L then L = K(β).

Since we are going to restate the hypothesis in the theorem many times, we give it a name

38 4 Cyclotomic and Kummer Extensions, Cubics and Quartics, Solution by Radicals

Definition (Kummer extension). A cyclic extension K ≤ L with |L : K| = m where char K - m and K contains a primitve mth root of unity is a Kummer extension.

We need the following to prove the theorem:

Lemma 4.7 (linear independence of group characters). Let φ1, . . . , φn be embeddings of a field K into a field L. Then there do not exist λ1, . . . , λn not all zero such that

λ1φ1(x) + ··· + λnφn(x) = 0

for all x ∈ K.

Proof. Example sheet 2 Q10. Proof of Theorem 4.6. Let Gal(G/K) = hσi which has order m. Observe that i m−1 {σ }i=0 are distinct maps L → L and we can apply linear independence of group characters: there exists α ∈ L such that

β = α + ζσ(α) + ··· + ζm−1σm−1(α) 6= 0 where ζ is a primitive mth root of unity. Observe that σ(β) = ζ−1β = β and so β ∈ K, the fixed field of Gal(L/K). Also σ(βm) = σ(β)m = βm. Let λ = βm ∈ K. Then

tm − λ = (t − β)(t − ζβ) ··· (t − ζm−1β) ∈ L[t] so K(β) is the splitting field of tm − λ over K (recall ζ ∈ K). i m−1 Observe also that {σ }i=0 are distinct K-automorphisms and so |K(β): K| ≥ m. Therefore L = K(β) = K(ζβ).

tm − λ is the minimal polynomial of β over K and hence is irreducible.

Definition (radical extension). A field extension K ≤ L is an extension by racidals if there exists

K = L0 ≤ L1 ≤ · · · ≤ Ln = L

such that each Li ≤ Li+1 is either cyclotomic or Kummer extension.

Definition (solubility by radicals). A polynomial f(t) ∈ K[t] is soluble by radicals if its splitting field lies in an extension by radicals.

39 4 Cyclotomic and Kummer Extensions, Cubics and Quartics, Solution by Radicals

4.3 Cubics In this section and the following one we assume char K 6= 2 for discussion about discriminant and char K 6= 3 for cubic Kummer extension. We have already seen that if f(t) is a monic irreducible cubic in K[t] with L its splitting field over K,

G = Gal(f) = Gal(L/K) is either A3 or S3 since irreducibility implies that the action on roots is transitive. Thus we have the following correspondence:

L 1 | | 3 K(∆) G ∩ A3 | | 1 or 2 KG where ∆2 = D(f) is the discriminant of f. But to see if we can solve f by radicals we want to make use of Theorem 4.6 and so we need to adjoin the appropriate roots of unity. So we get a bigger picture:

L(ω)

1 or 2

K(∆, ω) L 1 or 2 3 K(∆)

1 or 2 K where ω is a primitive cubic root of unity. From the Tower Law |L(ω): K(∆, ω)| = 3. Hence ∼ Gal(L(ω)/K(∆, ω)) = C3. We can now apply Theorem 4.6 to see that

L(ω) = K(∆, ω)(β) where β is a root of an irreducible polynomial t3 − λ ∈ K(∆, ω)[t]. In fact from the proof of Theorem 4.6 we see that

2 β = α1 + ωα2 + ω α3 where α1, α2, α3 are roots of f(t). Now all the extensions

K ≤ K(∆) ≤ K(∆, ω) ≤ L(ω) are either cyclotomic or Kummer. Thus f(t) is soluble by radicals. Let’s give a try to our theory. Given an irreducible cubic

3 2 f(t) = t + at + bt + c = (t − α1)(t − α2)(t − α3),

40 4 Cyclotomic and Kummer Extensions, Cubics and Quartics, Solution by Radicals

we have α1 + α2 + α3 = −a. But we don’t actually need that many parameters: 0 a 0 0 0 0 let α = αi + 3 so that α1 +α2 +α3 = 0 and the α ’s are roots of the polynomial g(t) = t3 + pt + q and most importantly,

0 0 0 K(α1, α2, α3) = K(α1, α2, α3) so the splitting field for g(t) is the same as that for f(t). Thus we could work with g(t) instead. The discriminant of g(t) is D(g) = −4p3 − 27q2. Set

0 0 2 0 β = α1 + ωα2 + ω α3 0 2 0 0 γ = α1 + ω α2 + ωα3 then

02 02 03 2 0 0 0 0 0 0 βγ = α1 + α2 + α3 + (ω + ω )(α1α2 + α1α3 + α2α3) 0 0 0 2 0 0 0 0 0 0 = (α1 + α2 + α3) − 3(α1α2 + α1α3 + α2α3) = −3p and so β3γ3 = −27p3. Also,

3 3 0 0 2 0 3 β + γ = (α1 + ωα2 + ω α3) 0 2 0 0 3 + (α1 + ω α2 + ωα3) 0 0 0 3 + (α1 + α2 + α3) | {z } =0 03 03 03 0 0 0 = 3(α1 + α2 + α3 ) + 18α1α2α3 = −27q

03 0 03 03 03 since α1 = −pα1 − q and so α1 + α2 + α3 = −3q. Thus after some messy algebra, we find that β3 and γ3 are roots of a quadratic t2 + 27qt − 27p3 and so are √ √ 27 3 −3p 27 3 −3√ − q ± −27q2 − 4p3 = − q ± D. 2 2 2 2 √ √ We can solve for β3 and γ3 in K( −3, D) = K(ω, ∆). From here we can get 3 3p β by adjoining a cubic root of β and set γ = − β . Finally we solve in L(ω) for 0 0 0 α1, α2, α3 1 α0 = (β + γ) 1 3 1 α0 = (ω2β + ωγ) 2 3 1 α0 = (ωβ + ω2γ) 3 3

41 4 Cyclotomic and Kummer Extensions, Cubics and Quartics, Solution by Radicals

4.4 Quartics 0 a As with the cubics, by making a substitution of the form αi = αi + 4 we may assume that the sum of the roots is zero and so the t3 term vanishes, leaving a general quartic polynomial of the form f(t) = t4 + bt2 + ct + d

= (t − α1)(t − α2)(t − α3)(t − α4) which we assume to be monic and irreducible. Let L = K(α1, . . . , α4) be the splitting field of f(t) over K. L 1 | | MG ∩ V4 E G | | K(∆) G ∩ A4 | | KG where M is the fixed field of the normal subgroup G ∩ V4 E G, which is an normal extension of K. By Fundamental Theorem of Galois Theory, ∼ Gal(M/K) = G/(G ∩ V4).

To determine this group concretely, we use a little knowledge from group theory: there is a group isomorphism ∼ S4/V4 = S3.

Let θ : S4 → S3 denote the quotient map, which has kernel precisely V4. There- fore θ|G : G → S3 induces an isomorphism ∼ G/ ker θ|G = G/(G ∩ V4) = im θ|G ≤ S3.

We therefore seek a cubic for which M is the splitting field. We introduce resolvent cubic: set

x = α1 + α2

y = α1 + α3

z = α1 + α4 and so 1 α = (x + y + z) 1 2 1 α = (x − y − z) 2 2 1 α = (−x + y − z) 3 2 1 α = (−x − y + z) 4 2

42 4 Cyclotomic and Kummer Extensions, Cubics and Quartics, Solution by Radicals

Thus K(α1, . . . , α4) = K(x, y, z).

Remebering that α1 + α2 + α3 + α4 = 0, 2 2 x = (α1 + α2) = −(α1 + α2)(α3 + α4) 2 2 y = (α1 + α3) = −(α1 + α3)(α2 + α4) 2 2 z = (α1 + α4) = −(α1 + α4)(α3 + α3) 2 2 These are all distinct since for example if y = z then y = ±z so α3 = α4 or α1 = α2. Now consider the resolvent cubic g(t) = (t − x2)(t − y2)(t − z2) ∈ K[t]. 2 2 2 x , y and z are permuted by G and are fixed by G ∩ V4. Thus K(x2, y2, z2) ≤ M = LG∩V4 . We claim that we actually have equality here: Proof. Check D(f) = D(g) — this will be addressed on example sheet — so K(∆) ≤ K(x2, y2, z2). Now observe that Gal(L/(K(x2, y2, z2)) = Gal(K(x, y, z)/K(x2, y2, z2)) and in K(x2, y2, z2) ≤ K(x, y2, z2) ≤ K(x, y, z2) ≤ K(x, y, z), each extension is of degree either 1 or 2 so |K(x, y, z): K(x2, y2, z2)| divides 8 so elements of Gal(L/K(x2, y2, z2)) have order dividing 8. But since 2 2 2 Gal(L/K(x , y , z )) ≤ G ∩ A4, by checking the order of elements in A4 we see that 2 2 2 Gal(L/K(x , y , z )) ≤ G ∩ V4. Therefore by Fundamental Theorem of Galois Theory M = K(x2, y2, z2).

Consider the coefficients of g(t): its roots x2, y2, z2 are permuted by G and so it has coefficients in K. We can actually write down these coefficients: x2 + y2 + z2 = −2b x2y2 + x2z2 + y2z2 = b2 − 4d xyz = −c x2y2z2 = c2 so g(t) = t3 + 2bt2 + (b2 − 4d)t − c2. We know how to solve cubics so we can solve for x2, y2, z2, from which we can 1 solve for x, y, z. Finally we use formula α1 = 2 (x + y + z) etc to recover the roots for the quartic. Whew, done!

43 4 Cyclotomic and Kummer Extensions, Cubics and Quartics, Solution by Radicals

Remark. In our map θ : S4 → S3, we have restriction ∼ θ|A4 : A4 → A3 = C3 and

θ|G∩A4 : G ∩ A4 → A3 so G ∩ A4/(G ∩ V4) ≤ A3 which is either 1 or A3, corresponding to M/K(∆). If the resolvent cubic is irreducible then it is isomorphic to A3. Otherwise it is the trivial group. Example. Let f(t) = t4 + 4t2 + 2. As an aside even if we didn’t study Galois theory we could solve it. Then the resolvent cubic g(t) = t3 + 8t2 + 8t. Note that c = 0 in f(t) so g(t) has no constant term and thus reducible. It follows that M = K(∆).

4.5 Solubility by Radicals Now suppose we have a Galois extension K ≤ L with

K = L0 ≤ L1 ≤ · · · ≤ Lm = L such that each Li ≤ Li+1 is either cyclotomic or Kummer extension. Let G = Gal(G/K). There is a corresponding chain of subgroups of G

G = G0 ≥ G1 ≥ · · · ≥ Gm = 1 with Gi = Gal(L/Ki) and, by Fundamental Theorem of Galois Theory, Li = Gi L . However each extension Li ≤ Li+1 is Galois and we know

Gi+1 = Gal(L/Li+1) E Gal(L/Li) = Gi and ∼ Gi/Gi+1 = Gal(Li+1/Li) which is abelian if Li ≤ Li+1 is cyclotomic and cyclic if Li ≤ Li+1 is Kummer. It is the perfect time to introduce some group theory:

Definition (soluble group). A group is soluble if there is a chain of sub- groups 1 E Gm E Gm−1 E ··· E G1 E G0 = G (∗)

with Gi/Gi+1 abelian.

Remark. Note that some authors use “cyclic” in the definition. While we will prove shortly that they are equivalent for a finite group G, in general they are different. Example.

1. S3 is soluble since 1 E h(123)i E S3.

44 4 Cyclotomic and Kummer Extensions, Cubics and Quartics, Solution by Radicals

2. S4 is soluble since 1 E V4 E A4 E S4. ∼ ∼ and A4/V4 = C3, S4/A4 = C2.

3. A5 is not soluble: we have proved in IA Groups and again in IB Groups, Rings and Modules that A5 is simple, and therefore any normal subgroup is 1 or A5. Then any chain of normal subgroups would have non-abelian quotients and thus A5 is not soluble.

Lemma 4.8. A finite group G is soluble if and only if

1 = Gm E Gm−1 E ··· E G1 E G0 = G (†)

with Gi/Gi+1 cyclic.

This says that we only have to consider cyclic extensions. Proof. The if part is easy. For the only if part, from Structural Theorem of Finitely Generated Abelian Groups, if A is abelian then there is a chain

0 = Ar E Ar−1 E ··· E A0 = A with Ar/Ar+1 cyclic. Thus if we have a chain (∗) with abelian factors Gi/Gi+1, we can refine it to one of the form(†) by Third Isomorphism Theorem.

Definition (derived subgroup). The derived subgroup G0 of a group G is the subgroup generated by all the commutators

−1 −1 g1g2g1 g2

for g1, g2 ∈ G.

Note that it is the subgroup generated by all such elements since it is not obvious that they are closed.

0 Lemma 4.9. Let K E G. Then G/K is abelian if and only if G ≤ K.

Proof. G/K is abelian if and only if

−1 −1 Kg1Kg2Kg1 Kg2 = K for all g1, g2 ∈ G, if and only if

−1 −1 g1g2g1 g2 ∈ K if and only if G0 ≤ K.

Remark. Some interesting asides: 1. In IID Representation Theory we will prove Burnside’s Theorem: if |G| = paqb with p, q distinct primes then G is soluble.

45 4 Cyclotomic and Kummer Extensions, Cubics and Quartics, Solution by Radicals

2. Feit-Thompson Theorem: if |G| is odd then G is soluble. 3. There is an analogue of Sylow’s Theorem due to Philip Hall: given a finite group G, for every m and n coprime and |G| = mn, there is a subgroup of order m if and only if G is soluble.

Definition (derived series). The derived series {G(m)} of G is defined in- ductively as

G(0) = G G(i+1) = (G(i))0

Thus (2) (1) (0) ··· E G E G E G = G with G(i)/G(i+1) abelian.

Lemma 4.10. Given a finite group G, G is soluble if and only if G(m) = 1 for some m.

Proof. If G(m) = 1 then the derived series gives a chain of the form (∗) as in the definition of solubility. Conversely, if there is a chain of the form (∗)

1 E Gm E ··· E G1 E G

(j) (m) with Gi/Gi+1 abelian then induction shows that G ≤ Gj and so G = 1.

Remark. The derived series is the fastest descending chain with abelian factors.

Lemma 4.11. 1. Let H ≤ G. If G is soluble then H is soluble.

2. Let H E G. Then G is soluble if and only if H and G/H are both soluble.

Proof.

1. G is soluble so there is some m such that G(m) = 1. But H(m) ≤ G(m) and so H is soluble.

2. Let H E G. Suppose G is soluble. Then by the previous line H is soluble. Let G(m) = 1, observed that

(G/H)0 = G0H/H ≤ G/H

and inductively (G/H)(j) = G(j)H/H ≤ G/H. Thus (G/H)(m) = H/H = 1 so G/H is soluble.

46 4 Cyclotomic and Kummer Extensions, Cubics and Quartics, Solution by Radicals

Conversely, suppose both H and G/H are soluble, i.e. H(r) = 1 and (G/H)(s) = H/H. But from above (G/H)(s) = G(s)H/H so G(s)H = H and thus G(s) ≤ H. Therefore G(r+s) ≤ H(r) = 1

so G is soluble.

Example. S5 is not soluble since its subgroup A5 is not soluble.

Theorem 4.12. Let K be a field with char K = 0 and f(t) ∈ K[t]. Then f(t) is soluble by radicals over K if and only if Gal(f) over K is soluble.

Remark. We don’t need to restrict to char K = 0. What we need to do for a particular f(t) is to avoid a finite number of bad characteristics (to avoid characteristics smaller than deg f(t)).

Corollary 4.13. If f(t) is a monic irreducible polynomial in K[t] with ∼ char K = 0 and Gal(f) = A5 or S5 then f(t) is not soluble by radicals.

Example. In the example after Theorem 3.6 on page 29, f(t) = t5−6t+3 ∈ Q[t] has Galois group S5 over Q: recall that it has three real roots and so complex conjugation gives a transposition. f(t) is irreducible and so 5 | | Gal(f)| and so there is a 5-cycle. Together they generate S5. Thus f(t) is not soluble by radicals. Proof of Theorem 4.12. Suppose f(t) is soluble by radicals. Then the splitting field of f(t) over K, L, lies in an extension of K by radicals:

K = L0 ≤ L1 ≤ · · · ≤ Lm with each Li ≤ Li+1 either cyclotomic or Kummer.

Lemma 4.14. If K ≤ N is an extension by radicals then there exists N ≤ N 0 with K ≤ N 0 an extension of radicals and a Galois extension.

Proof. Suppose we have a sequence as above and want to extend this into a Galois extension of the same form (assume char K = 0). By Primitive Element Theorem Lm = K(α1) for some α1. Let g(t) be the minimal polynomial of α1 over K with splitting field M. Thus M = K(α1, α2, . . . , αn) where α1, . . . , αn are roots of g(t). There are K-homomorphisms φi : M → M, α1 7→ αi extending K-homomorphisms K(α1) → K(αi) ≤ M. The tower

K ≤ φ1(K) ≤ φ2(L1) ≤ · · · ≤ φ1(Lm) = K(α1) has cyclotomic or Kummer extensions at each step as before. Consider

Lm = K(α1) ≤ φ2(L1)(α1) ≤ φ2(L2)(α1) ≤ · · · ≤ φ2(Lm)(α1) = K(α1, α2).

Consider the extension φ2(Lj)(α1) ≤ φ2(Lj+1)(α1) in this tower. There are two cases:

47 4 Cyclotomic and Kummer Extensions, Cubics and Quartics, Solution by Radicals

• if Lj ≤ Lj+1 is cyclotomic then all the roots of unity adjoined are now in Lm = K(α1) and so φ2(Lj)(αi) = φ2(Lj+1)(α1).

• if Lj ≤ Lj+1 is Kummer then we obtain Lj+1 by adjoining roots of an element of Lj and so we obtain φ2(Lj+1) by adjoining roots of an element in φ2(Lj). Hence we get from φ2(Lj)(α1) to φ2(Lj+1)(α1) by adjoining roots of an element of φ2(Lj). So this is a Kummer extension.

Now continue to get a suitable chain K(α1, α2) ≤ · · · ≤ K(α1, α2, α3) etc. Thus we get a suitable chain from K to K(α1, . . . , αn) = M. Observe that K ≤ M is Galois.

Assuming this lemma and so we may assume K ≤ Lm is Galois. By Funda- mental Theorem of Galois Theory there is a corresponding chain of subgroups Gal(Lm/K). Previous discussion shows that Gal(Lm/K) is soluble. So to sum up we have K ≤ L ≤ Lm with Lm Galois so by Fundamental Theorem of Galois Theory ∼ Gal(L/K) = Gal(Lm/K)/ Gal(Lm/L) But quotients of soluble group are soluble so Gal(L/K) is soluble. Conversely, assume char K = 0 and suppose G = Gal(f) over K is soluble. Let L be the splitting field of f(t) over K and so |G| = |L : K| = n. Set m = n! and let ζ be a primitive mth root of unity and consider L(ζ).

L(ζ) ≤n

K(ζ) L

n K Our proof is similar to that used for cubics. Observe that |L(ζ): K(ζ)| ≤ n: by Primitive Element Theorem L = K(α) for some α with minimal polynomial g(t) of degree n. Then L(ζ) = K(ζ)(α) and the minimal polynomial of α over K(ζ) divides g(t) and so has degree less than or equal to n. Note that Gal(L(ζ)/L) is abelian since the extension is cyclotomic. Then Gal(L(ζ)/K) is soluble since the subgroup Gal(L(ζ)/L is soluble and the quo- tient group Gal(L/K) =∼ Gal(L(ζ)/K)/ Gal(L(ζ)/L) is also soluble. Then the subgroup Gal(L(ζ)/K(ζ)) is soluble. Thus there is a chain of subgroups

1 = Gm E ··· E G1 E G0 E Gal(L(ζ)/K(ζ)) with Gi/Gi+1 cyclic (since for a finite group cyclic is equivalent to abelian in the definition of a soluble group). Now use Fundamental Theorem of Galois Theory to get a corresponding chain of fields

L(ζ) = Km ≥ · · · ≥ K1 ≥ K(ζ) with each Ki ≤ Ki+1 Galois with cyclic Galois group. By Theorem 4.6 all those extensions are Kummer (note all the extensions are of degree less than or equal to n and so we have the appropriate roots of unity, this is why we choose m = n!). Thus we have embedded L in an extension of K by radicals.

48 4 Cyclotomic and Kummer Extensions, Cubics and Quartics, Solution by Radicals

Example. 1. Recall previously we had f(t) = t4 + 4t2 + 2 which is irreducible by Eisen- 3 2 stein over Q. The resolvent√ cubic g(t) = t + 8t + 8t is reducible. It actually has roots 0, −4 ± 2 2. We thus have

p √ p √  L = K −4 + 2 2, −4 − 2 2

4 √ √ √ K(−4 + 2 2, −4 − 2 2) = K( 2)

2 K

so |L : K| = 8. The Galois group is a transitive subgroup of S4 of order 8. It is the dihedral group. p √ In particular the roots ± −2 ± 2 are two conjugate pairs so complex conjugation gives double transposition.

2. f(t) = t4 +2t+2 which is irreducible by Eisenstein over Q. A quick (hmm) calculation shows that the discriminant is 101 · 42 which is not a square. Thus the Galois group contains an odd permutation. The resolvent cubic g(t) = t3 −8t−4 is irreducible mod 5. Therefore there is a 3-cycle in the Galois group. We deduce that Gal(f) = S4.

3. Finally a quintic: f(t) = t5−t−1 is irreducible over Q since it is irreducible mod 5 and so the Galois group contains a 5-cycle and is a transitive sub- group of S5. Reduction mod 2, f(t) factorises as a product of irreducible cubic and an irreducible quadractic

f(t) = (t3 + t2 + 1)(t2 + t + 1)

so Gal(f) is generated by an element of cycle type (3, 2). Thus Gal(f) also contains an element of cycle type (3, 2). Then g3 is a transposition. A subgroup of S5 containing a 5-cycle and a transposition must be S5. Therefore f(t) is not solvable by radicals.

49 5 Final Thoughts

5 Final Thoughts

5.1 Algebraic Closure

Definition (algebracially closed). A field L is algebraically closed if any f(t) ∈ L[t] splits into a product of linear factors in L[t].

Remark. This is equivalent to saying that any f(t) ∈ L[t] has a root in L, or that any algebraic extension of L is L itself.

Definition (algebraic closure). An extension K ≤ L is an algebraic closure of K if K ≤ L is algebraic and L is algebraically closed.

Lemma 5.1. If K ≤ L is algebraic and every polynomial in K[t] splits completely over L then L is an algebraic closure of K.

Proof. We need to show that L is algebraically closed. Suppose L ≤ L(α) is a n n−1 finite extension and fα(t) = t + an−1t + ··· + a0 is the minimal polynomial of α over L. Let M = K(a0, . . . , an−1). Then M ≤ M(α) is a finite extension. But each ai is algebraic over K and so |M : K| < ∞. Hence |M(α): K| < ∞ by Tower Law and so α is algebraic over K. The minimal polynomial of α over K must split over L and so α ∈ L. Thus any algebraic extension of L is L itself.

Example (algebraic number). Let A = {α ∈ C : α algebraic over Q}. It is a subfield of C: if α and β are algebraic over Q then |Q(α, β): Q| < ∞. So if γ = α + β, α − β, αβ or α/β for β 6= 0 we get Q(γ) ≤ Q(α, β) and so |Q(γ): Q| < ∞ so γ is algebraic over Q and so γ ∈ A. Therefore A = Q is an algebraic closure of Q. Like construction of many other objects in this course, we want to prove the existence and uniqueness of algebraic closures. In general we need to appeal to Zorn’s Lemma. This will be covered in IID Logic and Set Theory, and it fact it is equivalent to both Axiom of Choice and Well-ordering Principle. We will do a quick ad hoc statement of Zorn’s Lemma here and relegate the main discussion to IID Logic and Set Theory.

Definition (partial order). (ζ, ≤) is a partial order on a set ζ is 1. reflexive: x ≤ x for all x ∈ ζ. 2. transitive: x ≤ y and y ≤ z implies x ≤ z. 3. antisymmetric: if x ≤ y and y ≤ x then x = y.

Definition (total order). A partial order (ζ, ≤) is a total order if it is total: for any x, y ∈ ζ, either x ≤ y or y ≤ x.

50 5 Final Thoughts

Definition (chain). A chain in a partially ordered set (ζ, ≤) is a totally ordered subset.

Theorem 5.2 (Zorn’s Lemma). Let (ζ, ≤) be a non-empty partially ordered set. Suppose that any chain has an upper bound. Then ζ has a maximal element.

Recall a result derived from Zorn’s Lemma in IB Groups, Rings and Modules:

Lemma 5.3. Let R be a ring. Then R has a maximal ideal.

Proof. Let ζ be the set of proper ideals of R. {0} is a proper ideal so ζ is non-empty. Let ζ be partially orderd by inclusion. Note that an ideal I E R is proper if and only if 1 ∈/ I. Any chain of proper ideals has an upper bound in ζ, namely the union of the chain, so by Zorn’s Lemma ζ has a maximal element, i.e. a maximal ideal of R.

Theorem 5.4 (existence of algebraic closure). Any field has an algebraic closure.

Proof. Let

ζ = {(f(t), j): f(t) irreducible monic in K[t], 1 ≤ j ≤ deg f}.

For each pair s = (f(t), j) ∈ ζ, introduce an indeterminate Xs = Xf,j. Consider the polynomial ring K[Xs : s ∈ ζ] and set

deg f ˜ Y f(t) = f(t) − (t − Xf,j) ∈ K[Xs : s ∈ ζ][t]. j=1

˜ Let I E K[Xs : s ∈ ζ] be the ideal generated by coefficients of all the f(t)’s, which we denote by af,` for 0 ≤ ` < deg f. Claim that I 6= K[Xs : s ∈ ζ]:

Proof. Suppose 1 ∈ I for contradiction. We thus have a relation

b1af1,`1 + ··· + bN afN ,`N = 1 ∈ K[Xs : s ∈ ζ]. (∗)

Let L be a splitting field for the product f1(t) ··· fN (t).For each i, fi splits over L so write deg fi Y fi(t) = (t − αij). j=1 Define a K-linear ring homomorphism which is identity on K

θ : K[Xs : s ∈ ζ] → L

Xfi,j 7→ αij

Xs 7→ 0 otherwise

51 5 Final Thoughts

This induces a map θ : K[Xs : s ∈ ζ][t] → L[t]. Then

deg fi ˜ Y θ(fi(t)) = θ(fi(t)) − θ(t − Xfi,j) j=1

deg fi Y = fi(t) − (t − αij) j=1 = 0

˜ so θ(afi,j) = 0 since afi,j are the coefficients of fi(t). But applying θ to (∗) shows 0 = 1. Absurd.

We have thus shown that I E K[Xs : s ∈ ζ] is proper. By Zorn’s Lemma, there is a maximal ideal P E K[Xs : s ∈ ζ] containing I. Set

L1 = K[Xs : s ∈ ζ]/P, which is a field. We thus have a field extension K ≤ L1. Claim that L1 is an algebraic closure of K:

Proof. We first show K ≤ L1 is algebraic. L1 is generated by the images xf,j ˜ of the Xf,j. However f(t) has coefficients in I and so its image in L1[t] is the zero polynomial. Thus in L1[t],

deg f Y f(t) = (t − xf,j) (†) j=1 and so f(xf,j) = 0. Thus the xf,j’s are algebraic. Any element of L1 involves only finitely many of the xf,j’s and so is algebraic over K. Moreover, from (†) any f(t) ∈ K[t] splits completely over L1. The claim thus follows from Lemma 5.1.

Theorem 5.5. Suppose θ : K → L is a ring homomorphism and L is algebraically closed. Suppose K ≤ M is an algebraic extension, then θ can be extended to a homomorphism φ : M → L, i.e. φ|K = θ.

Proof. Let

ζ = {(N, φ): K ≤ N ≤ M : φ : N → L homomorphism extending θ} and define a partial order ≤ on it by ( N1 ≤ N2 (N1, φ1) ≤ (N2, φ2) if φ2|N1 = φ1

(K, θ) ∈ ζ so it is non-empty. If there is an ascending chain

(N1, φ1) ≤ (N2, φ2) ≤ · · ·

52 5 Final Thoughts

S then set N = λ Nλ, which is a subfield of M and we can define a homo- morphism N → L as follow: if α ∈ N then α ∈ Nλ for some λ and we send α 7→ φλ(α). This is well-defined. Thus this is an upper bound for the chainin ζ. Zorn’s Lemma thus says that there is an maximal element of ζ, (N, φ). We now show that N = M. Given α ∈ M, it is algebraic over K and hence over N. Let fα(t) be its minimal polynomial over N. φ(fα(t)) ∈ L[t] and so splits completely since L is algebraically closed. Write

φ(fα(t)) = (t − β1) ··· (t − βr).

Since φ(fα(βj)) = 0 there is a map ∼ N(α) = N[t]/(fα(t)) → L

α 7→ β1 extending φ. Maximality of (N, φ) implies that N(α) = N. Thus α ∈ N and N = M. With this theorem we can finally show

Theorem 5.6 (uniqueness of algebraic closure). If K ≤ L1,K ≤ L2 are two algebraic closures of K, there exists an isomorphism φ : L1 → L2.

Proof. By the previous theorem there is a homomorphism φ : L1 → L2 extend- ing the embedding of K in L2. Since K ≤ L2 is algebraic, so is K ≤ φ(L1). But L1 is algebraically closed and so φ(L1) is algebraically closed. Thus L2 = φ(L1) and φ is an isomorphism.

5.2 Symmetric Polynomials & Invariant Theory In the build-up of Fundamental Theorem of Galois Theory we met Artin, which H says that if K ≤ L, H is a finite subgroup of AutK (L) and M = L then M ≤ L is a Galois extension and H = Gal(L/M).

Example. Let L = K(X1,...,Xn). Let Sn be the permutation group on the Xi’s. These permutations induce K-automorphisms of L. By Artin, M = Sn L ≤ L is Galois and Gal(L/M) = Sn. Thus Sn is a Galois group of some field extension.

Since we can regard any finite group G as a subgroup of some Sn, any finite group is a Galois group of some finite field extension

Q ≤ K.

Finding the field extension is the inverse Galois problem. Using notation from the previous example, let

n Y n n−1 n−2 n f(t) = (t − Xi) = t − s1t + s2t + ··· + (−1) sn ∈ M[t] i=1

53 5 Final Thoughts where

s1 = X1 + ··· + Xn X s1 = XiXj i

sn = X1X2 ··· Xn

Definition (elementary symmetric polynomial). The si’s are the elemen- tary symmetric polynomials.

Definition (algebraic independence). α1, . . . , αn are algebraically indepen- dent over K if the ring homomorphism

K[Y1,...,Yn] → K[α1, . . . , αn] ≤ L

Yi 7→ αi

is an isomorphism where K[Y1,...,Yn] is the polynomial ring in Y1,...,Yn.

Sn Theorem 5.7. The fixed field M = L equals to K(s1, . . . , sn) and the si’s are algebraically independent over K in L.

Proof. Certainly the si’s are fixed by Sn so M1 = K(s1, . . . , sn) ≤ M. Observe Qn that L is the splitting field for f(t) = i=1(t − Xi) over M1. But f(t) has degree n and so by a proposition we proved in example sheet, the degree of the splitting field over M1 is at most n!. Artin implies that |L : M| = |Sn| = n! so M1 = M. To show the algebraic independence of si’s, we make use of the idea of tran- scendence bases and transcendence degree: we may consider the algebraically independent subsets of L and partially order them by inclusion. Note that the union of a chain is algebraically independent and thus an upper bound of the chain. By Zorn’s Lemma there is a maximal algebraically independent subset, which is a transcendence basis. The transcendence degree of L over K is the cardinality of such a set, which is well-defined by Steinitz Exchange Lemma for infinite sets, and is denoted trdegK (L). Since L is algebraic over M, we have trdegK (L) = trdegK (M). If si’s were algebraically dependent then trdegK (M) < n as M would be algebraic over a subfield generated by fewer elements and trdegK K(α1, . . . , αn) ≤ n by induction. Absurd.

H Polynomial invariant theory studies K[X1,...,Xn] for a finite subgroup H ≤ Sn and as an aside, rather than confining ourselves to permutations of the variables, we may also consider H ≤ GL(V ) where V is the K-vector space generated by X1,...,Xn. Question. H 1. Is K[X1,...,Xn] finitely generated?

54 5 Final Thoughts

H 2. Is K[X1,...,Xn] isomormphic to any polynomial algebra?

We have shown above that the fixed field of K(X1,...,Xn) is K(s1, . . . , sn). A similar result holds for the ring:

Theorem 5.8. Sn K[X1,...,Xn] = K[s1, . . . , sn].

Definition (symmetric polynomial). The elements of K[s1, . . . , sn] are the symmetric polynomials.

Sn Proof. Let f(X1,...,Xn) ∈ K[X1,...,Xn] . The proof is by induction on the total degree of f. If the total degree is 0 then f is a constant polynomial and thus in K, so in K[s1, . . . , sn]. Suppose the total degree of f is positive. Let

θ : K[X1,...,Xn] → K[X1,...,Xn−1]

g(X1,...,Xn) 7→ g(Xn,...,Xn−1, 0) i.e. the projection of the first n − 1 coordinates. Then ker θ = (Xn). Since f(X1,...,Xn) is fixed by Sn, by slight abuse of notation θ(f(X1,...,Xn)) = f(X1,...,Xn−1) is fixed under the subgroup Sn−1, i.e. the subgroup that fixes n. Note that in particular

θ(sj(X1,...,Xn)) = sj(X1,...,Xn−1) for j ≤ n − 1 where sj is the jth elementary symmetric polynomial. Apply induction,

θ(f(X1,...,Xn)) = p(s1(X1,...,Xn−1), . . . , sn−1(X1,...,Xn−1)) where p is a polynomial. Rearrange,

θ(f(X1,...,Xn) − p(s1(X1,...,Xn), . . . , sn−1(X1,...,Xn))) = 0 so Xn divides the polynomial

f(X1,...,Xn) − p(s1(X1,...,Xn), . . . , sn−1(X1,...,Xn)) (∗) which is a symmetric polynomial since it is fixed under Sn. It follows that Xi divides (∗) for all i. However, K[X1,...,Xn] is a UFD and the Xi’s are coprime so the product X1 ··· Xn divides (∗). Write f(X1,...,Xn) = g(X1,...,Xn)X1 ··· Xn+p(s1(X1,...,Xn), . . . , sn−1(X1,...,Xn)).

Observe that the total degree of g(X1,...,Xn) is smaller that that of f(X1,...,Xn) and that g(X1,...,Xn) is fixed under Sn. Applying induction to g(X1,...,Xn) and so it is a polynomial in si’s. Thus f(X1,...,Xn) ∈ K[s1, . . . , sn].

An Example. K[X1,...,Xn] is generated by si’s and Y ∆(X1,...,Xn) = (Xi − Xj). i

Emmy Noether in 1920s consider other subgroups of Sn and showed that the invariant rings are Noetherian.

55 5 Final Thoughts

H Theorem 5.9 (Chevalley-Shephard-Todd). C[X1,...,Xn] where H ≤ GL(V ) finite is isomorphic to a polynomial algebra if and onlyif H is generated by pesudoreflections, whose 1-eigenspace has codimension 1.

56 Index

K-homomorphism, 10 normal, 13, 22 radical, 39 algebraic closure, 50 separable, 16 algebraic independence, 54 simple, 6 algebraically closed, 50 fixed field, 25 Artin’s Theorem, 26 Frobenius automorphism, 31 constructible number, 7 Galois group, 23, 28 cyclotomic polynomial, 33 inverse Galois problem, 53 derived series, 46 differentiation, 15 minimal polynomial, 5 discriminant, 29 norm, 19 elementary symmetric polynomial, separable, 15, 16 54 solubility, 39 soluble group, 44 field extension, 3 splitting field, 11 abelian, 36 splitting polynomial, 11 algebraic, 5 symmetric polynomial, 55 cyclic, 36 cyclotomic, 33 trace, 19 finite, 3 transcendence basis, 54 Galois, 23 Kummer, 39 Zorn’s Lemma, 51

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