Math 6320 Lecture Notes Lectures by Prof. David Zywina, Notes by Vivian Kuperberg Disclaimer: Professor David Zywina is responsible for the lectures behind these notes, but has nothing to do with the writing of the notes themselves, and is not responsible for any typos or mistaken proofs. All questions or concerns about the validity of the notes should be sent to Vivian Kuperberg. Contents 1 January 28th3 1.1 Galois Theory...................................3 2 February 2nd7 2.1 Galois Theory, continued.............................7 3 February 4th9 3.1 Proof of the fundamental theorem........................9 4 February 9th 12 4.1 More Thoughts on the fundamental theorem.................. 12 4.2 Application: Finite fields............................. 14 5 February 11th 15 5.1 Application: Fundamental Theorem of Algebra................. 15 5.2 Some General Structure Theorems....................... 16 6 February 18th 19 6.1 Last time..................................... 19 6.2 Solving for roots of polynomials of low degree................. 20 6.2.1 Degree 2.................................. 20 6.2.2 Degree 3.................................. 20 6.2.3 Degree 4.................................. 21 6.2.4 Degree ≥ 5................................ 22 7 February 23rd 23 7.1 Quintics...................................... 23 1 8 February 25th 25 8.1 Solvability and solvability............................ 25 8.2 Galois groups of polynomials of small degree.................. 26 9 March 3rd 28 9.1 Computing Galois groups over Q ........................ 28 10 March 8th 30 10.1 Last Time..................................... 30 10.2 Infinite Galois Theory.............................. 31 11 March 10th 32 11.1 A New Topic: Representation Theory of finite groups............. 32 12 March 15th 35 12.1 Last Time..................................... 35 12.2 Group Rings.................................... 35 13 March 17th 39 13.1 Representations and their Characters...................... 39 13.1.1 Interlude: Constructing New Representations............. 40 14 March 22nd 43 14.1 Complex Character Theory, continued..................... 43 15 March 24th 46 15.1 Character Tables................................. 46 16 April 5th 50 16.1 The last of the examples............................. 50 16.2 Frobenius Divisibility feat. maybe some Burnside............... 51 17 April 7th 53 17.1 Loose end..................................... 53 17.2 Frobenius and Burnside, cont'd......................... 53 18 April 12th: Transition to Homological Algebra 55 18.1 Hom........................................ 55 19 April 14th 58 19.1 Last time..................................... 58 19.2 Projective modules................................ 58 19.3 Injective Modules................................. 61 2 20 April 19th 62 20.1 Some more injective modules........................... 62 20.2 Tensor Products Revisited! D&Fx10.4..................... 63 21 April 21st 65 21.1 I'm Getting Tensor Every Day.......................... 65 22 April 26th 68 22.1 Homology/Cohomology.............................. 68 23 April 28th 70 23.1 Snake Lemma Continued, and More Homological Algebra........... 70 24 May 3rd 73 24.1 Cohomology Cont: Next Ext and More Tor................... 73 25 May 5th 78 25.1 Last time..................................... 78 26 May 10th 83 26.1 The Cohomology of Groups........................... 83 26.2 Central simple algebras.............................. 85 1 January 28th As review, look at the field theory sections, x 13.1, 13.2, 13.4, 13.5. 1.1 Galois Theory Let L be a field. Let Aut(L) be the set of field automorphisms, or structure-preserving ∼ bijections σ : L != L. Aut(L) is a group under composition. Let L=K be a field extension, i.e. a field L ⊇ K. (The book will usually use the notation K=F , but number theorists tend to use L=K.) We can then define a second group Aut(L=K), the set of automorphisms of L that fix K. Aut(L=K) is a subgroup of Aut(L). The question that Galois theory addresses is, what information does the group Aut(L=K) encode about the extension L=K? Example. 1. Aut(Q) = 1. σ(1) = 1, so all integers are fixed because addition is fixed, and then all rationals are fixed. 2. Aut(Fp) = 1, for a similar reason. 3. Aut(C=R) = f1; τg, where τ is complex conjugation. 4. f1; τg ⊆ Aut(C). Aut(C) has many elements assuming the axiom of choice. 3 4 Lemma 1.1.1. Let L=K be an algebraic extension. Take any α 2 L and let f(x) 2 K[x] be a non-zero polynomial with root α. Then σ(α) 2 L is a root of f for all σ 2 Aut(L=K). P i P i P i Proof. f(x) = i cix , with ci 2 K. 0 = f(α) = i cix , so 0 = σ(0) = i σ(ci)σ(α) = P i i ciσ(α) = f(σ(α)). p p p 3 3 3 Example. Lookp at σ 2 Aut(Q( 2=Q), with 2 2 R. Then σ( 2) has to mapp to a root of 3 3 x3 − 2 in Q( 2) ⊆ R, i.e. to itself, the only real root of x3 − 2. So Aut(Q( 2=Q) is trivial, because that fixation along with Q being fixed determines the trivial automorphism. 4 p p p p Example. Let σ 2 Aut(Fp(t)=Fp(t ). σ(t) must be a root of x − t = (x − t) , so σ(t) = t, and thus just as above, this automorphism group is trivial. 4 The first example is boring because there aren't enough roots in our field. The second is boring because the extension is inseparable; the polynomial doesn't have distinct roots when you increase the field. So we avoid both pitfalls by making the following definition. Definition 1.1.2. An algebraic extension L=K is Galois if it is normal and separable. Recall that an extension is normal if it is the splitting field of some polynomials in K[x], and an extension is separable if every element is the root of a separable polynomial in K[x], where, again, a separable polynomial is one that has distinct roots in the extension. For a Galois extension L=K, we define Gal(L=K) = Aut(L=K). The benefit of the new notation is that we call it Gal in the cases where we're sure that it'll be interesting. Now, fix L=K a finite Galois extension. 1. (Getting a group out of a field) Let K ⊆ F ⊆ L be a subfield. Then the extension L=F is Galois, so there's the group Gal(L=F ) ≤ Gal(L=K). 2. (Getting a field out of a group) Take a subgroup H ≤ Gal(L=K). Then let LH be the set of α 2 L that are fixed by H. Note that this is actually a subfield of L, because fixation is preserved by the field operations. Also, K ≤ LH , because H ≤ Gal(L=K). This is in fact a bijection between subgroups and subfields; it is a bijective correspondence, and the two directions are inverses of each other. Theorem 1.1.3 (Fundamental Theorem of Galois Theory). Let L=K be a finite Galois extension. The maps 4 F 7! Gal(L=F ) fF : K ≤ F ≤ L a fieldg fH : H ≤ Gal(L=K)g LH H [ are inverses. The maps are inclusion reversing: F1 ⊇ F2)Gal(L=F1) ⊆ Gal(L=F2), and H1 H2 H1 ⊇ H2)L ⊆ L . We will see that jGal(L=F )j = [L : F ], which is the definition of Galois in the book. p p Example. We take L=Q, with L = Q( 2; 3), the Galois splitting field of x2 − 2 and x3 − 3. jGal(L=Q)j should match [L : Q] =p 4. So we knowp we havep Gal(L=Qp( 3)) ≤ Gal(L=Qp) which should havep order 2,p i.e. it should be f1; σg.pσ( 3) = 3, so σ( 2) shouldn'tp be 2; thusp it sendsp 2p to − 2. Similarly, Gal(L=Q( 2)) ≤ Gal(L=Q) is f1; τg, where τ( 3) = − 3 and τ( 2) = 2. Then the group ∼ Gal(L=Q) is thus f1; σ; τ; στg. Each of these elements has order 2, so Gal(L=Q) = Z=2×Z=2. Okay, so we have this group. What are its subgroups? Well, the correspondence is illustrated by the diagrams below. 1 L p p p f1; σg f1; τg f1; στg Q( 3) Q( 2) Q( 6) Gal(L=Q) Q 4 p 4 p Example. Let L=Q be the splitting field of x4 − 2, in C. Let α = 2 2 R and let i = −1. Then the roots of x4 − 2 are ±α; ±iα, so L = Q(α; i). So there's some lattice of subfields, which we know contains L. 5 L 2 4 Q(α) Q(i) 4 2 Q Gal(L=Q) has order 8, so what are we missing? Well, Gal(L=Q(α)) has order 2; it's f1; τg, where τ(α) = α, so τ(i) = −i. Also, Gal(L=Q(i)) ⊆ Gal(L=Q) has order 4, so σ 2 Gal(L=Q(i)) is determined by σ(α); there are four options for this. There is a unique σ 2 Gal(L=Q) with σ(α) = iα and σ(i) = i. Note that σ(α) = iα, σ2(α) = σ(iα) = −α, σ3(α) = −iα and σ4(α) = α, so this group is cyclic. −1 −1 ∼ As it turns out, Gal(L=Q) = hσ; τi. One can check that τστ = σ , so Gal(L=Q) = D8, the dihedral group. So we have 1 f1; τg f1; σ2τg f1; σ2g f1; στg f1; σ3τg f1; σ2; τ; σ2τg f1; σ; σ2; σ4g f1; στ; σ2; σ3τg Gal(L=Q) which leads to the analogous diagram for fields: 6 Q(α; 1) = L Q(α) Q(iα) Q(i; α2) Q(α + iα) Q(α − iα) Q(α2) Q(i) Q(iα2) Q 4 2 February 2nd HW 1: Dummit and Foote 14.1 # 7,8; 14.2 # 1,3,5,12, 17 & 23, 18 & 21; 14.3 # 8.