NOTES FOR DRAGOS: MATH 210 CLASS 12, THURS. FEB. 22


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1 NOTES FOR DRAGOS: MATH 210 CLASS 12, THURS. FEB. 22 RAVI VAKIL Hi Dragos The class is in 381T, 1:15 2:30. This is the very end of Galois theory; you ll also start commutative ring theory. Tell them: midterm back on Tuesday. They can for their grade if they are feeling anxious. Here s an outline of what I did last day, and some proposals about what you can do. Feel free to just do what you feel happy about; don t feel the need to rush. 1. THINGS I DID LAST DAY 1.1. Solvability. First, a fun remark (thanks to Carl Erickson): odds of a quartic being solvable by ruler and compass are a little better than one in six. [DF 14.7 p. 606] Recall the statement from last time: Proposition. Let F be a field of characteristic not dividing n which contains the nth roots of unity. Easy. Then the extension F ( n a) for a F is cyclic over F of degree dividing n. Hard. Any cyclic extension of degree n is of the form F ( n a) for some a F. I proved the first half. I now want to prove the second half. [Let me do the prime case. As soon as you add one, you get the rest, so it is a splitting field. In other words, you get the degree of the extension.] The hard direction is analogous to the quadratic formula. Here is a reminder of how we solved the cubic. Cubic. e A 3 S 3. Q(a, b, c), Q(e 1, e 2, e 3, (a b)(b c)(c a) = δ). δ 2 =???. Hence can solve it. Let y = a + ωb + ω 2 c, z = a + ω 2 b + ωc. Note that the cubes are fixed by A 3. And y 3 + z 3 and y 3 z 3 are fixed by S 3. So solve for them. Then take cube roots to find y and z. (Can they be taken independently?) Then we know a, b and c. Exercise: x 3 3x + 1 solve explicitly. Solve for the cosine of 20. Now here is the proof of the hard direction. [DF p. 607] Suppose E/F Cyclic with Galois group Z/n where n is not necessarily prime. 1
2 For α E and any primitive nth root of unity ζ, define the Lagrange resolvent (α, ζ) K by (α, ζ) = α + ζσ(α) + ζ 2 σ 2 (α) + + ζ n 1 σ n 1 (α). If we apply the automorphism σ to (α, ζ), we obtain σ(α, ζ) = ζ 1 (α, ζ). Thus (α, ζ) n is fixed by Gal(E/F ) for any α E. Better: vandermonde. By the linear independence of the automorphisms, there is an element α E with a := (α, ζ) 0. Also, σ i (α, ζ) = ζ 1 (α, ζ) so σ i doesn t fix (α, ζ) for any i < n. Hence this element can t lie in any proper subfield of E, so E = F (α, ζ). Thus F ( n a) = F ((α, ζ)) = K Composite extensions and simple extensions. [DF 14.4] Proposition. Suppose K/F is a Galois extension and F /F is any extension. Then KF /F is a Galois extension, with Galois group Gal(KF /F ) = Gal(K/K F ) isomorphic to a subgroup of Gal(K/F ). Pictorially: KF K F K F F Immediate Corollary. If F /F is finite, then [KF : F ] = [K : F ][F : F ]/[K F : F ]. Warning: K/F Galois needed for this! Proof: For Galois: it is the splitting field of a separable polynomial. We gt a map φ : Gal(KF /F ) Gal(K/F ). It is a homomorphism. The kernel is trivial (any element is trivial on K and F ). So we want to see that it is surjective. Let H be the image. Then K H contains K F. On the other hand, the composite K H F is fixed by Gal(KF /F ), so K H F = F, so K H F, hence K H K F, hence equal, so H = Gal(K/K F ). Proposition. Let K 1 and K 2 be Galois extensions of F. Then (1) The intersection K 1 K 2 is Galois over F, and (2) The composite K 1 K 2 is Galois over F. The Galois group is isomorphic to the subgroup H = {(σ, τ) : σ K1 K 2 = τ K1 K 2 } of the direct product Gal(K 1 /F ) Gal(K 2 /F ) consisting of elements whose restrictions to the intersection K 1 K 2 are equal. K 1 K 2 K 1 K 2 K 1 K 2 F 2
3 Immediate Corollary. Let K 1 and K 2 be Galois extensions of a field F with K 1 K 2 = F. Then the big Galois group is the product of the smaller ones. Proof. (1) separable + normal (every irreducible polynomial with root splits completely). (2) Galois: Generated by separable elements, and is normal (splitting field). Consider φ : Gal(K 1 K 2 /F ) Gal(K 1 /F ) Gal(K 2 /F ). It is a homomorphism. It is injective. Image lies inside the set H described above. But they both have the same cardinality. Remark. Let E/F be any finite separable extension. Then E is contained in an extension K which is Galois over F and is minimal in the sense that in a fixed algebraic closure of K any other Galois extension of F containing E contains K. This is called the Galois closure. This is rather obvious: take the splitting field of everything in E. But also: take the intersection of all the Galois extensions of F containing E in the algebraic closure Primitive element theorem. An extension K of F is called simple if K = F (θ) for some element θ ( atomic ). In this case, θ is called a primitive element for K. Primitive element theorem. If K/F is finite and separable, then K/F is simple. In particular, any finite extension of fields of characteristic 0 is simple. Proof in infinite fields (characteristic 0) is easy, using the fundamental theorem of Galois theory, and infinitude of fields. General proof shortly. Finite fields: okay too, take a primitive p n 1th root of unity. See book for another proof, using: Let K/F be a finite extension. Then K = F (θ) iff there exist only finitely many subfields of K containing F. Remark: we need separability! Consider the field F p (x p, y p )/F p (x, y). Then F p (x p, y p, x+ cy) is a subfield, and each different. [page 576] YOU CAN START HERE. 2. CYCLOTOMIC EXTENSIONS AND ABELIAN EXTENSIONS [THIS IS IN DUMMIT AND FOOTE 14.5, see also 13.6] Reminder. Let µ n be the group of nth roots of unity over Q. Isomorphic to Z/n. Comment: why give it a different name? Define the nth cyclotomic polynomial Φ n (x) as the monic polynomial whose roots are the primitive nth roots of unity. It is degree φ(d). Exercise: Show that x n 1 = d n Φ d(x). [DF p. 533, possibly put on homework, but it is easy.] Perhaps give examples. Φ p is easy. Φ 4 = x 2 + 1, Φ 6 = x 2 x
4 Note: it is always a monic polynomial with integer coefficients. Reason: use exercise: divide out the rest, so use induction. Q(ζ) includes all the primitive roots of unity. Its degree is the degree of the minimal polynomial φ(n). However, it has at least φ(n) automorphisms. Conclusion: Theorem: The cyclotomic polynomial is irreducible, and the Galois group is a group of order φ(n). It is in fact (Z/n). Example: n = 5. (Z/5) is cyclic of order 4. Hence we can solve for the fifth root of unity in radicals; we can even construct it. Let s do that. Note ζ = 0. We need something preserved by a degree 2 automorphism. ζ ζ 1 is it. Hence α = ζ + ζ 1 is it, otherwise known as 2 cos(2π/5). α 2 = ζ 2 +2+ζ 2. α 2 +α = 1. Hence α = ( 1± 5)/2. (Which one? Interesting question! What s the meaning of the other root?) How to solve for ζ? Exercise: Find a generator of the quadratic extension for a general prime p. Note: we can construct a pgon where p is prime iff p 1 is a power of 2. These must be p = 2 2s + 1 (called Fermat primes), exercise. First few: 3, 5, 17, 257, Exercise. Given a prime decomposition, the Galois group of the primitive roots of unity is the product of what you get with the various prime powers. See p. 579, where this is done. Exercise. Show that a regular ngone can be constructed by straightedge and compass iff n = 2 k p 1 p r where the p i are distinct Fermat primes. Remark: These are all abelian extensions. Theorem (see book): if G is any finite abelian group. then there is a subfield K of a cyclotomic field with Gal(K/Q) = G. Converse is the KroneckerWeber theorem: if K is a finite abelian extension of Q. Then K is contained in a cyclotomic extension of Q. THAT S BASICALLY IT FOR GALOIS THEORY. 3. COMMUTATIVE RING THEORY The geometry of rings. All my rings will commutative, with unit. Noetherian rings. DF p. 637 Define Noetherian ring p Behaves well under quotients. 4
5 Theorem 2. TFAE: (a) R Noetherian, (b) every nonempty set of ideals of R contains a maximal element under inclusion. (c) Every ideal of R is finitely generated. Example: Any PID. Not Noetherian: k[x 1,... ]. How about k[x, y]? That s not a PID, so but it is Noetherian: State and prove the Hilbert basis theorem. Corollary: the polynomial ring over a field, or over the integers, or over any Noetherian ring, and any quotient thereof, is Noetherian. In particular, any finitely generated k algebra. HOPEFULLY YOU CAN SAFELY GET HERE. But it doesn t matter too much. Affine algebraic sets. Over an algebraically closed field k. Z(S) where S is a set of polynomials. (V or Z.) Note: Z(S) = Z((S)). It reverses inclusions. Intersections and unions. Zero ideal and everything. Thus we have a map: Z: ideals to affine algebraic sts. Proposition: each affine algebraic set is the intersection of a finite number of hypersurfaces in A n. Note: this is a geometric property that is a consequence of an algebraic property. We also get a map I from algebraic sets to ideals. Examples: an axis. A line. A point. y 2 x 3. Properties: revses inclusion; union of sets gives intersection of ideals. I( ), I(A n ). I I(Z(I)) (explain), A Z(I(A)). If V is an affine algebraic set, then V = Z(I(V )) (explain). This begs a question: what ideals arise in this way? Definition: if V is an affine algebraic set, the quotient ring k[a n ]/I(V ) is called the coordinate ring of V, denoted by k[v ]. The algebraic functions. Define an algebraic map (or polynomial map or regular map) if we can find polynomials. This induces maps of coordinate rings in the other direction. In fact: there is a bijective correspondence: Theorem 6 on p Do some examples. Corollary 7: a map of affine algebraic sets is an algebraic map iff it induces a map of rings. Important nonobvious consequence: we don t need to know anything about the embedding. YOU DEFINITELY SHOULDN T GET PAST HERE!!! 15.2 Radical and affine varieties 5
6 Define radical of I. nilradical. Radical ideal. Note/Proposition: rad I contains I. (radi)/i is the nilradical of R/I. I is radical iff R/I has no nonzero nilpotents. (I like this: prime ideals mean R/I is domain. Maximal: R/I is a field. Radical: R/I has no nonzero nilpotents.) Hence: prime ideals are radical. Proposition: the radical of a proper ideal is the intersection of all prime ideals containing it. Hence the nilradical is the intersection of all the prime ideals in R. Hilbert s Nullstellensatz, one version: radical ideals correspond to affine algebraic sets. Stated p. 652, proved in next section. Discuss: why we might expect this. We need algebraically closed field x in R[x]. Pictures of rings. Remark: I still have a picture of other ideals. Zariski topology. Irreducible. Proposition. Irreducible means prime ideal. Hence we can picture prime ideals of C[x 1, x 2 ]. We know what they all are. Proposition. Each nonempty affine algebraic set V may be written uniquely as a union of irreducible components. 6
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