Galois Theory, summary

Size: px

Start display at page:

Download "Galois Theory, summary"

Emerald Flynn
5 years ago
Views:

1 Galois Theory, summary Chapter UFD, definition. Any two elements have gcd 11.2 PID. Every PID is a UFD. There are UFD s which are not PID s (example F [x, y]) ED. Every ED is a PID (and hence a UFD). Chapter 15: Algebraic extensions of fields 15.1 Irreducible polynomials and Eisenstein s criterion. [Basic properties: F [x] is ED. The units are the non-zero elements of F. If p(x) is irreducible in F [x] then F [x]/(p(x)) is a field and conversely] Let f(x) F [x] have degree 2 or 3. Then f(x) is irreducible in F [x] if and only if it does not have a root in F , 6 Let f(x) Z[x]. Then f(x) is reducible over Z if and only if it is reducible over Q Let f(x) Z[x] be monic. If f(x) has a root a Q then a Z and a divides the constant coefficient Eisenstein Adjunction of roots. Let E F be a field extension, then E is a vector space over F. Denote (E : F ) = dim E Tower Theorem If p(x) F [x] is irreducible then there is a field E F in which p(x) has a root Let p(x) F [x] be irreducible, and u = a root of p(x) in E. Then (i) F (u) := the smallest subfield of E containing F and u. This is equal to {f(u) : f(x) F [x]} =: F [u] (ii) If p(x) has degree n then F (u) has basis {1, u, u 2,..., u n 1 } as vector space over F. Hence (F (u) : F ) = n Algebraic extensions. DEF 13 α E is algebraic over F, E F is an algebraic extension If u is algebraic over F then it has a (unique) minimal polynomial. Properties: If (E : F ) < then E is algebraic over F.

2 Let E = F (u 1,..., u r ). If each u i is algebraic over F then (E : F ) < and E is algebraic over F Let E F be a field extension. The set K := {u E : u is algebraic over F } is a subfield of F and it is algebraic over F Let E F be an algebraic extension. If σ : E E is an embedding then σ is onto, hence an isomorphism of E [we have not used algebraic closure seriously] 15. Lemma 4.2* Let σ : F L an embedding (F, L fields). Let E = F (α) where α is algebraic over F, and let p(x) be the minimal polynomial of α over F. Let β L. Then there is a homomorphism η : E L extending σ with η(α) = β β is a root of p σ (x) Corollary Let L F and α, β L algebraic over F. If α, β have the same minimal polynomial in F [x] then there is an isomorphism η : F (α) F (β) with η F = Id. useful to see whether two fields might be isomorphic Ch. 16 Normal and separable extensions 16.1 Splitting fields Definition /2 A polynomial f(x) F [x] has a splitting field E and (E : F ) < Any two splitting fields are isomorphic. To prove this, we showed more generally the following: Let σ : F 1 F 2 be an embedding and f(x) F 1 [x]. Let K = a SF of f(x) and E = a SF of f σ (x). Then there is an isomorphism η : K E extending σ Normal extensions DEF 19 E F is normal. 2.1* Let E F be a finite (algebraic) extension. Then E F is normal E is the SF of some f(x) F [x] Separable extensions DEF 20 If f(x) is irreducible, then f(x) is separable Suppose α is a root of f(x). then (x α) 2 divides f(x) if and only if f (α) = Assume F [x] is irreducible. Then f(x) has a multiple root if and only if f (x) = Assume f(x) F [x] is irreducible. (a) If char(f ) = 0 then f(x) is separable.

3 (b) If char(f ) = p > 0 then f(x) is not separable if and only if f(x) = g(x p ) for some g(x) F [x]. DEF 21 E F algebraic. α E is separable over F. The extension E F is separable. [Remark 5.1 ] Let E F be an algebraic extension. Then E F is separable if char(f ) = 0. As well if F is a finite field. 5.2 Assume E F is a finite and separable extension. Then E = F (θ) for some θ E. [we did not prove this, except when E is finite.] 5.3 Let E F be finite. Then equivalent: (a) E = F (α) for some α E; (b) There are only finitely many intermediate fields K with F K E. Ch 17 Galois Theory Let E F be a finite separable extension. DEF 23 G(E F ) := {σ Aut(E) : σ F = Id}. When E F is in addition normal, call E F a Galois extension, and call G(E F ) the Galois group. If E =SF of a polynomial f(x) F [x] then call G(E F ) the Galois group of f(x) FTGT Let E F be a Galois extension, with G = G(E F ). Then (1) There is a bijection, inclusion reversing intermediate fields subgroups of G where K G(E K) and for the converse H E H, the fixed field of H. Moreover (E : K) = G(E K) (2) The extension K F is Galois G(E K) G. If so then G(K F ) = G/G(E K). Part of the proof of (2) was /4 Let E K F. Then K F is normal For every σ G(E F ) we have σ(k) K. [Example 2.2a] Let E F be a Galois extension, the SF of f(x) F [x]. Assume f(x) has distinct roots α 1,..., α n. Then G(E F ) is isomorphic to a subgroup of S n, permuting the α i. If f(x) is irreducible then the action is transitive.

4 [That is, for each i, j there is some g G with g(α i ) = α j. ] Conversely if (E : F ) = n and the action is transitive then f(x) is irreducible. Ch 16.4 Finite fields Assume F has characteristic p. The map σ : F F given as σ(x) = x p is a homomorphisms, and embedding. If F is finite, it is an isomorphism The multiplicative group of a finite field is cyclic If E F and E is finite then E = F (α). Any field has a prime field, which is either isomorphic to Q (characteristic zero), or to a field Z p (characteristic p). If F is finite then it has characteristic p and it contains Z p. So it has size p n where n = (F : Z p ) Let F be a field with p n elements. Then F is the SF of the polynomial x pn x Z p [x]. Hence any two finite fields of the same size are isomorphic The set of roots of x pn x over characteristic p is a field. FTGT for finite fields: Let (E : Z p ) = n. E is a Galois extension and G(E Z p ) = σ, is cyclic of order n (with σ as in 4.0). The intermediate fields are in 1-1 correspondence with divisors of n Suppose F is a finite field. For each n there is an irreducible polynomial in F [x] of degree n. Method to count irreducible polynomials over finite fields. Ch. 18 Applications 18.1 Cyclotomic extensions Φ n (x) = ω (x ω), where ω runs through the primitive n-th roots of 1. That is ω = e 2πik/n with (k, n) = 1 and 1 k < n. This has degree = φ(n) the Euler function Let ω be some primitive n-th root of 1. Then E := Q(ω) is the SF of Φ n (x) and also of x n 1. (E : Q) = φ(n) = G(E Q) and G(E Q) = Z n, it is abelian Cyclic extensions

5 A Galois extension is cyclic if the Galois group is cyclic Assume E F is finite separable, and F contains a primitive n-th root of 1. Then equivalent: (1) E F is a cyclic Galois extension of degree n. (2) E = F (α) and α is a root of an irreducible polynomial x n b F [x] Polynomials solvable by radicals DEF E F is a radical extension DEF char(f ) = 0, let f(x) F [x] solvable by radicals f(x) is solvable by radicals if and only if the Galois group G(E F ) is solvable. A group G is solvable if there is a sequence G = G 0 > G 1 >... > G r = {e} such that G i+1 G i and the factor group is abelian. S 3 and S 4 are solvable. For n 5, the groups A n and S n are not solvable. Properties: (1) If G is solvable and H is a subgroup then H is solvable. (2) if G is solvable and N G then G/N is solvable. (3) If N G and if N and G/N are solvable then G is solvable Let E be the splitting field of x n a F [x]. Then G = G(E F ) is solvable Let f(x) Q[x] be irreducible of degree p where p is prime, and E its SF. If f has precisely two non-real roots in C then G(E Q) = S p. With this one can see that for example x 5 6x + 3 is not solvable by radicals Ruler and Compass constructions DEF 32 [cf book ] Construct subsets P 0 P 1... R 2 inductively. P i+1 is the union of P i together with the points constructed as follows. (i) The intersection of lines p 1 q 1 and p 2 q 2 is in P i+1 for p 1, p 2, q 1, q 2 P i ; (ii) The intersections of circles with centres p 1, p 2 through q 1, q 2 is in P i+1 if p 1, p 2 and q 1, q 2 P i. (iii) The intersections of a line with a circle is in P i+1 where the line goes through two points in P i and the circle has centre in P i and goes through a point in P i. DEF 33 (a) A point X is constructible from P 0 if X P i for some i. (b) A line l is constructible from P 0 if it passes through two distinct points in some P i. c) A circle C is constructible from P 0 if its centre and one point on it lie in P i for

6 some i. Say that a point X, or a line L or a circle C is constructible if it is constructible from P 0 := Q Q. Say that a number u R is constructible if (u, 0) is constructible from P 0 = Q Q Let u R be constructible from Q. Then there is a subfield F of R containing u such that (F : Q) = 2 m for some m Let K be the subset of R consisting of numbers constructible. Then K is a subfield of R such that for all u 0 in K, also u K. Impossibility proofs (1) The angle π/3 cannot be trisected by ruler and compass. (2) One cannot square a circle by ruler and compass. (3) One cannot double a cube by ruler and compass. (4) A regular n-gon is constructible by ruler and compass if and only if φ(n) is a power of 2.

Finite Fields. [Parts from Chapter 16. Also applications of FTGT]

Finite Fields. [Parts from Chapter 16. Also applications of FTGT] Finite Fields [Parts from Chapter 16. Also applications of FTGT] Lemma [Ch 16, 4.6] Assume F is a finite field. Then the multiplicative group F := F \ {0} is cyclic. Proof Recall from basic group theory