Algebra Prelim Notes

Algebra Prelim Notes Eric Staron Summer 2007 1 Groups Define C G (A) = {g G gag 1 = a for all a A} to be the centralizer of A in G. In particular, this is the subset of G which commuted with every element of A. The centralizer is a subgroup of G. Define Z(G) = {g G gx = xg for all x G} to be the center of G. Since Z(G) = C G (G), Z(G) is a subgroup of G. Define N G (A) = {g G gag 1 = A} to be the normalizer of A in G. N G (A) is a subgroup of G, and C G (A) N G (A). Define G s = {g G gs = s} to be the stabilizer of s in G, where G is a group acting on set S. G s G. The kernel of an action is defined a {g G gs = s for all s S}. Legrange s Theorem If G is a finite group and H is a subgroup of G, then H G. Furthermore, [G : H] = G / H. Cauchy s Theorem If G is finite and prime p divides the order of G, then G has an element of order p. If G = p, p prime, G is cyclic. G is a Simple group if the only normal subgroups are 1 and G. If G is an abelian group order n and m n, then G has a subgroup of order m. If K G, then KH G for all H G. Isomorphism Theorems for Groups 1. If φ : G H is a homomorphism of groups, then ker(φ) G and G/ ker(φ) = φ(g). 2. Let A and B be subgroups of G. If A N G (B), then AB is a subgroup of G, B G, A B A, and AB/B = A/A B. 3. Let H and K be normal subgroups of G with H K. Then K/H G/H and (G/H)/(K/H) = G/K. 1

1 GROUPS 2 Correspondence Theorem Let G be a group and N G. Then there is a bijection from the set of subgroups of A of G which contain N and subgroups of G/N. A group is solvable is there is a chain of subgroups 1 = G 0 G 1 G s = G such that G i+1 /G i is abelian. If N and G/N are solvable, then G is solvable. In S n, a transposition is a 2-cycle. The alternating group A n is the subset of S n consisting of only even permutations, ie an even number of transpositions. Cycles of odd length are even permutations. Group Actions, D&F ch 4 Recall our definition of stabilizer and kernel. An action is called faithful if its kernel is the identity. Let G act on A. Then the following is an equivalence relation: a b a = gb for some g G. For each a A, the number of elements in the equivalence class containing a is G : G a, the index of the stabilizer of a. The equivalence class {ga g G} is called the orbit of G containing A. An action is transitive if it only has one orbit, ie for all a,b there is a g s.t. a = gb. Cayley s Theorem Every group G, where G = n, is isomorphic to a subgroup of S n. If G = n, and H is a subgroup of G s.t. [G : H] = p, there p is the smallest prime dividing n, then H is normal in G. (proof is an old prelim problem) Consider G acting on A by conjugation, g a = gag 1. 2 elements a nad b are conjugate if there is a g G s.t. gag 1 = b. Orbits acting by conjugation are called conjugacy classes. The number of conjugates of a subset A is the index [G : G A ] = [G : N G (A)]. number of conjugates of an element s is [G : G s ] = [G : N G (s)] = [G : C G (s)]. The The Class Equation Let G be a group and g 1,..., g n be representatives of the distinct conjugacy classes if G not contained in Z(G). Then G = Z(G) + [G : C G (g i )]. If G is abelian, the class equation is meaningless, since everything is in the center. If p is prime and G = p α, then Z(P ) 1. If G = p 2, then G is abelian, and G = Z p Z p or G = Z + p 2. If G/Z(G) is cyclic, then G is abelian. Two elements on S n are conjugate iff they have the same cycle type. Recall: 1. A group of order p α is a p-group

1 GROUPS 3 2. If G = p α m, and p m, then a subgroup of order p α is a Sylow p-subgroup. 3. The number of Sylow p-subgroups will be denoted n p. Sylow s Theorem 1. There exists a sylow p-subgroup. 2. If P is a sylow p-subgroup and Q is a p-subgroup, there is a g G such that Q gp g 1. In particular, any two sylow p-subgroups are conjugates. 3. n p 1 mod (p). n p = [G, N G (P )], thus n p m. A n is simple for all n 5. Direct and Semidirect Products Fundamental Theorem of Finitely Generated Abelian Groups Let G be a finitely generated abelian group. Then 1. G = Z r Z n1 Z ns, where r 0 and n i+1 n i. 2. The expression described above is unique. In the above notation, r is the free rank or Betti number of G, and n 1,..., n s are the invariant factors of G. The above decomposition is called the invariant factor decomposition. This can be used to classify all finite abelian groups. The largest invariant factor is atleast the product of each distinct prime. If n is the product of distinct primes, then up to isomorphism the only abelian group of order n is the cyclic group of order n, Z n. Let G be a group, x, y G, and A, B G be nonempty. Define the commutator of x and y as [x, y] = xyx 1 y 1. Define G = [x, y] x, y G, the group generated by commutators, to be the commutator subgroup of G. Let G be a group, x, y G, H G. Then 1. H G iff [H, G] G. 2. G/G is abelian. 3. G/G is the largest abelian quotient of G, ie of H G and G/H is abelian, then G H. Conversely, if G H, then H G and G/H is abelian.

2 RING THEORY (CH 7-9 D&F) 4 If H and K are subgroups of G, then the number of ways of writing each element of the set HK in the form hk is H K. If H and K are normal in G, and H K = 1, then HK = H K. HK is the internal direct product of H and K, H K is the external direct product of H and K. Semidirect Product Now we assume H G, H G, and H K = 1. Note HK is a subgroup, since H is normal. Also, since the intersection is trivial, each element of HK has a unique representation hk. So we define multiplication (h 1 k 1 )(h 2 k 2 ) = (h 1 k 1 h 2 k1 1 )(k 1 k 2 ) HK. The group described above is the semidirect product and is denoted H K. p-groups Suppose G = p a. 1. Z(G) 1. 2. If H is a nontrivial subgroup of G then K intersects the center nontrivially. In particular, every normal subgroup of order p is contained in the center. 3. If H is a normal subgroup of G then H contains a subgroup of order p b that is normal in G. In particular, G has a normal subgroup of order p b for every b {0, 1,..., a}. 4. Every proper subgroup H of G is a proper subgroup of its normalizer in G, N G (H). 5. Every maximal subgroup of G is of index p and is normal in G. skipping most ch 6 stuff now for the moment 2 Ring Theory (ch 7-9 D&F) (R, +, ) such that (R, +) is an abelian group, is associative, and distributive laws hold. R is a division ring if every element of R is a unit, ie has a multiplicative inverse. The set of units is denoted R. A commutative division ring is a field. a R is a zero divisor if b s.t. ab = 0. R is an integral domain if it has no zero divisors. A finite domain is a field. If I is an integral domain, the units of R[x] are the units of R and R[x] is an integral domain. A ring homomorphism is such that the expected addition and multiplicative rules are satisfied. If φ : R S is a ring homomorphism, then Im(φ) is a subring of S, and ker(φ) is

2 RING THEORY (CH 7-9 D&F) 5 an IDEAL of R. (much like groups and normal subgroups.) An ideal is a subring such that I is a subring and ri I, Ir I for all r R. Isomorphism Theorems for Rings 1. If φ : R S is a ring homomorphism, then the kernel of φ is an ideal of R, I(φ) is a subring of S, and R/ ker(φ) = Im(φ) AND for any ideal I, R R/I is a surjective ring homomorphism with kernel I. 2. A a subring, B an ideal, then A + B is a subring of R, and A B is an ideal of A, and (A + B)/B = A/(A B). 3. I,J ideals s.t. I J, then J/I is an ideal of R/I and (R/I)/(J/I) = R/J. 4. (Lattice Iso Theorem) I an ideal of R. A A/I sets up a correspondence between subrings of R containing I and subrings of R/I. Also, A ideal of R iff A/I ideal of R/I. (A) is the smallest ideal containing A. If R is commutative, then (A) = RA = AR = { a i r a, a i A, r i R}. An ideal generated by one element, (a), is called principal. NOTE, (2, x) Z[x] is not principal. I = R iff I contains a unit. R is a field iff the only ideals are 0 and R. All non-zero field homomorphisms are injective. (consider kernel) An ideal M is maximal if M R and the only ideals containing M are M and R. Every proper ideal is contained on a maximal ideal. M is maximal iff R/M is a field. To show whether or not an ideal is maximal (or principle), consider R/I. P is called a prime ideal if P R and if ab P a P or b P. P is a prime ideal iff R/P is an integral domain (no zero divisors). If R is commutative, every maximal ideal is a prime ideal. R is a field iff 0 is a maximal ideal. Chinese Remainder Theorem Let A 1,..., A k be ideals in R. The map R R/A 1 R/A k is a ring homomorphism with ker = A 1 A k. The map is surjective, and A 1 A k = A 1... A k. Euclidean Domains A function N : R Z 0 with N(0) = 0 is a norm. In N(a) > 0 if a 0, N is a positive norm. R is a

2 RING THEORY (CH 7-9 D&F) 6 textbfintegral domain is there is a norm N on R such that for any two elements a and b of R, b 0, there is q and r s.t. a = bq + r such that N(r) < N(b)orr = 0. q is the quotient, r is the remainder. Examples of Euclidean Domains All fields (N(r) = 0, q = ab 1 ), Z(N(r) = r ), F a field, F[x] is an ED (the norm is the degree of the poly). Every ideal of a ED is principal, ie all EDs and PIDs. Z[x] has non-principal ideals [(2, x)], so Z[x] is not a PID, so it is not a ED. If R is a ED and a, b R, then if gcd(a, b) = d, then (d) is the ideal generated by a and b. In particular, there exists x, y R such that d = xa + yb. PIDs Z[ 5] not a PID [(3, 1 + 5)]. Z[(1 + 19/2)] is a PID, but not an ED. Let R be a PID, a, b R, and (a, b) = (d). Then d = gdc(a, b), and d = ax + by for some x, y R. Also, d is unique up to unit. In a PID, all prime ideals are maximal (recall that maximal ideals are always prime). If R[x] is a PID, then R is a field. UFDs Every PID is a UFD. If R is an integral domain (no zero divisors), then r R (non-unit) is said to be irreducible in R if whenever r = ab with a, b R, then either a or b is a unit. p is said to be prime if the ideal (p) is a prime ideal, ie p is not a unit and p ab p a or p b. 2 elements are associates if the differ by a unit. In an integral domain prime elements are always irreducible. Conversely, consider 3 in Z[ 5]. 3 (2 + 5)(2 5), but neither is in the ideal (3). So in general irreducible does not imply prime. However, in a PID irreducible and prime are the same (ie Z and F [x], F a field.). If we can find an element in a R such that a is irreducible but NOT prime, then we

2 RING THEORY (CH 7-9 D&F) 7 have shown R is not a PID, or ED. A unique factorization domain, UFD, is an integral domain where every non-zero non-unit element r R has the following properties: 1. r can be written as the product of irreducibles p i of R, 2. the decomposition above is unique up to unit. If R is a UFD, then R[x] is a UFD. Z[2i] is not a UFD (4 = 2 2 = 2i 2i). Also, Z[ 5, 6 = 2 3 = (1 + 5)(1 5). In a UFD, prime iff irreducible. In summary, Fields ED PID UFD Integral Domains Z, F [x] Z[(1 + 19)/2] Z[x] Z[ 5] Polynomial Rings If R is an ID, then the units of R[x] are the units of R, and R[x] is an ID. If I is an ideal of R, let (I) = I[x]. Then R[x]/I = (R/I)[x]. R[x 1,..., x n 1, x n ] = R[x 1,..., x n 1 ][x n ]. Recall that if F is a field, F [x] is a ED. Q[x, y] is NOT an ED or PID, but IS a UFD. Gauss Lemma If R is a UFD with field of fractions F, and p(x) R[x]. If p(x) irreducible in F [x], then irreducible in R[x]. corollary If the gcd of the coefficients of p(x) is 1, then p(x) irreducible in R[x] implies it is irreducible in F [x]. R is a UFD iff R[x] is a UFD. In particular, if R is a UFD, then R[x 1,..., x n ] is a UFD. Irreducibility If F is a field, then (x-a) is a factor of p(x) iff p(a) = 0. In p(x) is degree 2 or 3, p(x) is reducible iff p(x) has a root. If p(x) is a monic polynomial with integer coefficients and p(d) 0 for all integers d dividing the constant term, then p(x n) has no roots in Q. In general, if r/s is a root, then s a n and r a 0. Ex x 3 3x 1 is irreducible in Z[x]. x 3 + x + 1 irreducible in Z/2Z. Need Gauss and previous statements.

3 MODULES (CH 10,12 D&F) 8 For degree 4 or more, need new stuff. If p(x) cannot be factored in (R/I)[x], then p(x) cannot be factored in R[x]. In particular, we can mod out by some number. Eisenstein s Criterion P a prime ideal, R an integral domain, f(x) = a n x n + + a 0. If there is prime p s.t. p a i, i = 0, 1,..., n 1, p 2 a 0, then f(x) irreducible in R[x]. We should say a i P, but a 0 / P 2, but when in Z[x] the 1st statement is equivalent. EXAMPLE: Note that f(x) = x 4 + 1 is not a case for E. criteria, but g(x) = f(x + 1) = x 4 + 4x 3 + 6x 2 + 4x 2 is a case, and it follows f(x) is irreducible. Maximal ideals of F [x] are generated by irreducible polynomials. Of note, F [x]/(f(x)) is a field iff f(x) is irreducible. If g(x) = f a 1 1... f an n, then F [x]/(g(x)) = F [x]/(f a 1 1 ) F [x]/(f an n ). A commutative ring is Noetherian if every ideal of R is finitely generated. Hilbert Basis Theorem If R is Noetherian, so it R[x]. Every ideal of F [x 1,..., x n ] is finitely generated. 3 Modules (ch 10,12 D&F) A module over R is a set M with the rules: 1. (M,+) is an abelian group 2. the action of R on M (ie R M M), denoted rm, is distributive as expected. Given a ring R, R is a module of R. Given a field F, F n is a submodule. Likewise, R n is a submodule. If S is a subring of R, and M an R-module, then M is an S-module as well. If M is annihilated by I, ie am = 0 for all m M, a I, then M is an R/I-module. A is a Z-module iff A is an abelian group. An R-algebra is a ring A together with a ring homomorphism f : R A s.t. f(r) is contained in the center of A. If A is an R-algebra, then ra = ar = f(r)a. A R-algebra homomorphism is a ring homomorphism φ : A B s.t. φ(ra) = rφ(a) Let R be a ring and M,N be R-module homomorphisms. φ : M N is an R-module homomorphism if φ(x + y) = φ(x) + φ(y) and φ(rx) = rφ(x). Hom R (M, N) is the set of

3 MODULES (CH 10,12 D&F) 9 all R-module homomorphisms from M into N. Ring homomorphisms are not always module isomorphisms, likewise is true for the converse. x 2x is module, not ring. x x 2 is a ring, not a module. If R is a field, R-mod homomorphisms are linear transformations. φ is an R-mod homomorphism iff φ(rx + y) = rφ(x) + φ(y). Hom R (M, N) is an abelian group under addition. If R is commutative, then Hom R (M, N) is an R-module. End(M) = Hom R (M, M). Let R be a ring, M an R-module, N an R-submodule of M. Then M/N makes sense as an R-module, where π : M M/N, π(x) = x + N is R-module homomorphism with kernel N. Sum of submodules is submodule. The submodule of M generated by A is denoted RA, where RA = {r 1 a 1 + +r m a m r i R, a i A}. A free module is a module with a free basis. E = {e 1,..., e n } is a free basis if E is a generating set and if r 1 e 1 + + r n e n = 0, then r 1 = r 2 =... = r n = 0. Universal Property If M is an R-module and φ : A M is any map of sets, then there is a unique R-module homomorphism Φ : F (A) M s.t. Φ(a) = φ(a). Tensor Products, Exactness, Projective, Injective, Flat Modules Modules over a PID A left R-module M id said to be Noetherian R-Module or satisfies the ascending chain condition if M 1 M 2... is an increasing chain of submodules of M, then there is an m s.t. M i = M m for all i m. A ring is a Noetherian Ring if it is Noetherian as a noetherian module over itself. TFAE 1. M is Noetherian R-Module. 2. Every nonempty set of submodules of M contains a maximal element under inclusion. 3. Every submodule of M is finitely generated. If R is a PID, then every nonempty set of ideals of R has a maximal element and R is noetherian (follows from above). T or(m) = {x M rx = 0 for some nonzero r R}. Ann(M) = {r R rx = 0 for all m M}

3 MODULES (CH 10,12 D&F) 10 Let R be a PID, M a free R-Module of finite rank n, N a submodule of M. Then N is free of rank m, m n. there is basis y 1,..., y n of M s.t. a 1 y 1,..., a m y m is a basis of N where a 1 is an elt of R and a 1 a 2... a m. PID MODULE THRM Let R be a PID, M a f.g. R-Module. 1. M is isomorphic to the direct sum of finitely many cyclic submodules, ie M = R r R/(a 1 ) R/(a m ) where each a i is not a unit and a i a i+1. 2. M is torsion free iff M is free 3. T or(m) = R/(a 1 ) R/(a m ). In the notation above, r is called the free rank or Betti number of M, and each a i is can invariant factor of M. Generalized thrm R a PID, M a f.g. R-module. Then M = R r R/(p α 1 1 ) R/(p αt t ), where p α 1 i are powers of (not necessarily distinct) primes. The prime powers are called elementary divisors. Let R be a PID, p prime in R, the field F = R/(p). Then 1. If M = R r, then M/pM = F r. 2. If M = R/(a), a R, a 0, then M/pM = F if p a, 0 otherwise. 3. If M = R/(a 1 ) R/(a m ), each a i divides p, then M/pM = F k. Thus, if R is a PID, two finitely generated R-Modules are isomorphic iff they have the same free rank and invariant factors iff they have teh same free rank and same list of elementary divisors. If R is a PID, and M is a finitely generated R-module, then the elementary divisors of M are the prime power factors of the invariant factors of M. Rational Canonical Form Recall, if F is a field, F [x] is a PID. Thus there is lots we can say about modules over F [x]. If A is an n n matrix, the characteristic polynomial of A is c T (x) = det(xi A). The minimum polynomial on A is the smallest degree polynomial which is zero when evaluated at A. Important Stuff

4 FIELD AND GALIOS THEORY 11 The minimal polynomial is the largest invariant factor. All invariant factors divide the min poly. The product of the invariant factors is the characteristic polynomial. (Cayley-Hamilton Theorem) the minimal polynomial divides teh characteristic polynomial. A characteristic polynomial divides a power of the minimal polynomial...in particular the min poly and char poly have the same roots. If x k + b k 1 x k 1 + + b 0 is a monic polynomial, then the companion matrix is: 0 0......... b 0 1 0......... b 1 0 1......... b 2 0 0.......... 0 0...... 1 b k 1 A matrix is said to be in rational canonical form if it is the direct sum of companion matrices for monic polys a 1 (x),..., a m (x) of degrees at least 1 and a 1 (x)... a m (x). The polys are the invariant factors. This is a block matrix. A matrix is similar to its Rational Canonical Form. It follows that 2 matrices are similar (ie A and B are similar is there is a P s.t. A 1 AP = B) iff they have the same Rational Canonical Form. The Smith Normal Form of A is the matrix obtained by taking xi A and using row and column operations so that 1 s and invariant factors lie of the diagonal. Jordan Canonical Form A k k matrix with λ along the diagonal and 1 s on the subdiagonal is the k k elementary Jordan Matrix with eigenvalue λ. A matrix is in Jordan Canonical Form if it is the block diagonal matrix with jordan blocks along the diagonal. A matrix A is similar to a matrix in Jordan Canonical Form. Computation?? 4 Field and Galios Theory The characteristic of a field, ch(f), is the smallest positive integer p such that p 1 F = 0 if p exists, is zero otherwise. The characteristic is either prime or zero. The prime subfield

4 FIELD AND GALIOS THEORY 12 of F is the subfield generated by the multiplicative identity 1. It is isomorphic to Q or F p. If K is a field containing the subfield F, then K is said to be an extension field if F, denoted K/F, K over F. Every field F is an extension of its prime subfield. F is sometimes called the base field. The degree of a field extension K/F, denoted [K : F ], is teh dimension of K as a vector space over F. The field F/(p(x)) is a field containing an isomorphic copy of F and having a root of p(x). ex, R/(x 2 + 1) = C. Let K be an extension of F, and let α, β, K be a collection of elements in K/ Then the smallest subfield of K containing both F and the elements above is the field generated by α, β,.... If K = F (α), then K is said to be a simple extension over F, and α is the primitive element. (eventually we show every finite extension of a field of characteristic zero is simple). Let F be a field and p(x) F [x] be an irreducible polynomial. If K is an extension of F containing an element α s.t. p(α) = 0, then F (α) = F [x]/(p(x)). If p(x) is of degree n, then F (α) = {a 0 + a 1 α + + a + n 1α n 1 a i F } K. An element α K is said to be algebraic over F if α is the root of some nonzero poly in F [x]. Otherwise, α is transcendental over F. K/F is algebraic if every element of K is algebraic over F. Let α be algebraic over F. Then there is a unique monic irreducible poly m α (x) F [x] which has α as a root. A poly f(x) F [x] has α as a root iff m(x) f(x). The polynomial just described is called the minimal polynomial for α. If α algebraic over F, then F (α) = F [x]/m α (x). [F (α) : F ] = deg(α). α is algebraic over F iff F (α)/f is finite. If F K L, then [L : F ] = [L : K][K : F ]. F (x, y) = (F (x))(y). If a and b are algebraic over F, then so are a ± b, ab, a/b(b 0). The composite field of K 1 and K 2, denoted K 1 K 2 is the smallest subfield of K containing both K 1 and K 2. [K 1 K 2 : F ] [K 1 : F ][K 2 : F ] with equality iff the F-basis of K 1 and K 2 are linearly independent. If [K 1 : F ] = m, [K 2, F ] = n, where m and n are relatively prime, then equality above holds. The extension K of F is called the splitting field if f(x) F [x] if f(x) factors completely into linear factors (splits completely) in K[x] and f(x) does no split completely in a proper subfield of K containing F.

4 FIELD AND GALIOS THEORY 13 Degree n polynomial f(x) has precisely n roots in F iff f splits completely in F [x]. Given f(x), there is a K s.t. K is the splitting field of f(x). If K is an algebraic extension of F and K is the splitting field of f(x), then K is a normal extension of F. Example: S.F. of x 2 2 in Q is Q( 2). Since the roots of x 3 2 are 3 2, and 3 ( 1±i 2 ) 3. 2 Thus the splitting field is Q( 3, 3 2). If f(x) is degree n, the splitting field of f(x) has degree at most n! over F. A generator of the cyclic group of all n th roots of unity is called a primative n th root of unity. The splitting field of x n 1 over Q is the field Q(ζ n ) and is called the cyclotomic field of n th roots of unity. For prime p, [Q(ζ p ) : Q] = p 1. In general, [Q(ζ p ) : Q] = φ(n). The splitting field of x p 2 is Q( p 2, ζ p ), which has degree p(p 1). Any two splitting fields for a polynomial f(x) F [x] over a field F are isomorphic. F is the algebraic closure of F if F is algebraic over F and every f(x) F [x] splits completely over F. K is said to be algebraically closed if every polynomial with coefficients in K has a root in K. The algebraic closure is always algebraically closed. C is algebraically closed. Over a splitting field for f(x), we have f(x) = (x α 1 ) n 1... (x α k ) n k, where each αi is distinct. In n i > 1, α i is a multiple root. If n i = 1, then α i is a simple root. A polynomial is called separable if it has no multiple roots. Otherwise, it is inseparable. x 2 2 is separable over Q. (x 2 2) n, n 2 is inseparable. x 2 t F 2 (t) is irreducible, but t is the only root, with multiplicity 2, so is not irreducible. A polynomial f(x) has a multiple root α iff α is a root of the derivative of f(x). x pn x F p [x] is separable. Every irreducible polynomial over a field of characteristic 0 is separable. In such a field, a polynomial is separable iff it is the product of distinct irreducible polynomials. In a field of characteristic p, (a + b) p = a p + b p. The map described is the Frobenius endomorphism of F. Every irreducible polynomial over a finite field is separable. A polynomial in F[x] is separable iff it is the product of distince irreducible polynomials in F [x]. A filed K is separable over F if every element of K is the root of a separable polynomial over F (ie the minimal polynomial over F of every element of K is separable.)

4 FIELD AND GALIOS THEORY 14 Every finite extension of Q or a finite field is separable. Galois Theory Let K be a field. The prime field of K is generated by 1 K, and for any automorphism σ, σα = α for all α in the prime field. Any automorphism of a field fixes its prime subfield. Aut(K/F ) is the collection of automorphisms of K which fix F. Aut(K) is a group under composition, and Aut(K/F ) is a subgroup. Of note, Aut(K/F ) permutes the roots of irreducible polynomials (if α K is algebraic over F, then σα is a root of the min poly of α. Aut(K/F ) is completely determined by what it does to its generators. If H is a subgroup of the group of automorphisms of K, the subgroup of K fixed by H is called the fixed field of H. If F 1 F 2 K, then Aut(K/F 2 ) Aut(K/F 1 ). H 1 H 2 Aut(K) associated with F 1, F 2, then F 2 F 1. Let E be a splitting field over F of the polynomial f(x) F [x]. Then Aut(E/F ) [E : F ], with equality if f(x) is separable over F. Let K/F be a finite extension. Then K is said to be Galois over F and K/F is a Galois Extension if Aut(K/F ) = [K : F ]. If K is Galois, then Aut(K/F ) is called the Galois Group of K/F, and is denoted Gal(K/F ). If K is the splitting field over F of a separable polynomial f(x), then K/F is Galois. (The converse is also true). We say if f(x) is separable over F, then the Galois Group of f(x) over F is the splitting field of f(x) over F. Any quadratic extension K of a field F (F not of char=2) is Galois. The group of automorphisms of a field extension F(x) is completely determined by where it sends x (x must be sent into F(x)(ie, Q[ 3 2])). Fundamental Theorem of Galois Theory Let G = {σ 1 = 1, σ 2,..., σ n } be a subgroup of automorphisms of a field K and let F be the fixed field. Then [K : F ] = n = G.

4 FIELD AND GALIOS THEORY 15 Let K/F be any finite extension. Then Aut(K/F ) [K : F ] with equality iff F is the fixed field of K. Let G be a finite set of automorphisms of a field K and lat F be the fixed field. Then every automorphism of K fixing F is contained in G, ie Aut(K/F ) = G, so K/F is Galois, with Galois group G. Thrm The extension K/F is Galois iff K is the splitting field of a separable polynomial over F. Let K/F be a Galois Extension. If α K, the elements σα for σ Gal(K/F ) are conjugates of α over F. If E is a subfield of K containing F, the field σ(e) is called the conjugate field of E over F. Now we have four characterizations of Galois Extensions K/F. 1. Splitting fields of separable polynomials over F 2. Fields where F is precisely the set of elements fixed by Aut(K/F). (In general, the set of fixed elements may be larger than F) 3. Fields with [K : F ] = Aut(K/F ) 4. Finite, Normal, and Separable extensions. The Fundamental Theorem of Galois Theory Let K/F be a Galois extension and G = Gal(K/F ). Then there is a bijection: {Subfields E of K containing F} {Subgroups H of G} given by E {the elements of G fixing E} and {the fixed field of H} H. Here is the list of consequences: 1. If E 1, E 2 correspond to H 1, H 2, then E 1 E 2 iff H 2 H 1. (know diagram) 2. [K : E] = H and [E : F ] = G : H, the index of H in G. 3. K/E is always Galois, and Gal(K/E) = H. 4. E is Galois over F iff H is normal in G. Gal(E/F ) = G/H. 5. E 1 E 2 corresponds to H 1, H 2, and E 1 E 2 corresponds to H 1 H 2. F p n is Galois over F. All finite fields are isomorphic to F p n. F p n is the splitting field over F p of the polynomial x pn x, with cyclic Galois Group order n. The subfields of F p n are all Galois over F and in 1-1 correspondence with the divisors d on n. F p n is a simple extension. x pn x is the product of all the distinct irreducible polynomials in F p [x] of degree d where d runs thru the divisors of n. Stop at p. 591