Math 201C Homework. Edward Burkard. g 1 (u) v + f 2(u) g 2 (u) v2 + + f n(u) a 2,k u k v a 1,k u k v + k=0. k=0 d

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1 Math 201C Homework Edward Burkard 5.1. Field Extensions. 5. Fields and Galois Theory Exercise If v is algebraic over K(u) for some u F and v is transcendental over K, then u is algebraic over K(v). Let η K(u)[x] be a nonzero polynomial such that η(v) = 0. Then 0 = η(v) = f 0(u) g 0 (u) + f 1(u) g 1 (u) v + f 2(u) g 2 (u) v2 + + f n(u) g n (u) vn (1) where f i, g i K[x] and g i (u) 0 for all i. Now let s multiply (1) by where h i (x) = f i (x) n k i n g i (u) to get: i=0 0 = h 0 (u) + h 1 (u)v + h 2 (u)v h n (u)v n, (2) g k (u). Since K[x] is a ring, h i K[x] for all i. Let d = max 0 i n {deg h i}, then h i (x) = where a i,k = 0 if k > deg h i. Then substituting this into (2) we have: Let ψ(x) = 0 = = = a 0,k u k + n i=0 a 1,k u k v + a i,k u k v i ( n ) a i,k v i i=0 u k a 2,k u k v a n,k u k v n a i,k x k, ( n ) a i,k v i x k, then ψ K(v)[x] and ψ(u) = 0. In order to conclude that u is algebraic over K(v) i=0 we need to show that ψ is not a constant polynomial. Suppose that ψ is a constant polynomial, i.e. for all k > 0 we have n a i,k v i = 0. Recall that, since v is trancendental over K, the set {1, v, v 2,...} is linearly independent over K. This implies that a i,k = 0 for all k > 0 (otherwise there would be a finite linear combination of powers of v which is zero, hence v would be algebraic over K). Thus each h i above is of the form h i (x) = a i,0 (a constant polynomial), namely h i K. But then (2) implies that v is algebraic over K, a contradiction unless each a i,0 = 0. However, this is also a contradiction since a i,k = 0 for all i, k implies that η is the zero polynomial. i=0 Thus ψ K(v)[x] is a nonconstant polynomial such that ψ(u) = 0. Therefore u is algebraic over K(v). Exercise If u F is algebraic of odd degree over K, the so is u 2 and K(u) = K(u 2 ). Exercise If x n a K[x] is irreducible and u F is a root of x n a and m divides n, then prove that the degree of u m over K is n m. What is the irreducible polynomial for um over K? 1

2 2 Since n m we have that f(x) = x n m a K[x]. Then: f(u m ) = (u m ) n m a = u n a = 0 hence u m is a root of f. Now I clam that f is irreducible and hence is the irreducible of u m. If f were reducible, then we would have that f = f 1 f 2 and hence that f 1 (x m )f 2 (x m ) = f(x m ) = x n a is reducible, a contradiction to our assumptions. Thus f is irreducible and it follows that [K(u m ) : K] = deg h = n m by theorem 1.6. Exercise If F is algebraic over K and D is an integral domain such that K D F, then D is a field. Exercise (a) Consider the extension Q(u) of Q generated by a real root u of x 3 6x 2 + 9x + 3. (Why is this irreducible?) Express each of the following elements in terms of the basis {1, u, u 2 } : u 4, u 5, 3u 5 u 4 + 2, (u + 1) 1, (u 2 6u + 8) 1. (b) Do the same with respect to the basis {1, u, u 2, u 3, u 4 } of Q(u) where u is a real root of x 5 + 2x + 2 and the elements in questions are: (u 2 + 2)(u 3 + 3u), u 1, u 4 (u 4 + 3u 2 + 7u + 5), (u + 2)(u 2 + 3) 1. (a) Let p = 3. Since p 1 and p 6, 9, 3 and p 2 3, by Eisenstein s Criterion, f is irreducible in Q[x]. Let u be a root of the polynomial and notice that u 3 = 6u 2 9u 3. Thus: u 4 = 6u 3 9u 2 3u = 6(6u 2 9u 3) 9u 2 3u = 27u 2 57u 18 u 5 = 27u 3 57u 2 18u = 27(6u 2 9u 3) 57u 2 18u = 162u 2 243u 81 57u 2 18u = 105u 2 261u 81 (b) Exercise (a) If F = Q ( 2, 3 ), find [F : Q] and a basis of F over Q. (b) Do the same for F = Q(i, 3, ω), where i C, i 2 = 1, and ω is a complex (nonreal) cube root of 1. Exercise In the field K(x), let u = x3 x+1. Show that K(x) is a simple extension of the field K(u). What is [K(x) : K(u)]? Exercise In the field C, Q(i) and Q( 2) are isomorpic as vector spaces, but not as fields.

3 3 Exercise Find an irreducible polynomial f of degree 2 over the field Z 2. Adjoin a root u of f to Z 2 to obtain a field Z 2 (u) of order 4. Use the same method to construct a field of order The Fundamental Theorem. Exercise What is the Galois group of Q( 2, 3, 4) over Q? Exercise (a) If 0 d Q, the Q( d) is Galois over Q. (b) C is Galois over R. Exercise Let f g K(x) with f g K(x). ( ) [ (a) x is algebraic over K f g and K(x) : K (b) If E K is an intermediate field, then [K(x) : E] is finite. / K and f, g relatively prime in K[x] and consider the extension of K by ( )] f g = max{deg f, deg g}. (c) The assignment x f ϕ(x) g induces a homomorphism σ : K(x) K(x) such that ψ(x) ϕ( f g ) ψ( f g ). σ is a K automorphism if K(x) if and only if max{deg f, deg g} = 1. (d) Aut K K(x) consists of all those automorphisms induced (as in (c)) by the assignment where a, b, c, d K and ad bc 0. x ax + b cx + d Exercise Assume chark = 0 and let G be the subgroup of Aut K K(x) that is generated by the automorphism induced by x x + 1 K. Then G is an infinite cyclic group. Determine the fixed field E of G. What is [K(x) : E]? Exercise (a) If K is an infinite field, the K(x) is Galois over K. (b) If K is finite, the K(x) is not Galois over K. Exercise If K is an infinite field, then the only closed subgroups of Aut K K(x) are itself and its finite subgroups. Exercise In the extension of Q by Q(x), the intermediate field Q(x 2 ) is closed, but Q(x 3 ) is not. Exercise If D is an intermediate field of the extension such that D is Galois over K, F is Galois over D, and every σ Aut K E is extendible to F, then F is Galois over K.

4 4 Exercise In the extension of an infinite field K by K(x, y), the intermediate field K(x) is Galois over K, but not stable (relative to K(x, y) and K) Splitting Fields, Algebraic Closure, and Normality. Exercise No finite field K is algebraically closed. Exercise (Lagrange s Theorem on Natural Irrationalities). If L and M are intermediate fields such that L is a finite dimensional Galois extension of K, then LM is finite dimensional and Galois over M and Aut M LM = Aut L M L. Exercise Let E be an intermediate field. (a) If F is algebraic Galois over K, then F is algebraic Galois over E. (b) If F is Galois over E, E is Galois over K and F is a splitting field over E of a family of polynomials in K[x], then F is Galois over K. Exercise Let F be an algebraic closure of the field Q of rational numbers and let E F be a splitting field over Q of the set S = {x 2 + a a Q} so that E is algebraic and Galois over Q. (a) E = Q(X) where X = { p p = 1 and p is a prime integer }. (b) If σ Aut Q E, then σ 2 = 1 E. Therefore, the group Aut Q E is actually a vector space over Z 2. (c) Aut Q E is infinite and not denumerable. (d) If B is a basis of Aut Q E over Z 2, then B is infinite and not denumerable. (e) Aut Q E has an infinite nondenumerable number of subgroups of index 2. (f) The set of extension fields of Q contained in E of dimension 2 over Q is denumerable. (g) The set of closed subgroups of index 2 in Aut Q E is denumerable. (h) [E : Q] ℵ 0, whence [E : Q] < Aut Q E. Exercise If an intermediate field E is normal over K, then E is stable (relative to F and K). Exercise Let F be normal over K and E an intermediate field. Then E is normal over K if and only if E is stable. Furthermore Aut K F /E = Aut K E. Exercise Part (ii) or (ii) of the Fundamental Theorem is equivalent to: an intermediate field E is normal over K if and only if the corresponding subgroup E is normal in G = Aut K F in which case G/E = Aut K E. Exercise If F is normal over an intermediate field E and E is normal over K, then F need not be normal over K.

5 5 Exercise Let F be algebraic over K. F is normal over K if and only if for every K-monomorphism of fields σ : F N, where N is any normal extension of K containing F, σ(f ) = F so that σ is a K-automorphism of F. Exercise If F is algebraic over K and every element of F belongs to an intermediate field that is normal over K, then F is normal over K. Exercise If [F : K] = 2, then F is normal over K The Galois Group of a Polynomial. Exercise Suppose f K[x] splits in F as f = (x u 1 ) n1 (x u k ) n k (u i distinct; n i 1). Let v 0,..., v k be the coefficients of the polynomial g(x u 1 )(x u 2 ) (x u k ) and let E = K(v 0,..., v k ). Then (a) F is a splitting field of g over E. (b) F is Galois over E. (c) Aut E F = Aut K F. Exercise Let f be a separable cubic with Galois group S 3 and roots u 1, u 2, u 3 F. Then the distinct intermediate fields of the extension of K by F are F, K( ), K(u 1 ), K(u 2 ), K(u 3 ), K. The corresponding subgroups of the Galois group are 1, A 3, T 1, T 2, T 3, and S 3 where T i = {(1), (jk) j i k}. Exercise If chark 2 and f K[x] is a cubic whose discriminant is a square in K, the f is either irreducible or factors completely in K. Exercise Let f be an (irreducible) separable quartic over K and u a root of f. There is no field properly between k and K(u) if and only if the Galois group is either A 4 or S 4. Exercise Let x 4 + ax 2 + b K[x] (with chark 2) be irreducible with Galois group G. (a) If b is a square in K, then G = V. (b) If b is not a square in K and b(a 2 4b) is a square in K, then G = Z 4. (c) If neither b nor b(a 2 4b) is a square in K, then G = D 4. Exercise Determine the Galois groups of the following polynomials over the fields indicated: (a) x 4 5 over Q; over Q( 5); over Q( 5i). (b) (x 3 2)(x 2 3)(x 2 5)(x 2 7) over Q. (c) x 3 x 1 over Q; over Q( 23i). (d) x 3 10 over Q; over Q( 2). (e) x 4 + 3x 3 + 3x 2 over Q. (f) x 5 6x + 3 over Q. (g) x 3 2 over Q.

6 6 (h) (x 3 2)(x 2 5) over Q. (i) x 4 4x over Q. (j) x 4 + 2x 2 + x + 3 over Q. Exercise Determine all the subgroups of the Galois group and all of the intermediate fields of the splitting field (over Q) of the polynomial (x 3 2)(x 2 3) Q[x]. Exercise Let K be a subfield of the real numbers and f K[x] an irreducible quartic. If f has exactly two real roots, the Galois group of f is S 4 or D 4. Exercise Assume that f(x) K[x] has distinct roots u 1, u 2,..., u n in the splitting field F and let G = Aut K F < S n be the Galois group of f. Let y 1,..., y n be indeterminates and define g(x) = σ S n (x (u σ(1) y 1 + u σ(2) y u σ(n) y n )). (a) Show that g(x) = (x (u 1 y σ(1) + u 2 y σ(2) + + u n y σ(n) )). σ S n (b) Show that g(x) K[y 1,..., y n, x]. (c) Suppose g(x) factors as g 1 (x)g 2 (x) g n (x) with g i K(y 1,..., y n )[x] monic irreducible. If x u σ(i) y i is i a factor of g 1 (x), then show that g 1 (x) = τ G ( x i u τσ(i) y i ). (d) If K = Q, f Z[x] is monic, and p is a prime, let f Z p [x] be the polynomial obtained from f by reducing the coefficients of f( mod p). Assume f has distinct roots u 1,..., u n in some splitting field F over Z p. Show that g(x) = ( x ) u i y τ(i) F [x, y 1,..., y n ]. i τ S n (e) Show that x x 5 9x x 3 37x 2 29x 15 Q[x] has Galois group S 6. (f) the Galois group of x 5 x 1 Q[x] is S 5. Exercise Here is a method for constructing a polynomial f Q[x] with Galois group S n for a given n > 3. It depends on the fact that there exist irreducible polynomials of every degree in Z p [x] (p prime). First choose f 1, f 2, f 3 Z[x] such that (i) deg f 1 = n and f 1 Z 2 [x] is irreducible. (ii) deg f 2 = n and f 2 Z 3 [x] factors in Z 3 [x] as gh with g an irreducible of degree n 1 and h linear. (iii) deg f 3 = n and f 3 Z 5 [x] factors as gh or gh 1 h 2 with g an irreducible quadratic in Z 5 [x] and h or h 1 h 2 irreducible polynomials of odd degree in Z 5 [x]. (a) Let f = 15f f 2 + 6f 3. Then f f 1 ( mod 2), f f 2 ( mod 3), and f f 3 ( mod 5). (b) The Galois group G of f is transitive. (c) G contains a cycle of the type ζ = (i 1 i 2...i n 1 ) and element σλ where σ is a transposition and λ a product of cycles of odd order. therefore σ G, whence (i k i n ) G for some k (1 k n 1) by Exercise I.6.3 and transitivity. (d) G = S n.

7 5.5. Finite Fields. Exercise If K = p n, then every element of K as a unique p t h root in K. Clearly the unique p th root of 0 is 0. Now consider the multiplicative group (K, ) which has order p n 1. Thus for all u K, we have that u pn 1 = 1 K so that u pn = u and hence: ( ) p u pn 1 = u p n = u so that u pn 1 is a p th root of u. Thus every element in F has a p th root. Now let v, w be two p th roots of u. By lemma 5.5, we have that the map φ : u u p is injective, thus: so that w = v. Thus every element has a unique p th root. φ(w) = w p = u = v p = φ(v) 7 Exercise If the roots of a monic polynomial f K[x] (in some splitting field of f over K) are distinct and form a field, then chark = p and f = x pn x for some n 1. Let f K[x] be a monic polynomial, then deg f <. Let R = {u f(u) = 0} be the set of roots in some splitting field of f. If R is a field, then R = p n for some prime p and n N and since all the roots are distinct, deg f = p n. Thus chark = p. Since R is a finite field of order p n, by proposition 5.6, R is a splitting field of g(x) = x pn x. Since both f and g are monic and have the same roots, it must be that f = g. Exercise (a) Construct a field with 9 elements and give its addition and multiplication tables. (b) Do the same for a field of 25 elements. (a) Consider the following irreducible quadratic in Z 3 [x]: f(x) = x (f is irreducible since f(0) = 1, f(1) = 2, f(2) = = 2). Let u be a root of f. Then Z 3 (u) is a field containing 3 2 = 9 elements. (b) Consider the polynomial f(x) = x 2 2 in Z 5 [x]. Since: f(0) = 2 = 3, f(1) = 1 = 4, f(2) = 4 2 = 2, f(3) = 9 2 = 7 2, f(4) = 16 2 = 12 = 2, none of which are zero, f is irreducible and hence if u is a root of f, then Z 5 (u) is a field of 5 2 = 25 elements. Exercise If K = q and (n, q) = 1 and F is a splitting field of x n 1 K over K, then [F : K] is the least positive integer k such that n (q k 1). Let u be a root of x n 1, then u n = 1. Moreover, since F is a finite dimensional extension of K (because F is the splitting field of a single polynomial) let [F : K] = k <. Since K = q and [F : K] = k, it follows that F = K [F :K] = q k. Thus u also satisfies u qk 1 = 1, so that u n = u qk 1, hence n (q k 1). If there were an integer m smaller k such that n (q k 1), then that would imply that u qm 1 = 1 so that q k = q m, an obvious contradiction. Exercise If K = q and f K[x] is irreducible, then f divides x qn x if and only if deg f divides n. Exercise If K = p r and F = p n, then r n and Aut K F is cyclic with generator φ given by u u pr. Since F = p n we have that [F : Z p ] = n so that [F : K] <, and hence, by propositon 5.10, the Galois group Aut K F is cyclic. The map φ is a Z p -monomorphism by lemma 5.5. Moreover, since F = p n it follows that φ n = id F. If φ Exercise If n 3, then x 2n + x + 1 is reducible over Z 2.

8 8 Exercise Every element in a finite field may be written as the sum of two squares. Recall: For a finite group G and A, B G, if A + B > G, then G = AB. Let F be a finite field and suppose that charf = p and F = p n. Define φ : F F by x x 2. If p = 2 then φ is an isomorphism by lemma 5.5, so that every element in F is clearly a sum of two squares (one being 0 2 ). Now if p > 2, for all x, y F, if x 2 = y 2, then x = ±y. Thus it must be that im φ p Let m = p2 + 1 and choose 2 2 m distinct elements A = {x 2 1,..., x 2 m} F. Given any u F, each x 2 i u is distinct, so B = {x2 1 u,..., x 2 m u} is a set of m elements, thus A + B = p n + 1 > p n = F, thus by the above comment F = A + B. In particular, for 0 F : 0 = x 2 i + x 2 j u so that u = x 2 i + x 2 j. Since u was arbitrary, we are done. Exercise Let F be an algebraic closure of Z p (p prime). (a) F is algebraic Galois over Z p. (b) The map φ : F F given by u u p is a nonidentity Z p -automorphism of F. (c) The subgroup H = φ is a proper subgroup of Aut Zp F whose fixed field is Z p, which is also the fixed field of Aut Zp F by (a). Exercise If K is finite and F is an algebraic closure of K, then Aut K F Aut K F (except 1 F ) has infinte order. is abelian. Every element of 5.6. Separability. Exercise Exercise Exercise Exercise Exercise Exercise Exercise

9 9 Exercise Exercise Cyclic Extensions. Exercise Exercise Exercise Exercise Exercise Cyclotomic Extensions Radical Extensions.

AN INTRODUCTION TO GALOIS THEORY

AN INTRODUCTION TO GALOIS THEORY STEVEN DALE CUTKOSKY In these notes we consider the problem of constructing the roots of a polynomial. Suppose that F is a subfield of the complex numbers, and f(x) is