Math 201C Homework Edward Burkard 5.1. Field Extensions. 5. Fields and Galois Theory Exercise 5.1.7. 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 2 + + 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 2 + + 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 5.1.8. If u F is algebraic of odd degree over K, the so is u 2 and K(u) = K(u 2 ). Exercise 5.1.9. 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 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 5.1.10. If F is algebraic over K and D is an integral domain such that K D F, then D is a field. Exercise 5.1.13. (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 5.1.14. (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 5.1.15. 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 5.1.16. In the field C, Q(i) and Q( 2) are isomorpic as vector spaces, but not as fields.
3 Exercise 5.1.17. 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 8. 5.2. The Fundamental Theorem. Exercise 5.2.4. What is the Galois group of Q( 2, 3, 4) over Q? Exercise 5.2.5. (a) If 0 d Q, the Q( d) is Galois over Q. (b) C is Galois over R. Exercise 5.2.6. 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 5.2.8. 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 5.2.9. (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 5.2.10. If K is an infinite field, then the only closed subgroups of Aut K K(x) are itself and its finite subgroups. Exercise 5.2.11. In the extension of Q by Q(x), the intermediate field Q(x 2 ) is closed, but Q(x 3 ) is not. Exercise 5.2.12. 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 Exercise 5.2.13. 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). 5.3. Splitting Fields, Algebraic Closure, and Normality. Exercise 5.3.8. No finite field K is algebraically closed. Exercise 5.3.14 (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 5.3.15. 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 5.3.16. 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 5.3.17. If an intermediate field E is normal over K, then E is stable (relative to F and K). Exercise 5.3.18. 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 5.3.19. 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 5.3.20. If F is normal over an intermediate field E and E is normal over K, then F need not be normal over K.
5 Exercise 5.3.21. 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 5.3.22. 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 5.3.23. If [F : K] = 2, then F is normal over K. 5.4. The Galois Group of a Polynomial. Exercise 5.4.1. 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 5.4.3. 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 5.4.5. 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 5.4.8. 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 5.4.9. 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 5.4.10. 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 (h) (x 3 2)(x 2 5) over Q. (i) x 4 4x 2 + 5 over Q. (j) x 4 + 2x 2 + x + 3 over Q. Exercise 5.4.11. 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 5.4.12. 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 5.4.13. 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 2 + + 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 6 + 22x 5 9x 4 + 12x 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 5.4.14. 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 1 + 10f 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.
5.5. Finite Fields. Exercise 5.5.3. 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 5.5.4. 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 5.5.5. (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 2 + 1 (f is irreducible since f(0) = 1, f(1) = 2, f(2) = 4 + 1 = 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 5.5.6. 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 5.5.7. If K = q and f K[x] is irreducible, then f divides x qn x if and only if deg f divides n. Exercise 5.5.8. 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 5.5.9. If n 3, then x 2n + x + 1 is reducible over Z 2.
8 Exercise 5.5.10. 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 p2 + 1. 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 5.5.11. 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 5.5.12. 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 5.6.1. Exercise 5.6.2. Exercise 5.6.3. Exercise 5.6.6. Exercise 5.6.7. Exercise 5.6.8. Exercise 5.6.9.
9 Exercise 5.6.10. Exercise 5.6.11. 5.7. Cyclic Extensions. Exercise 5.7.6. Exercise 5.7.7. Exercise 5.7.8. Exercise 5.7.9. Exercise 5.7.10. 5.8. Cyclotomic Extensions. 5.9. Radical Extensions.