The third exam will be on Monday, April 9, 013. The syllabus for Exam III is sections 1 3 of Chapter 10. Some of the main examples and facts from this material are listed below. If F is an extension field of a field K, then the Galois group of F over K, denoted gal(f : K) is the group, under composition, of field automorphism of F which fix K, i.e. σ gal(f : K) if σ : F F is a field isomorphism and σ(a) = a for all a K. The Galois group of a polynomial f(x) K[x] is the Galois group of E over K where E is a splitting field for f(x) over K. If F is an extension field of K, f(x) K[x], u F, and σ gal(f : K), then σ(f(u)) = f(σ(u)). In particular, if u is a root of f(x), then σ(u) is also a root. As a consequence of the previous item, if F is the splitting field of f(x) K[x], then the Galois group of F over K is isomorphic to a subgroup of the group of permutations of the roots of f(x). In particular, if deg f(x) = n then gal(f : K) is isomorphic to a subgroup of the symmetric group S n. (Theorem 3, Page 417.) Let K be a field, let f(x) be a polynomial of positive degree in K[x], and let F be a splitting field for f(x) over K. If no irreducible factor of f(x) has repeated roots in F, then (Corollary, Page 419.) gal(f : K) = [F : K]. If F is a finite field with char(f ) = p, the map φ : F F defined by φ(x) = x p, for all x F, is an automorphism of F called the Frobenius automorphism of F. Let K be a finite field with K = p r, where p = char(k). Let F be an extension field of K with [F : K] = m, and let φ be the Frobenius automorphism of F. The gal(f : K) is a cyclic group of order m, generated by φ r. A polynomial f(x) K[x] has no multiple roots if and only if gcd(f(x), f (x)) = 1. A polynomial f(x) K[x] is separable if its irreducible factors have only simple roots. An algebraic extension F of K is called separable if the minimal polynomial of each element of F is separable. If F is a finite separable extension of K, then F is a simple extension, i.e. F = K(u) for some u F. If K is an extension field of a field F, know the Galois correspondence between fields between F and K and subgroups of Gal F (K). If E is a field between K and F (i.e., F E K) and if H gal(k : F ), then the correspondence is given by: E gal(k : E), H K H = {a K : σ(a) = a for all σ H}. K is said to be an Galois extension of F (or K is Galois over F ) is K is finite-dimensional over F and K gal(k:f ) = F. That is, the only elements of K fixed by all σ gal(k : F ) are the elements of F. 1
An extension K of F is a Galois extension if and only if K is the splitting field of a separable polynomial over F. If the characteristic of F is 0, then K is a Galois extension of F if and only if K is a splitting field of a polynomial f(x) F [x]. Know the Fundamental Theorem of Galois Theory, which states briefly that if F is a splitting field of a separable polynomial over the field K, then the Galois correspondence is bijective and preserves normality (of subgroups and subfields). More precisely: Fundamental Theorem of Galois Theory Let F be the splitting field of a separable polynomial over the field K, and let G = gal(f : K). 1. For an intermediate field K E F, the Galois group gal(f : E) is a subgroup of gal(f : K) and it has fixed field F gal(f :E) = E. Furthermore, gal(f : E) = [F : E] and [E : K] = [gal(f : K) : gal(f : E)].. For a subgroup H gal(f : K), its fixed field K F H F has Galois group gal(f : F H ) = H. Furthermore, [F : F H ] = H and [F H : K] = [gal(f : K) : H]. 3. An intermediate field K E F is a splitting field of a separable polynomial over K if and only if the corresponding group gal(e : K) is a normal subgroup of gal(f : K), and in this case gal(e : K) = gal(f : K)/ gal(f : E). If E n denotes the splitting field of x n 1 over a field K of characteristic 0, then gal(e n : K) is isomorphic to a subgroup of Z n and hence is an abelian group. (Theorem 1, Page 438.) Know what are roots of unity and primitive roots of unity. If K is a field of characteristic 0 that contains a primitive n th root of unity, and if u is a root of x n a K[x] in some extension field of K, then F = K(u) is a splitting field of x n 1 over K and gal(f : K) is abelian. (Lemma 4, Page 438.) Know what it means for an equation f(x) = 0 to be solvable by radicals. Know what it means for a group to be solvable: G is solvable if there is a chain of subgroups G = G 0 G 1 G G n 1 G n = {e} such that (1) G 1 is a normal subgroup of G i 1 for 1 i n, and () G i 1 /G i is abelian for all i. S n is not solvable if n 5. Know Galois s criterion for solvability by radicals: f(x) = 0 is solvable by radicals if and only if the Galois group of f(x) is a solvable group.
Know how to use Galois criterion to show that there are rational polynomials of degree 5 that are not solvable by radicals. The n th cyclotomic polynomial is the polynomial Φ n (x) = (k, n)=1, 1 k<n (x α k ) where α = e πi/n is a complex primitive n th root of unity. Thus, Φ n (x) is the complex polynomial which has all primitive n th roots of unity as roots. Properties of Φ n (x) are: 1. deg(φ n (x)) = φ(n);. x n 1 = d n Φ d(x); 3. Φ n (x) is monic, with integer coefficients; 4. Φ n (x) is irreducible over Q for every positive integer n. Know how to use the above properties to compute Φ n (x). For every positive integer n, the Galois group of the n th cyclotomic polynomial Φ n (x) over Q is isomorphic to Z n, the group of units of the ring Z n. Review Exercises Here are some sample exercises such as might be given on the exam. 1. If E = Q(e πi/8 ), compute gal(e : Q). Solution. Since e πi/8 is a primitive 8-th root of unity it follows that E is the splitting field of x 8 1 and gal(e : Q) = Z 8 = Z Z.. If E = Q(e πi/6 ), compute gal(e : Q). Solution. Since e πi/6 is a primitive 6-th root of unity it follows that E is the splitting field of x 6 1 and gal(e : Q) = Z 6 = Z. 3. If E = Q(i, 7), compute gal(e : Q). Solution. This is the splitting field of (x +1)(x 7) and gal(e : Q) = Z Z. A similar example was done in class. 4. If E = Q( 4 ), show that gal(e : Q) = Z. Solution. Since the minimal polynomial of 4 is x 4 it follows that any automorphism σ of E must take 4 to another root of x 4 in E. That is, σ( 4 ) = ± 4. Thus, there are at most automorphisms of E over Q. But each of these is realized. Namely, σ 1 (x) = x and σ (a + b 4 + c 4 4 + d 4 8) = a b 4 + c 4 4 d 4 8 are two automorphisms. Hence gal(e : Q) = Z. 5. Let K = Q(, 3, 5). Determine [K : Q], prove that K is a splitting field of a polynomial over Q, and determine gal(k : Q) 3
Solution. There is a chain of field extensions: Q Q( ) Q(, 3) Q(, 3, 5) = K. Since each successive extension is obtained by adjoining a square root, it follows from the tower theorem that [K : Q] = = 8. K is the splitting field of the polynomial (x )(x 3)(x 5). Analogous to adjoining square roots, gal(k : Q) = Z Z Z. 6. Let K be the splitting field over Q of the polynomial f(x) = (x x 1)(x x 7). Determine gal(k : Q), and find all intermediate fields explicitly. Solution. The roots of f(x) are 1 ± and 1 ± 4. Since all of these are included in the quadratic extension Q( ) it follows that K = Q( ) is the splitting field of f(x) over Q. Since [K : Q] = it follows that gal(k : Q) = Z. Since [K : Q] = there are no intermediate fields other than K and Q. 7. Let F be a splitting field of a separable polynomial over a field K such that gal(f : K) = Z 1. How many intermediate fields L are there such that (a) [L : K] = 4, (b) [L : K] = 9, (c) gal(f : L) = Z 6? Solution. Let G = gal(f : K). Then, by the Fundamental Theorem of Galois Theory, any intermediate field L has the form L = F H for some subgroup H of G and [F : L] = H while [L : K] = [G : H]. Hence: (a) [L : K] = 4 if and only if [G : H] = 4 if and only if H = 1/4 = 3. There is exactly one subgroup of Z 1 of order 3 and hence there is exactly one subfield L of F with [L : K] = 4. (b) Since 9 does not divide 1, there is no subfield L with [L : K] = 9. (c) There is exactly one subgroup of Z 1 isomorphic to Z 6. Hence there is exactly on subfield L of F containing K such that gal(f : L) = Z 6. 8. Let f(x) = x 3 + 5 Q[x]. (a) Show that the splitting field of f(x) is K = Q( 3, 3 5), and compute [K : Q]. Solution. The roots of f(x) are 3 5, 3 5ω, and 3 5ω, where ω = e πi/3 = 1 + 3i is a primitive cube root of unity in C. Thus, the splitting field of f(x) is K 1 = Q( 3 5, 3 5ω, 3 5ω ). Note that ω = 1 + 3i = 1 + 3 K = Q( 3, 3 5), and clearly, 3 5 K = Q( 3, 3 5). Thus, each of the roots of f(x), namely, 3 5, 3 5ω, and 3 5ω are in K. Hence K 1 K. To show the reverse inclusion, it is sufficient to show that each of the generators of K, namely, 3 5 and 3, is in K 1. That 3 5 is in K 1 is clear since 3 5 is one of the generators of K 1. Moreover, ω = 3 5ω/( 3 5) K 1, and similarly ω K 1. Hence 3 = ω ω K 1. Thus, K K 1 and hence K = K 1. The degree of the extension is calculated by the tower formula: [K : Q] = [K : Q( 3 5)][Q( 3 5) : Q] = 3 = 6. 4
(b) Find the Galois group gal(k : Q). Solution. Since gal(k : Q) is isomorphic to a subgroup of S 3 of order [K : Q] = 6, it follows that gal(k : Q) = S 3. (c) How many different fields L (other than Q and K) satisfy Q L K? That is, how many intermediate fields are there between K and Q? Solution. By the Fundamental Theorem of Galois Theory, there is a one-to-one correspondence between intermediate fields of K and subgroups of gal(k : Q) = S 3. But the subgroups of S 3 are {(1)}, (1 ), (1 3), ( 3), (1 3), and S 3. Since Q corresponds to S 3, and {(1)} corresponds to K, it follows that there are exactly 4 additional intermediate fields. (d) For each intermediate field L, compute the degree [L : Q] and write L in the form L = Q(α) for an appropriate α K. Solution. It is only necessary to find 4 distinct intermediate fields, since we know that that is the number of intermediate fields, according to the fundamental theorem. Each of the roots of the polynomial f(x) generates a subfield of K of degree deg f(x) = 3 over Q. Hence, 3 of the 4 intermediate fields are: L 1 = Q( 3 5) = Q( 3 5), L = Q( 3 5ω) = Q( 3 5ω), and L 3 = Q( 3 5ω ) = Q( 3 5ω ). The fourth intermediate field is obtained from the other generator of K, namely 3, so that L 4 = Q( 3). The degrees are: 9. Let F L K be fields. Prove or disprove: [L 1 : Q] = [L : Q] = [L 3 : Q] = 3 and [L 4 : Q] =. (a) If K is Galois over F, then K is Galois over L. Solution. This is item (4) in the Main Theorem of Galois Theory. Page 49 (b) If K is Galois over F, then L is Galois over F. Solution. This is false. Q( 3, ω) where ω is a primitive cube root of unity is the splitting field of x 3 over Q and hence it is a Galois extension. However, the intermediate field L = Q( 3 ) is not a Galois extension of Q since gal(q( 3 ) : Q) is the identity group, and hence it has fixed field Q( 3 ) = Q. (c) If L is Galois over F, and K is Galois over L, then K is Galois over F. Solution. This is also false. Consider the chain of subfields Q Q( ) Q( 4 ). Q( 4 ) is Galois over Q( ) since it is the splitting field of the polynomial x Q( )[x], while Q( ) is Galois over Q since it is the splitting field of the polynomial x Q[x]. However, Q( 4 ) is not Galois over Q since the polynomial x 4 has a root in Q( 4 ) but does not split in this field since the roots ±i 4 are not in this field. (See Theorem 3, part (b), Page 47.) 5
10. Let w = 4 and K = Q(w, i). Then K is a splitting field over Q of the polynomial f(x) = x 4. Label the roots of f(x) as r 1 = w, r = w, r 3 = wi, and r 4 = wi. Give an example (with justification) of a permutation of these roots which does not correspond to an element of the Galois group gal(k : Q). Solution. The 3-cycle (r, r 3, r 4 ) cannot come from an automorphism σ gal(k : Q). Suppose there is such a σ. Since σ must permute the roots of x 4, we must have σ(w) = w. But if this is true, then σ( w) = σ(w) = w. Thus, σ( w) cannot be wi for any σ that fixes w. 6