10.1 Galois Groups and Separability
1. First ε 
gal(E : F) because ε(a) = a for all a F. If σ, τ gal(E : F) then σ(a) = a for all a F, so a = σ−1(a) for all a; hence (since σ−1 is an automorphism) σ−1 gal(E : F). Finally στ(a) = σ[τ(a)] = σ(a) = a, so στ gal(E : F).
gal(E : F) because ε(a) = a for all a F. If σ, τ gal(E : F) then σ(a) = a for all a F, so a = σ−1(a) for all a; hence (since σ−1 is an automorphism) σ−1 gal(E : F). Finally στ(a) = σ[τ(a)] = σ(a) = a, so στ gal(E : F).
3. Let σ(ui) = τ(ui) for all i where σ, τ gal(E : F). If 
, write 
, ai F. Then
gal(E : F). If , write , ai F. Then
As 
was arbitrary, this shows that σ = τ.
was arbitrary, this shows that σ = τ.
5. Let σ : E → E be an automorphism. Then if 
, σ(n) = n · σ(1) = n. Thus 
for all 
131 Hence σ gal 
, σ(n) = n · σ(1) = n. Thus for all 131 Hence σ gal
7. Put u = e2πi/6. Then u satisfies x6 − 1 = (x2 − 1)(x2 + x + 1)(x2 − x + 1), and the only roots in 
are the sixth roots of unity 1, u, u2, u3, u4, u5. But u3 = − 1 so u satisfies x3 + 1 = (x + 1)(x2 − x + 1). Hence u is a root of m = x2 − x + 1 . Since m is irreducible over 
, it is the minimal polynomial of u. The other root of m is u5 (because (u5)3 = (u3)5 = (− 1)5 = − 1), so Theorem 1 gives 
where σ(u) = u5. Note that σ2 = ε because σ2(u) = σ[σ(u)] = σ(u5) = [σ(u)]5 = u25 = u. Hence 
are the sixth roots of unity 1, u, u2, u3, u4, u5. But u3 = − 1 so u satisfies x3 + 1 = (x + 1)(x2 − x + 1). Hence u is a root of m = x2 − x + 1 . Since m is irreducible over , it is the minimal polynomial of u. The other root of m is u5 (because (u5)3 = (u3)5 = (− 1)5 = − 1), so Theorem 1 gives where σ(u) = u5. Note that σ2 = ε because σ2(u) = σ[σ(u)] = σ(u5) = [σ(u)]5 = u25 = u. Hence
9. The minimal polynomial of i is x2 + 1, and that of 
is x2 − 3. By Lemma 2, there exists a 
-isomorphism 
with σ0(i) = − i. This extends to an automorphism σ of 
satisfying 
. Thus 
, σ(i) = − i and 
. Similarly, there exists 
with τ(i) = i and 
. Note that
is x2 − 3. By Lemma 2, there exists a -isomorphism with σ0(i) = − i. This extends to an automorphism σ of satisfying . Thus , σ(i) = − i and . Similarly, there exists with τ(i) = i and . Note that
σ(i) {i, − i} and 
so 
by Theorem 2. Now observe that
{i, − i} and so by Theorem 2. Now observe that
στ(i) = − i = τσ(i) and 
Hence τσ = στ by Theorem 2. Since o(σ) = 2 = o(τ), it follows that {ε, σ, τ, στ} ≅ C2 × C2. But 
and 
so 
and so
10. a. Write 
. Then u satisfies x4 − 2, irreducible by the Eisenstein criterion. The roots of x4 − 2 in 
are u, −u, iu and −iu, and the only ones in 
are u and −u. Thus Theorem 1 gives 
so 
where σ(u) = − u.
. Then u satisfies x4 − 2, irreducible by the Eisenstein criterion. The roots of x4 − 2 in are u, −u, iu and −iu, and the only ones in are u and −u. Thus Theorem 1 gives so where σ(u) = − u.
11. Choose u E, u ∉ F. Thus [F(u) : F] ≠ 1 so, since [E : F] = 2 is prime, E = F(u) by Theorem 5 §6.2 (the multiplication theorem). If m is the minimal polynomial of u over F, then °m = 2. Thus since one root u lies in E, so does the other (their sum is the negative of the coefficient of x in m). Now the result is clear by Theorem 1.
E, u ∉ F. Thus [F(u) : F] ≠ 1 so, since [E : F] = 2 is prime, E = F(u) by Theorem 5 §6.2 (the multiplication theorem). If m is the minimal polynomial of u over F, then °m = 2. Thus since one root u lies in E, so does the other (their sum is the negative of the coefficient of x in m). Now the result is clear by Theorem 1.
13. Write 
. Now x4 − 2 is irreducible (Eisenstein) so x4 − 2 and x2 + 1 are the minimal polynomials of u and i. The roots are {u, − u, iu, − iu} and {i, − i}, and all are in E so |G| ≤ 8 by Theorem 1 where 
. By Lemma 1 there exists 
satisfying σ0(u) = iu. Extend it to an automorphism σ of 
where σ(i) = i. Thus σ(u) = iu and σ(i) = i, and so σ2(u) = − u, σ3(u) = − iu, σ4(u) = u. Thus o(σ) = 4. Next let 
have τ0(i) = − i, extend τ0 to an automorphism τ of 
where τ(u) = u. Then τ2 = ε so o(τ) = 2. Finally
. Now x4 − 2 is irreducible (Eisenstein) so x4 − 2 and x2 + 1 are the minimal polynomials of u and i. The roots are {u, − u, iu, − iu} and {i, − i}, and all are in E so |G| ≤ 8 by Theorem 1 where . By Lemma 1 there exists satisfying σ0(u) = iu. Extend it to an automorphism σ of where σ(i) = i. Thus σ(u) = iu and σ(i) = i, and so σ2(u) = − u, σ3(u) = − iu, σ4(u) = u. Thus o(σ) = 4. Next let have τ0(i) = − i, extend τ0 to an automorphism τ of where τ(u) = u. Then τ2 = ε so o(τ) = 2. Finally
Thus στσ = τ, whence 
σ, τ 
≅ D4. Thus | σ, τ |= 8 so, since σ, τ ⊆ G and |G| = 8, G = σ, τ ≅ D4.
σ, τ ≅ D4. Thus | σ, τ |= 8 so, since σ, τ ⊆ G and |G| = 8, G = σ, τ ≅ D4.
15. a. If 
and 
, then 
is 0 or 
. So a = 1 is none of these, whence 
by the proof of Theorem 6.
and , then is 0 or . So a = 1 is none of these, whence by the proof of Theorem 6.
16. a. If 
and 
, then 
is 0 or 
. So a = 1 is neither of these, whence 
by the proof of Theorem 6.
and , then is 0 or . So a = 1 is neither of these, whence by the proof of Theorem 6.
17. If 
then 
for some 
. If σ ≠ ε in gal(
then 
and 
. Now let 
. If u ≠ σ(u) let u < σ(u). If u < a < σ(u), 
, then σ(u) < σ(a) = a, a contradiction. A similar argument diminates σ(u) < u.
then for some . If σ ≠ ε in gal( then and . Now let . If u ≠ σ(u) let u < σ(u). If u < a < σ(u), , then σ(u) < σ(a) = a, a contradiction. A similar argument diminates σ(u) < u.
19. We proceed by induction on n. If n = 1 then E = F(u1) = {f(u1) 
f F[x]} . Hence σ(f(u1)) = f(σ(u1)) = f(τ(u1)) = τ(f(u1)) for all f, as required. In general, write K = F(u1, u2, . . ., un−1) so that E = K(un) . By induction, σ = τ on K, so σ, τ gal(K : F) . Since σ(un) = τ(un) the result follows from the case n = 1.
f F[x]} . Hence σ(f(u1)) = f(σ(u1)) = f(τ(u1)) = τ(f(u1)) for all f, as required. In general, write K = F(u1, u2, . . ., un−1) so that E = K(un) . By induction, σ = τ on K, so σ, τ gal(K : F) . Since σ(un) = τ(un) the result follows from the case n = 1.
21. Let σ gal (F(t) : F). If 
, f, g relatively prime in F[t], then 
, so if 
, σ(λ(t)) = λ(σ(t)). Thus σ is determined completely by f and g. Clearly f ≠ 0, g ≠ 0. Now suppose 
. Then 
. Suppose
gal (F(t) : F). If , f, g relatively prime in F[t], then , so if , σ(λ(t)) = λ(σ(t)). Thus σ is determined completely by f and g. Clearly f ≠ 0, g ≠ 0. Now suppose . Then . Suppose
Then
Suppose n > m. Then
Since anfn is the only term not involving g, it follows that g|anfn. Hence g|an since f, g are relatively prime, so °g = 0. Similarly n < m implies that g|bmtfm so, again, °g ≤ 1 . Thus either °g ≤ 1 or n = m. But if n = m we have t(b0gn + b1fgn−1 + 
+ bnfn) = a0gn + + anfn. This yields g|(an − bnt)fn and °g ≤ 1 in all cases. Now observe that 
so 
. Hence 
and the same type of argument implies °f ≤ 1. Hence we have proved that 
so that σ = σM where 
. Now observe that σM[σN(λ(x))] = σNMλ(x) holds for all λ(t) F(t). In particular if σ gal(F(t) : F) and σ = σM, and if σ−1 = σN, we have 
where 
. It follows that c′′x2 + d′′x = a′′x + b′′ so c′′ = b′′ = 0, a′′ = d′′ ≠ 0 so 
and so 
. Thus the map M 
σM from GL2(F) → gal (F(t) : F) is an onto group homomorphism. Moreover the above shows that the kernel is Z.
+ bnfn) = a0gn + + anfn. This yields g|(an − bnt)fn and °g ≤ 1 in all cases. Now observe that so . Hence and the same type of argument implies °f ≤ 1. Hence we have proved that so that σ = σM where . Now observe that σM[σN(λ(x))] = σNMλ(x) holds for all λ(t) F(t). In particular if σ gal(F(t) : F) and σ = σM, and if σ−1 = σN, we have where . It follows that c′′x2 + d′′x = a′′x + b′′ so c′′ = b′′ = 0, a′′ = d′′ ≠ 0 so and so . Thus the map M σM from GL2(F) → gal (F(t) : F) is an onto group homomorphism. Moreover the above shows that the kernel is Z.
22. a. (3) ⇒ (1). Let E ⊇ F be a field and suppose f has a repeated root u in E. Then f(u) = 0 = f′(u) by Theorem 3 §6.4. But (3) implies 1 = fg + f′h in F[x] with g, h F[x], and this is valid in E[x]. But then, taking x = u gives 1 = 0 in E, a contradiction.
F[x], and this is valid in E[x]. But then, taking x = u gives 1 = 0 in E, a contradiction.
23. Here f′ = nxn−1 − 1. Write d = gcd (f, f′); we must show d = 1 by the preceding exercise. Let E ⊇ F be a splitting field for f. If d ≠ 1 then °d > 1 so d has a root u in E. Thus f(u) = 0 = f′(u) so un = u and nun−1 = 1. Hence (n − 1)u = 0, a contradiction if char F = 0 . But if char F = p it implies p|(n − 1), contrary to hypothesis.
25. If E ⊇ F and f F[x] is separable over F, let q be any irreducible factor of f in E[x]. We must show that q has no repeated root in E. Now f = p1p2 pr in F[x] where the pi are irreducible and (by hypothesis) separable. But f = p1p2 pr in E[x] so q|pi for some i. Hence any repeated root of q in some splitting field of f would be a repeated root of pi, contrary to assumption. So q is separable.
F[x] is separable over F, let q be any irreducible factor of f in E[x]. We must show that q has no repeated root in E. Now f = p1p2 pr in F[x] where the pi are irreducible and (by hypothesis) separable. But f = p1p2 pr in E[x] so q|pi for some i. Hence any repeated root of q in some splitting field of f would be a repeated root of pi, contrary to assumption. So q is separable.
27. Suppose f = xp − a is not a power of a linear polynomial in F[x]; we must show it is irreducible. If u is a root of f in an extension E ⊇ F, then up = a so f = xp − up = (x − u)p in E[x] because the characteristic is p. Then u ∉ F and F(u) is a splitting field of f over F. Let q be an irreducible factor of f in F[x]. Then q|(x − u)p in E[x] so q = (x − u)t. Then t > 1 because u ∉ F so q′ = 0 by Lemma 5. Hence q = g(xp) by Theorem 4. Thus, the factorization of f into irreducibles in F[x] takes the form
Since def f = p, it follows that f = gi(xp) = qi for some i, so f is irreducible.
28.
a. (3) ⇒ (1). If E ⊇ F is algebraic and u E, the minimal polynomial of u is separable by (3), and so E ⊇ F is a separable extension.
E, the minimal polynomial of u is separable by (3), and so E ⊇ F is a separable extension.
c. Let E ⊇ F be algebraic, F perfect. If K ⊇ E is algebraic, then K ⊇ F is algebraic by Corollary 1, Theorem 6 §6.2, hence separable by hypothesis.
29.
a. If F is perfect, and a F, let E be the splitting field of f = xp − a. If u E is a root of f then up = a, so f = xp − a = xp − up = (x − u)p. Let q be an irreducible factor of f in F[x]. Then q = (x − u)m. But E ⊇ F is finite, hence separable by hypothesis. Thus q has distinct roots, whence m = 1 and q = x − u. But then u F and up = a, as required.
F, let E be the splitting field of f = xp − a. If u E is a root of f then up = a, so f = xp − a = xp − up = (x − u)p. Let q be an irreducible factor of f in F[x]. Then q = (x − u)m. But E ⊇ F is finite, hence separable by hypothesis. Thus q has distinct roots, whence m = 1 and q = x − u. But then u F and up = a, as required.
Conversely, let q be irreducible in F[x]. If q is not separable then q′ = 0 so q = f(xp) for some f F[x] by Theorem 4, say 
. By hypothesis, let 
, bi F. Then
F[x] by Theorem 4, say . By hypothesis, let , bi F. Then
This contradicts the irreducibility of q.
30.
a. Let q be the minimal polynomial of u over F. If K = F(up) let m K[x] be the minimal polynomial of u over K. Then q K[x] and q(u) = 0, so m|q. But q has distinct roots by hypothesis, so m has distinct roots. On the other hand, xp − up K[x] and xp − up = (x − u)p in E[x]. Hence m|(x − u)p so m = (x − u)r. Since m has distinct roots, r = 1 and so u K.
K[x] be the minimal polynomial of u over K. Then q K[x] and q(u) = 0, so m|q. But q has distinct roots by hypothesis, so m has distinct roots. On the other hand, xp − up K[x] and xp − up = (x − u)p in E[x]. Hence m|(x − u)p so m = (x − u)r. Since m has distinct roots, r = 1 and so u K.
c. Extend 
to a basis 
of E over F. Suppose 
, say 
. Write 
, so 
, whence 
. Thus 
spans E and so is F -independent because dim E = n (Theorem 7 §6.1). But then 
is F-independent.
to a basis of E over F. Suppose , say . Write , so , whence . Thus spans E and so is F -independent because dim E = n (Theorem 7 §6.1). But then is F-independent.
31. If E ⊇ F is separable, the result is Exercise 26. If E ⊇ K and K ⊇ F are separable, we may assume char F = p by the Corollary to Theorem 4. By the preceding exercise, E = K(Ep) and K = F(Kp). If u E then 
where 
and ui E. But 
where aij F and 
for all i, j. Thus 
. Thus E ⊇ F is separable by the preceding exercise.
E then where and ui E. But where aij F and for all i, j. Thus . Thus E ⊇ F is separable by the preceding exercise.
32.
a. Let p and q be the minimal polynomials of u over F and K respectively. Then p K[x] and p(u) = 0, so q|p. Since p has distinct roots in some splitting field L ⊇ K, q is separable over K.
K[x] and p(u) = 0, so q|p. Since p has distinct roots in some splitting field L ⊇ K, q is separable over K.
c. Let u, 
. Then F(u) ⊇ F is separable and, since 
is separable over F(u) by (a), 
is also separable. Hence 
is separable by the preceding exercise. Since 
and uv are in 
, they are separable over F, so S is a field. Clearly S ⊇ F since the minimal polynomial of a F is x − a, and S ⊇ F is separable by the definition of S. If E ⊇ K ⊇ F and K ⊇ F is separable, then each u K is separable over F; that is u K. Hence K ⊆ S.
. Then F(u) ⊇ F is separable and, since is separable over F(u) by (a), is also separable. Hence is separable by the preceding exercise. Since and uv are in , they are separable over F, so S is a field. Clearly S ⊇ F since the minimal polynomial of a F is x − a, and S ⊇ F is separable by the definition of S. If E ⊇ K ⊇ F and K ⊇ F is separable, then each u K is separable over F; that is u K. Hence K ⊆ S.
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