6.3 Splitting Fields
1.
a. x3 + 1 = (x + 1)(x2 − x + 1), and the roots of x2 − x + 1 are 
and 
so 
But 
so it follows that 
Since 
is a root of the polynomial x2 + 3, and since x3 + 3 is irreducible over 
(no root), then it is the minimal polynomial of 
Hence 
and so But so it follows that Since is a root of the polynomial x2 + 3, and since x3 + 3 is irreducible over (no root), then it is the minimal polynomial of Hence
c. f = (x2 − 7)(x2 + 1) so 
. Thus,
. Thus,
Moreover, 
because x2 − 2 is irreducible over 
and 
because x2 + 1 has no root in 
Hence 
by the multiplication theorem.
because x2 − 2 is irreducible over and because x2 + 1 has no root in Hence by the multiplication theorem.
2.
a. f = (x2 − 2x − 2)(x2 − 5) so the roots are 
and 
. Hence 
is the splitting field.
and . Hence is the splitting field.
4.
a. f = (x + 1)(x2 + x + 1) and x2 + x + 1 is irreducible over 
. If u is a root then the other is given by 
; 
. Thus 
and f = (x + 1)(x + u)(x + 1 + u).
. If u is a root then the other is given by ; . Thus and f = (x + 1)(x + u)(x + 1 + u).
c. f is irreducible over 
. If u is a root then, by long division, f = (x + u)g where g = x2 + (1 + u)x + (u + u2). We claim that g also splits in 
, so 
. We try possibilities to get g(u2) = 0. Then the other root 
satisfies 
, so 
. Thus
. If u is a root then, by long division, f = (x + u)g where g = x2 + (1 + u)x + (u + u2). We claim that g also splits in , so . We try possibilities to get g(u2) = 0. Then the other root satisfies , so . Thus
e. f = (x2 − 2)(x2 + 1) = (x2 + 1)2 over 
. Now x2 + 1 is irreducible over 
(no root in 
If u is a root then x2 + 1 = x2 − u2 = (x − u)(x + u). Thus 
and f = (x − u)2(x + u)2.
. Now x2 + 1 is irreducible over (no root in If u is a root then x2 + 1 = x2 − u2 = (x − u)(x + u). Thus and f = (x − u)2(x + u)2.
5. The roots are 
and 
. So 
is the splitting field for both.
and . So is the splitting field for both.
6.
a. If 
were the splitting field of 
then 
. Thus 
would be algebraic, contradicting the fact that π or e is transcendental.
were the splitting field of then . Thus would be algebraic, contradicting the fact that π or e is transcendental.
7. We have f = a(x − u1) 
(x − um) in E[x]. If p is a monic, irreducible factor of g in E[x] then p|f so p = x − ui for some i. Thus
(x − um) in E[x]. If p is a monic, irreducible factor of g in E[x] then p|f so p = x − ui for some i. Thus
by the unique factorization theorem, as required.
9. If gcd (f, g) = 1 let 1 = fh + gk; h, k in F[x]. If E ⊇ F is an extension containing an element u such that f(u) = 0 = g(u), substitution gives
a contradiction. Conversely, let d = gcd (f, g). If d ≠ 1 then deg d ≥ 1 so let E ⊇ F be a field containing a root u of d. Then d|f and d|g means f(u) = 0 = g(u), contrary to hypothesis.
11. We have E = F(u1, . . ., un) where f = a(x − u1) (x − un), a 
F, ui E. Since L ⊇ F, E = L(u1, . . ., un), and a L, ui E, show f splits in E over L.
(x − un), a F, ui E. Since L ⊇ F, E = L(u1, . . ., un), and a L, ui E, show f splits in E over L.
13. The roots of xp − 1 are 
(Theorem 6, Appendix A) so 
is the splitting field. We have xp − 1 = (x − 1)Φp where
(Theorem 6, Appendix A) so is the splitting field. We have xp − 1 = (x − 1)Φp where
is the pth cyclotomic polynomial. Then Φp is irreducible over 
by Example 13 §4.2, and so is the minimal polynomial of 
(
is a root of Φp). Hence 
by Theorem 4 §6.2.
by Example 13 §4.2, and so is the minimal polynomial of ( is a root of Φp). Hence by Theorem 4 §6.2.
15. Let K ⊇ E be a field in which f has a root 
. Write deg f = m and g = n so that [F(u) : F] = n and 
are relatively prime. By Exercise 21 §6.2, 
. Now let p be the minimal polynomial of 
over F(u); we show f = p. We have 
so p|f in F(u)[x], and so it is enough to show that deg p = m. We have 
deg p, so
. Write deg f = m and g = n so that [F(u) : F] = n and are relatively prime. By Exercise 21 §6.2, . Now let p be the minimal polynomial of over F(u); we show f = p. We have so p|f in F(u)[x], and so it is enough to show that deg p = m. We have deg p, so
by the multiplication theorem. Thus p = m as required.
17. The map f → fσ is clearly onto (since σ is onto). If 
then
then
so fσ = 0 implies σ(ai) = 0 for all i, which in turn implies ai = 0 for all i, whence f = 0. Thus fσ is one-to-one if we can show it is a ring homomorphism. If 
,
,
Similarly
Hence the map is a ring isomorphism (clearly 1σ = 1). Finally, if a F let g = a be the constant polynomial. Then g 
gσ means a σ(a). So the map extends σ.
F let g = a be the constant polynomial. Then g gσ means a σ(a). So the map extends σ.
19. If π were algebraic over 
then π would satisfy a nonzero polynomial f in 
. But f splits in 
because 
is algebraically closed, so this would imply that 
, a contradiction.
then π would satisfy a nonzero polynomial f in . But f splits in because is algebraically closed, so this would imply that , a contradiction.
20.
a. We show 
. Clearly 
is algebraic. We must show that if u E is algebraic over 
then u A. Since u is algebraic over A, we show that u A implies u is transcendental over A. We have E = A(π) so this follows from Exercise 31 §6.2 if we can show that π is transcendental over A. But if π were algebraic over A it would be algebraic over 
, contrary to the preceding exercise.
. Clearly is algebraic. We must show that if u E is algebraic over then u A. Since u is algebraic over A, we show that u A implies u is transcendental over A. We have E = A(π) so this follows from Exercise 31 §6.2 if we can show that π is transcendental over A. But if π were algebraic over A it would be algebraic over , contrary to the preceding exercise.
21. (1) ⇒ (2). If E ⊇ F is the splitting field of f F[x], then E = F(u1, . . ., um) where the ui are the roots of f in E, so [E : F] is finite by Theorem 6 §6.2. If p F[x] is irreducible with a root u E, let 
be a root in some field K ⊇ E. Then p is the minimal polynomial of both u and 
so let 
be an isomorphism (Theorem 4 §6.2). Now E is a splitting field of f over F(u), and 
is a splitting field of f over 
so Theorem 3 shows 
via an isomorphism extending σ. Hence 
, so
F[x], then E = F(u1, . . ., um) where the ui are the roots of f in E, so [E : F] is finite by Theorem 6 §6.2. If p F[x] is irreducible with a root u E, let be a root in some field K ⊇ E. Then p is the minimal polynomial of both u and so let be an isomorphism (Theorem 4 §6.2). Now E is a splitting field of f over F(u), and is a splitting field of f over so Theorem 3 shows via an isomorphism extending σ. Hence , so
Since E is an F-subspace of 
, this shows 
, and so 
(2) ⇒ (1). Since E ⊇ F is finite, let E = F(u1, . . ., un). Each ui is algebraic over F, say with minimal polynomial pi F[x]. Since pi splits in E by (2), E is the splitting field of f = p1p2 pn.
, this shows , and so (2) ⇒ (1). Since E ⊇ F is finite, let E = F(u1, . . ., un). Each ui is algebraic over F, say with minimal polynomial pi F[x]. Since pi splits in E by (2), E is the splitting field of f = p1p2 pn.
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