6.2 Algebraic Extensions
1.
a. 
whence (u2 − 8)2 = 60 . Hence u4 − 16u + 4 = 0.
whence (u2 − 8)2 = 60 . Hence u4 − 16u + 4 = 0.
c. 
so 
that is 
Thus (u4 + 7)2 = 12u4, so u8 + 2u4 + 49 = 0.
so that is Thus (u4 + 7)2 = 12u4, so u8 + 2u4 + 49 = 0.
2.
a. 
so (u2 − 5)2 = 24, u4 − 10u2 + 1 = 0. We claim m = x4 − 10x2 + 1 is the minimal polynomial; it suffices to prove it is irreducible over 
. It has no root in 
so suppose
so (u2 − 5)2 = 24, u4 − 10u2 + 1 = 0. We claim m = x4 − 10x2 + 1 is the minimal polynomial; it suffices to prove it is irreducible over . It has no root in so suppose
Then
Then c = − a and b = d = ± 1 . Hence 2b − a2 = − 10, so a2 = 10 + 2b = 12 or 8, a contradiction.
c. 
so (u2 − 1)2 = 3; u4 − 2u2 − 2 = 0. The minimal polynomial is x4 − 2x2 − 2 because it is irreducible by Eisenstein's criterion.
so (u2 − 1)2 = 3; u4 − 2u2 − 2 = 0. The minimal polynomial is x4 − 2x2 − 2 because it is irreducible by Eisenstein's criterion.
3.
a. A lgebraic; 
is a root of x2 − π F[x] where 
is a root of x2 − π F[x] where
c. Transcendental. For if f(π2) = 0 where 
then g(π) = 0 where g(x) = f(x2) ≠ 0. This would imply π itself is algebraic over 
, a contradiction.
then g(π) = 0 where g(x) = f(x2) ≠ 0. This would imply π itself is algebraic over , a contradiction.
4.
a. (u − 1)2 = i2 = − 1, so u2 − 2u + 2 = 0. We claim m(x) = x2 − 2x + 2inF[x] is the minimal polynomial. The roots of m in 
are 1 ± i and neither is in 
. Hence m(x) is irreducible in F[x].
are 1 ± i and neither is in . Hence m(x) is irreducible in F[x].
5. Clearly 
so 
. Since 
(because 
this means 
so 
. Thus 
Alternatively, let u = a + bi. Then b ≠ 0 (as 
) so 
. Hence 
, so 
.
so . Since (because this means so . Thus Alternatively, let u = a + bi. Then b ≠ 0 (as ) so . Hence , so .
7.
a. 
so 
. We claim
so . We claim
is the minimal polynomial. Its roots in 
are 
and neither is in 
. So it is irreducible in 
, as required.
are and neither is in . So it is irreducible in , as required.
9. If 
with f F[x], then 
, whence 
. Always 
. Conversely, if 
then 
so
with f F[x], then , whence . Always . Conversely, if then so
with ai F, n ≥ 1, by Theorem 4. Take f = a0 + a1x + 
+ an−1xn−1.
+ an−1xn−1.
11. If [F(u) : F] is finite then u is algebraic over F by Theorem 1 because u F(u). The converse follows from Theorem 4.
12.
a. If 
then u is a root of x3 − 2, which is irreducible over 
by the Eisenstein criterion. So the basis is {1, u, u2} by Theorem 4.
then u is a root of x3 − 2, which is irreducible over by the Eisenstein criterion. So the basis is {1, u, u2} by Theorem 4.
c. If 
then u satisfies x3 − 3, which is irreducible over 
by Eisenstein. Thus {1, u, u2} is a basis of 
on 
. We have 
satisfies x2 − 3 L[x]. This is irreducible because 
. [If 
then 
divides 
Thus 
is a basis of 
on L. By Theorem 5, 
is a basis of E over 
then u satisfies x3 − 3, which is irreducible over by Eisenstein. Thus {1, u, u2} is a basis of on . We have satisfies x2 − 3 L[x]. This is irreducible because . [If then divides Thus is a basis of on L. By Theorem 5, is a basis of E over
e. We have 
, so 
. Now 
is a 
-basis of 
. We have 
[If 
then b ≠ 0, and hence a ≠ 0]. Hence x2 − 5 is irreducible in L[x]. Thus 
is an L-basis of 
. Finally 
is a 
-basis of E by Theorem 5.
, so . Now is a -basis of . We have [If then b ≠ 0, and hence a ≠ 0]. Hence x2 − 5 is irreducible in L[x]. Thus is an L-basis of . Finally is a -basis of E by Theorem 5.
13.
a. Put 
. Then 
so u is a root of 
. The roots of m in 
are 
and these are not in F because 
means 
, a, 
, and b ≠ 0 and 
and 
. Thus 
shows 
, a contradiction.] Thus m is irreducible over F, so [F(u) : F] = 2. But F(u) = E.
. Then so u is a root of . The roots of m in are and these are not in F because means , a, , and b ≠ 0 and and . Thus shows , a contradiction.] Thus m is irreducible over F, so [F(u) : F] = 2. But F(u) = E.
c. Put 
so (u − i)2 = 3 and u is a root of m = x2 − 2ix − 4 F[x]. The roots of m in 
are u and 
, and neither is in F because 
means 
. Thus m is F-irreducible so E = F(u) has degree 2 over F; that is [E : F] = 2.
so (u − i)2 = 3 and u is a root of m = x2 − 2ix − 4 F[x]. The roots of m in are u and , and neither is in F because means . Thus m is F-irreducible so E = F(u) has degree 2 over F; that is [E : F] = 2.
15. If u E, u ∉ F, we have E ⊇ F(u) ⊇ F, so [F(u) : F] divides [E : F] by Theorem 5. Since [F(u) : F] ≠ 1 because u ∉ F, we have [F(u) : F] = [E : F] because [E : F] is prime. Thus F(u) = E by Theorem 8 §6.1.
17. If F(u) ⊇ L ⊇ F write p = [F(u) : F] = [F(u) : L][L : F]. Thus [L : F] = 1 or p; so L = F or [L : F] = [F(u) : F], whence L = F(u) by Theorem 8 §6.1.
19. Let 
. Then 
means f(u) = 0 where 
has degree 1 or 2. Since 
, °f = 2, say f = ax2 + bx + c. Thus 
and 
so 
. Now f(u) = 0 so (clearing denominators) we may assume a, b and c are integers. Thus 
where 
. Thus 
. If d = p2e, 
, p a prime, then 
. Thus we can continue until 
where m is square-free.
. Then means f(u) = 0 where has degree 1 or 2. Since , °f = 2, say f = ax2 + bx + c. Thus and so . Now f(u) = 0 so (clearing denominators) we may assume a, b and c are integers. Thus where . Thus . If d = p2e, , p a prime, then . Thus we can continue until where m is square-free.
20.
a. E(u) ⊇ E ⊇ F gives E(u) ⊇ F finite because u is algebraic over E. Let m be the minimal polynomial of u over F, so [F(u) : F] = °m. Now u is also algebraic over E because m(u) E[x], so let p E(x) be the minimal polynomial of u over E. Since m(u) = 0, m E[u], Theorem 3 gives p|m. Thus [E(u) : E] = °p ≤ °m = [F(u) : F].
21.
a. Write L = F(u) so 
. Now [L : F] = m by hypothesis so, since 
, it suffices to show 
. Let p and m be the minimum polynomials of 
over L and F respectively. Then p|m by Theorem 3, so
. Now [L : F] = m by hypothesis so, since , it suffices to show . Let p and m be the minimum polynomials of over L and F respectively. Then p|m by Theorem 3, so
as required.
c. No. If 
, 
, 
and 
, then m = 2 and n = 2 are not relatively prime, but 
by Example 15.
, , and , then m = 2 and n = 2 are not relatively prime, but by Example 15.
23. If 
then 
, f, 
, gcd (f, g) = 1. Then h(π) = 0 where h(x) = 2g2(x) − f2(x), and this is a contradiction if h ≠ 0 in 
. But h = 0 means 2g2 = f2 so, since gcd(f, g) = 1, f|2. Thus f = ± 1, ±2, g2(x) = ± 1, 
. This forces °g = 0; 
, g = ± 1, 
. Thus g = ± 1, 
, a contradiction. Thus h ≠ 0 and 
has led to a contradiction. So 
then , f, , gcd (f, g) = 1. Then h(π) = 0 where h(x) = 2g2(x) − f2(x), and this is a contradiction if h ≠ 0 in . But h = 0 means 2g2 = f2 so, since gcd(f, g) = 1, f|2. Thus f = ± 1, ±2, g2(x) = ± 1, . This forces °g = 0; , g = ± 1, . Thus g = ± 1, , a contradiction. Thus h ≠ 0 and has led to a contradiction. So
25. Write K = F(u1, . . ., uk, . . ., un) and L = F(u1, . . ., uk). Then L ⊆ K becaue ui K for 1 ≤ i ≤ k, and so L(uk+1, . . ., un) ⊆ K because ui K for k + 1 ≤ i ≤ n. On the other hand F ⊆ L(uk+1, . . ., un) and every ui L(uk+1, . . ., un) because ui L if 1 ≤ i ≤ k. So K ⊆ L(uk+1, . . ., un).
26.
a. If u2 is algebraic over F, let f(u2) = 0, f(x) ≠ 0 in F[x]. Hence g(u) = 0 where g(x) = f(x2) ≠ 0. Thus u is algebraic over F.
27. Let 
be a family of subfields of E and write
be a family of subfields of E and write
Then 1 
F because 1 Fi for all i. If a, b F then a, b Fi for all i, so a + b, a − b, ab are in Fi. Thus a + b, a − b, ab are in F, so F is a subring of E. Finally, if a ≠ 0 in F let ai be its inverse in Fi. Thus aai = 1 = aa−1 where a−1 is the inverse in E, so ai = a−1. Thus a−1 Fi for all i, so a−1 F. Hence F is a field.
F because 1 Fi for all i. If a, b F then a, b Fi for all i, so a + b, a − b, ab are in Fi. Thus a + b, a − b, ab are in F, so F is a subring of E. Finally, if a ≠ 0 in F let ai be its inverse in Fi. Thus aai = 1 = aa−1 where a−1 is the inverse in E, so ai = a−1. Thus a−1 Fi for all i, so a−1 F. Hence F is a field.
29. Yes. Take 
, 
and 
. Then 
by Exercise 23 and 
is algebraic over 
, being algebraic over 
. On the other hand, π is transcendental over 
. For if f(π) = 0, 
, write
, and . Then by Exercise 23 and is algebraic over , being algebraic over . On the other hand, π is transcendental over . For if f(π) = 0, , write
where either g(x) ≠ 0 or h(x) ≠ 0. Thus 
. If h(π) = 0 then g(π) = 0, g(x) ≠ 0, a contradiction. If h(π) ≠ 0 then 
, contrary to Exercise 23.
. If h(π) = 0 then g(π) = 0, g(x) ≠ 0, a contradiction. If h(π) ≠ 0 then , contrary to Exercise 23.
31.
a. We show F(u) = Q where
Since u is transcendental over F, f(u) ≠ 0 whenever f ≠ 0. Thus Q is a subfield of E containing F and u, so F(u) ⊆ Q. But any subfield of E containing F and u must contain Q, so F(u) ⊇ Q. Thus F(u) = Q.
c. Put 
where a = f(u), g = g(u) as in (a). If h ≠ 0 in F[x] satisfies 
then, clearing denominators leads to a polynomial f(x, y) ≠ 0 in F[x, y] such that f(a, b) = 0. Write f(a, b) = ∑ ifi(a)bi. Then fi(a) = 0 for all i because b is transcendental; whence fi = 0 for all i because a is transcendental. So f(x, y) = 0; contrary to assumption.
where a = f(u), g = g(u) as in (a). If h ≠ 0 in F[x] satisfies then, clearing denominators leads to a polynomial f(x, y) ≠ 0 in F[x, y] such that f(a, b) = 0. Write f(a, b) = ∑ ifi(a)bi. Then fi(a) = 0 for all i because b is transcendental; whence fi = 0 for all i because a is transcendental. So f(x, y) = 0; contrary to assumption.
32.
a. Write 
and 
. Clearly 
We have 
, so
and . Clearly We have , so
Substituting 
leads to 
. Hence 
is in 
so, since p ≠ q, 
. Then 
, so 
.
leads to . Hence is in so, since p ≠ q, . Then , so .
c. We have 
, so [u2 − (p + q)]2 = 4pq. This gives u4 − 2(p + q)u2 + (p + q)2 = 4pq, so u4 − 2(p + q)u2 + (p − q)2 = 0 . Thus f(u) = 0 where f = x4 − 2(p + q)x2 + (p − q)2. If m is the minimal polynomial of u over 
, this shows m|f. But 
by (b), so we have m = f (as required) because both are monic.
, so [u2 − (p + q)]2 = 4pq. This gives u4 − 2(p + q)u2 + (p + q)2 = 4pq, so u4 − 2(p + q)u2 + (p − q)2 = 0 . Thus f(u) = 0 where f = x4 − 2(p + q)x2 + (p − q)2. If m is the minimal polynomial of u over , this shows m|f. But by (b), so we have m = f (as required) because both are monic.
33. We show that if u is algebraic over A then u is algebraic over F, so u A contrary to hypothesis. If f(u) = 0 where f ≠ 0 in A[x], let
Then 
is a finite extension of F by Theorem 6. Moreover f L[x] so L(u) ⊇ L is a finite extension by Theorem 4. Thus L(u) ⊇ F is finite by Theorem 5, so u L(u) implies u is algebraic over F.
is a finite extension of F by Theorem 6. Moreover f L[x] so L(u) ⊇ L is a finite extension by Theorem 4. Thus L(u) ⊇ F is finite by Theorem 5, so u L(u) implies u is algebraic over F.
35.
a. Let 0 ≠ u R. Then u−1 exists in E and we must show it is in R. Now u is algebraic over F because E ⊇ F is an algebraic extension, so let f(u) = 0, 0 ≠ f F[x], say
If a0 = 0 we can cancel u to reduce the degree. So we may assume a0 ≠ 0. Clearly n ≥ 1. If n = 1, a1u + a0 = 0 gives u F so u−1 F ⊆ R. If n > 1 then u(anun−1 + + a1) = − a0 so
+ a1) = − a0 so
c. If x and y are indeterminants over 
let 
denote the field of fractions of the integral domain 
Then define 
where u2 = x and 
If R = span
then E ⊇ R ⊇ F, R is an F -space, but R is not closed under multiplication; in fact uv ∉ R. For if uv R then uv = a + bu + cv where 
and so (since the characteristic is 2)
let denote the field of fractions of the integral domain Then define where u2 = x and If R = span then E ⊇ R ⊇ F, R is an F -space, but R is not closed under multiplication; in fact uv ∉ R. For if uv R then uv = a + bu + cv where and so (since the characteristic is 2)
a contradiction because x and y are indeterminates over F.
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