Math 121 Homework 5 Solutions

Math 121 Homework 5 Solutions

Problem 2, Section 14.1. Let : C - C be the complex conjugation, defined by (a + bi) = a - bi. Prove that is an automorphism of C. First Solution. Every element of C may be written uniquely as a + bi. To see that is a ring homomorphism, we must check that

((a + bi) + (c + di)) = (a + bi) + (c + di)

and ((a + bi)(c + di)) = (a + bi) (c + di).

The first identity is easy to check. The second means

ac - bd - (ad + bc)i = (a - bi)(c - di),

which is also easy to check. Second Solution. Note that C = R(i) is the splitting field of the irreducible polynomial x2 + 1. This has roots i, -i an it follows from Theorem 8 on page 519 that there is an isomorphism : C - C over R such that (i) = -i. Now (a + bi) = (a) + (b) (i) = a - bi, that is, is complex conjugation. This proves that complex conjugation is an automorphism.

Problem 3, Section 14.1. Determine the fixed field of complex conjugation on C. Solution: Any complex number may be written uniquely as a + bi with a, b R. If (a + bi) = a + bi then a - bi = a + bi which is equivalent to b = 0 and so a + bi = a R. Therefore the fixed field of complex conjugation is R.

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Problem 4, Section 14.1. Prove that Q 2 and Q 3 are not isomor-

phic.

Solution. If there is an isomorphism :

2

= a + b 3satisfies

a+b 3

2

=

2.

Q

2 - Q

3 and then

That is, a2 + 3b2 + 2ab 3 = 2.

Note that 1 and 3 are linearly independent over Q, so 2ab = 0, that is,

either a = 0 or b = 0. Thus either a2 = 2 or b2 = 2/3. Neither of these

equations may be solved with a or b in Q, which is a contradiction.

Problem 1,Section 14.2. Determine the minimal polynomial over Q for the element 2 + 5.

Solution. Let f (x) be the minimal polynomial for = 2 + 5. If is any root of f (x) then by Theorem 8 on page 519 there is an isomorphism : Q() - Q() such that () = . By Theorem 27 on page 541, we may extend to an automorphism of the splitting field E of f over Q. Now consider 2 . Since 2 is aroot of the polynomial x2 - 2 over Q, so is 2 . Therefore 2 = ? 2, and similarly 5 = ? 5. Therefore

= 2 + 5 = ? 2 ? 5.

Thus the possible roots of 2+ 5 are among the four element set ? 2 ? 5

and so f divides the polynomial

x- 2- 5 x- 2+ 5 x+ 2- 5 x+ 2+ 5 .

This equals

x4 - 14x2 + 9.

To show that this is the minimal polynomial, it is sufficient to show that it

is irreducible over Q. None of its roots are in Q, so if itis reducible, it splits as two quadratic factors. One of these will have 2 + 5 as a root. Hence

one of the quadratic factors is one of

x - 2 - 5 x - 2 - 5 x- 2- 5

x - 2 + 5 = x2 - 22x - 3,

x + 2 - 5 = x2 - 25x + 3,

x + 2 + 5 = x2 - 2 10 - 7.

It is easy to see that none of these are in Q[x], so x4 - 14x2 + 9 is irreducible over Q, and this is the minimal polynomial.

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Section 14.2, Problem 5. Prove that the Galois group of xp - 2 over Q ab

for p a prime is isomorphic to the group of matrices 0 1 for a, b Fp,

a = 0.

Solution. The polynomial is irreducible by Eisenstein's criterion. So its

splitting field is generated by the roots of xp - 2 which are i (0 i < p)

where = 21/p and = e2i/p. Thus the splitting field is Q(, i). Note that

Q()/Q has degree (p) = p - 1, while Q()/Q has degree p. These degrees

are coprime, so [Q(, i) : Q] = p(p - 1) by Corollary 22 on page 529.

Let m Z be prime to p. We will show that there exists Gal(K/Q)

such that

m() = m() = m

.

(1)

Since K/Q is Galois, K/Q() is also Galois and has degree p - 1. Indeed, K = Q(, ) so the effect of Gal(K/Q()) is determined by () which must be a primitive p-th root of unity. There are p - 1 elements of the Galois group and exactly p - 1 such p-th roots of unity, so Gal(K/Q()) must be transitive on these. Therefore we can find m as required.

Now we will show that Gal(K/Q) contains an automorphism such that

() = , () = .

(2)

Indeed, and are roots of the same irreducible polynomial xp - 2,

so we may find an automorphism 1 such that 1() = . Now 1() is a primitive p-th root of unity, say 1() = k, p k. Then m(k) = for some . (So if k is the image of k in F?p , m = 1/k.) Then m1 has the desired effect (2).

The elements mk and kmm have the same effect, km, m.

So mkm-1 = km. now the "affine group" G of matrices

ab 01

for

a, b Fp, a = 0 also has generators

m

tm =

1,

s=

11 1

,

so

sk =

1k 1

.

subject to the same relations tmskt-m1 = skm. Therefore there is an isomorphism Gal(K/Q) G in which m tm and s.

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Section Q 2,

14.2, Problem 6. Let K F3 = Q -2 . Prove that

= Q 8 2, Gal(K/F1)

i =

and let F1 = Q(i), F2 = Z8, Gal(K/F2) = D8 and

Gal(K/F3) = Q8.

The book devotes several pages to this example. However we will solve this from scratch.

Solution. First let usobserve that K is Galois and of degree 16 over Q. To begin with, = 8 2 is the root of the polynomial x2 - 2 which is

irreducible by Eisenstein's criterion, so [Q() : Q] = 8. Now i / Q() since

Q() R. Therefore [Q(, i) : Q] = 2 and so [Q(, i) : Q] = [Q(, i) :

Q()][Q() : Q] = 16. The field Q(, i) contains the primitive 8-th roots of

unity since it contains

2=

4

and

hence

8

=

1 (1 + i) 2

=

e2i/8.

Hence

it

contains all the roots of

7

x8 - 2 = (x - k8).

k=0

It is thus a splitting field for this polynomial, and Galois over Q.

Lemma 1. Let = k8 be any root of x8 - 2. Then there exist elements , Gal(K/Q) such that

() = , (i) = i,

and

() = , (i) = -i.

Proof. There exists : Q() - Q() such that by Theorem 8 on page 519. By Theorem 27 on page 541, this extends to an automorphism of the splitting field K.

We show that i / Q(). Otherwise -1(i) Q() R and -1(i) is a root of x2 + 1 = 0, that is, -1(i) = ?i which is not real, so this is a contradiction.

Since Q() has degree 8 over Q and i / Q(), Q(, i) has degree 16 over Q and hence Q(, i) = K. Since K is Galois over Q, it is Galois over Q() and | Gal(K/Q())| = 2. Let be the generator, so that () = and (i) = -i. Then and both take to , and one takes i to i, the other i to -i. This gives us and .

We may nowhandle the three cases. Each of Gal(K/Q(i)), Gal K/Q 2 and Gal K/Q -2 has order 8, which is the degree of the extension.

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First, the Lemma gives us an element Gal(K/Q) such that () = 8 and (i) = i. This is then in Gal(K/Q(i)). We find that 2 = - 2 since

2 = 4 and - 2 = (8)4. Therefore

1

1

(8) =

(1 + i) 2

=

(1 -2

+

i)

=

-8.

We may now calculate that 2() = -28 = -i and 2(8) = 8 so 4() = - and 4(i) = i. We see that has order 8 and therefore Gal(K/Q(i)) is

cyclic of order 8.

Next, let us show that Gal K/Q 2

= D8. The Lemma gives us two

elements , Gal(K/Q) such that

() = i, (i) = i,

and

() = , (i) = -i.

Since 2 = 4, both of these fix 2 and so they are in Gal K/Q 2 . It

is easy to see that has order 4 and has order 2, and -1 = -1, so they

generate a dihedral group of order 8. Finally, let us show that Gal K/Q -2

= Q8. The Lemma gives us

two elements ? and such that

?() = 8 ?(i) = -i

,

and

() = -8 1 (i) = -i

.

We will show that ?, satisfy

?2 = 2, ?4 = 1, ?-1 = ?-1.

These relations define the quaternion group Q8. Since -2 = i4, ? sends

-2 to (-i)(8)4 = -2 and is in Gal K/Q -2 ; similarly is also

in

this

Galois

group.

Using

8

=

1 -2

(-1

+

i)

we

also

calculate

?(8) = (8) = 38.

Using this we calculate

?2() = - ?2(i) = i

and has the same effect. Thus ?2 = 2 and ?, have order 4. The relation ?-1 = ?-1 may now be checked.

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