MATH 202 - ALGEBRA IV - HW 3 SOLUTIONS

CHRIS LEBAILLY, MIGUEL-ANGEL MANRIQUE, WEI YUAN

1. Section 14.2

3.

Proposition. Let K be the splitting field of the separable polynomial (x22)(x23)(x25) Q[x]. ThenGal(K/Q) = Z/2Z Z/2Z Z/2Z and the set of all of subfields of K containing Q is

S = {Q,Q(

2),Q(

3),Q(

5),Q(

6),Q(

10),Q(

30),Q(

15),Q(

2,

3),

Q(

2,

5),Q(

3,

5),Q(

2,

15),Q(

3,

10),Q(

5,

6),Q(

6,

10),K}.

Proof. It follows that K = Q(

2,

3,

5) = Q(

2,

3)(

5). The field Q(

2,

3) has degree 4 over Q beExample 4 of 14.1 in [1]; since 5 6 Q(2,3) and 5 is a root of x2 5, the degree of Q(2,3)(5)over Q(

2,

3) is equal to 2; therefore

[K : Q] = [Q(

2,

3)(

5) : Q]

= [Q(

2,

3)(

5) : Q(

2,

3)][Q(

2,

3) : Q]= (2)(4)

= 8.

Hence, by Theorem 14 of 14.2 in [1], # Gal(K/Q) = 8.Any automorphism Gal(K/Q) must map the generators 2, 3 and 5 to a root of their respective

minimal polynomials. Thus, because is defined by where it maps the generators

2,

3 and

5, we maycharacterize by

:

2 7 23 7 35 7 5.

Since # Gal(K/Q) = 8, these eight distinct possibilities must be elements of Gal(K/Q). Furthermore, for1, 2, 3 Gal(K/Q) such that

1 :

2 7 23 7 35 7 5

, 2 :

2 7 23 7 35 7 5

and 3 :

2 7 23 7 35 7 5

,

it follows that

Gal(K/Q) = 1 2 3 = Z/2Z Z/2Z Z/2Z.There are 7 elements of order 2 (all non-identity elements), and their fixed fields are the 7 bi-quadratic

fields given in the statement. There are 7 subgroups of order 4, and their fixed fields are the 7 quadraticfields listed. Corresponding to 1 and to G we have K and Q. These subfields of K exhaust the possibilitiesby the bijective correspondence of Theorem 14 of 14.2 in [1] to complete the proof.

Date: 06/01/2011.

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6.

Proposition. Let K = Q( 8

2, i) and let F1 = Q(i), F2 = Q(

2) and F1 = Q(2). Then Gal(K/F1) = Z8,

Gal(K/F2) = D8 and Gal(K/F3) = Q8.Proof. Let be a primitive eighth root of unity. By the discussion in 14.2 of [1],

Gal(Q( 8

2, i)/Q) =, : 8 = 2 = 1, = 3

(1)

for the automorphisms and of K such that

:

8

2 7 82i 7 i 7 5

and :

8

2 7 82i 7 i 7 7.

From these relations, it follows that F1 is the fixed field of H1 = , F2 is the fixed field of H2 =2,

and F3 is the fixed field of H3 =

2, 2

; therefore, by Corollary 11 of 14.2 in [1], Gal(K/Fi) = Hi for all

i {1, 2, 3}.Also by the relations in (3), 8 = 1 and so H1 is a group of order 8 containing an element of order 8.

Thus H1 is isomorphic to the cyclic group Z8. Similarly, we have the equations (2)4 = 1, 2 = 1 and

2 = () = (3) = 1. Hence

H2 =2, : (2)4 = 2 = 1, = 1

.

Since these generators and relations uniquely define the dihedral group of order 8, H2 = D8. Lastly, therelations in (3) also imply that (2)4 = 1, (3)4 = 1, 2(3) = (3)12 and (2)2 = 4 = (3)2, wefind that

H3 =2, 3 : (2)4 = (3)4 = 1, 2(3) = (3)12, (2)2 = (3)2

.

Therefore H3 = Q8. 10.

Proposition. Let K be the splitting field over Q of x8 3 Q[x] and let be a primitive eighth rootof unity. The Galois group Gal(K/Q) is the group of automorphisms a,b : K K indexed by (a, b) (Z/8Z) (Z/8Z), where a,b fixes Q and

a,b :

{8

3 7 a 83 7 b. (2)

This identification yields an isomorphism

Gal(K/Q) = (Z/8Z)o (Z/8Z) (semi-direct product),where the multiplicative group (Z/8Z) acts on the additive group Z/8Z by multiplication.

Proof. The eight distinct roots of x8 3 are given by a 83 for a {0, 1, . . . , 7}. Thus K = Q(, 83). Notethat by the discussion in 14.2 of [1], Q() = Q(2, i). We claim that 2 6 Q( 83). Granting this, we seethat

[Q(

2,8

3) : Q] = 2 [Q( 8

3) : Q] = 16.Then since i cannot be embedded in R, we have that K is the degree 2 extension of Q(

2, 8

3) obtained byadjoining i, and hence [K : Q] = 2 16 = 32.

There are various ways to prove the claim that

2 6 Q( 83); perhaps the simplest is to use Eisensteinscriterion with the polynomial x8 3 over the ring Z[2] with the maximal ideal (3); this shows that thepolynomial is irreducible over the UFD Z[

2], hence irreducible over Q(

2) by Gausss Lemma, and hence

[Q(

2, 8

3) : Q] = 16, proving the claim.For any Gal(K/Q), the generators 83 and must be sent to roots of their respective minimal

polynomials. That is, is of the form a,b as in (4). Since there are exactly 32 pairs (a, b) and [K : Q] = 32,we see that each a,b does indeed yield an element of Gal(K/Q).

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It remains to determine the group structure of Gal(K/Q). We calculate that

a,b c,d( 8

3) = a,b(c 8

3) = a+bc8

3.

a,b c,d() = a,b(d) = bd.Therefore a,b c,d = a+bc,bd, which gives the semi-direct product as desired:

Gal(K/Q) = (Z/8Z)o (Z/8Z).

12.

Proposition. Let K be the splitting field over Q of x4 14x2 + 9 Q[x] and let 1 =

7 + 2

10. ThenK = Q(1), and Gal(K/Q) = Z/2Z Z/2Z, the Klein 4-group.Proof. Let 2 =

7 210. A straightforward computation using the quadratic formula shows that 1

and 2 are the distinct roots of x4 14x2 + 9. So K = Q(1, 2). Since 12 =

49 40 = 3, we see that2 Q(1); so in fact K = Q(1).

We claim that [K : Q] = 4, i.e. that x4 14x2 + 9 is irreducible. One can prove this directly by exploringthe possible factorizations (the factorizations will be of the form (x2 + ax 3)(x2 ax 3) or the samewith (3, 3) replaced by (1, 9); now solve for a and show that a is not rational). Here is an alternate method.Since we know how the polynomial factors in K, and neither 1 nor 2 is rational, a factorization occursonly if (x+1)(x2) has rational coefficients. The constant term is indeed rational (namely, 3), but thex coefficient is 1 2, and we calculate

(1 + 2)2 = 20, (1 2)2 = 8,

so in particular neither of 1 2 is rational.In any case, regardless of which method we chose, we see that the polynomial is irreducible and [K : Q] = 4.

To figure out the structure of the Galois group, one method is to note that by our calculations above, wesee that K contains

20 = 2

5 and

8 = 2

2, and also it clearly contains 21 = 7 + 2

10. Hence

K = Q(

2,

5) is a biquadratic extension with Galois group (Z/2Z) (Z/2Z).Another way to see this if we did not do the calculation above with (1 2)2 is to directly write down

the automorphisms. We know that Gal(K/Q) acts transitively on the roots, so the Galois elements aredetermined by the possible images of 1; we have the identity and the elements:

1(1) = 2; 2(1) = 1; 3(1) = 2.Since 2 = 3/1, it is easy to compute the action of these automorphisms on 2:

1(2) = 1; 2(2) = 2; 3(2) = 1.With these formulas it is easy to check that 2i = 1 for each i = 1, 2, 3. Hence Gal(K/Q) is the Klein 4-group.

17. This is similar to the proof of the next exercise.

18. With notation as in the previous problem define the trace of from K to F to be

TrK/F () =

(),

a sum of Galois conjugates of .

(a) Prove that TrK/F () F .Proof. The sum here is taken as follows. Let L be a Galois extension of F containing K, andwrite G = Gal(L/F ). Let H G be the subgroup corresponding to K. Then the sum runs overrepresentatives for the left cosets of H in G. These are the embeddings of K in L (equivalently,of K in F ) by the Fundamental Theorem of Galois Theory.

To begin, note that the trace is well-defined, i.e. () depends only on the left H-coset of .Since H fixes K we see that if H = H, then 1() = h() = for some h H. Therefore,() = ().

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To see that TrK/F () F , let S be a set of coset representatives for H, and notice that if Gal(L/F ) then S is still a set of coset representatives for H. In other words, the action of Gby left multiplication simply permutes the left H-cosets. It follows that (TrK/F ()) = TrK/F ()for all Gal(L/F ), and since the only elements fixed by all the automorphisms of L/F are thosein F , we have TrK/F () F .

(b) Prove that TrK/F (+ ) = TrK/F () + TrK/F (), so that the trace is an additive map from K toF .

Proof. By the definition of trace,

TrK/F (+ ) =

(+ )

=

(() + ())

=

() +

()

= TrK/F () + TrK/F (),

so the trace map is additive. (c) Let K = F (

D) be a quadratic extension of F . Show that TrK/F (a+ b

D) = 2a.

Proof. The quadratic extension K/F is a Galois extension with Gal(K/F ) = {1, }, where (a +bD) = a bD. As such,

TrK/F (a+ bD) = a+ b

D + (a+ b

D)

= a+ bD + a b

D

= 2a.

(d) Let m(x) be as in the previous problem. Prove that TrK/F () = nd ad1.

Proof. Begin by noticing that

(x ()) = xn (

())xn1 +

= xn TrK/F ()xn1 + .

Also,

(m(x))nd = (xd + ad1xd1 + + a0)nd

= xn +n

dad1xn1 + + a

nd0 .

In the following problem we prove that(x ()) = (m(x))

nd , so equating the two xn1 terms

gives the desired result.

20. With notation as in the previous problems (beginning with 17) show more generally that(x()) =

(m(x))nd .

Proof. Let f(x) =(x ()). Let H G be the subgroup corresponding to F (), and observe that it

follows from F F () K that H H . We have the following diagram of fields, with certain extensionslabelled by their Galois groups:

4

LH

H

G

K

F ()

F

Now we claim thatm(x) =

H

(x ()),

where the product runs over the left cosets of H in G. Indeed, it is clear that is a root of the producton the right (taking = 1), and that this product is in F [x] by the argument of #19 part (a). Sincedegm = [F () : F ] = [G : H

], we have the desired equality.Lets return to the product f(x) =

(x ()), with the product running over all left H-cosets. We

have that() = () 1 H H = H .

Since H H , each left H -coset is a disjoint union of left H-cosets. The number of H-cosets in this disjointunion is

[H : H] = [G : H]/[G : H ] = n/d,where n = [K : F ] and d = [F () : F ] = degm.

Therefore, in the product defining f(x), each distinct root () is repeated nd times, and hence f(x) =

(m(x))nd as desired.

References

[1] Dummit and Foote, Abstract Algebra, Third Algebra.

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