Assignment 05

Question 1

Let

Show that its cubic resolvent is

Solution

Let the roots of in a splitting field be . The cubic resolvent is built from the three pairings of the four roots:

The resolvent cubic is the monic polynomial having as its roots:

Now use Vieta’s formulas for :

A standard symmetric-polynomial computation gives

Therefore

That is exactly the cubic resolvent.

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Question 2

Find the Galois groups of

Solution

We use the standard quartic criterion:

• if the quartic resolvent is irreducible over and the discriminant is not a square, the Galois group is ;
• if the resolvent is irreducible and the discriminant is a square, the Galois group is .

(i)

Here , , . By Question 1, the cubic resolvent is

By the rational root theorem, any rational root must divide , so the only possibilities are . But

So is irreducible over .

The discriminant of is

which is not a square in .

Hence the Galois group is

(ii)

Now , , . The cubic resolvent is

Reduce modulo :

Checking shows that this cubic has no root in , so it is irreducible over . Therefore is irreducible over .

Its discriminant is

which is a square in .

Therefore

So the answers are:

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Question 3

Let be an irreducible quartic over a field , let be its splitting field, and let be its cubic resolvent.

Solution

Let the roots of be . The three roots of the cubic resolvent are

The Galois group acts on the four roots, hence also on the three pairings above.

(i) If the cubic resolvent has exactly one root in

Then fixes one of the three pairings, so it preserves a decomposition of the four roots into two pairs. This forces to be a transitive subgroup of the symmetry group of a square, hence

(ii) If the cubic resolvent splits completely over

Then all three pairings are fixed by . That leaves only the Klein four group acting on the roots, so

(iii) In case (i), the discriminant decides between and

If the discriminant is a square in , then . In the one-root case, the only transitive subgroup left is . If the discriminant is not a square, then is not contained in , so the only possibility is .

So, in the one-root case:

This is the standard quartic classification.

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Question 4

Let

be irreducible, let be its splitting field, and let be the discriminant of . Prove the following:

1. If is a square in , then .
2. If is not a square, but is a square in , then .
3. If neither nor is a square in , then .

Solution

The cubic resolvent of is obtained from Question 1 by setting :

So the resolvent roots are

(1) is a square

Then , so the resolvent splits completely over . By Question 3(ii), the Galois group is the Klein four group:

(2) is not a square, but is a square

Now the resolvent has exactly one rational root, namely . So by Question 3(i), the Galois group is either or .

The extra condition is exactly the cyclic case for this even quartic: it means the two quadratic subextensions glue together into a single cyclic tower of degree . Hence

(3) Neither nor is a square

Again the resolvent has only the one rational root , so the group is either cyclic or dihedral. Since the cyclic condition fails, the only possibility is the dihedral group of order :

So the classification is exactly:

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Question 5

For an integer , let be a primitive -th root of unity. If is odd, show that

Solution

Since , we immediately have

For the reverse inclusion, use that is odd. Then is an integer, and

Hence

So

Therefore

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Question 6

Show that

where .

Solution

Let

be the compositum of the two cyclotomic fields.

Since and , we have

Therefore

so

For the reverse inclusion, write

These two integers are coprime. Hence there exist integers such that

Now set . Then

Thus , so

Combining the two inclusions gives

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Question 7

Prove that

is a Galois extension. Find its Galois group.

Solution

Let

Then

so

Now consider the polynomial of over :

So is a root of

Step 1: is irreducible over

If had a rational root, then by the rational root theorem it would be among , but none of these are roots. So there is no linear factor.

If factored over into two quadratics, because has no or term we may write

for some . Expanding gives

Comparing coefficients with yields

If , then and , impossible over because would be roots of .

If , then , also impossible in .

So is irreducible, hence it is the minimal polynomial of .

Step 2: The field contains all roots of

The roots of are

We already have and . Let

Then

and also

Since and , we get

Thus contains all four roots of the minimal polynomial of . Therefore is the splitting field of over .

Since the extension is generated by a splitting field of a separable polynomial over , it is Galois.

Step 3: Determine the Galois group

Because is irreducible of degree , we have

So the Galois group has order .

Now define an automorphism by sending

This is allowed because is also a root of the minimal polynomial and lies in the field.

We compute

Then

Hence

So has order .

Therefore the Galois group is cyclic of order :

So the extension is Galois and its Galois group is