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Assignment 04

10 min read

Question 1

Let be a perfect field of characteristic and let

be the Frobenius map.

Is an isomorphism?

Solution

Yes. The Frobenius map is always a ring homomorphism in characteristic , because

It is always injective on a field: if , then .

Since is perfect, every element of is a -th power. So is surjective.

Therefore is bijective, hence an automorphism of .

So the answer is

Question 2

Let be a field of characteristic and let be an irreducible polynomial over . Suppose is perfect.

The handwritten sheet appears to intend the standard statement that is reducible for every . That is what is proved below.

Solution

Let

Because , the derivative of is zero, so by the chain rule,

Now use the key fact about perfect fields:

  • every irreducible polynomial over a perfect field is separable;
  • a separable irreducible polynomial must have nonzero derivative.

So if were irreducible over , it would have to be separable, but its derivative is zero. That is impossible.

Hence is not irreducible.

Therefore, for every ,

Question 3

Let be a field of characteristic and let be a finite extension. Let .

Prove that is separable over if and only if

Solution

Set

Then .

If is separable, then

The extension is separable because is separable over .

On the other hand, is a root of

so is purely inseparable.

But a finite extension that is both separable and purely inseparable must be trivial. Therefore

If , then is separable

Assume and suppose, for contradiction, that is not separable over .

Then the minimal polynomial has zero derivative, so in characteristic it has the form

for some polynomial .

In particular, is a root of the lower-degree polynomial , so the degree of over is strictly smaller than the degree of over .

That contradicts .

So must be separable over .

Therefore

Question 4

Let and let be algebraically closed. Find infinitely many intermediate subfields of

Solution

Let

For each , define

Then

so each is an intermediate field:

Now show that there are infinitely many distinct such fields.

Suppose for two scalars . Then

Since , this implies . Then

Hence , so .

But satisfies

while

So , a contradiction.

Therefore whenever .

Since is algebraically closed, it is infinite, so the family

gives infinitely many intermediate subfields.

Thus the answer is:

Question 5

Let be a finite Galois extension. Let and let be its minimal polynomial over .

Suppose

Prove that

and explain why the are called the Galois conjugates of .

Solution

Because is Galois, it is normal and separable.

Step 1: Every is a root of the minimal polynomial

Let be the minimal polynomial of over . If , then fixes every coefficient of , so applying to the equation

gives

Hence each is a root of .

Step 2: Every root of the minimal polynomial is a Galois conjugate

Let be any root of in . Then the -embedding

exists because has the same minimal polynomial as .

Since is normal and separable, this embedding extends to an automorphism of over . So for some .

Thus the roots of in are exactly the elements .

Step 3: Factorization

Because is separable, has no repeated roots. Therefore it factors as the product of its distinct roots:

The numbers are called the Galois conjugates of because they are exactly the orbit of under the action of the Galois group.

So the result is

Question 6

Let be a finite field of characteristic and .

Show that for any , there exist such that

Solution

Write .

Case 1:

In characteristic , the Frobenius map is an automorphism of the finite field . So every is a square:

for some .

Then take .

So the claim holds.

Case 2: is odd

Let be a root of in an algebraic closure of , and set

Then is a quadratic extension, and every element of can be written uniquely as with .

Consider the norm map

Since is cyclic of order , the norm map is surjective onto , which has order .

So for any , there exists such that

But

Hence

For , simply take .

Therefore every element of is a sum of two squares:

Question 7

Find all monic irreducible polynomials of degree and over and .

Solution

We list them field by field.

Over

Degree 2

A monic quadratic over is irreducible iff it has no root in .

The only monic irreducible quadratic is

Degree 3

A monic cubic over is irreducible iff it has no root in .

The monic irreducible cubics are

So over the irreducible monic polynomials are:

and

Over

Degree 2

A monic quadratic over is irreducible iff it has no root in .

The monic irreducible quadratics are

Degree 3

A monic cubic over is irreducible iff it has no root in .

The monic irreducible cubics are

So over the irreducible monic polynomials of degrees and are exactly the ones listed above.

Question 8

Let be prime and let . Find the intermediate subfields of

corresponding to the subgroups of

Solution

The extension is Galois, and its Galois group is cyclic of order , generated by the Frobenius automorphism

Since , its subgroups are in one-to-one correspondence with the divisors of .

For each divisor , the unique subgroup of order is generated by , and its fixed field is

Thus the intermediate fields are exactly

So the correspondence is:

Question 9

Let be a monic irreducible polynomial of degree , and let be a root of .

(i) Prove that are also roots of .

(ii) Show that for integers ,

(iii) Conclude that

are all the roots of .

Solution

(i) Frobenius preserves roots

Because the coefficients of lie in , they are fixed by the Frobenius map. If

then applying the Frobenius map gives

Repeating this argument shows that is also a root for every .

(ii) Equality of powers

Since is irreducible of degree , we have

Hence .

The Frobenius automorphism on has order exactly . Therefore the orbit of under Frobenius has length .

So two Frobenius powers of are equal if and only if they differ by a multiple of :

(iii) All roots are obtained

From (i), the elements

are roots of .

From (ii), they are pairwise distinct. Since has degree , it has exactly roots in its splitting field.

Therefore these are all the roots of :

Question 10

Find the splitting field over of the following polynomials:

(i) , .

(ii) , .

Solution

(i) over

In , we have , so

But in characteristic ,

So the polynomial already splits over itself.

Hence the splitting field is

(ii) over

The nonzero elements of form a cyclic group of order . Since , the equation

has all its roots in .

Therefore splits completely over , so its splitting field is just

Question 11

Find the intermediate subfields of the following extensions and identify the Galois groups.

Solution

The handwriting in the sheet is most clearly readable for the following two extensions.

(i)

Let

The field is generated by two independent square roots, so it is the splitting field of

Hence is Galois.

The automorphisms are determined by independent sign changes:

So

The three nontrivial proper subgroups of this Klein four group correspond to the three quadratic subfields:

So the intermediate fields are exactly

(ii)

Let

There are two independent involutions:

They generate a group of order acting on over .

Thus

The three quadratic intermediate fields are the fixed fields of the three subgroups of order :

So the full list of intermediate fields is

If the second handwritten subitem intended a different quadratic-compositum field, the same method applies: compute the independent sign-change automorphisms and read off the fixed fields from the subgroup lattice.

Question 12

Let and be finite Galois extensions. Show that is finite Galois and that

is an injective group homomorphism.

Solution

Let be the compositum of and inside a fixed algebraic closure of .

Step 1: is finite

Since and are finite,

So is finite.

Step 2: is Galois

Because and are Galois, they are both normal and separable.

The compositum of separable extensions is separable, and the compositum of normal extensions is normal. Therefore is separable and normal.

Hence is Galois.

Step 3: Define the restriction map

For , define

This is a group homomorphism because restriction respects composition:

Step 4: Injectivity

If , then fixes both and pointwise. Since is generated by and , it follows that fixes all of . So .

Therefore is injective.

Hence

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