Lecture 19 - Feb 19

Recap

Let be a field.

Case 1:

Every irreducible polynomial over is separable.

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Case 2:

Let be irreducible.

If , then is separable.

If , then

Let be the Frobenius map,

If is an isomorphism (i.e. surjective), then for each there exists such that

Hence

Thus is not irreducible — contradiction.

Therefore:

If the Frobenius map is an isomorphism, every irreducible polynomial is separable.

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Conclusion

Let be a field with .

If the Frobenius map

is an isomorphism, then:

1. Every irreducible polynomial over is separable.
2. The product of two distinct irreducible polynomials is separable.

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Proposition

Let be a field with .

Let be the Frobenius morphism.

If , then for every ,

is irreducible over .

Sketch of Proof

Suppose

Let be a root of in .

Then

In ,

Since is a UFD, any divisor has the form

Write with .

Then

From coefficients, we obtain , hence

contradiction.

Thus is irreducible.

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Perfect Fields

Definition

A field is perfect if every irreducible polynomial over is separable.

Examples

1. Every field of characteristic .
2. A field of characteristic where Frobenius is an isomorphism.
3. Every algebraically closed field.

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Primitive Element Theorem

Let be a finite extension.

1. is simple if there are only finitely many intermediate fields.
2. If is separable, then is simple.

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Lemma (Finite Groups)

Let be a finite group of order .

Suppose for every divisor ,

Then is cyclic.

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Proof Sketch

Let

If , choose .

Then has order .

Since

we get .

Hence consists precisely of the generators of , so

Summing over all divisors of ,

Thus equality holds and .

Hence has an element of order and is cyclic.

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Fact

If , then is a root of