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Lecture 15 - Feb 12

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Continuation of Theorem (Normal Extensions)

We complete the proof of:

normal

is a splitting field

every irreducible polynomial over having one root in splits in .

(iii) (i)

Let

be a -algebra homomorphism.

Take . Since is algebraic, is algebraic over .

Let

be its minimal polynomial.

By assumption, if has one root in , then it splits completely in .

Since is a root, all roots of lie in .

Now is also a root of .

Hence

Thus

Since is algebraic, is surjective, so

Therefore is normal.

Proposition

Suppose

  • is normal,
  • is any algebraic extension.

Then:

  1. is normal.
  2. If both and are normal, then
    and are also normal.

Proof of (i)

Let

be an -algebra homomorphism.

Every element of is a finite sum of elements of the form

Then

Since fixes ,

Because is normal,

Hence

Since is algebraic, is surjective.

Thus

So is normal.

Proof of (ii)

Let

be a -algebra homomorphism.

If , then and .

Since and are normal,

Thus

Hence

By algebraicity, is surjective.

Therefore is normal.

Example

Let

Let be a root of .

Then the roots of are

Take

If is a -algebra homomorphism, then

for some .

Since

we get

are all conjugates.

Hence

is normal.

Separable Extension

Definition

Let be an algebraic extension.

Fix an embedding

into an algebraic closure .

Let

The separable degree of is defined by

Independence of Choice

The definition appears to depend on:

  • the embedding ,
  • the algebraic closure .

We show it is independent of these choices.

Claim

Let

be two embeddings into algebraic closures.

Then there is a bijection between the sets of embeddings

Sketch of Argument

We have:

The map

is an isomorphism.

Since both closures are algebraic over these subfields, this isomorphism extends to

Thus embeddings correspond under conjugation by .

Hence the separable degree does not depend on the chosen embedding or algebraic closure.

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