Lecture 13 - Feb 9

Recap

If is finite, then

If is finite, then

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Theorem

is a Galois extension

there exists such that

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Splitting Field

Let be a field and let be a family of polynomials over , i.e.,

A field is called a splitting field of if:

1. Every polynomial in splits completely into linear factors in .

2. is generated by the roots of polynomials in .

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Existence of Splitting Field

Let be an algebraic closure of .

For each , contains all roots of .

Define to be the subfield of generated by all roots of polynomials in .

Then:

• is a splitting field of .
• is algebraic.

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Examples

(i)

Splitting field:

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(ii)

The roots are:

where is a primitive cube root of unity.

Splitting field:

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(iii)

Splitting field:

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(iv)

Factorization:

and further:

Splitting field:

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Proposition

Let be a field and a family of polynomials.

Let and be two splitting fields of .

Then for any -algebra homomorphism

we have

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Proof (Sketch)

Step 1: Single Polynomial

Assume .

Let

in .

Applying :

in .

Since is a splitting field of , the roots of in are exactly

Thus

Since

we get

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Step 2: General Case

If is infinite, take a finite subfamily and apply Step 1.

Write:

Since for each finite ,

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Corollary

If and are two splitting fields of a family , then

as -extensions.

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Example

Let .

Then

for some .

Then is the splitting field of

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Further Example

Let

Case 1: reducible

If

and is reducible over ,

then the splitting field is .

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Case 2: irreducible quadratic

Suppose

Then the roots are

Splitting field:

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Case 3: irreducible cubic

Suppose

in .

If ,

then is the splitting field.

If not, then

is the splitting field.