Lecture 04 - Jan 15

Example: Degree of

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Example: Algebraic Closure of in

Let

and let

Then

Question 1

Is

For every ,

since it satisfies

Hence

But

Therefore,

so

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Question 2

Is countable?

Every algebraic number is a root of some polynomial

For fixed degree , the set of such polynomials is in bijection with

which is countable.

Hence the set of all polynomials with rational coefficients is countable.

Each polynomial has finitely many roots.

Therefore the set of all algebraic numbers is a countable union of finite sets, hence countable.

Thus is countable.

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Example

Let

Then

Since

we have

Also

so

A minimal polynomial computation gives

leading to

Thus

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Example: ,

Let

Then

Since

we get

Now is algebraic over , and

Hence

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Simple Extension

Let

Take the ordered basis

for .

One checks that

generate the same field.

Since

and

we conclude

Thus the extension is simple.

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Field Generated by a Set

Let .

• The smallest field containing and is denoted

• The smallest ring containing and is denoted

Explicitly,

The field is the field of fractions of .

If

we say is generated by over .

If the elements of are algebraic over , then is algebraic over .

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Composite Field

Let and be field extensions.

The composite field of and over is the smallest field containing both and .

It is denoted by

Equivalently,

Any element of can be written as

where

and the denominator is nonzero.

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Proposition

If and are algebraic extensions, then

is algebraic (and similarly is algebraic).