Lecture 08 - Jan 30

Algebraic Closure

Let be a field.

An algebraic closure of is a field such that:

3. Every non-constant polynomial over has a root in .

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Algebraically Closed Field

A field is called algebraically closed if every non-constant polynomial over has a root in .

Observation

If is algebraically closed, then

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Proposition

Let , where is algebraically closed.

Define

Then

Proof Sketch

• If are algebraic over , then

  are algebraic over , so is a field.

• Clearly,

  so it is an algebraic extension of .

• Let be non-constant. Since is algebraically closed, has a root .

  Because satisfies a polynomial in , it is algebraic over , hence

Thus every polynomial over has a root in , so

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Proposition

Let be a field and its algebraic closure.

Then is algebraically closed.

Proof Sketch

Let

be a non-constant polynomial.

Let be a root of in some extension field of .

Then:

• is algebraic over .
• Since is algebraic, is algebraic over .
• Let be the minimal polynomial of over .

Because contains all roots of polynomials over ,

Hence every polynomial over has a root in , so it is algebraically closed.

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Polynomial Rings in Many Variables

Let be a set.

Define

to be the polynomial ring in variables indexed by .

An element if there exists a finite subset

such that

Equivalently,

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Theorem: Existence of Algebraic Closure

Let be a field.

Then an algebraic closure exists.

Proof Idea

We construct a field containing in which every polynomial over has a root.

1. Let

2. Consider the polynomial ring

   where each corresponds to a variable .

3. For each , consider the polynomial

4. Let be the ideal generated by all these elements:

5. One checks that is a proper ideal.

6. Let be a maximal ideal containing .

7. Define

Then:

• is a field,
• ,
• every polynomial over has a root in .

Finally, take the algebraic elements over inside to obtain .