statement theorem
Existence of an algebraic closure
Statement
Every field has an algebraic closure.
Remarks
One proof: take a polynomial ring over with one variable for each root of each monic irreducible polynomial over . A maximal ideal of that contains the relations gives a field in which every polynomial over splits. Repeating the step gives fields , and their union is an algebraic closure of . The proof uses Zorn's lemma, through the existence of maximal ideals. The weaker ultrafilter lemma is enough. For or , the complex numbers give a direct proof.
Explanations
L1 human
Every field sits inside an algebraically closed field that is algebraic over it. The general proof needs a form of the axiom of choice, because a field can have very many polynomials.
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What this needs
To read this node, read these 30 nodes in this order. The foundations come first, and the list ends with this node.
- no evidence level
- Group concept
- Ring definition
- Commutative ring definition
- Ideal definition
- Polynomial ring definition
- Subring definition
- Unit of a ring definition
- Zero divisor definition
- Maximal ideal definition
- Field definition
- Polynomial evaluation definition
- Integral domain definition
- Existence of maximal ideals statement
- Root of a polynomial definition
- Subfield definition
- Irreducible element definition
- Field extension definition
- Split polynomial definition
- Irreducible polynomial definition
- Degree of a field extension definition
- Algebraic element definition
- Intermediate field definition
- Algebraically closed field definition
- Finite extension definition
- Algebraic extension definition
- Tower law statement
- Finite extensions are algebraic statement
- Algebraic closure definition
- An algebraic extension of an algebraic extension is algebraic statement
- Existence of an algebraic closure statement
Sources
- Web https://en.wikipedia.org/wiki/Algebraic_closure, Existence of an algebraic closure and splitting fields
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