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# algebraically closed

A field $K$ is *algebraically closed* if every non-constant polynomial in $K[X]$ has a root in $K$.

An extension field $L$ of $K$ is an *algebraic closure* of $K$ if $L$ is algebraically closed and every element of $L$ is algebraic over $K$. Using the axiom of choice, one can show that any field has an algebraic closure. Moreover, any two algebraic closures of a field are isomorphic as fields, but not necessarily canonically isomorphic.

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algebraic closure

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## Mathematics Subject Classification

12F05*no label found*

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## Recent Activity

Jul 5

new correction: Error in proof of Proposition 2 by alex2907

Jun 24

new question: A good question by Ron Castillo

Jun 23

new question: A trascendental number. by Ron Castillo

Jun 19

new question: Banach lattice valued Bochner integrals by math ias

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new question: young tableau and young projectors by zmth

Jun 11

new question: binomial coefficients: is this a known relation? by pfb

new correction: Error in proof of Proposition 2 by alex2907

Jun 24

new question: A good question by Ron Castillo

Jun 23

new question: A trascendental number. by Ron Castillo

Jun 19

new question: Banach lattice valued Bochner integrals by math ias

Jun 13

new question: young tableau and young projectors by zmth

Jun 11

new question: binomial coefficients: is this a known relation? by pfb

## Attached Articles

## Corrections

The word Polynomial linked to Sextic Equation by iddo ✓

merge and expand by mps ✓

merge? by mps ✓

missing "of" by niel ✓

merge and expand by mps ✓

merge? by mps ✓

missing "of" by niel ✓