Algebraically Closed Fields

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Pathway
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Algebraically Closed Fields

Postby Pathway » Mon May 26, 2008 3:53 pm UTC

A post in another thread led to the Wikipedia article on algebraically closed fields, which got me thinking. I now have a question.

The article states that no finite field is algebraically closed. It also mentions that for infinite fields, R in particular is not algebraically closed. It also provides a proof that no subfield of R is algebraically closed.

It does not, however, either:
1) state that for any field F which is not algebraically closed, no subfield of F is algebraically closed , or
2) provide a counterexample to statement 1).

I don't know how to prove 1), and I have no clue how to find a counterexample for 2). Can anyone shed light on this?
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Re: Algebraically Closed Fields

Postby jestingrabbit » Mon May 26, 2008 4:27 pm UTC

Take [imath]\mathbb{C}[/imath] and extend it to [imath]\mathbb{C}(t)[/imath] then take the algebraic closure of that, call it [imath]\mathbb{K}[/imath]. [imath]\mathbb{K}[/imath] is algebraically closed, and has [imath]\mathbb{C}[/imath] as a subfield. So 1) is false and 2) is true.
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Re: Algebraically Closed Fields

Postby Ended » Mon May 26, 2008 6:41 pm UTC

Just to expand a bit:

C(t) is a field which is not algebraically closed. E.g. the polynomial x2+t in C(t)[x] has no root in C(t).

But C(t) has C as an algebraically closed subfield. So C(t) is a counter-example to 1).
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