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Must an algebraically closed field necessarily be an algebraic closure of some field?

In other words, does there exist an algebraically closed field that is NOT an algebraic closure of ANY field?

(This question popped up in my mind while I was reading the definition of an 'algebraic closure' of a field -- a field extension E of F is an algebraic closure of F if 1) E is an algebraically closed field and 2) E is an algebraic extension of F )

Thanks!

stoneaa
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    any algebraically closed field is the algebraic closure of itself. – Sasha Jun 14 '21 at 07:42
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    The perhaps more interesting question is whether any algebraically closed field is the algebraic closure of some proper subfield. I could swear I've seen that question on MO before, but I can neither find it again nor recall the answer right now. – Thorgott Jun 14 '21 at 09:57

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A trivial answer is that any algebraically closed field is the algebraic closure of itself. However, every algebraically closed field also contains a subfield, which is not algebraically closed - its prime field. Indeed, $\Bbb Q$ is not algebraically closed, as well as $\Bbb F_p$.

Reference: No finite field is algebraically closed

Perhaps the following question is more interesting for you: is there a field extension $L/K$, where $L$ is algebraically closed, while $L/K$ is not an algebraic extension?

This has a positive answer, see here:

Algebraically closed field extension can be not algebraic?

Dietrich Burde
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