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A Lemma stated:

Let $F$ be a finite field, of prime characteristic $p$, and with algebraic closure $\overline{F}$. The polynomial $x^{p^{n}} - x$ has $p^{n}$ distinct zeros in $\overline{F}$.

The first line of the proof goes like this:

Since $\overline{F}$ is algebraically closed, $x^{p^{n}} - x$ factors into $p^{n}$ linear factors. So all that is left to show is that each factor does not appear more that once.

My question is how do we know that $\overline{F}$ is closed?

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    Isn't that the *definition*? That $\overline{F}$ is the intersection of all algebraically closed fields containing $F$; and that the intersection of algebraically closed fields is algebraically closed?2012-08-03
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    Or it's defined as an algebraic extension that is algebraically closed.2012-08-03
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    I was confusing it with the algebraic closure of $F$ *in some extension field $E$*. Since there is no such $E$ in this context, your definition must be the case. Clearly I should re-skim this chapter before going forward. Thanks!2012-08-03
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    You are certainly correct. We used Zorn's Lemma to prove that at least one such algebraically closed extension exists, and then appropriately defined the minimal one as you describe. Thanks again.2012-08-03
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    Check out Dummit and Foote, p.543, Prop 29 if that's what you're asking.2012-08-03
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    By definition, an algebraic closure of a field $K$ is an algebraically closed extension $\overline{K}$ of $K$ which is algebraic over $K$. So it's a matter of definition.2012-08-03
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    @AsafKaragila: I usually don't disagree with you, but here I do. Since there is no "universal extension field" of$~F$, I cannot see any setting in which such an intersection can be reasonably defined. Note also that while all algebraic closures of $F$ are isomorphic, they are _not canonically_ isomorphic; your description would sort-of define a canonical algebraic extension, but this does not exist. The other descriptions given in the comments do make sense.2015-03-07
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    @Marc: Of course, this is not a canonical construction. But pick any sufficiently large algebraically closed field, and just consider this construction working in that field. While this construction (and any construction of algebraic closures) lack canonicity, there is a very good reason why we say "the" algebraic closure.2015-03-07
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    The definition I had while asking this question (3 years ago), was different than the one most people use. The book I was using supplied a more direct concreted definition using roots of polynomials, but then proceeded to work with the abstract one instead. I found this confusing, because although I could clearly see that $\overline{F}$ *ought* to be algebraically closed, it was not yet established.2015-03-08

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The comments already stated as much, but I'm posting an answer to get this question out of the unanswered queue.

The algebraic closure of any field is algebraically closed by definition. Being algebraically closed is the key defining property of the algebraic closure. Details depend on what definition you use, but defining it as an algebraic field extension which is algebraically closed should be quite common and makes this property clear.

(Note however that this algebraic closure will no longer be finite.)

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    The definition used at the time was a different ( but equivalent one ). I agree the one you mention is more natural and now the one I think of [if ever I find myself in the world of algebra :) ]. At the time of asking I didn't know they were equivalent.2014-02-19