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I have a question on the proof of the following proposition in Dummit and Foote (3rd edition, Proposition 31, p.544).


This section is what I entirely quote from the book.

Proposition. Let $K$ be an algebraically closed field and let $F$ be a subfield of $K$. Then the collection of elements $\overline{F}$ of $K$ that are algebraic over $F$ is an algebraic closure of $F$. An algebraic closure is unique up to isomorphism.

Proof. By definition $\overline{F}$ is an algebraic extension of $F$. Every polynomial $f(x) \in F[x]$ splits completely over $K$ into linear factors $x - \alpha$ (the same is true for every polynomial even in $K[x]$). But each $\alpha$ is a root of $f(x)$, so is algebraic over $F$, hence is an element of $\overline{F}$. It follows that all the linear factors $x - \alpha$ have coefficients in $\overline{F}$, i.e., $f(x)$ splits completely in $\overline{F}[x]$ and $\overline{F}$ is an algebraic closure of $F$.

......


It seems like the authors showed that every polynomial in $F[x]$ splits over $\overline{F}$, but in order to show that $\overline{F}$ is algebraic closure of $F$, shouldn't we show that $\overline{F}$ is algebraically closed? (That is, shouldn't we take $f(x)$ from $\overline{F}[x]$ instead of $F[x]$?) I don't know what I am missing, but I thought the proof should have been proceeded as follows:

Fix any $f(x) \in \overline{F}[x]$, which splits over $L$. Let $\alpha \in L$ be any root of $f$. Then $\alpha$ is algebraic over $\overline{F}$ and since $\overline{F}$ is algebraic extension of $F$, it follows that $\alpha$ is algebraic over $F$ so that $\alpha \in \overline{F}$.

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    $\begingroup$ Perhaps it depends what definition of "alegbraic closure of $F$" you're using. You could be using "smallest algebraically closed field containing $F$", in which case you're correct. Or you could be using "smallest field containing all roots of polynomials in $F$", in which case the proposition's proof is fine (and your comment would be part of the proof of a lemma that algebraic closures are always algebraically closed). $\endgroup$ Commented Jun 18, 2013 at 17:07
  • $\begingroup$ @GregMartin That's an answer. Can you copy that as an answer? $\endgroup$
    – Gil
    Commented Jun 18, 2013 at 17:58
  • $\begingroup$ @GilYoungCheong, notice that in prop. 29 D&F already proved that an algebraic closure is algebraically closed... $\endgroup$
    – DonAntonio
    Commented Jun 18, 2013 at 18:52
  • $\begingroup$ @DonAntonio Thank you. I guess this happens when I do not stick to one book for reading (and I usually can't). I should have looked at the section more carefully! $\endgroup$
    – Gil
    Commented Jun 19, 2013 at 1:23

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Perhaps it depends what definition of "alegbraic closure of $F$" you're using. You could be using "smallest algebraically closed field containing $F$", in which case you're correct. Or you could be using "smallest field containing all roots of polynomials in $F$", in which case the proposition's proof is fine (and your comment would be part of the proof of a lemma that algebraic closures are always algebraically closed).

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