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Suppose $E/F$ is an algebraic extension, where every polynomial over $F$ has a root in $E$. It's not clear to me why $E$ is actually algebraically closed.

I attempted the following, but I don't think it's correct:

I let $f$ be an irreducible polynomial in $E[X]$. I let $\alpha$ be a root in some extension, so $f=m_{\alpha,E}$. Since $\alpha$ is algebraic over $E$, it is also algebraic over $F$, let $m_{\alpha,F}$ be it's minimal polynomial. I now let $K$ be a splitting field of $m_{\alpha,F}$, which is a finite extension since each root has finite degree over $F$.

If $m_{\alpha,F}$ is separable, then $K/F$ is also separable, so as a finite, separable extension, we can write $K=F(\beta)$ for some primitive element $\beta$. By assumption, $m_{\alpha,F}$ has a root in $E$, call it $r$. Then we can embed $F(\beta)$ into $r$ by mapping $\beta$ to $r$. It follows that $m_{\alpha,F}$ splits in $E$. Since $f\mid m_{\alpha,F}$, we must also have the $f$ is split in $E$.

But what happens if $m_{\alpha,F}$ is not separable? In such case, $F$ must have characteristic $p$. I know we can express $m_{\alpha,F}=g(X^{p^k})$ for some irreducible, separable polynomial $g(X)\in F[X]$. But I'm not sure what follows after that.

NB: I say $E$ is algebraically closed if every nonconstant polynomial in $E[X]$ has a root in $E$.

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marked as duplicate by Brahadeesh, YuiTo Cheng, user3417, José Carlos Santos, Yanior Weg Jul 8 at 16:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ There are (unfortunately) two slightly different definitions of algebraically closed. Which one are you using? $\endgroup$ – Eric Towers Mar 21 '14 at 22:43
  • $\begingroup$ @EricTowers I say $E$ is algebraically closed if every nonconstant polynomial in $E[X]$ has a root in $E$. $\endgroup$ – Nastassja Mar 21 '14 at 22:44
  • $\begingroup$ You cannot map $\beta$ to $r$ in general: a primitive element for the splitting field of a polynomial need not be a root of the polynomial. $\endgroup$ – Jan Ladislav Dussek Mar 21 '14 at 23:04
  • $\begingroup$ Your title and first sentence don't match. Does every polynomial in $F[x]$ have a root in $F$ or in $E$? $\endgroup$ – Greg Martin Mar 21 '14 at 23:08
  • $\begingroup$ @GregMartin I'm sorry, that was a typo, every polynomial in $F[X]$ has a root in $E$. $\endgroup$ – Nastassja Mar 21 '14 at 23:09
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This is a recent qual question at my school. :)

If $F$ is not perfect, say it has characteristic $p$. Let $F_{perf}$ be the extension of $F$ obtained by adjoining a root for $x^{p^n}-a$ for every $n \in \mathbb N, a \in F$. Since these polynomials are purely inseparable, $F_{perf}$ embeds (via a unique isomorphism) into $E$, and we may identify $F_{perf}$ with its isomorphic copy in $E$.

Two facts for you to verify: $F_{perf}$ is a perfect field, and every element of $F_{perf}$ is a root of $x^{p^n}-a$ for some $n \in \mathbb N$ and $a \in F$.

Now let $f$ be a polynomial with coefficients in $F_{perf}$, say $f(x) = x^k + a_{k-1} x^{k-1} + \cdots + a_0$. Then for sufficiently large $n$, $a_i^{p^n} \in F$ for all $i$. So $$f^{p^n}(x) = x^{k p^n} + a_{k-1}^{p^n} x^{(k-1)p^n} + \cdots + a_0^{p^n} \in F[x],$$ which has a root in $E$ by assumption. But $f$ and $f^{p^n}$ have the same roots, hence $f$ has a root in $E$. So without loss of generality, we may assume $F = F_{perf}$, and Bruno's answer gets us the rest of the way.

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  • $\begingroup$ Ah, so you want to take a perfect closure of $F$. May I ask, why is it that each $x^{p^n}-a$ is purely inseparable, and why is that needed to embed $F_{perf}$ into $E$? $\endgroup$ – Nastassja Mar 22 '14 at 1:05
  • $\begingroup$ Purely inseparable means that the polynomial has only one root. If $\alpha$ is a root of $x^{p^n}-a$, then $x^{p^n}-a = x^{p^n}-\alpha^{p^n} = (x-\alpha)^{p^n}$. Since each of these polynomials has only one root, they all split in $E$, which can't be said, a priori, for a polynomial with more than one root. $\endgroup$ – Dustan Levenstein Mar 22 '14 at 1:08
  • $\begingroup$ Thanks, I had not heard that terminology for polynomials before. $\endgroup$ – Nastassja Mar 22 '14 at 1:10
  • $\begingroup$ Done! Thanks both for your help. $\endgroup$ – Nastassja Mar 22 '14 at 1:55
  • $\begingroup$ lol, Bruno, thanks! :P $\endgroup$ – Dustan Levenstein Mar 22 '14 at 6:22
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Note: feel free to ignore the warzone in the comments; it's not really relevant anymore.

If $F$ is perfect, we can proceed like this. Let $f$ be a polynomial with coefficients in $F$. Let $K/F$ be a splitting field for $f$. Then $K=F(\alpha)$ for some $\alpha \in K$. Let $g$ be the minimal polynomial of $\alpha$ over $F$. Then $g$ has a root in $E$ by assumption, hence $E$ contains a copy of $F(\alpha)$, i.e. a splitting field for $f$.

Thus every $f\in F[X]$ splits in $E$. Now I claim that $E$ is algebraically closed. Let $E'/E$ be an algebraic extension and let $\beta \in E'$. By transitivity, $\beta$ is algebraic over $F$; let $h(X)$ be its minimal polynomial over $F$. By the above, $h$ splits in $E$, and therefore $\beta \in E$. Thus $E$ is algebraically closed.

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    $\begingroup$ Thank you. But how does the assumption on $E$ imply $f$ splits in $E$? If $f$ has a root $\beta\in E$, $f=(X-\beta)g(X)$ for $g\in E[X]$. If $\alpha=\beta$, we are done, otherwise $g(\alpha)=0$...not sure what follows next? $\endgroup$ – Nastassja Mar 21 '14 at 23:14
  • $\begingroup$ @Nastassja Oops, I misread the assumption. My bad. I'm not immediately sure how to prove it (and I'm actually not sure that it's true - where did you get this statement?). $\endgroup$ – Bruno Joyal Mar 21 '14 at 23:35
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    $\begingroup$ @BrunoJoyal, where and how did you prove that every non-constant pol. in $\;E[x]\;$ (!!) has a root in $\;E\;$ ? You don't even mess with any pol. over $\;E\;$ ! $\endgroup$ – DonAntonio Mar 22 '14 at 0:19
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    $\begingroup$ @BrunoJoyal, or I'm missing some rather simple point here, or you are...or something: in order to prove some field $\;B\;$ is an alg. clos. of another field $\;A\;$ you must prove THAT $\;B\;$ is both algebraic over $\;A\;$ and alg. closed. For the third time, where did you prove $\;E\;$ is alg. closed (the only interesting thing to do since we're given it is alg. over $\;F\;$) ? $\endgroup$ – DonAntonio Mar 22 '14 at 0:24
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    $\begingroup$ Ah! Now I get your point, @BrunoJoyal ....finally! It must be my two remaining living neurons are on vacation in Bahamas...Yes, you're right: if we're given (or if we prove) that $\;E\;$ is s.t. that any non-constant pol. in $\;F\;$ splits there then we're done. $\endgroup$ – DonAntonio Mar 22 '14 at 0:34

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