Let $E$ be a (possibly infinite) field extension of $F$, and let $\Omega$ be an algebraic closure of $E$. I'm trying to prove that if $\sigma(E)=E$ for all $F$-algebra embeddings $\sigma \colon E\to \Omega$, then $E/F$ is normal.

The finite case is easy: Suppose $E=F(\alpha_1,\ldots,\alpha_n)$. Consider the splitting field $K$ of the polynomial $f:=m_{\alpha_1}\cdots m_{\alpha_n}$, where $m_\alpha=m_{\alpha,F}$ denotes the minimal polynomial of $\alpha$ over $F$. Let $\alpha=\alpha_i$ be any one of $\alpha_1,\ldots,\alpha_n$, and let $\beta$ be another root of $m_\alpha$. Then there's an $F$-algebra isomorphism $\tau\colon F(\alpha)\to F(\beta)$ taking $\alpha$ to $\beta$. But since $\tau$ fixes $F$, it takes $f$ to itself. Therefore, $\tau$ extends to an automorphism $\sigma$ of $K$. But $\sigma|_E$ is an $F$-algebra embedding of $E$ into $\Omega$, meaning $\sigma(E)=E$. So, $\beta=\tau(\alpha)=\sigma(\alpha)\in E$. Hence, all the roots of $f$ are in $E$. Thus, $K\subseteq E$. Since $E\subseteq K$ by definition, it follows that $E=K$. Hence, $E$ is normal.

How do I generalize this to the infinite case? (Note that this isn't homework. I'm doing some extra reading on my own.)

  • $\begingroup$ It really doesn't matter whether it is or not homework: people will still try to help you, but perhaps you won't be getting as many fully solved answers...which is something people really willing to learn mathematics should try to avoid. $\endgroup$ – DonAntonio May 1 '13 at 23:00
  • $\begingroup$ Hints are fine. $\endgroup$ – Avi Steiner May 1 '13 at 23:09

I think the infinite case is not that big a problem after you proved what you say you already did: let $\,p(x)\in F[x]\,$ be irreducible non-constant s.t. $\,p(w)=0\,$ for some $\;w\in E\;$ , and let $\,w_1:=w\,,\,w_2\,,\,...\,w_n\,$ be all its conjugates in $\,\Omega\,$ .

Now we have an isomorphism $\,F(w_1)\to F(w_2)\,$ , and this isomorphism can be extended to an embedding $\,E\to \Omega\,$ (if you're not sure about this you can check, for example, theorem 2.8 , chap.5.2, in Lang's "Algebra") , and since this embedding is an $\,F$-automorphism of $\,E\,$ then in fact $\,w_2\in E$...

Complete the argument above to show all the roots of $\,p(x)\,$ are in $\,E\,$ and thus $\,E/F\,$ is normal

  • $\begingroup$ Does Dummit & Foote have a similar theorem? I couldn't find anything when I looked. $\endgroup$ – Avi Steiner May 1 '13 at 23:51
  • $\begingroup$ In the 3rd. Edition of D&F, try Theorem 8 in chapter 13.1 (page 519) and, in particular, theorem 27 in chapter 13.4 (page 541). Lang's book ("Algebra") is a very good reference book, though I would never use it to teach and/or learn from it for beginners. I like though the way he proves this stuff, using Zorn's Lemma in a very beautiful way. Try to read it in your university's library or even perhaps in Google books - it is in page 233 of the revised third edition of Springer Verlag - $\endgroup$ – DonAntonio May 2 '13 at 8:42
  • $\begingroup$ I'll check out Lang. D&F's theorems 8 and 27 only cover the finite case, and I couldn't figure out how to generalize to the infinite. $\endgroup$ – Avi Steiner May 2 '13 at 17:50
  • $\begingroup$ I figured it out on my own! Thanks @DonAntonio. $\endgroup$ – Avi Steiner May 2 '13 at 18:39
  • $\begingroup$ Hey @AviSteiner, I'd love to see your solution, even as a new answer to your own question. It's the usual thing here...and good for you! $\endgroup$ – DonAntonio May 2 '13 at 21:41

The only part of the infinite case the was giving me trouble was, using the notation in DonAntonio's answer, showing that the isomorphism $F(w_1)\to F(w_2)$ extends to an embedding $E\to\Omega$. To prove this, I'll prove something slightly more general:

Theorem. Let $E$ be the splitting field of a family $\mathcal{F}$ of polynomials over a field $F$, let $\tau\colon F\to F'$ be an isomorphism, and let $E'$ be the splitting field of $\mathcal{F}':=\tau(\mathcal{F})$. Then there exists an isomorphism $\sigma\colon E\to E'$ extending $\tau$.

Proof. Consider the set $\mathcal{E}$ of isomorphisms $L\subseteq E\to L'\subseteq E'$ extending $\tau$---this is non-empty since it contains $\tau$ itself. Equip $\mathcal{E}$ with the partial order $\leq$ given by $\sigma_1 \leq \sigma_2$ iff $\sigma_2$ extends $\sigma_1$. Any totally ordered subset $\{\varphi_i\colon L_i\to L_i'\}_{i\in I}$ of $\mathcal{E}$ is clearly bounded above by the isomorphism $\varphi\colon \bigcup_{i\in I} L_i\to \bigcup_{i\in I} L_i'$ in $\mathcal{E}$ given by $\varphi(\alpha)=\varphi_i(\alpha)$ if $\alpha\in L_i$. Therefore, by Zorn's Lemming, the set $\mathcal{E}$ has a maximal element $\sigma\colon L\to L'$.

We show that $L=E$. To see this, suppose not. Then $L$ doesn't contain all the roots of some $f\in \mathcal{F}$, and the splitting field $M$ of $f$ over $L$ strictly contains $L$. So, by the finite version of this theorem, $\sigma$ extends to an isomorphism $\eta\colon M\to M'$, where $M'$ is the splitting field of $\sigma(f)$ over $L'$. But then $\sigma <\eta$, contradicting the fact that $\sigma$ is maximal. Hence, $L=E$.

Finally, we show that $L'=E'$. Let $\beta$ be a root $\beta$ of some $g\in \mathcal{F}'$. By definition, $g=\tau(f)$ for some $f\in \mathcal{F}$, meaning $\sigma(\alpha)=\tau(\alpha)=\beta$ for some $\alpha\in E$. Therefore, $\beta\in L'$. Hence, $L'\supseteq E'$. Thus, since $L'\subseteq E'$ by construction, $L'=E'$. q.e.d.


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