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Prove that a finite-dimensional extension field $K$ of $F$ is normal if and only if it has this property: Whenever $L$ is an extension field of $K$ and $\sigma :K\rightarrow L$ an injective homomorphism such that $\sigma(c) = c$ for every $c\in F$, then $\sigma(K)\subseteq K$.

First off, one thing I don't understand is why do they put $\sigma(K)\subseteq K$ instead of $\sigma(K) = K$? Unless I'm missing something $\sigma(K)\subset K$ (strict inclusion) is impossible if $\sigma$ is injective.

As to the main part of my question, I've proved this tautology in the forward direction (after significant effort), but I've been unsuccessful in proving the other direction.

By contrapositive seems like the most likely to succeed, so I assumed $K$ wasn't a normal extension of $F$ and thus there exists some irreducible polynomial $f(x)\in F[x]$ which has some but not all its roots in $K$. What I need to do now is find some field extension $L$ of $K$ and some injective hom $\sigma : K\rightarrow L$ which fixes $F$ but sends an element of $K-F$ to an element of $L-K$. This is tough, what makes the most sense to me is choosing $L$ to be the splitting field of $f(x)$, so the part of $f(x)$ which factors in $K[x]$ has to have at least one of its roots sent outside $K$, I'm thinking maybe I should send them to roots of $f(x)$ which only exist in $L$, but I just can't figure out how to construct a homomorphism from this.

I tried looking at the example of $x^3 - 2$ with $K = \mathbb{Q}(\sqrt[3]{2})$ and $L = \mathbb{Q}(\sqrt[3]{2},\sqrt[3]{2}\omega,\sqrt[3]{2}\bar{\omega}$) to get some intuition but it seems too trivial, is there a better example to help my understanding? Anyways hopefully someone can help me with this, thanks.

Addition: In the other direction I made extensive use of an injective hom $\sigma:K\rightarrow L$ implying the existence of an injective hom $\phi:K[x] \rightarrow L[x]$.

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What's your definition of normal? I might be able to guess from the fourth paragraph, but note that $f$ needs to be irreducible for your characterization there to hold. – Dylan Moreland Jul 4 '12 at 4:58
Ah yes sorry forgot to say that, $f(x)$ is irreducible in $F[x]$. Normal means if it is irreducible in $F[x]$ and has one root in $K[x]$ then it has all roots in $K[x]$ – Thoth Jul 4 '12 at 5:00
Note: in the contrapositive, you are not trying to find $\sigma$ such that $\sigma(K)\supset K$. The negation of "$\sigma(K)\subset K$" is not $\sigma(K)\supset K$; the negation is $\sigma(K)\not\subset K$. So all you need is to find some map $\sigma K\to L$ and some element $k\in K$ such that $\sigma(k)\notin K$. Perhaps... take one of the roots of your $f$, and make sure you map it to a root that is not in $K$? To construct such a homomorphism, remember the results about extending homomorphisms. – Arturo Magidin Jul 4 '12 at 5:08
@Arturo Magidin: you're right, thanks for pointing that out. – Thoth Jul 4 '12 at 5:26
up vote 1 down vote accepted

I think you're on the right track. What is written below mostly fleshes out and gives references for the plan given by Prof Magidin in the comments.

Let $\alpha$ be a root of $f$ in $K$, and view $K$ as being embedded in an algebraic closure $F^a$ of $F$. There is some root $\beta \in F^a$ of $f$ which is not contained in $K$, and there is an isomorphism $F(\alpha) \to F(\beta)$ over $F$ [As both of these fields are isomorphic to $F[X]/(f)$; see Proposition 2.1a of Milne for more discussion.]. Moreover, since $K/F$ is algebraic this extends to an embedding $\sigma\colon K \to F^a$ [See Corollary 2.8b of Milne for a sharper version of this claim.], and we've constructed this such that $\sigma(\alpha) \notin K$.

Regarding $\sigma(K) \subset K$ versus $\sigma(K) = K$: the answers to this question show that if $K/F$ is algebraic and $\sigma\colon K \to K$ is the identity on $F$ then $\sigma$ is surjective, so there is no difference.

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Let me know if the relation to the referenced propositions in Milne is unclear. I couldn't really find an online reference that lined up perfectly. Also, I have no qualms about using algebraic closures when they're aren't necessary, but you might! – Dylan Moreland Jul 4 '12 at 6:24
Also, we don't need finite-dimensionality anywhere — algebraicity is really the key — though it does make some of the proofs of the associated lemmas easier. – Dylan Moreland Jul 4 '12 at 6:33

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