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 $\phi:K[x] \rightarrow L[x]$.