Let $K \prec F$ and $\tau : F \rightarrow$ is a homomorphism where $\tau(u) = u \forall u\in K$ then $\tau \in \text{Gal}_K F$ I'm having some difficulty going about trying to solve this problem. Here is the full problem statement.

Prove that, if $[F:K]$ is finite and $\tau : F \rightarrow F$ is a homomorphism such that $\tau(u) = u$ for all $u \in K$, then $\tau \in \text{Gal}_K F$. 

I know that I need to show that that $\tau$ is an isomorphism and I have a hunch that I need to use the fact that $\tau$ fixes all elements of $K$ and that $[F:K]$ is finite, and therefore that $F = K(S)$ for some $S \subseteq F$ where for all $x \in S$ that $x$ is algebraic over $K$. Furthermore, I know that a $K$-automorphism is completely determined by where it sends generating elements. I'm confused as to how to piece these pieces of information together and whether I'm even close.
Thanks!

It's given that $\tau$ fixes $K$, so $\tau$ is not the zero map.

It follows that the kernel of $\tau$ is $(0)$ (the only ideal of a field other than the field itself is the zero ideal).

Thus, $\tau$ is injective.

Regarding $F$ as a vector space over $K$, the map $\tau$ is a $K$-linear map (since $\tau$ is a ring homomorphism, and $\tau$ fixes $K$).

Hence, since $\tau$ is injective, and $[F:K]$ is finite, $\tau$ must be bijective.

It follows that $\tau$ is an automorphism of $F$, hence, since $\tau$ fixes $K$, we get $\tau \in \text{Gal}_K F$.
