I have been going through some problems in field theory recently, and problem that I came across was the following:
"Give an example of (or show it is not possible to have) a field extension $E/F$ that is finite and a ring homomorphism $\varphi : E \longrightarrow E$ that is the identity on $F$ but is not an isomorphism."
Among the examples I tried were the map $\varphi$ that sends the imaginary part of a complex number to zero and the Frobenius Endomorphism on $\Bbb{F}_p(t)$. The first one did not work out because the map $\varphi$ was not a ring homomorphism, and the second about the Frobenius Endomorphism does not work too because $t$ is transcendental over $\Bbb{F}_p$.
So then I realised that suppose we have such a ring homomrphism from $E$ to $E$ that is the identity on $F$. Then actually $\varphi$ gives us an $F$ - linear transformation $T$ from $E$ to itself if we just look at addition now. The part I am unsure is that I want to say that the kernel of $T$ is actually trivial. I can't actually relate this to the statement that $E$ and $F$ have no proper ideals because the kernel of $T$ is neither an ideal of $E$ nor $F$. How to get around this?
The idea I am trying to use is that if the kernel is trivial, then because $E$ is finite dimensional as a vector space over $F$ by the Rank - Nullity Theorem we have the $\varphi$ is always surjective, so it is not possible to have such a $\varphi$.