# Ring homomorphism between two field extensions that is the identity over ground field

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 $\varphi$.

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The kernel of a ring map is always an ideal of the source ($E$ in this case). – Keenan Kidwell Apr 5 '12 at 1:11
@KeenanKidwell I know that the kernel of $\phi$ as a ring map is trivial. However how does this mean then that the kernel of $\phi$ as an $F$ - vector space linear transformation is trivial? – user38268 Apr 5 '12 at 1:15
@Benjamin: Because the two notions of kernel coincide here. This is something that needs checking if you defined kernel in terms of category theory but it's entirely obvious from the concrete definition. – Zhen Lin Apr 6 '12 at 23:51

Right, you are correct. Namely, since $\phi$ fixes $F$ you know that $\phi$ is actually $F$-algebra map. The fact that $E$ is a field and $\phi$ a ring map tells you that $E$ is necessarily injective, but since $\phi$ is an endomorphism of a finite dimensional $F$-space which is injective it is necessarily surjective (by the Rank-Nullity theorem) and thus an isomorphism.
How can I tell that the kernel of $\phi$ is trivial? Like I said it is neither an ideal of $E$ nor an ideal of $F$, so how do I know this? – user38268 Apr 5 '12 at 1:11
You are thinking entirely too hard. We can view $\phi$ in three different ways: as a set map (forget that it satisfies any extra special properties), as a ring map, and as a liner map. Now, regardless of which of the last two we are thinking about, they both have the same underlying thought of the map as a set map. Thus, something defined entirely while thinking only about it as a set map should be the same in both. The kernel when thinking about our map as a ring map and as a linear map is just the preimage of zero--an entirely set map associated concept. Thus, the kernels are the same thing. – Alex Youcis Apr 5 '12 at 1:23
Said, more simply let $X=\ker_{\mathbf{Ring}}\phi$ and $Y=\ker_{\mathbf{Vect}_F}(\phi)$. Then, $X=\phi^{-1}(\{0\})$ and $Y=\phi^{-1}(\{0\})$ (both by DEFINITION). Now, of course we see that $X=Y$. And since $X$ is an ideal so is $Y$ because it's $X$....I don't know how to say it any better, I'm sorry. – Alex Youcis Apr 5 '12 at 1:24