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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).2012-04-05
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    @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?2012-04-05
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    @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.2012-04-06

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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.

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    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?2012-04-05
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    What do you mean? You have that $E$ is a ring and that $\phi:E\to E$ is a ring map. You always know that the kernel of a ring map is an ideal from where I think you can finish (since you seem to know that ideals of fields are either the full space or trivial).2012-04-05
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    Yes I know the kernel of the ring map $\phi$ is trivial. However does this then mean the kernel of $\phi$ as an $F$ - linear transformation is trivial?2012-04-05
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    They are the same thing. The kernel of both a ring map and a linear transformation is the preimage of the additive identity element in the target space. Even if you forgot this you should have recalled that "both" kernels are trivial if and only if the map is injective--and since injectivity is only a property of the underlying set map, it doesn't matter if you are looking at it as a ring map or linear map, etc.2012-04-05
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    I am not seeing this at all. Let us look at the axiom for a subspace being closed under multiplication from the outside: If $a \in \ker T$, and $c \in F$ then $ac \in \ker T$. Where does this talk about that $\ker T$ is closed under multiplication from elements in $E$? This is the crucial bit that makes me not convinced it is an ideal.2012-04-05
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    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.2012-04-05
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    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.2012-04-05
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    I get what you saying now exactly. The fact that $\phi$ is injective iff it's kernel is trival holds whether we are talking about the kernel of a ring map or the kernel of a linear transformation. Since it is injective already as a ring map, it is injective as a linear transformation. Done.2012-04-05
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    Actually this has showed me in algebra that it can get confusing which underlying "structure" one is talking about.2012-04-05
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    Yes, that is a common thing to be careful about--very careful.2012-04-05