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Can we find a field $K$ and an endomorphism $f$ of $K$, such that $f$ is not trivial and $f$ is not surjective? In other words, can we find an endomorphism of $K$ which is not an automorphism?

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Take a look at this: en.wikipedia.org/wiki/Transcendence_degree#Applications –  M Turgeon Sep 25 '12 at 1:53
    
If I'm not mistaken, you're looking for the cohopfian objects of the category of fields. (This has no real content, it's just an observation on terminology) –  lentic catachresis Sep 25 '12 at 2:52

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up vote 6 down vote accepted

Consider the Frobenius endomorphism of fields of characteristic $p$ given by

$$ x \to x^{p}$$

This is not always an automorphism. For example, the image of rational function field $\mathbb{F}_{p}(t)$ under the Frobenius endomorphism does not contain $t$.

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Of course, this is exactly the difference between a perfect and a non-perfect field. –  M Turgeon Sep 25 '12 at 1:56

Yes, lots; for example, the endomorphisms $F(x) \to F(x)$ fixing $F$ are precisely given by extending $x \mapsto \frac{p(x)}{q(x)}$ where $p, q$ are two nonzero polynomials of total degree at least $1$. Assuming WLOG that $\gcd(p, q) = 1$, this map is an automorphism if and only if $p = ax + b, q = cx + d$ where $ad - bc \neq 0$ (exercise).

On the other hand, if $K$ is a finite extension of its prime subfield, then any endomorphism of $K$ is an automorphism (exercise). These are precisely the number fields and the finite fields.

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