# The endomorphism of field

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) – Bruno Stonek Sep 25 '12 at 2:52

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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Aren't the endomorphisms $F(x)\to F(x)$ given by extending any map of the form $x\mapsto R(x)$, where $R(x)$ is any nonconstant rational function (i.e., not just the ratio of linear polynomials)? – user134824 Mar 27 '15 at 21:31
@user134824: yes. That's what I said. – Qiaochu Yuan Mar 28 '15 at 3:08