In my algebra book I found the following exercise:

Show a field homomorphism $f:\mathbb K\longrightarrow \mathbb K$ from a field $\mathbb K$ to itself is the zero map or an isomorphism?

We have two possibilities $f(1)=0$ or $f(1)\neq 0$. If $f(1)=0$ then:

$$f(x)=f(x\cdot 1)=f(x)f(1)=0$$

for every $x\in \mathbb K$ and therefore $f$ is the zero map. Now, if $f(1)\neq 0$ happens then we can show $f$ is injective as follows: Suppose $f(x)=f(y)$ for some $x\in\mathbb K$ and $y\in\mathbb K$. If we had $x\neq y$ then $u:=x-y\neq 0$ is invertible hence we get the following absurd:

$$0=0f(u^{-1})=f(u)f(u^{-1})=f(uu^{-1})=f(1)\neq 0.$$

But, how about the surjectivity of $f$? I'm starting to think that result is not true.


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    $\begingroup$ You can see that $f$ sends idempotents to idempotents. So $f(1)=0$ or $f(1)=1$, as $0$ and $1$ are the unique idemoptents of a field. $\endgroup$ – Murphy Feb 25 at 22:07
  • $\begingroup$ The kernel of a ring morphism is an ideal. So a ring morphism is injective if and only if the kernel is zero. $\endgroup$ – Murphy Feb 25 at 22:08
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    $\begingroup$ math.stackexchange.com/questions/826019/… $\endgroup$ – Murphy Feb 25 at 22:10
  • $\begingroup$ Thanks for the tips concerning idempotents @FrankMurphy. I wasn't supposed to use ideals, of course, in that case it would be very easy to show $f$ is injective. $\endgroup$ – PtF Feb 25 at 22:12

You are correct in suspecting that $f$ may not be surjective. For instance, consider the map $f:\Bbb R(t)\hookrightarrow \Bbb R(t)$ $$f(p)=p(t^2)$$


It won't necessarily be surjective. But it will be an isomorphism onto its image.

Consider $\Bbb Q(x_1, x_2, \ldots)$ where each $x_i$ is transcendental and is algebraically independent of $x_j$ for all $j \neq i$. Then the field homomorphism induced by $x_i \rightarrow x_{i+i}$ is an isomorphism onto its image but it is not surjective (because $x_1$ is not in its image).

  • $\begingroup$ Yes. Edited to fix. $\endgroup$ – Robert Shore Feb 25 at 22:17

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