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.
Thanks.
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).