# Field homomorphism $f:\mathbb K\longrightarrow \mathbb K$ is zero map or an isomorphism?

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.

• 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. – Murphy Feb 25 at 22:07
• The kernel of a ring morphism is an ideal. So a ring morphism is injective if and only if the kernel is zero. – Murphy Feb 25 at 22:08
• math.stackexchange.com/questions/826019/… – Murphy Feb 25 at 22:10
• 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. – PtF Feb 25 at 22:12

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