If $f$ is not injective then $F$ is finite.

Question

I have to prove or confute the follow proposition:

Let $F$ a Field and $f:\mathbb{Z} \to F$ an homomorphism of rings such that $f(1_\mathbb{Z})=1_F$. Show that if $f$ isn't injective then $F$ is finite.

My work:

Let $f$ an homomorphism of rings.

We know that $Ker f$ is an ideal of $\mathbb{Z}$ and $Ker f= n \mathbb{Z}$ with $n \in \mathbb{N}$ uniquely determined. This $n$ is the characteristic of the field $F$.

Such that $f$ is not injective $Ker f \neq \{0\}$ then $n \neq 0$ then $F$ is not finite.

There is an error in the last "then". How can i proof this exercise?

Thx.

Answer 1

Let $F=\overline{\Bbb F_2}$ be the algebraic closure of the two-element field. Then $f$ as in the problem statement isn't injective (has kernel $2\Bbb Z$), but $F$ is infnite.