I have to prove or confute the follow proposition:
Let $F$ a field and $ f: $$\mathbb Z \rightarrow F$ an homomorphism of rings such that $f(1_\mathbb Z ) =1_F$.
If $f$ isn't injective $\Rightarrow$ $F$ is finite
This is my work :
$f$ is an homomorphism of rings $\Rightarrow Ker f $ is an ideal of $\mathbb Z \Rightarrow $ $Ker f = n \mathbb Z $
$n \in \mathbb N$ is uniquely determined and is the characteristic of the field $F$
Such that $f$ is not injective $\Rightarrow Ker f \neq (0)$ then $n \neq 0$
Now I can't conclude that $F$ is infinite neither that is finite.
If I take $\mathbb Z_2(x)$ I know that is a field infinite with characteristic 2, so I found an infinite field with finite characteristic.
Is it correct? I forgot something important to say?
How can I prove that $\mathbb Z_2 (x) $ is an infinite field?
The field $\Bbb Z_2(x)$ is an infinite field since $1,x,x^2,x^3\ldots\in\Bbb Z_2(x)$. And this proves that the statement is false.