## $\mathbb{Z}\ncong F$ where $F$ is a field and $f\colon\mathbb{Z}\to F$ is an onto morphism

I must claim that $$\mathbb{Z}\ncong F$$ where $$F$$ is a field and $$f\colon\mathbb{Z}\to F$$ is an onto morphism.

Let $$F$$ a field ed $$f\colon\mathbb{Z}\to F$$ an onto morphism. We know that $$\ker f$$ is an ideal of $$\mathbb{Z}$$, then $$\ker f=(n)=n\mathbb{Z}$$ for same $$n\in\mathbb{Z}$$. For the first isomorphism theorem we have that $$\mathbb{Z}_n:=\mathbb{Z}/n\mathbb{Z}=\mathbb{Z}/\ker f\cong F.$$

Now, if $$n=0$$, the canonical projection is $$\pi\colon\mathbb{Z}\to \mathbb{Z}$$, but the only non-zero morphism from $$\mathbb{Z}$$ to $$\mathbb{Z}$$ is $$id_{\mathbb{Z}}$$ which is, in particular, injective. Since $$\ker f=n\mathbb{Z}$$, then $$\tilde{f}\colon\mathbb{Z}\to F$$ is injective, but for the first isomorphism theorem $$f=\tilde{f}\circ\pi$$, then $$f$$ is injective, moreover, for hypotesis, $$f$$ is onto, then $$f$$ is an isomorphism. But this is absurd, because $$\mathbb{Z}$$ is not a field.

Correct?

Thanks!