$\mathbb{Z}\ncong F$ 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!