Krdytkyu

Suppose $ch(F)=0$ and let $phi: mathbb{Z} to F$ with $phi:n mapsto n cdot 1_F$. Since the characteristic is $0$, we have $ker phi = {0 }$. By the First Isomorphism Theorem, we have $mathbb{Z} / { 0}cong text{Im}(phi)$. But the ring on the right is just $mathbb{Z}$, and the ring on the left is the prime "subfield" of $F$. Could someone point out the problem please?

Your $phi$ is the morphism of the initial object in the category of unital rings to the ring $F$. If the codomain is a field of prime characteristic, then the image of the morphism is not just a subring, but a subfield. This is due to the fact that the prime non-zero ideals of $Bbb Z$ are also maximal. If not, then it's just a subring.

The image of $varphi$ is not the prime subfield of $F$, which is isomorphic to $mathbb{Q}$. The problem is that $phi$ is a morphism of rings, and not a morphism of fields. Therefore, you could think of $operatorname{im}(varphi)$ as being a sort of "prime subring" associated to $F$, but it is not a field. In general, for any field of characteristic 0, there is a unique field homomorphism $psi:mathbb{Q} to F$ whose image is the prime subfield associated to $F$, and which is isomorphic to $mathbb{Q}$.