Prime subfield when $ch(F) = 0$; shouldn't it be prime subring isomorphic to $\mathbb{Z}$?

Question

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?

Answer 1

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.

Answer 2

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 $\mathbb{Q}$.