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

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.
• Sorry I'm having trouble understanding your answer. In $\mathbb{Z} / \{ 0\}\cong \text{Im}(\phi)$, is the ring on the left not $\mathbb{Z}$, or is the ring on the right not the prime subfield? – Ovi Nov 21 '18 at 2:31
• $\operatorname{im}\phi$ is not a subfield. – Saucy O'Path Nov 21 '18 at 2:31
• Maybe my definition of prime subfield is wrong; as I understand it, the prime subfield of $F$ is the set of all finite sums of $1$, or $\{ n \cdot 1_F| n \in \mathbb{Z} \}$. I think that $\text {Im}(\phi) = \{ n \cdot 1_F| n \in \mathbb{Z} \}$. So why am I getting that this object is not a field, when it should be? – Ovi Nov 21 '18 at 2:34
• Your definition of prime subfield is not correct. The prime subfield of a field is its smallest subfield, which contains not only all the integer multiples of $1_F$, but in the case of a field of characteristic 0, all its rational multiples as well. – Sir Jective Nov 21 '18 at 2:43
• @Ovi note that it says field generated by $1_F$, so it's the set of sums and products and quotients that you obtain by starting with $1_F$. – jgon Nov 21 '18 at 2:48
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 $$\mathbb{Q}$$.