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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?

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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.

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  • $\begingroup$ 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? $\endgroup$ – Ovi Nov 21 '18 at 2:31
  • $\begingroup$ $\operatorname{im}\phi$ is not a subfield. $\endgroup$ – Saucy O'Path Nov 21 '18 at 2:31
  • $\begingroup$ 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? $\endgroup$ – Ovi Nov 21 '18 at 2:34
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    $\begingroup$ 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. $\endgroup$ – Sir Jective Nov 21 '18 at 2:43
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    $\begingroup$ @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 $1_F$. $\endgroup$ – jgon Nov 21 '18 at 2:48
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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 $\mathbb{Q}$.

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