How do we know that $\mathbb{Q}$ is the initial field of characteristic $0$?

Question

For any field $F$, I can see why any two morphisms $\mathbb{Q} \rightarrow F$ must be equal. If $F$ has characteristic $0$, how do we furthermore know that there exists a homomorphism $\mathbb{Q} \rightarrow F$?

Answer 1

Well, $1_{\mathbb Q}$ must go to $1_F$. You show inductively that $n\in\mathbb N$ has to go to $n\cdot1_F$, where this latter is defined as the sum of $1_F$ with itself $n$ times. You extend to a map $\iota\colon\mathbb Z\to F$, and you see as you go along that there are really no choices be made at any stage. Then you show that this $\iota$ is a homomorphism of rings. And you notice that the kernel of $\iota$ is trivial, by your hypothesis that the characteristic of $F$ is zero. Now you can define your homomorphism $\mathbb Q\to F$, by sending $m/n$ to $\iota(m)/\iota(n)$. Of course you have to show that this doesn’t depend on your representation of an arbitrary $\lambda\in\mathbb Q$ as a fraction of integers. Finally, you show that this is an extension of $\iota$ to a homomorphism of rings, $\mathbb Q\to F$. There’s a lot to prove, some of the inductions are very tedious, and the whole process is truly tiresome. But everybody has to do it once.