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

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 $\mathbb{Q} \rightarrow F$?

• Can't you just choose any $a \in F^{\#}$, send $1 \mapsto a$, and extend accordingly? It seems that the only issue here might be well-definition, but this ought to follow pretty quickly from how it's defined on $\mathbb{Z}$. – user61527 Jul 24 '14 at 4:13
• Every field of characteristic 0 has a copy of Q embedded in it. Any field map Q to F must send 1 to 1, and then the rest is determined. – Ragib Zaman Jul 24 '14 at 4:15
• $F$ is a $\Bbb Q$ vector space and field embeddings are linear. Hence $f:\Bbb Q\to F$ is determined via $f(m/n)={m\over n}f(1)$. Existence is easy, just sent $1_{\Bbb Q}\to 1_F$ which is required of a field homomorphism anyways. – Adam Hughes Jul 24 '14 at 4:16
• I think that this was asked at least several times before. – Asaf Karagila Jul 24 '14 at 5:20

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.