1
$\begingroup$

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 $\mathbb{Q} \rightarrow F$?

$\endgroup$
  • $\begingroup$ 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 $\mathbb{Z}$. $\endgroup$ – user61527 Jul 24 '14 at 4:13
  • $\begingroup$ 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. $\endgroup$ – Ragib Zaman Jul 24 '14 at 4:15
  • 2
    $\begingroup$ $F$ 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. $\endgroup$ – Adam Hughes Jul 24 '14 at 4:16
  • $\begingroup$ I think that this was asked at least several times before. $\endgroup$ – Asaf Karagila Jul 24 '14 at 5:20
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.