0
$\begingroup$

Let K be a number field and $O_K$ its ring of integers.

Trying to understand proof of "The ring of integers of a number field is a Dedekind domain", in the part:

Every element of the finitely generated abelian group $O_K /p$ is killed by $a_0$ , so $O_K /p$ is a finite set. What does it mean?

Thank you.

enter image description here

$\endgroup$
  • 1
    $\begingroup$ You quoted two sentences. What exactly do you not understand? $\endgroup$ – Bernard Jul 22 '19 at 20:16
  • $\begingroup$ Please check the sentence in bold, how does $a_0$ kill the elements in $O_K/p$ and why $O_K/p$ is a finite set.Thank you. $\endgroup$ – PerelMan Jul 22 '19 at 20:18
3
$\begingroup$

$a_0$ is a nonzero integer, and $a_0(\mathcal O_K/\mathfrak p)=0$, that is $\mathfrak{p}\supseteq|a_0|\mathcal{O}_K$ and so $\mathcal O_K/\mathfrak p$ is a quotient of $\mathcal O_K/|a_0|\mathcal O_K$. Therefore $|\mathcal O_K/\mathfrak p|\le|\mathcal O_K/|a_0|\mathcal O_K|$. But $|\mathcal O_K/|a_0|\mathcal O_K|=|a_0|^n<\infty$, where $n=|K:\Bbb Q|$ since as an Abelian group, $\mathcal O_K$ is free of rank $n$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you. I don't know if $|\mathcal O_K/|a_0|\mathcal O_K|=|a_0|^n$ is always true? how do we get that result? Thanks $\endgroup$ – PerelMan Jul 22 '19 at 23:08
1
$\begingroup$

$a_0$ kills the elements of $\mathcal O_K/\mathfrak p$ simply because the author has just explained why $a_0 \in \mathfrak p$.

On the other hand, $\mathcal O_K/\mathfrak p$ is a finitely generated abelian group, and the above remark shows it is torsion, which implies finiteness, by the Structure theorem for finitely generated abelian groups.

| cite | improve this answer | |
$\endgroup$

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.