1
$\begingroup$

I am studying the theorem that every Dedekind domain is integrally closed. Here's the setup.

Let $D$ be a Dedekind domain, $F$ its field of fractions and let $u$ be an element of $F$ that is $D$-integral (there exists a monic polynomial $f(x)=x^n+a_1x^{n-1}+...+a_n$ such that $f(u)=0$). Then, there exists a faithful $D[u]$ submodule of $F$ that is finitely generated as $D$ module. This submodule is $M=D1+Du+...+Du^{n-1}$.

I now wish to prove that this fractional ideal $M$ satisfies $M^2=M$, but I am having difficulties.

Would you help me, please? Thank you in advance.

$\endgroup$
  • $\begingroup$ Maybe $M^2=D1M+DuM+...+Du^n-1M$ is included in $M$ since $M$ is regarded as a $D$-module which contains $1$. Also, $M$ is included in $M^2$ and hence $M=M^2$. $\endgroup$ – User1999 Jun 30 '17 at 18:33
  • $\begingroup$ @Eric Wofsey I need to prove $M=M^2$ because from this it follows that $M=D$ since $M$ is a fractional ideal and hence invertible. $\endgroup$ – User1999 Jun 30 '17 at 18:48
2
$\begingroup$

Just note that $M$ is none other than the ring $D[u]$ itself, since any higher power of $u$ can be reduced to an element of $M$ using $f(u)=0$. Thus $M$ is closed under multiplication.

$\endgroup$
  • $\begingroup$ Thank you for your answer! Well, I understand why $M$ is a subring of $D[u]$ but I am not clear why it is the ring $D[u]$ itself. So, if $x$ is any element of $D[u]$ why is it in $M$? $\endgroup$ – User1999 Jun 30 '17 at 19:37

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.