# Why the faithful $D$-submodule of $F$ which is finitely generated as $D$ module, $M=M^2$?

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.

• 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 $M=M^2$. – User1999 Jun 30 '17 at 18:33
• @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. – User1999 Jun 30 '17 at 18:48

## 1 Answer

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.

• 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 $M$? – User1999 Jun 30 '17 at 19:37