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Let $A$ be an integral domain, $K$ its field of fractions. An $A$-submodule $M$ of $K$ is a fractional ideal of $A$ if $xM\subset A$ for some $x\neq 0$ in $A$.

Proposition 9.7. Let $A$ be a local domain. Then $A$ is a discrete valuation ring $\iff$ every non-zero fractional ideal of $A$ is invertible.

Proof. $\Rightarrow$. Let $x$ be a generator of the maximal ideal $m$ of $A$ , and let $M \neq 0$ be a fractional ideal. Then there exists $y\in A$ such that $yM\subset A$ :thus $yM$ is an integral ideal , say $(x^r)$, and therefore $M=(x^{r-s}) $ where $S=v(y)$ $\cdots$

What puzzle me is that $M=(x^{r-s})$ . We need to show that $M$ is an ideal. By definition of fractional ideal. $M$ is a $A$-module, thus an ideal of $A$ ($a\in A,m\in M, am\in M$). Every fractional ideal is an ideal of $A$ Is this right? I think $M$ could contain $A$ (thus no an ideal of $A$) or I have a misunderstanding of the definition of fractional ideal.

I also have no idea to prove $M=(x^{r-s})$. I want to show $v(M)=v(yMy^{-1})=v(x^r)-v(y)$, but there is no definition of $v(M)$.

Zeldovich Yakov
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  • No. That's not the goal. The goal is tha $M$ is invertible. This of course differs from $M$ being an integral ideal. – xbh Apr 22 '21 at 11:03
  • A better notation could be $Ax^{r-s}$, which would make you less easier to confuse this with integral ideals. – xbh Apr 22 '21 at 11:09
  • I think $M=(x^{r-s})$ means $M$ is a $A$-module generated by $(x^{r-s})$. But how can I prove this? – Zeldovich Yakov Apr 22 '21 at 11:45

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