Given a Noetherian integral domain $A$ and a finitely generated torsion $A$-module $M$, we can define the divisor, or content of $M$ to be $\mathrm{div}(M)= \sum_{\mathrm{ht}(P)=1} \ell(M_P) [P]$, where the sum ranges over all height one prime ideals of $A$, and $M_P$ is the localization of $M$ at $P$, which is an $A_P$-module of finite length $\ell(M_P)$. The above sum makes sense because the only nonzero terms correspond to associated prime ideals of $M$ with height $1$.

Now for the question, as in the title. Let $a,b\in A$ be such that $\mathrm{div}(A/aA)=\mathrm{div}(A/bA)$. Can we deduce that $a,b$ differ only by a unit of $A$?

This is trivially true if $A$ is a UFD (factorial domain) and maybe also if $A$ is integrally closed, but is it true for every Noetherian integral domain? The question should be easy to settle once one has studied (which I have not yet done) Chapter 7 of Bourbaki's Commutative Algebra, for example.

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    $\begingroup$ The answer is negative for non-integrally closed domains. Try $A=k[[x^2,x^3]]$, $a=x^2, $b=x^2+x^3$. $\endgroup$ – Mohan May 5 '16 at 2:06
  • $\begingroup$ Thanks, but.. are you sure? $\endgroup$ – Luca Ghidelli May 5 '16 at 13:16
  • $\begingroup$ As sure as one can be. $\endgroup$ – Mohan May 5 '16 at 13:23
  • $\begingroup$ Oh sure. I developed a fever these days, so I'm a bit slow at understanding (and that's why I'm asking other people to solve my problems, ehe!). For some reason I was convinced that the first element of your example had multiplicity 1, whereas the second had 2. And I was dumbly trying to match them the wrong way. Thank you again. $\endgroup$ – Luca Ghidelli May 5 '16 at 18:27

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