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This is not a duplicate, but an attempt to give an alternative to the answer given in this question. The proof I found there confuses me a bit, and I think it can be proven in a more formal and clear way, but I wanted to make sure that I am not doing any mistake. The statement is:

In a Dedekind domain, we can simplify common factors in a quotient of fractional ideals: $$ IK/JK \cong I/J $$

For any (commutative unitary) ring $A$, $A$-module $M$ and $\mathfrak{p}$ a (non-zero) prime ideal of $A$, we have the following isomorphisms:

  • $M_{\mathfrak{p}}\cong M\otimes_{A} A_{\mathfrak{p}}$
  • $M/\mathfrak{p}M \cong M\otimes_{A} A/\mathfrak{p}$
  • $\operatorname{Frac}(A/\mathfrak{p})\cong A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}$ (note that in Dedekind domains $A/\mathfrak{p}$ is already a field).
  • Chinese Reminder Theorem.

Lemma: for any non-zero fractional ideal $I$ we have $ A/\mathfrak{p} \cong I/\mathfrak{p}I $.

$$ I/\mathfrak{p}I \cong I\otimes_{A} A/\mathfrak{p} \cong I\otimes_{A}A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}} \cong I\otimes_{A} A_{\mathfrak{p}} \otimes_{A_{\mathfrak{p}}} A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}} \cong I_{\mathfrak{p}}\otimes_{A_{\mathfrak{p}}} A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}} \cong I_{\mathfrak{p}}/\mathfrak{p}I_{\mathfrak{p}} $$

In this way we reduce our global problem to a local problem. Now we can argue over the discrete valuation ring $A_{\mathfrak{p}}$, on which $\mathfrak{p}=(f)$ for some $f\in A_{\mathfrak{p}}$. But then, if $v_{\mathfrak{p}}(I)=n \in \mathbb{Z}$, we have

$$ I_{\mathfrak{p}}/\mathfrak{p}I_{\mathfrak{p}} \cong (f)^n/ (f)^{n+1} $$

and the isomorphism $A_{\mathfrak{p}}/(f) \cong (f)^n/ (f)^{n+1}$ induced by multiplication times $f^n$ is clear. So we have shown

$$ A/\mathfrak{p} \cong I/\mathfrak{p}I $$

Proof of the statement in yellow:

If $I=A$ we are done by the argument on the prime factors of $J$ and using the Chinese Reminder Theorem (in this case $J$ must be an integral ideal for the quotient to make sense). So we have shown $$ K/JK \cong A/J $$

Finally, if $I$ is another fractional ideal (containing $J$, again, so that the quotient makes sense), it follows from the correspondence theorem for submodules of a quotient module that $$IK/JK\cong I/J $$

Is this proof correct?

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  • $\begingroup$ You are right that my answer is not satisfactory ; I will correct it as soon as I have some free time. However, your approach doesn't seem to work, since locally isomorphic modules doesn't have to be isomorphic (see here). $\endgroup$ – Watson Mar 13 '18 at 22:28
  • $\begingroup$ Thanks a lot for your feedback @Watson ! But I am not really using that, right? I am only taking advantage of the isomorphism between the quotient ring and the residue field (at least that the impression I have, but again, I am not sure of this proof, that's the reason to post it as proof verification) $\endgroup$ – Pedro Mar 13 '18 at 22:38
  • $\begingroup$ And in any case I think this argument would work if I argued directly by localizing because indeed it isn't true that modules with isomorphic stalks are isomorphic but if you have a globally defined map and it induces an isomorphism on every stalk then it is an isomorphism, right? Because you can check that kernel and cokernel are zero locally $\endgroup$ – Pedro Mar 13 '18 at 22:42
  • $\begingroup$ While I'm reading again (one month later…) your question, you might see my edit to the answer here. $\endgroup$ – Watson Apr 8 '18 at 16:27
  • $\begingroup$ Your lemma seems to be fine. But I'm not sure how you can conclude directly at the very end, since nor $I$ nor $J$ might be contained in $A$ (typically, you probably need a little argument to be able to apply the case $I=A$, don't you?). $\endgroup$ – Watson Apr 8 '18 at 18:32

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