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This is from Conrad's notes.

  1. I agree that we can promote $\frak{a}$ and $\frak{b}$ from fractional ideals to integral ideals by multiplying by some element in $A$. i.e. For each prime ideal $\frak{p}$ with a negative exponent in the prime factorization, there is some principal ideal $(p)$ that $\frak{p}$ divides. Then multiplying by powers of $p$ gets rid of this negative exponent.

  2. However, I don't agree that we can "divide" to assume that $\frak{a}$ and $\frak{b}$ are coprime. My objection is that you can't divide by nonprincipal ideals. If you were to be able to do that, every ideal would be isomorphic as $A$-modules.

  3. If we assume that $\frak{a}$ and $\frak{b}$ are coprime integral ideals, we can say $\mathfrak{a}+\mathfrak{b}\cong A\oplus \frak{ab}$. Now, surely you can't use this when $\mathfrak{a}=\mathfrak{b}$. In particular, I am not sure about the isomorphism in Example 9.

Am I missing something?

daruma
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    Somewhat related. Wish I could address your doubts right away, but I need to spend more time on what's going on here. Actually K. Conrad spends quite a bit of time roaming our site, you may get a first hand account as an answer! – Jyrki Lahtonen May 04 '23 at 19:26
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    Hmm. I suspect his point is that $\mathfrak{a}_0\mathfrak{c}$ is principal, say $(\alpha)$. Then as $A$-modules $\mathfrak{a}\simeq \alpha\mathfrak{a}=\mathfrak{c}$. In other words, "division" is by a principal ideal, which is, as you pointed out, the unproblematic case. Or may be I'm missing something. – Jyrki Lahtonen May 04 '23 at 19:30
  • @JyrkiLahtonen Actually, that makes sense. Fractional ideals in the same ideal class groups are isomorphic by multiplying by an element (i.e. principal ideal). The thought of being free to move between different representatives of the same ideal class is what I guess I had been missing. – daruma May 05 '23 at 00:41

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