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Thank you all! This is very interesting and helpful. @Wojowu, I have been unable to pull up Weinberger's original paper, but it's not the case that it specifies a Euclidean function, correct? (Wishful thinking, probably...)
Thank you! Do you have advice on how to show that that $M$ suffices? e.g. in $K=\mathbb{Q}[\sqrt{-5}]$ where $\frak{a}=\langle 2,1+\sqrt{-5}\rangle$ gives a nontrivial element of the class group $Cl(\mathcal{O}_K)\cong\mathbb{Z}/2\mathbb{Z}$, I'm having some trouble seeing the way to view $\frak{a}\oplus\frak{a}$ as free (either as a $\mathcal{O}_K$-module or $\mathbb{Z}$-module), and more generally how to see direct-summing of ideals as being equivalent in some way to their product.
Yes, I'd say that this solution falls into "in terms of $P$", thanks! To really get a sense of what direct-summing $M$ does, I was also hoping to be able to think about what the module $P\oplus M$ "looks like", separately from the exact sequence context (since that seems to be the most natural way to actually prove the freeness of $P\oplus M$) --- is there a nice way to think about the isomorphism between $P\oplus M$ and the free module $FP$ on $P$? I'm having trouble picturing exactly what that might be, i.e. how to carry $p\oplus\sum r_ip_i$ to a $FP$-element.
