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Suppose there's a projective $R$-module $P$ (non-free). We know that there is another $R$-module $M$ such that $P\oplus M$ is free over $R$. Is there a way to write down such an $M$ in terms of $P$?

If this is not always tractable, is it possible in certain specialized circumstances? The setting that comes to mind is where $P$ is a non-principal ideal of the ring of integers of a number field (as Wikipedia says this is an instance of $P$ being projective but not free).

Thanks!

(This was originally posted on math.stackexchange but I realized that here might be the better place to ask, apologies if not.)

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    $\begingroup$ I can imagine it's one of those "I'll know it when I see it", but do you have a rigorous definition of "in terms of $P$"? For example, one can take $M$ to be the kernel of the projection to the module $P$ from the free $R$-module on the set $P$. Does that count? $\endgroup$
    – LSpice
    Commented Jun 20, 2020 at 23:20
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    $\begingroup$ 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. $\endgroup$
    – zjs
    Commented Jun 21, 2020 at 2:09
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    $\begingroup$ The problem with a question like "how to carry $p \oplus m$ to an element of $F P$" is that, in the generality of a totally arbitrary (ring $R$ and) projective $P$, we just have no way of peeking into the structure of $P$ other than by using the universal property; and, in terms of the universal property, the answer is that we use a splitting $P \to F P$ of $M \hookrightarrow F P \twoheadrightarrow P$ to map $P \oplus M \to F P$, and I think that's all that can be said! Obviously @StevenLandsburg's answer is more interesting. $\endgroup$
    – LSpice
    Commented Jun 21, 2020 at 3:17

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If you want something completely general, LSpice's comment is the answer.

For the special case of an ideal $P$ in the ring $R$ of integers of a number field (or more generally if $R$ is a Dedekind domain) you can take $M=\{x\in K|xI\subset R\}$, where $K$ is the fraction field of $R$. The keyword to Google is fractional ideal.

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  • $\begingroup$ 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. $\endgroup$
    – zjs
    Commented Jun 21, 2020 at 2:15
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    $\begingroup$ In a Dedekind ring $R$ the direct sum of two ideals is related to their product by the formula $P\oplus Q\approx PQ\oplus R$. This is quite standard material, and any commutative algebra textbook should give you all the details. $\endgroup$
    – Steven Landsburg
    Commented Jun 21, 2020 at 2:21

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