# Second summand to make projective module free

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.)

• 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? – LSpice Jun 20 '20 at 23:20
• 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. – zjs Jun 21 '20 at 2:09
• 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. – LSpice Jun 21 '20 at 3:17

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.
• 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. – zjs Jun 21 '20 at 2:15
• 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. – Steven Landsburg Jun 21 '20 at 2:21