A projective ideal in an integral domain is finitely generated

Let $$R$$ be an integral domain and $$I$$ a projective ideal. Then $$I$$ is finitely generated as an $$R$$-module.

This seems like it should be easy but I don't know what to argue, or where to bring in the integral domain condition.

I tried localising at a prime $$P$$ containing $$I$$, and we get that $$I_P\subseteq PR_P\subseteq R_P$$ is a projective $$R_P$$ module. Then if $$0\to I_P\to R_P\to R_P/I_P\to 0$$ is short exact, since projective modules are flat, we can tensor with $$I_P$$, and the last term is $$0$$. But there is a result that if $$M$$ is a flat module over a ring then if $$J$$ is an ideal, then $$J\otimes M\cong JM$$. Hence, $$I_P^2\cong I_P\otimes I_P\cong R_P\otimes I_P\cong I_P$$.

But if $$I$$, hence $$I_P$$ were finitely generated, then wouldn't that mean $$I\subseteq I_P=0$$ by Nakayama?

• Are you sure that $I_p^2\simeq I_p\otimes I_p$? – Ehsaan Apr 26 at 22:14
• $I_P^{2}$ is not the same thing as $I_P \otimes_{R_P} I_P$ . When you tensor the short exact sequence $0 \to I_P \to R_P \to R_P / I_P \to 0$ with $I_P$, you get $0 \to I_P \to I_P \to 0 \to 0$, and so unsurprisingly $I_P = I_P$. – Adam Higgins Apr 26 at 22:15
• Localisation seems like a good idea. For every prime $p$, the ideal $I_p$ is a projective ideal for the local ring $R_p$, but every projective over a local ring is free --- so $I_p$ is a free ideal, so it has to be principal, in particular it is finitely-generated. Thus $I$ is locally finitely-generated, which implies $I$ is finitely-generated doesn't it? Maybe the domain hypothesis is used to get $R\subseteq R_p$? – Ehsaan Apr 26 at 22:15
• But If a module $M$ over a ring is flat, then for every ideal of the ring we have an isomorphism $M\otimes I\cong IM$. Applying this to $M=I$ we get $I^2\cong I\otimes I$ – George Apr 26 at 22:19
• @Ehsaan I do see that projectives over local rings are free, although it looks to be a deep result by Kaplansky (and is not covered by my lecture course - only the finitely generated projectives over local rings are free is covered). Perhaps there is a simpler way? Regardless, you're right and it works. However, I am still confused about the $I^2=I\otimes I$ business, since $I$ is projective hence flat and the result I quoted is from Liu's algebraic curves – George Apr 26 at 22:24

Your approach is wrong since $$R_P/I_P\otimes I_P$$ is typically not $$0$$. For instance, if $$I$$ is principal, then it is isomorphic to $$R$$ as a module, so $$R_P/I_P\otimes I_P\cong R_P/I_P\otimes R_P$$.
Here is a hint for what you can do instead. Take a set $$S$$ of generators of $$I$$, which gives a surjective homomorphism $$f:R^{\oplus S}\to I$$. Since $$I$$ is projective, this maps splits via some map $$g:I\to R^{\oplus S}$$. Now use the fact that $$R$$ is a domain and $$I$$ is an ideal to show that the image of $$g$$ is nonzero on only finitely many coordinates, and so there is a finite subset $$S_0\subseteq S$$ which still generates $$I$$.
For any $$a,b\in I$$, note that $$bg(a)=g(ab)=ag(b).$$ If $$a$$ and $$b$$ are nonzero, this implies that $$g(a)$$ and $$g(b)$$ are nonzero on exactly the same (finite) set of coordinates, since multiplying by $$a$$ or $$b$$ cannot change whether a coordinate is $$0$$ (here we use the fact that $$R$$ is a domain). So there is a finite set $$S_0\subseteq S$$ which is the set of coordinates on which $$g(a)$$ is nonzero for all nonzero $$a\in I$$. We then see that $$f$$ is still surjective when restricted to $$R^{\oplus S_0}$$, since $$f\circ g=1_I$$ and the image of $$g$$ is contained in $$R^{\oplus S_0}$$. Thus $$S_0$$ generates $$I$$.
• I made the crucial mistake of implicitly believing that $1\in I$, since I thought I could pass any element from $I$ across the tensor to annihilate $R/I$. – George Apr 27 at 1:40