A projective ideal in an integral domain is finitely generated

Question

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?

Answer 1

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 $I$.

More details are hidden below.

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 $I$.