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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?

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  • $\begingroup$ Are you sure that $I_p^2\simeq I_p\otimes I_p$? $\endgroup$ – Ehsaan Apr 26 at 22:14
  • $\begingroup$ $I_P^{2}$ 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 $I_P = I_P$. $\endgroup$ – Adam Higgins Apr 26 at 22:15
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    $\begingroup$ 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 $R\subseteq R_p$? $\endgroup$ – Ehsaan Apr 26 at 22:15
  • $\begingroup$ 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 $I^2\cong I\otimes I$ $\endgroup$ – George Apr 26 at 22:19
  • $\begingroup$ @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 $\endgroup$ – George Apr 26 at 22:24
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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 $I$.

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  • $\begingroup$ 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 $R/I$. $\endgroup$ – George Apr 27 at 1:40
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Projective ideals in integral domains are invertible, hence finitely generated.

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