Let $R$ be a Noetherian ring and $I,J$ proper ideals that are isomorphic as $R$-modules. Can we conclude that the projective dimensions of $R/I,R/J$ are equal?
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There is a short exact sequence $0\to I\to R\to R/I\to 0$, so the projective dimension of $R/I$ is one more than that of $I$. Similarly, the projective dimension of $R/J$ is one more than that of $J$. Now, the projective dimension of a module depends only on its isomorphism class, so if $I$ and $J$ are isomorphic, they have the same projective dimension.
Mariano Suárez-Álvarez
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From the short exact sequence you don't get that the projective dimension of the cokernel is one more than the dimension of the kernel: consider the case $I=R$. It is at most one more. – egreg Jan 07 '14 at 22:15
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No.
For example, take $R=\mathbb{Z}\times\mathbb{Z}$, $I=0\times\mathbb{Z}$, and $J=0\times 2\mathbb{Z}$. Then $R/I$ has projective dimension zero, and $R/J$ has projective dimension one.
Though Mariano's answer shows that the only counterexamples have one of $R/I$ and $R/J$ projective and the other of projective dimension one.
Jeremy Rickard
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