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Let $R$ be a regular ring i.e. a commutative Noetherian ring whose localisations at every prime ideal is regular local ring. Then every finitely generated $R$-module has finite projective dimension, however not every $R$-module may have finite projective dimension.

My question is: Let $M$ be an $R$-module of finite projective dimension. Then, does every submodule also have finite projective dimension ? If this is not true in general, what if we also assume $R$ is an integral domain ?

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  • $\begingroup$ First I'd like to ask why deleted the question about the finiteness of projective dimension of ideals in the same class of rings. $\endgroup$
    – user26857
    Mar 26, 2019 at 15:36
  • $\begingroup$ @user26857: because going by Mohan's suggestion, I should read it in a book and I found it in Lam's Lectures on Rings and Modules $\endgroup$
    – user521337
    Mar 26, 2019 at 16:32
  • $\begingroup$ Found what? An example of ideal of infinite projective dimension? $\endgroup$
    – user26857
    Mar 26, 2019 at 20:52
  • $\begingroup$ @user26857: No no, found that every ideal does indeed have finite projective dimension, in fact as is proved in the book by Lam, being regular is equivalent to the property that every finitely generated module over the ring has finite projective dimension $\endgroup$
    – user521337
    Mar 26, 2019 at 21:49

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For any ring $R$, if there is an $R$-module $N$ of infinite projective dimension then there is a projective $R$-module (which certainly has finite projective dimension!) with a submodule of infinite projective dimension.

Indeed, choose an epimorphism from a projective module $P$ to $N$, and let $K$ be the kernel, so we have a short exact sequence $$0\longrightarrow K\longrightarrow P\longrightarrow N\longrightarrow0.$$

Since $N$ has infinite projective dimension and $P$ is projective, $K$ also has infinite projective dimension.

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