Given any natural number $n$, does there a ring $A$ and an $A$-module $M$ such that projective dimension of $M$ is $n$?

I am think this statement should be true but I don’t know how to find such $A$ and $M$ for arbitrary $n$. I don’t explicitly need $A$ and $M$ (could be very hard). Can anyone just tell me that whether this statement is true or not?

  • $\begingroup$ You may want to look at some papers of Paul Roberts. e.g. this or this. $\endgroup$ – user5325 May 1 at 5:21

Yes. For a simple example, let $k$ be any nonzero ring, let $A=k[x_1,\dots,x_n]$, and let $M=A/(x_1,\dots,x_n)$. Then $M$ has projective dimension $n$. To verify this, you can explicitly write down a free resolution of $M$ of length $n$ (the Koszul complex of the sequence $(x_1,\dots,x_n)$) to bound the projective dimension above by $n$, and then use that resolution to compute that $\operatorname{Ext}_A^n(M,M)$ is nontrivial to bound the projective dimension below by $n$.

  • $\begingroup$ Ok thanks, this is what I wanted. $\endgroup$ – Sunny Rathore May 1 at 5:33

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