# Module of arbitrary projective dimension

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?

• You may want to look at some papers of Paul Roberts. e.g. this or this. – 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 $$n$$.

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