Projective dimension of module over local ring

Question

This question arose reading the well known article by Buchsbaum Lectures on regular local rings.

He states without proof that, given $(R,m)$ a local ring and an $R$-module $M$ over $R$, we have the inequality: $$\mathrm{projdim}(M)\leq \mathrm{projdim}(R/m).$$

I think this is rather simple, but I can't figure out how to produce a proof. If we take a projective resolution $P_\bullet$ of $M$, how can I tell that its length is less or equal to $\mathrm{projdim}(R/m)$?

Thank you for help, I think I need only some hints and I'll fill the details.

Edit. I would like to point that in Buchsbaum's article $R$ and $M$ are apparently taken without any other condition than $R$ being local (and noetherian).

Answer 1

You want to prove that the projective dimension of the module $R/m$ is equal to the global dimension of the ring $R$. This is done in textbooks, like C. Weibel's Introduction to homological algebra or in Cartan-Eilenberg's Homological Algebra.

(This is for noetherian local rings)