# Projective dimension of graded modules

Short version:

Why is the projective dimension of a graded module the same as the projective dimension of its underlying ungraded module?

Longer version:

Let $$G$$ be a commutative group, let $$R$$ be a $$G$$-graded commutative ring, and let $$M$$ be a $$G$$-graded $$R$$-module. The category of $$G$$-graded $$R$$-modules has enough projectives, and so we can define the projective dimension $${\rm pd}(M)$$ of $$M$$ as the infimum of the lengths of all projective resolutions of $$M$$ in the category of $$G$$-graded $$R$$-modules.

Let $$U(M)$$ denote the ungraded $$R$$-module underlying $$M$$. Then, $$U(M)$$ also has a projective dimension $${\rm pd}(U(M))$$, defined in the category of (ungraded) $$R$$-modules.

At several places in the literature one finds the statement that these two projective dimensions coincide, i.e., that $${\rm pd}(M)={\rm pd}(U(M)).$$ The reason for this is always given by the fact that a $$G$$-graded $$R$$-module is projective if and only if its underlying $$R$$-module is so. But this seems to yield only the inequality $${\rm pd}(M)\geq{\rm pd}(U(M)).$$

If we wish to show the converse, then we may consider a projective resolution $$0\rightarrow P_n\rightarrow P_{n-1}\rightarrow\cdots\rightarrow P_1\rightarrow P_0\rightarrow U(M)\rightarrow 0$$ in the category of $$R$$-modules and try to get from this a projective resolution of the same length in the category of $$G$$-graded $$R$$-modules. But as the $$P_i$$ need not be obtained from $$G$$-graded $$R$$-modules it is not immediately clear how to proceed. So:

Is the above equality true? And if so, how do we prove it?

Note 1: Of course this question (and hopefully also its answer) can be generalised to arbitrary coarsenings, but for the moment the forgetful functor $$U$$ will suffice.

Note 2: There seems to be something more general going on, for the same claim is found in the literature about the weak dimension; here, one should note that a $$G$$-graded $$R$$-module is flat if and only if its underlying $$R$$-module is so.

For $$M$$ as is in your question, consider a truncated resolution of projective $$G$$-graded modules $$Q_{n-1}\xrightarrow{f}Q_{n-2}\to\dots\to Q_1\to Q_0\to M.$$ Then $$U(Q_{n-1})\xrightarrow{f}U(Q_{n-2})\to\dots\to U(Q_1)\to U(Q_0)\to U(M)$$ is a truncated projective resolution of $$U(M)$$. If $$\mathrm{pd}\, U(M)\leq n$$, then Schanuel's lemma implies that $$U(\ker f)$$ is projective. Thus, $$\ker f$$ is projective as a $$G$$-graded module and $$\mathrm{pd}\, M\leq n$$.
This argument works for any exact functor $$U$$ such that $$X$$ is projective if and only if $$U(X)$$ is projective (in the appropriate categories).
• Minor technical addition: When generalising this to arbitrary abelian categories (as suggested in the last sentence), one has to suppose that $M$ has a projective resolution. Dec 11, 2018 at 9:42