**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.