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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.

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1 Answer 1

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The other inequality follows from Schanuel's lemma.

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).

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  • $\begingroup$ @FredRohrer You are welcome! $\endgroup$ Dec 1, 2018 at 16:13
  • $\begingroup$ 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. $\endgroup$ Dec 11, 2018 at 9:42

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