5
$\begingroup$

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.

$\endgroup$
0
3
$\begingroup$

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).

$\endgroup$
2
  • $\begingroup$ @FredRohrer You are welcome! $\endgroup$ – Uriya First Dec 1 '18 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. $\endgroup$ – Fred Rohrer Dec 11 '18 at 9:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.