3
$\begingroup$

Let $G$ be an abelian group and let $R$ be a $G$-graded commutative ring, i.e., $R=\oplus_{g\in G} R_g$ with $R_gR_h\subseteq R_{g+h}$. Let $M$ be a $G$-graded $R$-module i.e. $M=\oplus_{g\in G}M_g$ and $R_g M_h\subseteq M_{g+h}$. We will say that a morphism of $G$-graded $R$-modules $f:M\rightarrow N$ has degree $h\in G$ if for all $g\in G$ one has that $f(M_g)\subseteq N_{g+h}$.

Q1: Is it always possible to find a resolution $F^{\cdot}$ of $M$ by Free $R$-modules which are $G$-graded and where the differentials have degree $0$ ?

Q2 If the answer to Q1 is no, then what are further assumptions that one can impose on $R$ and $G$ in order to guarantee the existence of such a resolution?

The motivation for $Q1$ and $Q2$ is this: If $M$ and $N$ are $G$-graded $R$-modules, is it possible to put a natural $G$-graded structure on $Tor_R^i(M,N)$?

Q3: What is a good reference for graded homological algebra?

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ For Q1: it is enough to show that every graded module $M$ is an epimorphic image of a graded free one by an homogeneous map: just pick an homogeneous set of generators of $M$ and proceed as in the ungraded case. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Feb 19, 2012 at 22:29
  • 1
    $\begingroup$ It follows from that that there are enough projectives in the category of graded modules and homogeneous maps, so you can compute Tors there. The homology groups will be themselves in the category (because your $R$ is commutative) and in particular will be graded. The usual well-definedness of Tor then gives you a naturality result for the grading. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Feb 19, 2012 at 22:42
  • 2
    $\begingroup$ (By the way: your title mentions cohomology but you then refer to Tor... If you wanted Ext instead, then things are a bit more delicate because hom(M,N) is not G-graded in the obvious way unless M is finitely generated.) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Feb 19, 2012 at 22:44

1 Answer 1

Reset to default
2
$\begingroup$

In short, everything works fine in the graded case, i.e. there are projective and free resolutions and a graded tensor product. Eventually, since kernels and quotients of graded modules are graded, there is also a natural grading on $\operatorname{Tor}$.

In more detail: I don't know of a reference for graded homological algebra in the general case. However, Herzog, Bruns (Cohen-Macaulay Rings) treat the $\mathbb{Z}$-graded case and the formal properties carry over to the general case without difficulty. The existence of a free graded resolution with morphisms of degree zero, for example, can be found (in my edition) at the end of page 32 (somewhere after Theorem 1.5.8).

As mentioned by Mariano, more care has to be taken when considering the graded $\operatorname{Ext}$-functor. Details can again be found in Herzog-Bruns.

Graded rings and modules (without homological algebra) are treated in detail in

Nastasescu, Oystaeyen: Graded and filtered rings and modules.

There you learn for example in 3.3.7 that a graded module is a projective object in the graded category iff it is projective as an abstract module (i.e. when forgetting the grading) while the same is not true for injective modules (3.3.11). The definition of the graded tensor product is given in section 3.4.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks a lot Ralph for the reference on Cohen Macaulay rings. $\endgroup$
    – Hugo Chapdelaine
    Commented Feb 20, 2012 at 3:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .