Lemma 24.11.1. Let (\mathcal{C}, \mathcal{O}) be a ringed site. Let \mathcal{A} be a graded \mathcal{O}-algebra. The category \textit{Mod}(\mathcal{A}) is a Grothendieck abelian category.
24.11 Shift functors on sheaves of graded modules
We urge the reader to skip this section. It turns out that sheaves of graded modules over a graded algebra are an example of the phenomenon discussed in Differential Graded Algebra, Remark 22.25.7.
Let (\mathcal{C}, \mathcal{O}) be a ringed site. Let \mathcal{A} be a sheaf of graded algebras on (\mathcal{C}, \mathcal{O}). Let \mathcal{M} be a graded \mathcal{A}-module. Let k \in \mathbf{Z}. We define the kth shift of \mathcal{M}, denoted \mathcal{M}[k], to be the graded \mathcal{A}-module whose nth part is given by
(\mathcal{M}[k])^ n = \mathcal{M}^{n + k}
is the (n + k)th part of \mathcal{M}. As multiplication maps
(\mathcal{M}[k])^ n \times \mathcal{A}^ m \longrightarrow (\mathcal{M}[k])^{n + m}
we simply use the multiplication maps
\mathcal{M}^{n + k} \times \mathcal{A}^ m \longrightarrow \mathcal{M}^{n + m + k}
of \mathcal{M}. It is clear that we have defined a functor [k], that we have [k + l] = [k] \circ [l], and that we have
\mathop{\mathrm{Hom}}\nolimits _{\textit{Mod}^{gr}(\mathcal{A})}(\mathcal{L}, \mathcal{M}[k]) = \mathop{\mathrm{Hom}}\nolimits _{\textit{Mod}^{gr}(\mathcal{A})}(\mathcal{L}, \mathcal{M})[k]
(without the intervention of signs) functorially in \mathcal{M} and \mathcal{L}. Thus we see indeed that the graded category of graded \mathcal{A}-modules can be recovered from the ordinary category of graded \mathcal{A}-modules and the shift functors as discussed in Differential Graded Algebra, Remark 22.25.7.
Proof. By Lemma 24.4.2 and the definition of a Grothendieck abelian category (Injectives, Definition 19.10.1) it suffices to show that \textit{Mod}(\mathcal{A}) has a generator. We claim that
\mathcal{G} = \bigoplus \nolimits _{k, U} j_{U!}\mathcal{A}_ U[k]
is a generator where the sum is over all objects U of \mathcal{C} and k \in \mathbf{Z}. Indeed, given a graded \mathcal{A}-module \mathcal{M} if there are no nonzero maps from \mathcal{G} to \mathcal{M}, then we see that for all k and U we have
\mathop{\mathrm{Hom}}\nolimits _{\textit{Mod}(\mathcal{A})}(j_{U!}\mathcal{A}_ U[k], \mathcal{M}) = \mathop{\mathrm{Hom}}\nolimits _{\textit{Mod}(\mathcal{A}_ U)}(\mathcal{A}_ U[k], \mathcal{M}|_ U) = \Gamma (U, \mathcal{M}^{-k})
is equal to zero. Hence \mathcal{M} is zero. \square
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