The Stacks project

22.18 Injective modules over graded algebras

In this section we discuss injective graded modules over graded algebras analogous to More on Algebra, Section 15.55.

Let $R$ be a ring. Let $A$ be a $\mathbf{Z}$-graded algebra over $R$. Section 22.2 for our conventions. If $M$ is a graded $R$-module we set

\[ M^\vee = \bigoplus \nolimits _{n \in \mathbf{Z}} \mathop{\mathrm{Hom}}\nolimits _\mathbf {Z}(M^{-n}, \mathbf{Q}/\mathbf{Z}) = \bigoplus \nolimits _{n \in \mathbf{Z}} (M^{-n})^\vee \]

as a graded $R$-module (no signs in the actions of $R$ on the homogeneous parts). If $M$ has the structure of a left graded $A$-module, then we define a right graded $A$-module structure on $M^\vee $ by letting $a \in A^ m$ act by

\[ (M^{-n})^\vee \to (M^{-n - m})^\vee , \quad f \mapsto f \circ a \]

as in Section 22.13. If $M$ has the structure of a right graded $A$-module, then we define a left graded $A$-module structure on $M^\vee $ by letting $a \in A^ n$ act by

\[ (M^{-m})^\vee \to (M^{-m - n})^\vee , \quad f \mapsto (-1)^{nm}f \circ a \]

as in Section 22.13 (the sign is forced on us because we want to use the same formula for the case when working with differential graded modules — if you only care about graded modules, then you can omit the sign here). On the category of (left or right) graded $A$-modules the functor $M \mapsto M^\vee $ is exact (check on graded pieces). Moreover, there is an injective evaluation map

\[ ev : M \longrightarrow (M^\vee )^\vee , \quad ev^ n = (-1)^ n \text{ the evaluation map }M^ n \to ((M^ n)^\vee )^\vee \]

of graded $R$-modules, see More on Algebra, Item (17). This evaluation map is a left, resp. right $A$-module homomorphism if $M$ is a left, resp. right $A$-module, see Remarks 22.13.5 and 22.13.6. Finally, given $k \in \mathbf{Z}$ there is a canonical isomorphism

\[ M^\vee [-k] \longrightarrow (M[k])^\vee \]

of graded $R$-modules which uses a sign and which, if $M$ is a left, resp. right $A$-module, is an isomorphism of right, resp. left $A$-modules. See Remark 22.13.7.

We claim that $A^\vee $ is an injective object of the category $\text{Mod}_ A$ of graded right $A$-modules. Namely, given a graded right $A$-module $N$ we have

\[ \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}_ A}(N, A^\vee ) = \mathop{\mathrm{Hom}}\nolimits _{\text{Comp}(\mathbf{Z})}(N \otimes _ A A, \mathbf{Q}/\mathbf{Z})) = (N^0)^\vee \]

by Lemma 22.13.2 (applied to the case where all the differentials are zero). We conclude because the functor $N \mapsto (N^0)^\vee = (N^\vee )^0$ is exact.

Finally, for every graded right $A$-module $M$ we can choose a surjection of graded left $A$-modules

\[ \bigoplus \nolimits _{i \in I} A[k_ i] \to M^\vee \]

where $A[k_ i]$ denotes the shift of $A$ by $k_ i \in \mathbf{Z}$. We do this by choosing homogeneous generators for $M^\vee $. In this way we get an injection

\[ M \to (M^\vee )^\vee \to \prod A[k_ i]^\vee = \prod A^\vee [-k_ i] \]

Observe that the products in the formula above are products in the category of graded modules (in other words, take products in each degree and then take the direct sum of the pieces).

We conclude that

  1. the category of graded $A$-modules has enough injectives,

  2. for every $k \in \mathbf{Z}$ the module $A^\vee [k]$ is injective, and

  3. every $A$-module injects into a product in the category of graded modules of copies of shifts $A^\vee [k]$.


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0FQC. Beware of the difference between the letter 'O' and the digit '0'.