The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

22.12 Injective modules over algebras

In this section we discuss injective modules over algebras and over graded algebras. Thus it is the analogue of More on Algebra, Section 15.54 in the setting of this chapter.

Algebras and modules. Let $R$ be a ring and let $A$ be an $R$-algebra, see Section 22.2 for our conventions. For a right $A$-module $M$ we set

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

which we think of as a left $A$-module by the multiplication $(a f)(x) = f(xa)$. Namely, $((ab)f)(x) = f(xab) = (bf)(xa) = (a(bf))(x)$. Conversely, if $M$ is a left $A$-module, then $M^\vee $ is a right $A$-module. Since $\mathbf{Q}/\mathbf{Z}$ is an injective abelian group (More on Algebra, Lemma 15.53.1), the functor $M \mapsto M^\vee $ is exact (More on Algebra, Lemma 15.54.6). Moreover, the evaluation map $M \to (M^\vee )^\vee $ is injective for all modules $M$ (More on Algebra, Lemma 15.54.7).

We claim that $A^\vee $ is an injective right $A$-module. Namely, given a right $A$-module $N$ we have

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

and we conclude because the functor $N \mapsto N^\vee $ is exact. The second equality holds because

\[ \mathop{\mathrm{Hom}}\nolimits _\mathbf {Z}(N, \mathop{\mathrm{Hom}}\nolimits _\mathbf {Z}(A, \mathbf{Q}/\mathbf{Z})) = \mathop{\mathrm{Hom}}\nolimits _\mathbf {Z}(N \otimes _\mathbf {Z} A, \mathbf{Q}/\mathbf{Z}) \]

by Algebra, Lemma 10.11.8. Inside this module $A$-linearity exactly picks out the bilinear maps $\varphi : N \times A \to \mathbf{Q}/\mathbf{Z}$ which have the same value on $x \otimes a$ and $xa \otimes 1$, i.e., come from elements of $N^\vee $.

Finally, for every right $A$-module $M$ we can choose a surjection $\bigoplus _{i \in I} A \to M^\vee $ to get an injection $M \to (M^\vee )^\vee \to \prod _{i \in I} A^\vee $.

We conclude

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

  2. $A^\vee $ is an injective $A$-module, and

  3. every $A$-module injects into a product of copies of $A^\vee $.

Graded algebras and modules. Let $R$ be a ring. Let $A$ be a graded algebra over $R$. If $M$ is a graded $A$-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 with the $A$-module structure defined as above (for homogeneous elements). This again switches left and right modules. On the category of graded $A$-modules the functor $M \mapsto M^\vee $ is exact (check on graded pieces). Moreover, the evaluation map $M \to (M^\vee )^\vee $ is injective as before (because we can check this on the graded pieces).

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

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

and we conclude because the functor $N \mapsto (N^0)^\vee = (N^\vee )^0$ is exact. To see that the second equality holds we use the equalities

\[ \mathop{\mathrm{Hom}}\nolimits _\mathbf {Z}(N^ n, \mathop{\mathrm{Hom}}\nolimits _\mathbf {Z}(A^{-n}, \mathbf{Q}/\mathbf{Z})) = \mathop{\mathrm{Hom}}\nolimits _\mathbf {Z}(N^ n \otimes _\mathbf {Z} A^{-n}, \mathbf{Q}/\mathbf{Z}) \]

of Algebra, Lemma 10.11.8. Thus an element of $\mathop{\mathrm{Hom}}\nolimits _{\text{Mod}_ A}(N, A^\vee )$ corresponds to a family of $\mathbf{Z}$-bilinear maps $\psi _ n : N^ n \times A^{-n} \to \mathbf{Q}/\mathbf{Z}$ such that $\psi _ n(x, a) = \psi _0(xa, 1)$ for all $x \in N^ n$ and $a \in A^{-n}$. Moreover, $\psi _0(x, a) = \psi _0(xa, 1)$ for all $x \in N^0$, $a \in A^0$. It follows that the maps $\psi _ n$ are determined by $\psi _0$ and that $\psi _0(x, a) = \varphi (xa)$ for a unique element $\varphi \in (N^0)^\vee $.

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 $A^\vee [k]$.

If $(A, \text{d})$ is a differential graded algebra and $I$ is an object of $\text{Mod}_{(A, \text{d})}$ then we say $I$ is injective as a graded $A$-module to mean that $I$ is a injective object of the abelian category $\text{Mod}_ A$ of graded $A$-modules.

Lemma 22.12.1. Let $(A, \text{d})$ be a differential graded algebra. Let $I \to M$ be an injective homomorphism of differential graded $A$-modules. If $I$ is an injective object of the category of graded $A$-modules, then $I \to M$ is an admissible monomorphism.

Proof. This is immediate from the definitions. $\square$

Let $(A, \text{d})$ be a differential graded algebra. If $M$ is a left differential graded $A$-module, then we will endow $M^\vee $ (with its graded module structure as above) with a right differential graded module structure by setting

\[ \text{d}_{M^\vee }(f) = - (-1)^ n f \circ \text{d}_ M^{-n - 1} \quad \text{in }(M^\vee )^{n + 1} \]

for $f \in (M^\vee )^ n = \mathop{\mathrm{Hom}}\nolimits _\mathbf {Z}(M^{-n}, \mathbf{Q}/\mathbf{Z})$ and $\text{d}_ M^{-n - 1} : M^{-n - 1} \to M^{-n}$ the differential of $M$1. We will show by a computation that this works. Namely, if $a \in A^ m$, $x \in M^{-n - m - 1}$ and $f \in (M^\vee )^ n$, then we have

\begin{align*} \text{d}_{M^\vee }(f a)(x) & = - (-1)^{n + m} (f a)(\text{d}_ M(x)) \\ & = - (-1)^{n + m} f(a\text{d}_ M(x)) \\ & = -(-1)^ n f(\text{d}_ M(ax) - \text{d}(a)x) \\ & = -(-1)^ n[-(-1)^ n \text{d}_{M^\vee }(f)(ax) - (f\text{d}(a))(x)] \\ & = (\text{d}_{M^\vee }(f)a)(x) + (-1)^ n (f\text{d}(a))(x) \end{align*}

the third equality because $\text{d}_ M(ax) = \text{d}(a)x + (-1)^ m a\text{d}_ M(x)$. In other words we have $\text{d}_{M^\vee }(fa) = \text{d}_{M^\vee }(f)a + (-1)^ n f \text{d}(a)$ as desired.

If $M$ is a right differential graded module, then the sign rule above does not work. The problem seems to be that in defining the left $A$-module structure on $M^\vee $ our conventions for graded modules above defines $af$ to be the element of $(M^\vee )^{n + m}$ such that $(af)(x) = f(xa)$ for $f \in (M^\vee )^ n$, $a \in A^ m$ and $x \in M^{-n - m}$ which in some sense is the “wrong” thing to do if $m$ is odd. Anyway, instead of changing the sign rule for the module structure, we fix the problem by using

\[ \text{d}_{M^\vee }(f) = (-1)^ n f \circ \text{d}_ M^{-n - 1} \]

when $M$ is a right differential graded $A$-module. The computation for $a \in A^ m$, $x \in M^{-n - m - 1}$ and $f \in (M^\vee )^ n$ then becomes

\begin{align*} \text{d}_{M^\vee }(a f)(x) & = (-1)^{n + m} (f a)(\text{d}_ M(x)) \\ & = (-1)^{n + m} f(\text{d}_ M(x)a) \\ & = (-1)^{n + m} f(\text{d}_ M(ax) - (-1)^{m + n + 1} x\text{d}(a)) \\ & = (-1)^ m \text{d}_{M^\vee }(f)(ax) + f(x\text{d}(a)) \\ & = (-1)^ m (a\text{d}_{M^\vee }(f))(x) + (\text{d}(a)f)(x) \end{align*}

the third equality because $\text{d}_ M(xa) = \text{d}_ M(x)a + (-1)^{n + m + 1} x\text{d}(a)$. In other words, we have $\text{d}_{M^\vee }(af) = \text{d}(a) f + (-1)^ ma\text{d}_{M^\vee }(f)$ as desired.

We leave it to the reader to show that with the conventions above there is a natural evaluation map $M \to (M^\vee )^\vee $ in the category of differential graded modules if $M$ is either a differential graded left module or a differential graded right module. This works because the sign choices above cancel out and the differentials of $((M^\vee )^\vee $ are the natural maps $((M^ n)^\vee )^\vee \to ((M^{n + 1})^\vee )^\vee $.

Lemma 22.12.2. Let $(A, \text{d})$ be a differential graded algebra. If $M$ is a left differential graded $A$-module and $N$ is a right differential graded $A$-module, then

\[ \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}_{(A, \text{d})}}(N, M^\vee ) \]

is isomorphic to the set of sequences $(\psi _ n)$ of $\mathbf{Z}$-bilinear pairings

\[ \psi _ n : N^ n \times M^{-n} \longrightarrow \mathbf{Q}/\mathbf{Z} \]

such that $\psi _{n + m}(y, ax) = \psi _{n + m}(ya, x)$ for all $y \in N^ n$, $x \in M^{-m}$, and $a \in A^{m - n}$ and such that $\psi _{n + 1}(\text{d}(y), x) + (-1)^ n \psi _ n(y, \text{d}(x)) = 0$ for all $y \in N^ n$ and $x \in M^{-n - 1}$.

Proof. If $f \in \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}_{(A, \text{d})}}(N, M^\vee )$, then we map this to the sequence of pairings defined by $\psi _ n(y, x) = f(y)(x)$. It is a computation (omitted) to see that these pairings satisfy the conditions as in the lemma. For the converse, use Algebra, Lemma 10.11.8 to turn a sequence of pairings into a map $f : N \to M^\vee $. $\square$

Lemma 22.12.3. Let $(A, \text{d})$ be a differential graded algebra. Then we have

\[ \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}_{(A, \text{d})}}(M, A^\vee [k]) = \mathop{\mathrm{Ker}}(\text{d} : (M^\vee )^ k \to (M^\vee )^{k + 1}) \]

and

\[ \mathop{\mathrm{Hom}}\nolimits _{K(\text{Mod}_{(A, \text{d})})}(M, A^\vee [k]) = H^ k(M^\vee ) \]

for any differential graded $A$-module $M$.

Proof. This is clear from the discussion above. $\square$

[1] The sign rule is analogous to the one in Example 22.19.8, although there we are working with right modules and the same sign rule taken there does not work for left modules. Sigh!

Comments (5)

Comment #499 by m.o. on

After Lemma 22.12.1 and before Lemma 22.12.2, there is a formula to show that if is a right differential graded -module then is a left differential graded -module. However since has degree and not , two parts of the signs in the formla change, and I am wondering the conclusion might be that
is a left differential graded -module. On the other hand, if is a left differential graded -module then is a right differential graded -module. The reason of this (strange looking ?) phenomenon seems to come from the fact that the differential always act from the left, but I am not an expert in this field at all, and I am not quite sure these things. Do I misunderstand or overlook some important part ?

Comment #502 by on

Thank you very much for this comment. I agree the sign is wrong and I agree with your explanation. I corrected this here and I also tried to explain a little bit what you said.

Note: I cannot currently update the web server because we are making a change in the way things work but in a week or so this change will percolate to the webinterface. Thanks again.

Comment #506 by m.o. on

Thank you for your quick reply. After I read your tex file, I noticed the point where I made a mistake: let be a right differential graded -module, and let , , and . Then the correct left action of on would be given by . The reason of this definition is as follows. Denote by the differential graded module with the usual differential as before (i.e, the one which does not depend on the (left or right) -module structure). We are given a homomorphism of differential graded modules. Then we have , and this corresponds to a homomorphism of graded differential modules. Now we have the commutativity: , which is the origin of the sign if . Thus we have , and this corresponds to the desired map of differential graded modules. By the same way, the evaluation map will be given by if and . These sign changes agree with your suggestion that the definition of left action is wrong if and is odd. Am I right ? (I am very sorry for the undefined control sequence.)

Comment #507 by on

Yes, that is exactly right and this is what I was thinking too. But I decided we should not want to change the -action on in the graded case and we should not have different -actions depending on whether is a graded module or a differential graded module, so I suggest we leave it like it is now (I mean in the commit I linked to above).

Thank you very much once again (and please realize that everybody is an expert on this stuff in the sense that once the material is written out clearly, then we can all decide whether some formula is right or wrong).

Comment #513 by m.o. on

Now I understand the problem well, and I also think that we should leave it like it is now. It's fun to read the Stacks project. Thank you.


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