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.

## 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

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

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

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

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

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

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

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

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

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)

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

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

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

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

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.

**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

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

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

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

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

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

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

and

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

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

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