## 22.4 Differential graded modules

Our default in this chapter is right modules; we discuss left modules in Section 22.11.

Definition 22.4.1. Let $R$ be a ring. Let $(A, \text{d})$ be a differential graded algebra over $R$. A (right) differential graded module $M$ over $A$ is a right $A$-module $M$ which has a grading $M = \bigoplus M^ n$ and a differential $\text{d}$ such that $M^ n A^ m \subset M^{n + m}$, such that $\text{d}(M^ n) \subset M^{n + 1}$, and such that

$\text{d}(ma) = \text{d}(m)a + (-1)^ n m\text{d}(a)$

for $a \in A$ and $m \in M^ n$. A homomorphism of differential graded modules $f : M \to N$ is an $A$-module map compatible with gradings and differentials. The category of (right) differential graded $A$-modules is denoted $\text{Mod}_{(A, \text{d})}$.

Note that we can think of $M$ as a cochain complex $M^\bullet$ of (right) $R$-modules. Namely, for $r \in R$ we have $\text{d}(r) = 0$ and $r$ maps to a degree $0$ element of $A$, hence $\text{d}(mr) = \text{d}(m)r$.

The Leibniz rule relating differentials and multiplication on a differential graded $R$-module $M$ over a differential graded $R$-algebra $A$ exactly means that the multiplication map defines a map of cochain complexes

$\text{Tot}(M^\bullet \otimes _ R A^\bullet ) \to M^\bullet$

Here $A^\bullet$ and $M^\bullet$ denote the underlying cochain complexes of $A$ and $M$.

Lemma 22.4.2. Let $(A, d)$ be a differential graded algebra. The category $\text{Mod}_{(A, \text{d})}$ is abelian and has arbitrary limits and colimits.

Proof. Kernels and cokernels commute with taking underlying $A$-modules. Similarly for direct sums and colimits. In other words, these operations in $\text{Mod}_{(A, \text{d})}$ commute with the forgetful functor to the category of $A$-modules. This is not the case for products and limits. Namely, if $N_ i$, $i \in I$ is a family of differential graded $A$-modules, then the product $\prod N_ i$ in $\text{Mod}_{(A, \text{d})}$ is given by setting $(\prod N_ i)^ n = \prod N_ i^ n$ and $\prod N_ i = \bigoplus _ n (\prod N_ i)^ n$. Thus we see that the product does commute with the forgetful functor to the category of graded $A$-modules. A category with products and equalizers has limits, see Categories, Lemma 4.14.11. $\square$

Thus, if $(A, \text{d})$ is a differential graded algebra over $R$, then there is an exact functor

$\text{Mod}_{(A, \text{d})} \longrightarrow \text{Comp}(R)$

of abelian categories. For a differential graded module $M$ the cohomology groups $H^ n(M)$ are defined as the cohomology of the corresponding complex of $R$-modules. Therefore, a short exact sequence $0 \to K \to L \to M \to 0$ of differential graded modules gives rise to a long exact sequence

22.4.2.1
$$\label{dga-equation-les} H^ n(K) \to H^ n(L) \to H^ n(M) \to H^{n + 1}(K)$$

of cohomology modules, see Homology, Lemma 12.13.12.

Moreover, from now on we borrow all the terminology used for complexes of modules. For example, we say that a differential graded $A$-module $M$ is acyclic if $H^ k(M) = 0$ for all $k \in \mathbf{Z}$. We say that a homomorphism $M \to N$ of differential graded $A$-modules is a quasi-isomorphism if it induces isomorphisms $H^ k(M) \to H^ k(N)$ for all $k \in \mathbf{Z}$. And so on and so forth.

Definition 22.4.3. Let $(A, \text{d})$ be a differential graded algebra. Let $M$ be a differential graded module whose underlying complex of $R$-modules is $M^\bullet$. For any $k \in \mathbf{Z}$ we define the $k$-shifted module $M[k]$ as follows

1. the underlying complex of $R$-modules of $M[k]$ is $M^\bullet [k]$, i.e., we have $M[k]^ n = M^{n + k}$ and $\text{d}_{M[k]} = (-1)^ k\text{d}_ M$ and

2. as $A$-module the multiplication

$(M[k])^ n \times A^ m \longrightarrow (M[k])^{n + m}$

is equal to the given multiplication $M^{n + k} \times A^ m \to M^{n + k + m}$.

For a morphism $f : M \to N$ of differential graded $A$-modules we let $f[k] : M[k] \to N[k]$ be the map equal to $f$ on underlying $A$-modules. This defines a functor $[k] : \text{Mod}_{(A, \text{d})} \to \text{Mod}_{(A, \text{d})}$.

Let us check that with this choice the Leibniz rule is satisfied. Let $x \in M[k]^ n = M^{n + k}$ and $a \in A^ m$ and denoting $\cdot _{M[k]}$ the product in $M[k]$ then we see

\begin{align*} \text{d}_{M[k]}(x \cdot _{M[k]} a) & = (-1)^ k \text{d}_ M(xa) \\ & = (-1)^ k \text{d}_ M(x) a + (-1)^{k + n + k} x \text{d}(a) \\ & = \text{d}_{M[k]}(x) a + (-1)^ n x \text{d}(a) \\ & = \text{d}_{M[k]}(x) \cdot _{M[k]} a + (-1)^ n x \cdot _{M[k]} \text{d}(a) \end{align*}

This is what we want as $x$ has degree $n$ as a homogeneous element of $M[k]$. We also observe that with these choices we may think of the multiplication map as the map of complexes

$\text{Tot}(M^\bullet [k] \otimes _ R A^\bullet ) \to \text{Tot}(M^\bullet \otimes _ R A^\bullet )[k] \to M^\bullet [k]$

where the first arrow is More on Algebra, Section 15.72 (7) which in this case does not involve a sign. (In fact, we could have deduced that the Liebniz rule holds from this observation.)

The remarks in Homology, Section 12.14 apply. In particular, we will identify the cohomology groups of all shifts $M[k]$ without the intervention of signs.

At this point we have enough structure to talk about triangles, see Derived Categories, Definition 13.3.1. In fact, our next goal is to develop enough theory to be able to state and prove that the homotopy category of differential graded modules is a triangulated category. First we define the homotopy category.

Comment #5351 by Noah Olander on

Shouldn't a right $A$-module be the same as a map $\mathrm{Tot} (M^\bullet \otimes _R A^\bullet) \to M^\bullet$, not $A^\bullet \otimes _R M^\bullet$ as you have written? It's possible that I'm misremembering a sign convention but this seems not right.

Comment #7061 by Álvaro Sánchez on

I think that the actual definition for the $k$-shifted module doesn't satisfy Leibniz. Maybe the action of $A$ on $M[k]$ should be $A^n \cdot M^{m+k} \to M^{m+k+n}$, $(a,x) \mapsto (-1)^{|a|k}ax$? (sorry I write it for left $A$-modules).

Comment #7062 by on

Dear Álvaro Sánchez, thanks very much for this comment. So, this section is for right modules. In the section for left modules I put in a sign, but at the time I thought that for right modules one didn't need the sign. But now that I am redoing the calculation I guess you are right and one needs a sign in both cases to get the Leibniz rule to work out! OK, I will fix this the next time I go through all the comments (and I will also double check the signs in the left modules case). Thanks again!

Comment #7247 by on

It turns out that the sign was/is correct. And... since you were only complaing about the sign in the left module case, this means you weren't wrong either! Phew! It would've been a real mess if the sign didn't work out since it would probably mean there was some problem with the sign rules in More on Algebra, Section 15.72. I have added a check for the Leibniz rule so you can (when the website updates in a few weeks) plainly see that it works! See this commit.

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