# stacks/stacks-project

Fix a sign error in dga.tex

Thanks to Masahiro Ohno
http://stacks.math.columbia.edu/tag/04JD#comment-499
 @@ -1367,35 +1367,78 @@ \section{Injective modules over algebras} \noindent Let $(A, \text{d})$ be a differential graded algebra. If $M$ is a (right) differential graded $A$-module, then $M^\vee$ (as defined above for graded modules) is a differential graded module as well with differential given by the contragredient of $\text{d}_M$ up to sign. The sign rule (see Example \ref{example-dgm-dg-cat}) is to set {\bf 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 $$\text{d}_{M^\vee}(f) = - (-1)^n f \circ \text{d}_M^{-n - 1} \quad\text{in }(M^\vee)^{n + 1}$$ where $f \in (M^\vee)^n = \Hom_\mathbf{Z}(M^{-n}, \mathbf{Q}/\mathbf{Z})$ and $\text{d}_M^{-n - 1} : M^{-n - 1} \to M^{-n}$ is the differential of $M$. This does indeed work, namely, if $a \in A^m$, $x \in M^{-n - m - 1}$ and $f$ as above, then for $f \in (M^\vee)^n = \Hom_\mathbf{Z}(M^{-n}, \mathbf{Q}/\mathbf{Z})$ and $\text{d}_M^{-n - 1} : M^{-n - 1} \to M^{-n}$ the differential of $M$\footnote{The sign rule is analogous to the one in Example \ref{example-dgm-dg-cat}, although there we are working with right modules and the same sign rule taken there does not work for left modules. Sigh!}. 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. \medskip\noindent If $M$ is a {\bf 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} (a f)(\text{d}(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}(x)a) \\ (-1)^{n + m} f(\text{d}_M(ax) - (-1)^{m + n + 1} x\text{d}(a)) \\ & = - (-1)^{n + m} f(\text{d}(xa) - (-1)^{n + m}x\text{d}(a)) \\ (-1)^m \text{d}_{M^\vee}(f)(ax) + f(x\text{d}(a)) \\ & = (\text{d}(a) f)(x) + (-1)^m (a \text{d}_{M^\vee}(f))(x) (-1)^m (a\text{d}_{M^\vee}(f))(x) + (\text{d}(a)f)(x) \end{align*} in other words we have $\text{d}(af) = \text{d}(a)f + (-1)^m a \text{d}(f)$ 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. \medskip\noindent Similarly, if $M$ is a left differential graded module, then $M^\vee$ becomes a right differential graded module with the same sign rule for $\text{d}_{M^\vee}$. Computation omitted. 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 $((M^n)^\vee)^\vee \to ((M^{n + 1})^\vee)^\vee$. \begin{lemma} \label{lemma-map-into-dual} @@ -1649,7 +1692,7 @@ \section{I-resolutions} \item $I = \lim I/F_pI$, \item the inclusions $F_iI \to F_{i - 1}I$ are admissible monomorphisms, \item the quotients $F_{i - 1}I/F_iI$ are direct sums of the\ \item the quotients $F_{i - 1}I/F_iI$ are direct sums of the modules $A^\vee[k]$. \end{enumerate} In fact, condition (2) is a consequence of condition (3), see