I'm trying to understand K\"ahler differentials for CDGAs (commutative differential graded algebras). A few minor things have been confusing me all afternoon. Pointers and references are welcome.


Let $A$ be a CDGA over a field of characteristic zero with differential of degree $1$ and with no cohomology in positive degrees. (If we made the differential of degree $-1$, then we might call these connective CDGAs), let $M$ be a left dg-module over $A$, and let $I=ker(A \otimes A \rightarrow A)$. Then I'd like to verify that $\delta:A \rightarrow I/I^{2}, a \mapsto 1 \otimes a - a \otimes 1$ is a derivation, just like for classical (not dg) commutative algebras. 


For this to work, I seem to be forced to take the following convention for what I mean by derivation: $\delta(a \cdot b)=b \cdot \delta(a)+a\cdot \delta(b)$. On the other hand, some informal references I read give $\delta(a \cdot b)=\delta(a)\cdot b+a\cdot \delta(b)$, and use the convention that an action on the right $m \cdot x$ is defined by $(-1)^{|m||x|}x \cdot m$, where $x,m$ are homogeneous. Thus altogether
we would require $\delta(a \cdot b)=(-1)^{|a||b|+|b|}b \cdot \delta(a)+a\cdot \delta(b)$ for homogeneous $a,b$. But this seems to mess up the signs to make the given $\delta$ a derivation. 


So the question is: what is the right notion of derivation $\delta: A \rightarrow M$ of degree $0$, and more generally of degree $k$?


My apologies for the triviality of the question.



Sign convention for derivations in CDGAs
Yes, left modules for a graded-commutative ring can be routinely converted to right modules by defining $ma=(-1)^{|m||a|}am$, and then the "right" definition of degree zero derivation is


$\delta(ab)=a\delta(b)+\delta(a)b$.


More generally


$\delta(ab)=(-1)^{|\delta||a|}a\delta(b)+\delta(a)b$.


There really isn't a sign problem if you follow the Koszul rule faithfully.


Note that $ab\otimes 1-1\otimes ab=a(b\otimes 1-1\otimes b)+(a\otimes 1-1\otimes a)b$.


