If $A$ and $I$ are $k$-algebras, what is a '$k$-linear derivation $A \to I$?'
Solution 1:
We can actually define the $A$-module structure with $g$ or $g'$ and the leibniz rule will make sense, since $g(x) \theta(y) = g'(x) \theta(y)$. This fact follows from the fact that $I^2 = 0$. After all, $$g(x) \theta(y) - g'(x) \theta(y) = \theta(x) \theta(y) = 0$$ since $\theta(x) \theta(y) \in I^2$.
Sorry to answer my own question so quickly - I couldn't find a similar one online so I thought I shouldn't delete it.