I am reading how derivations of an algebra work. I am following the reference: Introduction to affine group schemes, Waterhouse. After having defined $\Omega_B=B^n$ for $B=k[x_1,...,x_n]$ and having shown that $\Omega_B$ satisfies Der$_k$($B,M)\cong \hom_B(\Omega_B,M)$, we now want to define $\Omega_A$ for $A$ a generic $k$-algebra (which is the then of the form $A=B/I$).
The book defines it as $\Omega_A=\Omega_B/I \cdot \Omega_B+B \cdot dI$, where $d: B \rightarrow \Omega_B$ is the map $dX_i=e_i$, $e_i$ the generator of the $i$-th $B$ in the direct sum of $\Omega_B$. I am not sure that $\Omega_B/I$ is. Does it mean that $I$ acts trivially and so $\Omega_B/I$ acts on $\Omega_B$ by multiplication of the representative? Is it $\Omega_B/I=\Omega_{B/I}$?
Hope someone can clarify this! Thanks!
It probably means $$ \Omega_A = \frac{\Omega_B}{I\cdot \Omega_B+B\cdot dI} $$ So you're taking the quotient of $\Omega_B=B^n$ by the submodule generated by everything of the form $df$ for $f\in I$ and everything of the form $fdx_i$ for $f\in I$ (otherwise it wouldn't even be a $B/I$-module).