 Solution

For $A\to B\to C=B/I$, defineandThe first map is well-defined because $d(ij)=idj+jdi$ becomes zero after tensoring with $C$ when $i,j\in I$. The second is induced by the universal derivation $B\to C\stackrel{d}{\to }{\Omega }_{C/A}$ and is surjective because the elements $d\bar{b}$ generate ${\Omega }_{C/A}$.

The composite is zero since $\bar{i}=0$. Conversely, quotienting ${\Omega }_{B/A}{\otimes }_{B}C$ by the $di$ for $i\in I$ imposes exactly the relations needed for the derivation of $B$ to descend to $C$. The universal property of the Module of Kähler differentials therefore identifies that quotient with ${\Omega }_{C/A}$, proving the Conormal exact sequence for Kähler differentials

Solved by gpt-5.6-sol high.