The conormal sequence is only right exact

For a homomorphism of commutative rings $A\to P$, an ideal $I\subseteq P$ and $B=P\mathrm{/}I$, the conormal sequence $I\mathrm{/}{I}^2⟶B{\otimes }_P{\mathrm{\Omega }}_{P\mathrm{/}A}⟶{\mathrm{\Omega }}_{B\mathrm{/}A}⟶0$ of Conormal exact sequence for an algebra quotient is exact at the middle and final terms only: the left arrow is not asserted to be injective, and it need not be. The kernel of that arrow is therefore genuine information about the pair $(I,P)$, not a defect of the construction.

The smallest witness is $P=k[x]$ with $k$ a field, $I=(x^2)$ and $B=k[x]\mathrm{/}(x^2)$. Then $I\mathrm{/}{I}^2=(x^2)\mathrm{/}(x^4)$, and the class $[x^3]$ is nonzero there, because $x^3\mathrm{\notin }(x^4)$ by degrees. Its image under the conormal map is $1\otimes \mathrm{d}(x^3)$. By Polynomial differentials are free the module ${\mathrm{\Omega }}_{k[x]\mathrm{/}k}$ is free on $\mathrm{d}x$ with $\mathrm{d}g=g^{\mathrm{\prime }}(x)\mathrm{d}x$, so $\mathrm{d}(x^3)=3x^2\mathrm{d}x$; after identifying $B{\otimes }_P{\mathrm{\Omega }}_{P\mathrm{/}k}\cong B\mathrm{d}x$ this becomes $3x^2\mathrm{d}x=0$, because $x^2=0$ in $B$. The class $[x^3]$ thus lies in the kernel of the left arrow, in every characteristic: for $p=3$ the coefficient $3$ is already zero in $k$, and otherwise the coefficient dies only after passing to $B$.

Two qualifications. First, the failure is not an artefact of a badly chosen presentation: it depends on the ideal $I$ and not on the number of generators used to describe it. Second, with additional regularity hypotheses on $I$ the left arrow can become injective, so that the sequence starts as a short exact sequence; no such hypothesis is part of the general statement, and the injectivity is never to be used on this page without an explicit regular hypothesis. The companion examples page records the witness above as a counterexample with the same computation.