Lemma 48.29.3. Let $f : X \to Y$ be a morphism of schemes. Let $r \geq 0$. Assume

1. $Y$ is Cohen-Macaulay (Properties, Definition 28.8.1),

2. $f$ factors as $X \to P \to Y$ where the first morphism is an immersion and the second is smooth and proper,

3. if $x \in X$ and $\dim (\mathcal{O}_{X, x}) \leq 1$, then $f$ is Koszul at $x$ (More on Morphisms, Definition 37.59.2), and

4. if $\xi$ is a generic point of an irreducible component of $X$, then we have $\text{trdeg}_{\kappa (f(\xi ))} \kappa (\xi ) = r$.

Then with $\omega _{Y/X} = H^{-r}(f^!\mathcal{O}_ Y)$ there is a map

$\wedge ^ r\Omega _{X/Y} \longrightarrow \omega _{Y/X}$

which is an isomorphism on the locus where $f$ is smooth.

Proof. Let $U \subset X$ be the open subscheme over which $f$ is a local complete intersection morphism. Since $f$ has relative dimension $r$ at all generic points by assumption (4) we see that the locally constant function of Lemma 48.29.2 is constant with value $r$ and we obtain a map

$\wedge ^ r\Omega _{X/Y}|_ U = \wedge ^ r \Omega _{U/Y} \longrightarrow \omega _{U/Y} = \omega _{X/Y}|_ U$

which is an isomorphism in the smooth points of $f$ (this locus is contained in $U$ because a smooth morphism is a local complete intersection morphism). By Lemma 48.21.5 and the assumption that $Y$ is Cohen-Macaulay the module $\omega _{X/Y}$ is $(S_2)$. Since $U$ contains all the points of codimension $1$ by condition (3) and using Divisors, Lemma 31.5.11 we see that $j_*\omega _{U/Y} = \omega _{X/Y}$. Hence the map over $U$ extends to $X$ and the proof is complete. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).