I'm studying the module of Kähler differentials from Eisenbud's Commutative Algebra, you can easily get a pdf by typing the name on Google. We have proposition 16.3:
If $\pi:S\to T$ is a surjective morphism of $R$-algebras, and $I=\ker(\pi)$ we have an exact sequence $$I/I^2\to T\otimes_S\Omega_{S/R}\to \Omega_{T/R}\to 0.$$
Now consider $B$ an essentially finitely generated $A$– algebra, i.e we can write $B\cong S_p$ where $S$ is a quotient of a polynomial ring with coefficients in $A$, $p\subseteq S$ prime. I want to show that $\Omega_{B/A}$ is a finitely generated $B$-module.
Since the right morphism is surjective, showing that $T\otimes_S\Omega_{S/R}$ is finitely generated is enough.
Consider first the case $B=A[x_1,\dots,x_n]$. What seems natural to me is taking $T=B$, and maybe $S=A$ or $S=B$. However I'm not sure if there is necessarily an epimorphism from $A$ to $B$, and $id_B=\pi$ just leads to $\Omega_{B/A}=\Omega_{B/A}$ since $I/I^2=0$.
Do you have ideas?
Best Answer
Yes, $\Omega_{B/A}$ is a finitely-generated $B$-module. Using the notation from your post, we first see that $\Omega_{S/A}$ is a finitely-generated $S$-module from applying the result listed in your post to a surjection $A[x_1,\cdots,x_n]\to S$ as $\Omega_{A[x_1,\cdots,x_n]/A}$ is the free $A[x_1,\cdots,x_n]$-module on the symbols $dx_i$ for $1\leq i\leq n$. Next, we use the fact that for a ring map $C\to D$ and any multiplicative subset $U\subset D$ we have $U^{-1}\Omega_{D/C}\cong \Omega_{U^{-1}D/C}$ applied to $A\to S$ with $U=S\setminus p$. As the localization of a finitely-generated module is again finitely generated over the localized ring, we have the result. (One proof of this last statement is that finite generation of the $R$-module $M$ is equivalent to a surjective map $R^{\oplus n}\to M$, and as localization is exact we get a surjective map $S^{-1}R^{\oplus n}\to S^{-1}M$ by localizing at the multiplicative set $S\subset R$.)