1
$\begingroup$

I'm reading Commutative Algebra with a View Toward Algebraic Geometry written by Eisenbud. In chapter 16(Module of Differentials) there is a proposition:

If $S$ is a finitely generated $R$-algebra, say $S=R[x_1,...,x_r]/I$, and if $I=(f_1,...,f_s)$, then $S \otimes_R\Omega_{R[x_1,...,x_r]/R}=\oplus_iSdx_i$.

My attempt: Let $T=R[x_1,...,x_r]$, we have the following map with $I$ as its kernel: $T\to S\to 0$. $\{d(x_i+I)\}_{I\in\{1,...,r\}}$ generate $\Omega_{S/R}$ therefore, $\Omega_{S/R}=\oplus_iSd(x_i+I)$. Using the conormal sequence, there is a surjective morphism $D\pi$ from $S \otimes_R\Omega_{R[x_1,...,x_r]/R}$ to $\Omega_{S/R}$. $D\pi$ is not necessarily an isomorphism...

$\endgroup$
4
  • $\begingroup$ What is $K$ in your second to last sentence? Next, how much of the compatibility of Kahler differentials with extension of scalars have you proven so far? $\endgroup$ – KReiser Oct 26 '20 at 3:00
  • $\begingroup$ @KReiser 🙏🏻 Thanks it was a typo I corrected it. You mean compatibility between $\Omega_{S/R}$ and $\Omega_{R[x_1,...,x_r]/R}$? I haven't shown it yet $\endgroup$ – user766483 Oct 26 '20 at 3:15
  • $\begingroup$ @KReiser $dx \in \Omega_{R[x_1,...,x_n]/R}$ is sent to $d(x+I) \in \Omega_{S/R}$ and elements in $\Omega_{S/R}$ are subjected to the same relations additionally we have $\forall i \in I: d(i)\mapsto0$ $\endgroup$ – user766483 Oct 26 '20 at 3:22
  • 1
    $\begingroup$ I remember being confused by this, too and I'm fairly certain this is a typo: it should be $S \otimes_{R[x_1, \ldots, x_r]}\Omega_{R[x_1,...,x_r]/R}=\oplus_i S \, dx_i$. I found it really annoying and confusing how he changed the roles of $R, S,$ and $T$ in this section, and apparently it confused the author, too! $\endgroup$ – Viktor Vaughn Dec 6 '20 at 20:42
0
$\begingroup$

If $A:=R[x_1,..,x_r]$ is a polynomial ring over $R$ in $r$ variables it follows there is an isomorphism of $A$-modules $\Omega_{A/R} \cong \oplus_i Adx_i$. The sum is a direct sum, hence $\Omega_{A/R}$ is a free $A$-module of rank $r$ on the elements $dx_i$. It follows $S\otimes_R \Omega_{A/R} \cong \oplus_i S\otimes_R Adx_i$ and $S\otimes_R A \neq S$ are non-isomorphic rings, hence it seems your formula is incorrect. If the elements $f_1,..,f_l$ generate your ideal $I$, and you want a formula for the $S$-module $\Omega_{S/R}$ the following holds:

Formula 1. $\Omega_{S/R} \cong \oplus_i Sdx_i/C,$

where the module $C$ is the left $S$-module generated by the differentials $df_i$ for all $i=1,..,l$. You may find details on this construction in a commutative algebra book where they discuss the module of derivations and differentials. There are "fundamental exact sequences" involving the module of differentials, and these sequences can be used to prove such results. Mastumuras book "Commutative Ring Theory" studies derivations and differentials and gives explicit proofs of properties of this construction. Formula 1 is proved.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy