Throughtout, assume all rings are commutative with identity, and all schemes are separated. If $ A \rightarrow B$ is an $A$-algebra, I am having a lot of trouble understanding precisely what is meant by the module $\Omega_{B/A}$. In particular, I am having trouble understanding precisely what the module structure is, and as a result am having trouble understanding how to globalise this to the sheaf of differentials on a scheme.

We begin by defining some morphisms:

Let $\mu: B \otimes_A B \rightarrow B$ be the multiplication map and let $I = \text{ker} \mu$ be the kernel.

Let $\lambda_{1}: B \rightarrow B\otimes_{A} B$ be the map defined by $\lambda(b) = b \otimes 1$ and similarly define $\lambda_{2}(b) = 1 \otimes b$.

It is often said loosely that $\Omega_{B/A}$ is "the $B$-module" $I/I^2$.

I want to be careful with precisely how this set inherits a $B$-module structure, because this is never made clear, and I can see a potential ambiguity arising when this is globalised. My understanding of the module structure is as follows:

$B \otimes_{A} B$ inherits the structure of a $B$-module via the morphism $\lambda_{1}$, or in other words multiplication on the left. We are able to show that $I$ is a $B$-submodule of $B \otimes_{A} B$. So $I$ is a $B$ module in this way, and we again see that $I^2$ is an $B$-submodule of $I$ so that $I/I^2$ is the resulting quotient module.

Now we try to globalise this. Let $X = \text{spec}B$ and $Y = \text{Spec}A$. The morphism of rings $\mu: B \otimes_{A} B \rightarrow B$ gives rise to the diagonal morphism $\Delta: X \rightarrow X \times_{Y} X$ and this is a closed immersion since we are only dealing with affine schemes. So we have a surjective morphism of sheaves, $$ \Delta^{\#}: \mathcal{O}_{X \times_{Y} X} \longrightarrow \Delta_{*}\mathcal{O}_{X}, $$ with kernel $\mathscr{I}$, which is the quasicoherent sheaf of ideals corresponding to the ideal $I$ of $B \otimes_{A} B$.

We want $\Delta^{*}(\mathscr{I}/\mathscr{I}^2)$ to be a quasicoherent sheaf of $\mathcal{O}_{X}$-modules, and this is where I get stuck. We should have \begin{align} \Delta^{*}(\mathscr{I}/\mathscr{I}^2) &= \Delta^{*}\big((I/I^2)^{\sim}\big) \\ & = \big( B \otimes_{B \otimes_{A} B} I / I^2 \big)^\sim \end{align} and I am not even sure how to handle such an object. Moreover, I don't see how the object $B \otimes_{B \otimes_{A} B} I / I^2$ as a $B$-module is related to the $B$-module $I/I^2$ given by the map of rings $\lambda_{1}$.

Can someone explain to me what is actually going on here, in explicit detail. So many books are loose with their language and just say things like "the $B$-module" when they are actually performing some change of rings, etc, and it results in me not having an idea how the sheaf of modules actually works.


If $f:R\to S$ is a surjective ring homomorphism with kernel $I$, then $I/I^2$ is canonically an $S$-module (take any section, as a map of sets, of $f$ to give the structure; in your case you can take $\lambda_1$ or $\lambda_2$). $I/I^2$ is called the conormal module of $f$.

If $M$ is an $R$-module, then $M\otimes_RS$ is canonically isomorphic to $M/IM$, so if $M$ is an $S$-module, then $M\otimes_RS=M/IM=M$.

I think this answers your questions.

  • $\begingroup$ Let me just clarify to make sure I have understood. Let $f: R \rightarrow S$ be a surjective morphism of rings with kernel $I$ and let $\lambda: S \rightarrow R$ be a section. Let $M$ be an $S$-module and denote by $f^{*}M$ the restriction of scalars to $R$. Then are you saying that $f^{*}M \otimes_{R} S = M$ as $S$-modules? Is there some way to state this in terms of the adjunction between extension and restriction of scalars? I am assuming this has to do with the unit or counit of that adjunction. $\endgroup$
    – Joe
    Feb 25 '19 at 17:44
  • $\begingroup$ Then are you saying that $f^∗M\otimes_RS=M$ as S-modules? Yes. Is there some way to state this in terms of the adjunction between extension and restriction of scalars? The counit $f^∗M\otimes_RS\to M$ is an isomorphism. $\endgroup$
    – A.G
    Feb 25 '19 at 21:42

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