# What is the module and sheaf of differentials (actually)?

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 $$M\otimes_RS=M/IM=M$$.
• 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. – Joe Feb 25 at 17:44
• 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. – A.G Feb 25 at 21:42