# Functoriality of the module of Kähler differentials

In Eisenbud's Commutative Algebra, at the start of Chapter 16, he describes the module of Kähler differentials: given a ring $$R$$ and an $$R$$-algebra $$S$$, we have the associated $$S$$-module $$\Omega_{S/R}$$. This comes equipped with an $$R$$-module homomorphism $$d: S \to \Omega_{S/R}$$, called the universal $$R$$-linear derivation, which satisfies an associated universal property.

He goes on to state that the module of Kähler differentials is functorial in the following sense: given a commutative diagram of rings $$\require{AMScd}$$ $$\begin{CD} R @>>> R'\\ @V{}VV @VVV\\ S @>>> S', \end{CD}$$ where $$S$$ is an $$R$$-algebra, and $$S'$$ is an $$R'$$-algebra, there is a commutative square of abelian groups

$$\require{AMScd}$$ $$\begin{CD} S @>>> S'\\ @V{d}VV @VV{d}V\\ \Omega_{S/R} @>>> \Omega_{S'/R'}, \end{CD}$$ where $$S \to S'$$ is the associated $$R$$-algebra homomorphism, $$\Omega_{S/R} \to \Omega_{S'/R'}$$ is an $$S$$-module homomorphism, and $$d$$ denotes the universal derivation in each context.

As Eisenbud notes, this is quite complicated to state. I am curious if this can be rephrased in a simpler way. My question can be stated concisely as follows:

1. As the module of Kähler differentials is functorial, we should be able to understand it as a functor of the form $$\Omega_{-/-}: \mathscr{C} \to \mathscr{D}$$. In this context, what are the categories $$\mathscr{C}$$ and $$\mathscr{D}$$?
2. Once question 1 is answered, how do you understand the universal property of the module of Kähler differentials in this categorical framework?
• It may help you to know that for a unital commutative $R$-algebra $S$, the module of Kähler differentials $\Omega_{S/R}$ is isomorphic to the first Hochschild homology module $HH_1(S, S)$ as $R$-modules. And $HH_n$ is a functor from the category of associative $R$-algebras to the category of $R$-modules. I'm not $100$% sure if this answers your first question, so I'll leave it here as a comment. – SeraPhim Aug 7 '20 at 15:13
• The domain category is the arrow category of the category of commutative rings. The codomain category is not as obvious – there are a few reasonable but different choices – but as Eisenbud suggests the arrow category of the category of abelian groups is one possibility. Another is the arrow category of the category of all modules over all rings. – Zhen Lin Aug 7 '20 at 15:27
• Kähler differentials are extension of k-forms from manifolds or smooth/Lie groups to rings, or to schemes, which is varieties (similar to vector fields?) over ring? – Charlie Chang Aug 7 '20 at 15:55
• See the comment to this answer: math.stackexchange.com/a/653036 – Jehu314 Aug 7 '20 at 16:01

$$\newcommand\Mod{\operatorname{Mod}}\newcommand\cRng{\operatorname{cRng}}\newcommand\Kappa{\mathrm{K}}\require{AMScd}$$Let $$A$$ be a commutative ring. Let $$\operatorname{Mod}_A$$ be the category such that:

• the objects are pairs $$(B,M)$$ where $$B$$ is an $$A$$-algebra and $$M$$ is a $$B$$-module;
• the morphisms are pairs $$(\varrho,\varphi):(B,M)\to(C,N)$$ where $$\varrho:B\to C$$ is an $$A$$-algebra homomorphism and $$\varphi:M\to N$$ is a $$B$$-module homomorphism (where $$N$$ is a $$B$$-module by scalar restriction trough $$\varrho$$).
Let $$\cRng_A$$ denote the category of (commutative, associative, unitary) $$A$$-algebras and $$\Xi_A:\Mod_A\to\cRng_A$$ be the functor such that:
• If $$(B,M)\in\Mod_A$$, then $$\Xi_A(B,M)$$ is the commutative $$A$$-algebra on the $$A$$-module $$B\times M$$ with multiplication defined for $$b,b'\in B$$ and $$x,x'\in M$$ by $$(b,x)(b',x')=(bb',xb'+bx')$$
• If $$(\varrho,\varphi):(B,M)\to(C,N)$$ in $$\Mod_A$$, then $$\Xi_A(\varrho,\varphi)$$ is the homomorphism of $$A$$-algebras $$\varrho\times\varphi:B\times M\to C\times N$$.
Let $$\Kappa_A:\cRng_A\to\Mod_A$$ the functor such that such that:
• if $$B$$ is an $$A$$-algebra $$B$$, then $$\Kappa_A(B)=(B,\Omega_A(B))$$.
• if $$\varrho:B\to C$$ is an homomorphism of $$A$$-algebras, then there exists an homomorphism of $$B$$-modules $$\Omega_A(\varrho)$$ making the following diagram commutative: $$\begin{CD} B@>>>C\\ @VdVV@VVdV\\ \Omega_A(B)@>>>\Omega_A(C) \end{CD}$$
Let $$B\in\cRng_A$$ and $$(C,N)\in\Mod_A$$.
• The forgetful functor $$\Mod_A\to\cRng_A$$ is a fibration.
• The forgetful functor $$\Mod_A\to\cRng_A$$ is a cofibration.
• $$[\varrho,\delta]:B\to C\times N$$ is an homomorphism of commutative $$A$$-algebras if and only if $$\varrho:B\to C$$ is an homomorphims of $$A$$-algebras and $$\delta:B\to N$$ is a derivation.
• We have an adjunction $$\Kappa_A:\cRng_A\rightleftarrows\Mod_A:\Xi_A$$.
• We have a bijection \begin{align} \hom_{\Mod_A}((B,\Omega_A(B)),(C,N))&\xrightarrow\sim\hom_{\cRng_A}(B,C\times N)\\ (\varrho,\varphi)&\mapsto[\varrho,\varphi\circ d] \end{align}
• We have an isomorphism of $$C$$-modules $$\Omega_A(B)\otimes_AC\cong\Omega_C(B\otimes_AC)$$.
• +1 Thank you for this detailed response! This is very helpful and clarifies some other confusions I had surrounding Kähler differentials and trivial square zero extensions. How would you phrase $\Omega_{-,-}$ as a functor in this framework though? It seems as though $K_A$ requires $\Omega_{-/A}$ in its definition. In particular (although we know it should be true), it's not obvious that what you've written isn't functorial in $A$, and I was curious how to make $\Omega_{B/A}$ functorial in both $B$ and $A$. – desiigner Aug 7 '20 at 16:28