5
$\begingroup$

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?
$\endgroup$
4
  • 2
    $\begingroup$ 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. $\endgroup$ – SeraPhim Aug 7 '20 at 15:13
  • 2
    $\begingroup$ 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. $\endgroup$ – Zhen Lin Aug 7 '20 at 15:27
  • $\begingroup$ 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? $\endgroup$ – Charlie Chang Aug 7 '20 at 15:55
  • $\begingroup$ See the comment to this answer: math.stackexchange.com/a/653036 $\endgroup$ – Jehu314 Aug 7 '20 at 16:01
2
$\begingroup$

$\newcommand\Mod{\operatorname{Mod}}\newcommand\cRng{\operatorname{cRng}}\newcommand\Kappa{\mathrm{K}}\require{AMScd}$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 $\Omega_A(B)\otimes_AC\cong\Omega_C(B\otimes_AC)$.
  • $\endgroup$
    1
    • 1
      $\begingroup$ +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 $A$. $\endgroup$ – desiigner Aug 7 '20 at 16:28

    Your Answer

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

    Not the answer you're looking for? Browse other questions tagged or ask your own question.