4
$\begingroup$

From nlab, the module of Kähler differentials over some category $\mathcal{C}$ is the free functor: $$\Omega: \mathcal{C} \to \mathsf{Mod_{\mathcal{C}}}$$ left-adjoint to the (forgetful) embedding: $$u: \mathsf{Mod}_{\mathcal{C}} \cong \mathsf{Ab}(\mathcal{C}) \hookrightarrow \mathcal{C}$$ with $\mathsf{Ab}(\mathcal{C})$ denoting abelian group objects of $\mathcal{C}$, and $\mathsf{Mod}_{\mathcal{C}}$ denoting the category of modules of $\mathcal{C}$. By this definition, we automatically get a bijection of hom-sets: $$\mathsf{Mod_{\mathcal{C}}}(\Omega(R), M) \cong \mathcal{C}(R, u(M))$$ Also, according to the same nlab page as above, $R$-derivations taking values in some $R$-module $M$ (with $R$ some object of $\mathcal{C}$) are morphisms: $$d: \Omega(R) \to M$$ in $\mathsf{Mod}_R \cong \mathsf{Mod}_{R/\mathcal{C}}$. Thus, they could be identified by $\mathcal{C}$-morphisms: $$X: R \to u(M)$$ because of the adjunction $\Omega \dashv u$.

At this point, I have two questions:

  1. When $\mathcal{C} = \mathsf{CRing}$, the category of commutative and unital rings, do we automatically get the Leibniz/product rule ? Why or why not ?
  2. If we do automatically get the Leibniz rule, then is it also the case in categories more general than $\mathsf{CRing}$ ?

Thank you.

$\endgroup$

1 Answer 1

2
$\begingroup$

1. The Leibniz rule follows immediately from the last description of derivations as morphisms of commutative rings X:R→u(M).

Indeed, u(M) is the square-zero extension of some R-module M' (in the traditional sense), i.e., u(M)=R⊕M'.

Now a morphism of commutative rings f:R→R⊕M' in the slice category C/R (not in C, as is claimed in the main post) necessarily has the form r↦(r,φ(r)), for some linear map φ. Since f is a homomorphism, we have f(1)=(1,φ(1))=(1,0), so φ(1)=0. Also $$(rr',φ(rm'+r'm))=f(rr',rm'+r'm)=f((r,m)(r',m'))=f(r,m)f(r',m')=(r,φ(m))(r',φ(m'))=(rr',rφ(m')+r'φ(m)),$$ so $$φ(rm'+r'm)=rφ(m')+r'φ(m).$$

This is precisely the Leibniz rule.

2. Yes, for instance, this is true for algebras over Fermat theories. See Carchedi and Roytenberg's Homological Algebra for Superalgebras of Differentiable Functions.

$\endgroup$
2
  • $\begingroup$ Why do we know that $u(M)$ is the square-zero extension of some $R$-module $M'$ ? $\endgroup$ Commented May 24, 2020 at 19:43
  • $\begingroup$ @DatMinhHa: This is explained in the link to the nLab that you yourself cited in the main post: ncatlab.org/nlab/show/… $\endgroup$ Commented May 24, 2020 at 19:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .