# “Formally unramified iff trivial Kähler differentials” using only universal properties?

For convenience we work with commutative rings instead of commutative algebras.

Fix a commutative ring $R$. Consider the functor $\mathsf{Mod}\longrightarrow \mathsf{CRing}$ defined by taking an $R$-module $M$ to $R\ltimes M$ (with dual number multiplication). Following the nlab, the module of Kähler differentials of $R$ is defined as the universal arrow from $R$ to this functor.

There's also the equivalence between the category of split zero-square extensions and the category of modules. These result in the natural isomorphisms ($\operatorname{dom}$ takes an arrow to its domain).

\begin{aligned}\mathsf{CRing}(R,\operatorname{dom}(A^{\prime}\twoheadrightarrow A)) & \cong{\substack{\text{split square-zero}\\ \text{extensions} } }(R\ltimes\Omega_{R}\twoheadrightarrow R,A^{\prime}\twoheadrightarrow A)\\ & \cong \mathsf{Mod}(\Omega_{R},\operatorname{Ker}(A^{\prime}\twoheadrightarrow A)). \end{aligned}

So $\Omega _R\cong \bf 0$ iff $R$ has exactly one arrow to the domain of every split square-zero extension.

On the other hand, we say $R$ is unramified if given a zero-square extension $A^\prime \twoheadrightarrow A$, the induced post-composition $$\mathsf{CRing}(R,A^\prime)\longrightarrow \mathsf{CRing}(R,A)$$ is injective. The diagrams of interest are below.

$$\require{AMScd} \begin{CD} A^\prime /I @<<< R\\ @AAA \\ A^\prime & \end{CD} \; \; \; \begin{CD} A^\prime /I @<<< R\\ @AAA @VVV\\ A^\prime @<<< I \end{CD}$$

Now, if $\Omega _R\cong \bf 0$ then there exists and is unique a diagonal arrow $R\to A^\prime$, which seems to easily give "unramified w.r.t split square-zero extensions". However, I see no way to deduce anything about the non-split ones.

The thing is, I feel a proof is very close: suppose $g,h$ are diagonal fillers $R\to A^\prime$ on the left below. Viewing them as module arrows this is equivalent to saying $g-h$ lands in $I$, i.e factors through $I\vartriangleleft A^\prime$. Finally $g=h$ iff this factorization $R\to I$ of $g-h$ must be zero. If only this could be given by the universal property...

Perhaps the correct way to proceed is using some universal property of the diagonal? (Not that it realizes the Kähler differentials.)

• As long as $A^' \to A$ is a split extension I don't see a problem with anything. It is true that formally unramified is equivalent to existence and uniqueness of lifts to split square zero extensions. – Saal Hardali Nov 25 '17 at 13:07
• The stacks project has a sketch of a proof at stacks.math.columbia.edu/tag/00UO – Vladimir Sotirov Nov 26 '17 at 0:52
• @VladimirSotirov that proof uses the construction of $\Omega_{R/S}$ as $J/J^2$ where $J = \mathop{\mathrm{Ker}}(S \otimes_R S \to S)$. I am hoping to understand how to move between the universal property of $\Omega \dashv \ltimes$ and the "at most one lift" criterion in a purely categorical fashion. The stacks project seems to have parts of this by considering "universal first order thickenings" but I am not able to see clearly what to do to remain with a purely categorical mechanism. – Arrow Nov 26 '17 at 1:00

Some time ago I worked out how to define Kähler differentials and derive their cotangent and conormal exact sequences in any protomodular category with pullbacks and pushouts. Based on the same idea, I think the following argument works to show that a morphism being unramified is equivalent to the vanihing of its module of relative Kähler differentials.

Given a morphism $X\xrightarrow{f} Y$, the category of Beck modules equipped with $f$-derivations has for its objects internal abelian group objects $Y\leftarrow M$ in $\mathcal C/Y$ equipped with an additional section $X\xrightarrow{d} M$ of $X\leftarrow M$ such that $d\circ f=0\circ f$ where $Y\xrightarrow{0}M$ is the zero section of the Beck module. The Beck module $Y\leftarrow\Omega_{Y/X}$ of Kähler differentials is the initial object of this category.

For example, a Beck module over a ring $R$ is precisely a square-zero split extension $R\leftarrow R\oplus M$ for $M$ an $R$-module, and a $k$-linear derivation amounts to an $k$-linear map $R\xrightarrow{d}M$ such that $d(ab)=d(a)b+ad(b)$ and $d(c)=0$ for $c$ in the image of $k\to R$.

In the case where $\mathcal C$ is a protomodular category (e.g. the category of rings), recall that the forgetful functor from Beck modules to morphisms equipped with a section has a left adjoint. For the category of rings, a pointed object over a ring $R$ is a split extension $R\leftarrow R\oplus A$ for an $R$-algebra $A$, which left adjoint sends to the Beck module $R\oplus (A/A^2)$ (i.e. the left adjoint sends $R$-algebras $A$ to $R$-modules $A/A^2$).

This allows us to generalize the construction of Kähler differentials to arbitrary protomodular categories as follows.

Define a weaker notion of a pointed object equipped with an $f$-prederivation, i.e. a morphism $Y\leftarrow P$ equipped with a "point" section $Y\xrightarrow{0}P$ and an additional section $X\xrightarrow{d} P$ of $X\leftarrow P$ such that $d\circ f=0\circ f$.

For example, a pointed object in the category of rings is a split extension $R\leftarrow R\oplus A$ for $A$ an $R$-algebra, and a $k$-linear pre-derivation amounts to an additive map $R\xrightarrow{d} A$ such that $d(ab)=d(a)b+ad(b)+d(a)d(B)$ and $d(c)=0$ for $c$ in the image of $k\to R$.

The pointed object $\Gamma_{Y/X}$ of Kähler pre-differentials is then an initial object in the category of morphisms $Y\leftarrow P$ equipped with a pair of sections that make a fork $X\xrightarrow{f}Y\overset d{\underset 0\rightrightarrows} P$.

It turns out that $\Gamma_{Y/X}$ is then given by the cokernel pair of $X\xrightarrow{f}Y$, and $\Omega_{Y/X}$ by its reflection in Beck modules. Hence in the category of rings, $\Gamma_{R/k}$ is $R\leftarrow R\otimes_kR$ and $\Omega_{R/k}$ is $R\leftarrow R\oplus(J/J^2)$ where $J=\ker(R\leftarrow R\otimes_k R)$.

Recall that in a protomodular category, being an abelian group object is a property of pointed objects, rather than a structure. Accordingly, we define a morphism $A'\xrightarrow{f} A$ in a protomodular category $\mathcal C$ to be a square-zero morphism if its kernel pair $K[f]\rightrightarrows A'$ equipped with the diagonal morphism $A'\xrightarrow{\Delta}K[f]$ is a Beck module over $A'$.

For example, $Y\leftarrow\Omega_{Y/X}$ is a square-zero morphism because its kernel pair is the pullback of a Beck module so still a Beck module.

Thus, we can define a morphism $X\to Y$ to be unramified, respectively etale, respectively smooth, if any square $\require{amsCD} \begin{CD} X @>>> A'\\ @VVV @VVV\\ Y@>>> A \end{CD}$ can be filled so that the resulting diagram commutes with at most one, respectively at least one, respectively exactly one morphism $Y\to A$.

We can now easily show that unramified is equivalent to $\Omega_{Y/X}=0$, i.e. to the fact that the zero section $Y\xrightarrow{0}\Omega_{Y/X}$ and the canonical $f$-derivation $Y\xrightarrow{d}\Omega_{Y/X}$ are equal.

First, unramified implies $\Omega_{Y/X}=0$ since there must be at most one fill in the square $\require{amsCD} \begin{CD} X @>>> \Omega_{Y/X}\\ @VVV @VVV\\ Y@= Y \end{CD}$ and these fills are precisely the $f$-derivations $Y\to\Omega_{Y/X}$.

Second, here's a sketch of a proof that $\Omega_{Y/X}=0$ implies $X\to Y$ is unramified. The pullback of the square-zero $A\leftarrow A'$ along $Y\to A$ is still square-zero, hence it suffices to show that there is at most one fill in any square $\require{amsCD} \begin{CD} X @>>> A\\ @VVV @VfVV\\ Y@= Y \end{CD}$. But a pair of fills $Y\overset a{\underset b\rightrightarrows}A$ is the data of a morphism $\Gamma_{Y/X}\to K[f]$ of pointed objects over $Y$. Since $A\xrightarrow{f}Y$ is square-zero, $K[f]$ is by assumption a Beck module, hence $\Gamma_{Y/X}\to K[f]$ factors as $\Gamma_{Y/X}\to\Omega_{Y/X}\to K[f]$. Finally, triviality of $\Omega_{Y/X}$ forces uniqueness of the morphism $\Omega_{Y/X}\to K[f]$.

• This is very cool. Thank you for sharing. What can be deduced from the cotangent and conormal sequences in this generality? – Arrow Nov 26 '17 at 7:03
• @Arrow I don't know whether you are still interested. My understanding is that, we should pass to cotangent complexes instead of Kähler differentials to discuss exact sequences. In fact, if you look at Lurie's Spectral Algebraic Geometry, the section of derived algebraic geometry gives a definition of absolute cotangent complexes via universal properties, and defines relative ones as the cofiber, then he verifies that this satisfies another universal property. This explains the cotangent exact sequence "neatly". – Yai0Phah Oct 15 '19 at 19:01
• @Arrow On the other hand, the conormal sequence is more magic. It is essentially computing the $H_1$ of the cotangent complex associated to a surjection. I have not come up or seen a computation which does not essentially depend on the first order infinitesimal neighborhood of the diagonal yet. (A remark: canonical proofs of Quillen's spectral sequence are also essentially considering first order infinitesimal neighborhoods of the diagonal) – Yai0Phah Oct 15 '19 at 19:07
• @Arrow Note that we can talk about surjections and simplicial ideals of simplicial commutative rings, while there are no counterparts at present in the world of commutative (namely, $\mathbb E_\infty$-) ring spectra, except the trivial one: surjections on $\pi_0$ and ideals of $\pi_0$. That might be an explanation why the conormal sequence could be more subtle. If it were done purly via universal properties, it could be easily migrated into world of commutative ring spectra. – Yai0Phah Oct 15 '19 at 19:14