Exalcomm

In algebra, Exalcomm is a functor classifying the extensions of a commutative algebra by a module. More precisely, the elements of Exalcommk(R,M) are isomorphism classes of commutative k-algebras E with a homomorphism onto the k-algebra R whose kernel is the R-module M (with all pairs of elements in M having product 0). Note that some authors use Exal as the same functor. There are similar functors Exal and Exan for non-commutative rings and algebras, and functors Exaltop, Exantop, and Exalcotop that take a topology into account.

"Exalcomm" is an abbreviation for "COMMutative ALgebra EXtension" (or rather for the corresponding French phrase). It was introduced by.

Exalcomm is one of the André–Quillen cohomology groups and one of the Lichtenbaum–Schlessinger functors.

Given homomorphisms of commutative rings A → B → C and a C-module L there is an exact sequence of A-modules
 * $$\begin{align}

0 \rightarrow\; &\operatorname{Der}_B(C,L)\rightarrow \operatorname{Der}_A(C,L)\rightarrow \operatorname{Der}_A(B,L) \rightarrow \\ &\operatorname{Exalcomm}_B(C,L)\rightarrow \operatorname{Exalcomm}_A(C,L)\rightarrow \operatorname{Exalcomm}_A(B,L) \end{align}$$ where DerA(B,L) is the module of derivations of the A-algebra B with values in L. This sequence can be extended further to the right using André–Quillen cohomology.

Square-zero extensions
In order to understand the construction of Exal, the notion of square-zero extensions must be defined. Fix a topos $$T$$ and let all algebras be algebras over it. Note that the topos of a point gives the special case of commutative rings, so the topos hypothesis can be ignored on a first reading.

Definition
In order to define the category $$\underline{\text{Exal}}$$ we need to define what a square-zero extension actually is. Given a surjective morphism of $$A$$-algebras $$p: E \to B$$ it is called a square-zero extension if the kernel $$I$$ of $$p$$ has the property $$I^2 = (0)$$ is the zero ideal.

Remark
Note that the kernel can be equipped with a $$B$$-module structure as follows: since $$p$$ is surjective, any $$b \in B$$ has a lift to a $$x\in E$$, so $$b \cdot m := x\cdot m$$ for $$m \in I$$. Since any lift differs by an element $$k \in I$$ in the kernel, and
 * $$(x + k)\cdot m = x\cdot m + k\cdot m = x\cdot m$$

because the ideal is square-zero, this module structure is well-defined.

From deformations over the dual numbers
Square-zero extensions are a generalization of deformations over the dual numbers. For example, a deformation over the dual numbers $$\begin{matrix} \text{Spec}\left( \frac{k[x,y]}{(y^2 - x^3 )} \right) & \to & \text{Spec}\left( \frac{k[x,y][\varepsilon]}{(y^2 - x^3 + \varepsilon)} \right) \\ \downarrow & & \downarrow \\ \text{Spec}(k) & \to & \text{Spec}(k[\varepsilon]) \end{matrix}$$ has the associated square-zero extension"$0 \to (\varepsilon) \to \frac{k[x,y][\varepsilon]}{(y^2 - x^3 + \varepsilon)} \to \frac{k[x,y]}{(y^2 - x^3 )} \to 0$"of $$k$$-algebras.

From more general deformations
But, because the idea of square zero-extensions is more general, deformations over $$k[\varepsilon_1,\varepsilon_2]$$ where $$\varepsilon_1\cdot \varepsilon_2 = 0$$ will give examples of square-zero extensions.

Trivial square-zero extension
For a $$B$$-module $$M$$, there is a trivial square-zero extension given by $$B \oplus M$$ where the product structure is given by
 * $$(b,m)\cdot (b',m') = (bb',bm' + b'm)$$

hence the associated square-zero extension is
 * $$0 \to M \to B\oplus M \to B \to 0$$

where the surjection is the projection map forgetting $$M$$.

Construction
The general abstract construction of Exal follows from first defining a category of extensions $$\underline{\text{Exal}}$$ over a topos $$T$$ (or just the category of commutative rings), then extracting a subcategory where a base ring $$A$$ $$\underline{\text{Exal}}_A$$ is fixed, and then using a functor $$\pi:\underline{\text{Exal}}_A(B,-) \to \text{B-Mod}$$ to get the module of commutative algebra extensions $$\text{Exal}_A(B,M)$$ for a fixed $$M \in \text{Ob}(\text{B-Mod})$$.

General Exal
For this fixed topos, let $$\underline{\text{Exal}}$$ be the category of pairs $$(A, p:E \to B)$$ where $$p:E\to B$$ is a surjective morphism of $$A$$-algebras such that the kernel $$I$$ is square-zero, where morphisms are defined as commutative diagrams between $$(A, p:E \to B) \to (A', p':E' \to B')$$. There is a functor
 * $$\pi: \underline{\text{Exal}} \to \text{Algmod}$$

sending a pair $$(A, p:E \to B)$$ to a pair $$(A\to B, I)$$ where $$I$$ is a $$B$$-module.

ExalA, ExalA(B, –)
Then, there is an overcategory denoted $$\underline{\text{Exal}}_A$$ (meaning there is a functor $$\underline{\text{Exal}}_A \to {\displaystyle {\underline {\text{Exal}}}}$$) where the objects are pairs $$(A, p:E \to B)$$, but the first ring $$A$$ is fixed, so morphisms are of the form
 * $$(A, p:E \to B) \to (A, p':E' \to B')$$

There is a further reduction to another overcategory $$\underline{\text{Exal}}_A(B,-)$$ where morphisms are of the form
 * $$(A, p:E \to B) \to (A, p':E' \to B)$$

ExalA(B,I&hairsp;)
Finally, the category $$\underline{\text{Exal}}_A(B,I)$$ has a fixed kernel of the square-zero extensions. Note that in $$\text{Algmod}$$, for a fixed $$A,B$$, there is the subcategory $$(A\to B, I)$$ where $$I$$ is a $$B$$-module, so it is equivalent to $$\text{B-Mod}$$. Hence, the image of $$\underline{\text{Exal}}_A(B,I)$$ under the functor $$\pi$$ lives in $$\text{B-Mod}$$.

The isomorphism classes of objects has the structure of a $$B$$-module since $$\underline{\text{Exal}}_A(B,I)$$ is a Picard stack, so the category can be turned into a module $$\text{Exal}_A(B,I)$$.

Structure of ExalA(B, I&hairsp;)
There are a few results on the structure of $$\underline{\text{Exal}}_A(B,I)$$ and $$\text{Exal}_A(B,I)$$ which are useful.

Automorphisms
The group of automorphisms of an object $$X \in \text{Ob}(\underline{\text{Exal}}_A(B,I) )$$ can be identified with the automorphisms of the trivial extension $$B\oplus M$$ (explicitly, we mean automorphisms $$B\oplus M \to B\oplus M$$ compatible with both the inclusion $$M\to B \oplus M$$ and projection $$B\oplus M \to B$$). These are classified by the derivations module $$\text{Der}_A(B,M)$$. Hence, the category $$\underline{\text{Exal}}_A(B,I)$$ is a torsor. In fact, this could also be interpreted as a Gerbe since this is a group acting on a stack.

Composition of extensions
There is another useful result about the categories $$\underline{\text{Exal}}_A(B,-)$$ describing the extensions of $$I\oplus J$$, there is an isomorphism"$\underline{\text{Exal}}_A(B,I\oplus J) \cong \underline{\text{Exal}}_A(B,I)\times \underline{\text{Exal}}_A(B,J)$|undefined"It can be interpreted as saying the square-zero extension from a deformation in two directions can be decomposed into a pair of square-zero extensions, each in the direction of one of the deformations.

Application
For example, the deformations given by infinitesimals $$\varepsilon_1,\varepsilon_2$$ where $$\varepsilon_1^2 = \varepsilon_1\varepsilon_2 = \varepsilon_2^2 = 0$$ gives the isomorphism $$\underline{\text{Exal}}_A(B,(\varepsilon_1) \oplus (\varepsilon_2)) \cong \underline{\text{Exal}}_A(B,(\varepsilon_1))\times \underline{\text{Exal}}_A(B,(\varepsilon_2))$$ where $$I$$ is the module of these two infinitesimals. In particular, when relating this to Kodaira-Spencer theory, and using the comparison with the contangent complex (given below) this means all such deformations are classified by"$H^1(X,T_X)\times H^1(X,T_X)$"hence they are just a pair of first order deformations paired together.

Relation with the cotangent complex
The cotangent complex contains all of the information about a deformation problem, and it is a fundamental theorem that given a morphism of rings $$A \to B$$ over a topos $$T$$ (note taking $$T$$ as the point topos shows this generalizes the construction for general rings), there is a functorial isomorphism"$\text{Exal}_A(B,M) \xrightarrow{\simeq} \text{Ext}_B^1(\mathbf{L}_{B/A}, M)$ (theorem III.1.2.3)"So, given a commutative square of ring morphisms $$\begin{matrix} A' & \to & B' \\ \downarrow & & \downarrow \\ A & \to & B \end{matrix}$$ over $$T$$ there is a square $$\begin{matrix} \text{Exal}_A(B,M) & \to & \text{Ext}^1_B(\mathbf{L}_{B/A}, M) \\ \downarrow & & \downarrow \\ \text{Exal}_{A'}(B',M) & \to & \text{Ext}^1_{B'}(\mathbf{L}_{B'/A'}, M) \end{matrix}$$ whose horizontal arrows are isomorphisms and $$M$$ has the structure of a $$B'$$-module from the ring morphism.