Derivations from Rings to Modules: A Categorical Perspective Exploration

Let $R$$R$ be a commutative ring, and let $M$$M$ be an $R$$R$-module.

A derivation of $R$$R$ into $M$$M$ is a map $d:R\to M$$d: R \to M$ such that for all $a,b\in R$$a, b \in R$,

$$d(a+b)=d(a)+d(b),d(ab)=ad(b)+bd(a).$$

We denote the set of all derivations from $R$$R$ to $M$$M$ by $\mathrm{Der}(R,M)$$\mathrm{Der}(R,M)$.

Proposition: $\mathrm{Der}(R,-)$$\mathrm{Der}(R,-)$ is a functor from $\mathrm{Mod}(R)$$\mathrm{Mod}(R)$ to $\mathrm{Mod}(R)$$\mathrm{Mod}(R)$.

Proof:

It is evident that $\mathrm{Der}(R,M)$$\mathrm{Der}(R,M)$ is an $R$$R$-module as well.

For any $\phi :M_1\to M_2$$\varphi: M_1 \to M_2$, the pushforward ${\phi }_{\ast }(d)=\phi \circ d\in \mathrm{Der}(R,M_2)$$\varphi_*(d) = \varphi \circ d \in \mathrm{Der}(R,M_2)$.

Readers may recall the concepts of bilinear map and alternating map here. They do look similar.

The commonality here is that both the bilinear map functor and the alternating map functor are representable.

That is,

$$\mathrm{Bil}(M\times N,-)\cong {\mathrm{Hom}}_{\mathrm{Mod}(R)}(M\otimes N,-)$$

and

$${\mathrm{Alt}}^k(M,-)\cong {\mathrm{Hom}}_{\mathrm{Mod}(R)}(\stackrel{k}{\bigwedge }M,-).$$

We can apply the same approach to derivations. Since whether bilinear or alternating, maps are just some functions that satisfy certain equations. Hence, we can use the free module and quotient map to construct the universal map for derivations.

This is what "free" means: you can construct "anything" you want.

We will see that

$${\mathrm{Der}}_S(R,-)\cong {\mathrm{Hom}}_{\mathrm{Mod}(S)}({\mathrm{\Omega }}_{R/S},-).$$

Let $S\subseteq R$$S \subseteq R$ be a subring. An $S$$S$-derivation is defined by

$$\mathrm{\forall }a,b\in R,s\in S,d(a+b)=d(a)+d(b),d(sa)=sd(a),d(ab)=ad(b)+d(a)b.$$

We denote by ${\mathrm{Der}}_S(R,M)$$\mathrm{Der}_S(R,M)$ the set of all $S$$S$-derivations.

In fact, an $S$$S$-derivation is nothing new compared to a normal derivation. We simply select derivations such that

$$S\subseteq \mathrm{Ker}(d).$$

Since the Leibniz rule tells us that

$$d(sa)=sd(a)+d(s)a=sd(a),$$

when $S=\mathbb{Z}$$S = \mathbb Z$, we recover $\mathrm{Der}(R,M)$$\mathrm{Der}(R,M)$.

Again, ${\mathrm{Der}}_S(R,-):\mathrm{Mod}(R)\to \mathrm{Mod}(R)$$\mathrm{Der}_S(R,-):\mathrm{Mod}(R)\to\mathrm{Mod}(R)$​ is a functor.

Remark The $i:S\to R$$i:S\to R$ make the $R-$$R-$module $M$$M$ becomes $S-$$S-$module.

Indeed, let $f:A\to B$$f:A\to B$ be a morphism in $\mathrm{C}\mathrm{R}\mathrm{i}\mathrm{n}\mathrm{g}$$\rm CRing$​.

A $B-$$B-$module is given by an ring homomorphism $g:B\to {\mathrm{End}}_{\mathrm{Ab}}(M)$$g:B\to\mathrm{End}_{\mathrm{Ab}}(M)$.

Then $f^{\ast }:\mathrm{Mod}(B)\to \mathrm{Mod}(A)$$f^*:\mathrm{Mod}(B)\to\mathrm{Mod}(A)$ define a functor.

For a $B-$$B-$module, $f^{\ast }(g)=g\circ f:A\to {\mathrm{End}}_{\mathrm{Ab}}(M)$$f^*(g)=g\circ f:A\to \mathrm{End}_{\mathrm{Ab}}(M)$ gives us a $A-$$A-$module.

For a $B-$$B-$module homomorphism $\phi :M\to N$$\varphi:M\to N$,

$$\begin{array}{c}{f}^{\ast }:\phi (a+b)\mapsto \phi (a+b)\\ f^{\ast }:\phi (g(r)a)\mapsto \phi (f^{\ast }(g)(r)a)=\phi (g(f(r))a)=gf(r)\phi (a)\end{array}$$

We will derive a new category, ${\mathrm{Der}}_S(R)$$\mathrm{Der}_S(R)$, via ${\mathrm{Der}}_S(R,-)$$\mathrm{Der}_S(R,-)$ as follows:

The objects of ${\mathrm{Der}}_{R/S}$$\mathrm{Der}_{R/S}$ are $S$$S$-derivations

$$d:R\to M.$$

The morphism between $d_1:R\to M_1$$d_1: R \to M_1$ and $d_2:R\to M_2$$d_2: R \to M_2$ is deduced by the $S$$S$-module homomorphism $\phi :M_1\to M_2$$\varphi: M_1 \to M_2$.

That is,

$${\phi }_{\ast }:d_1\mapsto \phi \circ d_1.$$

Definition: Let $d_{R/S}:A\to {\mathrm{\Omega }}_{R/S}$$d_{R/S}: A \to \Omega_{R/S}$ be the initial object of ${\mathrm{Der}}_{R/S}$$\mathrm{Der}_{R/S}$.

Proposition: The initial object $d_{R/S}:A\to {\mathrm{\Omega }}_{R/S}$$d_{R/S}: A \to \Omega_{R/S}$ exists.

Proof:

Let $F$$F$ be the free module generated by $\hat{d}(a)$$\hat{d}(a)$, and let $E$$E$ be the free $R-$$R-$module generated by

$$\hat{d}(s),\hat{d}(a+b)-\hat{d}(a)-\hat{d}(b),\hat{d}(ab)-(a\hat{d}(b)+\hat{d}(a)b).$$

Here, $\hat{d}(a)$$\hat{d}(a)$ are nothing but symbols!

Then we can define

$${\mathrm{\Omega }}_{R/S}:=F/E,$$

and the map

$$d_{R/S}:R\to {\mathrm{\Omega }}_{R/S},d_{R/S}(a)=\hat{d}(a).$$

We still need to prove that $d_{R/S}:R\to {\mathrm{\Omega }}_{R/S}$$d_{R/S}: R \to \Omega_{R/S}$ is the initial object of ${\mathrm{Der}}_{R/S}$$\mathrm{Der}_{R/S}$.

For all $d:R\to M\in {\mathrm{Der}}_S(R,M)$$d: R \to M \in \mathrm{Der}_S(R,M)$, there exists a unique $S$$S$-module homomorphism $f$$f$, such that

$$d=f\circ d_{R/S}.$$

In other words, the equation $d(a)=f(d_{R/S}(a))$$d(a) = f(d_{R/S}(a))$ uniquely determines $f$$f$​.

Here $f$$f$ is a $S-$$S-$module homomorphism since

$$\begin{array}{c}f(d_{R/S}(a)+d_{R/S}(b))=d(a+b)=d(a)+d(b)=f(d_{R/S}(a))+f(d_{R/S}(b))\\ f(s{d}_{R/S}(a))=d(sa)=sd(a)=sf(d_{R/S})\end{array}$$

The uniqueness is concluded by that $d_{R/S}$$d_{R/S}$ is an epimorphism. $◻$$\square$

Hence, giving an $S$$S$-derivation $D$$D$ from $R$$R$ to $M$$M$ is equivalent to giving an $R$$R$-module homomorphism $f$$f$ from ${\mathrm{\Omega }}_{R/S}$$\Omega_{R/S}$ to $M$$M$.

Therefore, the functor ${\mathrm{Der}}_S(R,-)$$\mathrm{Der}_S(R,-)$ is representable as well.