Differential Ring (4): a way to constrcut differential ring.

This blog can be viewed as a generalization of the previous one.

Math Essays: Differential Ring (3): An example from a monoid ring

Let $A$$A$ be a Ring. Consider the quotient map $\pi :A[X]\to A[X]/(X^2)$$\pi:A[X]\to A[X]/(X^2)$.

We can induce a derivation as follows:

$$\partial :a+bX\mapsto bX$$

It is obvious that $\partial$$\partial$ is a group homomorphism.

To verify the Leibniz Law:

$$\partial [(a+bX)(c+dX)]=\partial (ac+(ad+bc)X)=(ad+bc)X=\partial (a+bX)(c+dX)+(a+bX)\partial (c+dX)$$

Remark

The usual derivation on $A[X]$$A[X]$ is not a derivation on the quotient ring.

Since the Leibniz Law will not work any more.

$$\frac{d}{dx}[(a+bX)(c+dX)]=\frac{d}{dx}(ac+(ad+bc)X)=ad+bc\ne \frac{d}{dx}(a+bX)(c+dX)+(a+bX)\frac{d}{dx}(c+dX)$$

Remark

Let $A$$A$ be an integral domain, and consider a $A-$$A-$algebra $i:A\hookrightarrow B$$i:A\hookrightarrow B$.

Let $t\in B$$t\in B$ be a nilpotent elemt such that $t^2=0$$t^2=0$.

Then $A[t]\cong A[X]/(X^2)$$A[t]\cong A[X]/(X^2)$. You can define the derivation by $\partial :A[t]\to A[t],a+bt\to bt$$\partial:A[t]\to A[t],a+bt\to bt$.