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

Preliminary

Math Essays: Viewing Matrix Operations through Monoid Rings: A Rendezvous of Algebra and Graph Theory (wuyulanliulongblog.blogspot.com)

Math Essays: Differential Ring (1): Some general result and interesting application (wuyulanliulongblog.blogspot.com)

Let us consider the monoid generanted by

$$\begin{array}{c}U:=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\end{array}$$

That is,

$$M=(\{0,I,U\},\circ )$$

The monoid ring $\mathbb{R}[M]$$\mathbb R[M]$ is a subring of $2\times 2$$2\times 2$ matrix ring.

The elements of $\mathbb{R}[M]$$\mathbb R[M]$ are $aI+bU$$aI+bU$. Let us consider $d:\mathbb{R}[M]\to \mathbb{R}[M],aI+bU\mapsto bU.$$d:\mathbb R[M]\to\mathbb R[M],aI+bU\mapsto bU.$

We want to prove an interesting fact, $(\mathbb{R}[M],d)$$(\R[M],d)$ forms a differential ring.

Proof.

Obviously $d$$d$ is linear. We only need to check Leibniz Law.

$$d((a_{1}I+b_{1}U)(a_{2}I+b_{2}U))=d(a_1{a}_{2}I+(a_1{b}_2+a_2{b}_1)U)=(a_1{b}_2+a_2{b}_1)U$$

and

$$d(a_{1}I+b_{1}U)(a_2+b_{2}U)+(a_{1}I+b_{1}U)d(a_{2}I+b_{2}U)=a_2{b}_{1}U+a_1{b}_{2}U=(a_1{b}_2+a_2{b}_1)U$$

Readers may already oberve that

$$\mathbb{R}[M]\cong \mathbb{R}[X]/(X^2)$$

By the evaluation map $e{v}_U:\mathbb{R}[X]\to \mathbb{R}[M]$$ev_{U}:\R[X]\to \R[M]$ and first isomorphism theorem.

Hence we actually define a different derivation over $\mathbb{R}[X]/(X^2)$$\R[X]/(X^2)$

$$d:a+bX⟼bX$$