Draft

Monoid Ring

Let $M$$M$ be a monoid, the monoid ring $R[M]$$R[M]$ is defined as follows: the elements of $R[M]$$R[M]$ is $(r_m{)}_{m\in M}$$(r_m)_{m\in M}$

Here $r_m=0$$r_m=0$ all but finite many. That is, at most finite many $r_m\ne 0$$r_m\ne 0$.

As a $R$$R$-module

$$\begin{array}{}\text{(1)}& R[M]\cong R^{⨁M}\end{array}$$

Usually, we write the element as

$$\begin{array}{}\text{(2)}& f=\sum _{m\in M}{r}_{m}m\end{array}$$

Let

$$\begin{array}{}\text{(3)}& g=\sum _{m\in M}{r}_m^{\prime }m\end{array}$$

Define

$$\begin{array}{}\text{(4)}& \begin{array}{c}f\ast g=\sum _{m\in M}\left(\sum _{xy=m}{r}_x{r}_y^{\prime }\right)m\end{array}\end{array}$$

This makes $R[M]$$R[M]$ becomes an $R$$R$-Algebra.

We can view $f$$f$ as a function on $M$$M$, $f(m)=r_m$$f(m)=r_m$. (Recall that for a finite set $S$$S$, the free $R$$R$ module $F(S)\cong {\mathrm{Hom}}_{\mathrm{Set}}(S,R)$$F(S)\cong\mathrm{Hom}_{\mathrm{Set}}(S,R)$.)

Then it could be rewritten as $f\ast g(m)=\sum _{xy=m}f(x)g(y)$$f*g(m)=\sum_{xy=m}f(x)g(y)$

The construction of a monoid ring gives you a functor $R[-]:\mathrm{Mon}\to R-\mathrm{Alg}$$R[-]:\mathrm{Mon}\to R-\mathrm{Alg}$ .

Example, Let $M=\mathbb{N}\cong F(\{x\})$$M=\mathbb N\cong F(\{x\})$, where $F$$F$ is the free monoid functor. Then $R[\mathbb{N}]\cong R[X]$$R[\mathbb N]\cong R[X]$.

If $G$$G$ is a group, then $R[G]$$R[G]$ is the group ring.

$$\begin{array}{}\text{(5)}& f\ast g(x)=\sum _{u\in G}f(u)g(u^{-1}x)\end{array}$$

Now consider $L^1(S^1)$$L^1(S^1)$, then

$$\begin{array}{}\text{(6)}& f\ast g(\tau )={\int }_{S^1}f(x)g(x^{-1}\tau )d\tau \end{array}$$

Since $S^1\cong \mathbb{R}/\mathbb{Z}$$S^1\cong\mathbb R/\mathbb Z$

$$\begin{array}{}\text{(7)}& f\ast g(\tau )={\int }_0^{1}f(x)g(\tau -x)d\tau \end{array}$$