convolution is in the monoid ring Given a monoid ring $R[M]$ ($R$: ring, $M$:monoid). Why we have $$(f\ast g)(m)=\sum_{ab=m}f(a)g(b)\in R[M]$$ if $f,g\in R[M]$?
Is it because the set $$\{m\in M:\sum_{ab=m}f(a)g(b)\neq0\}$$ is finite? This seems a bit too easy.
Thank you.
 A: Write $f=\sum_i a_i m_i$ and $g=\sum_j b_j m_j$, for $a_i,b_j\in R, m_i,m_j\in M$. Then we have
$$ fg = (\sum_i a_i m_i)(\sum_j b_j m_j)=\sum_{i,j}a_i b_j m_i m_j.$$
Now if we want to know the coefficient of $m\in M$ in $fg$, we collect all the summands where $m_i m_j = m$, we hence find that the coefficient of $m$ in $fg$ is $\sum_{m_i m_j = m} a_i b_j$, which is a finite sum since there were only finitely many indices $i$ and $j$.
Edit: I see you define $R[M]=\{f\colon M\to R | f(m) = 0 \text{ for all but finitely many }m\}.$ I now define for given $m\in M$ the element $\chi_{m}\in R[M]$ by setting $\chi_m(m')=\delta_{m,m'}$ (Kroenecker delta). Hence if $f\in R[M]$, by definition there is a finite set of elements $m_i\in R[M]$ such that $f=\sum_{i} f(m_i)\chi_{m_i}$. Writing $a_i$ for $f(m_i)$ and just $m_i$ for $\chi_{m_i}$, we arrive at the way of writing elements of $R[M]$ presented above. 
A: 
Why we have [definition of convolution]?

If you write out two linear combinations of elements of the monoid, then multiply them by assuming the products distribute, you get exactly this formula. It's totally natural. It is an obvious consequence of trying to extend the multiplication in a monoid to a multiplication in a ring containing the monoid.
Have you ever multiplied two polynomials in $F[x]$  together? You are just doing convolution in the monoid ring over the infinite cyclic monoid generated by $x$.
