Multiplicative identity in a monoid ring.

Question

Let $R$ be a ring and $S$ a subset of $R$.

I want to prove that $1:S\rightarrow R: s \mapsto 1_R$ is the multiplicative identity in the ring $(R^{(s)},*,+,1,0)$ (with $R^{(S)}$ the subset of $R^S$ with functions of finite support). $*$ is defined as follows: $$ f*g:S \rightarrow: s\mapsto \sum_{a,b\in S; a*b=s}f(a)*g(b)$$.

This is what I have :

$$(f*1)(s) = \sum_{a,b\in S; a*b=s} f(a)*1(b) = \sum_{a,b\in S; a*b=s} f(a)*1_R = \sum_{a,b\in S; a*b=s} f(a) =\sum_{r \in f(S)} r \neq f(s)$$

Does somebody see where I went wrong?

Answer 1

(Also, you are assuming $S$ is a monoid, right?)

Does somebody see where I went wrong?

It happened in this line:

I want to prove that $1:S\rightarrow R: s \mapsto 1_R$ is the multiplicative identity in the ring $(R^{(s)},*,+,1,0)$

The function $s\mapsto 1_R$ does not have finite support if $S$ is infinite, so it is clearly not even a candidate to be the identity.

You'll have better luck proving that the map sending $s\mapsto 0_R$ if $s\neq 1_S$ and $1_S\mapsto 1_R$ is the identity of the monoid ring $R^{(S)}$.