Semigroup rings
Today I’ll give another great way to get rings: from semigroups.
Start with a semigroup 
. If it helps, think of a finite semigroup or a finitely-generated one, but this construction doesn’t much care. Now take one copy of the integers 
for each element 
of and direct sum them all together. There are two ways to think of an element of the resulting abelian group, as a function 
that sends all but finitely many elements of to zero, or as a “formal finite sum” 
where each 
is an integer and 
is “
” from the copy of 
corresponding to .
I’ll try to talk in terms of both pictures since some people find the one easier to understand and some the other. We can go back and forth by taking a valid function and using its nonzero values as the coefficients of a formal sum: 
. This sum is finite because most of the values of 
are zero. On the other hand, we can use the coefficients of a formal sum to define a valid function.
So we’ve got an abelian group here, but we want a ring. We use the semigroup multiplication to define the ring multiplication. In the formal sum picture, we define 
, and extend to sums the only way we can to make the multiplication satisfy the distributive law. In the function picture we define =\sum\limits_{xy=s}f(x)g(y)](https://s0.wp.com/latex.php?latex=%5Cleft%5Bfg%5Cright%5D%28s%29%3D%5Csum%5Climits_%7Bxy%3Ds%7Df%28x%29g%28y%29&bg=e6e6e6&fg=333333&s=0&c=20201002)
where we take the sum over all pairs 
of elements of whose product is . This takes the product of all nonzero components of and 
and collects the resulting terms whose indices multiply to the same element of the semigroup.
The ring we get is called the “semigroup ring” of , written ![\mathbb{Z}[S]](https://s0.wp.com/latex.php?latex=%5Cmathbb%7BZ%7D%5BS%5D&bg=e6e6e6&fg=333333&s=0&c=20201002)
. There are a number of easy variations on the same theme. If is actually a monoid we sometimes say “monoid ring”, and note that the ring has a unit given by the identity of the monoid. If is a group we usually say “group ring”. If in any of these cases we start with a commutative semigroup (monoid, group) we get a commutative ring.
So here’s the really important thing about semigroup rings. If we take any ring 
and forget its additive structure we’re left with a semigroup. If we take any semigroup homomorphism from to this “underlying semigroup” of we can uniquely extend it to a ring homomorphism from to . This is just like what we saw for free groups, and it’s just as important.
As a side note, I want to mention something about the multiplication in group rings. Since 
only if 
we can rewrite the product formula in the function case =\sum\limits_{x\in S}f(x)g(x^{-1}s)](https://s0.wp.com/latex.php?latex=%5Cleft%5Bfg%5Cright%5D%28s%29%3D%5Csum%5Climits_%7Bx%5Cin+S%7Df%28x%29g%28x%5E%7B-1%7Ds%29&bg=e6e6e6&fg=333333&s=0&c=20201002)
. This way of multiplying two functions on a group is called “convolution”, and it shows up all over the place.
April 12, 2007 -
Posted by John Armstrong |
Ring theory
The Unapologetic Mathematician 
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Is there a standard term for a “reduced semigroup ring” when the semigroup has a zero? For a semigroup with zero element, the “full semigroup ring” can be taken quotient of by the ideal generated by the semigroup zero.
I don’t know that there is such a standard term.
I was curious about “reduced semigroup rings” because matrix algebras are such. Consider a ring $R$ and a semigroup $S$ of $n^2+1$ elements $0$ and $(i,j)$, where $i$ and $j$ are taken from some $n$-element set, and the semigroup product is defined by $(i,j)(j,k)=(i,k)$, and $(i,j)(k,l)=0$ if $j\ne k$. Then the “reduced semigroup ring” of $S$ over $R$ is isomorphic to the ring of $n\times n$ matrices over $R$.
I need an example of semigroup ring
Well, take the positive integers (
) with addition as your semigroup. Then integer linear combinations of these form a ring:
Hm, the latex rendering there doesn’t seem to make a clear distinction between bold (ring generator) and non-bold (coefficient) numbers. I trust you can figure it out from context.