The Unapologetic Mathematician

Semigroup rings

Today I’ll give another great way to get rings: from semigroups.

Start with a semigroup $S$. 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 $\mathbb{Z}_s$ for each element $s$ of $S$ and direct sum them all together. There are two ways to think of an element of the resulting abelian group, as a function $f:S\rightarrow\mathbb{Z}$ that sends all but finitely many elements of $S$ to zero, or as a “formal finite sum” $c_1e_{s_1}+c_2e_{s_2}+...+c_ne_{s_n}$ where each $c_i$ is an integer and $e_s$ is “$1$” from the copy of $\mathbb{Z}$ corresponding to $s$.

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: $f=\sum\limits_{s\in S}f(s)e_s$. This sum is finite because most of the values of $f$ 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 $e_{s_1}e_{s_2}=e_{s_1s_2}$, and extend to sums the only way we can to make the multiplication satisfy the distributive law. In the function picture we define $\left[fg\right](s)=\sum\limits_{xy=s}f(x)g(y)$ where we take the sum over all pairs $(x,y)$ of elements of $S$ whose product is $s$. This takes the product of all nonzero components of $f$ and $g$ 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 $S$, written $\mathbb{Z}[S]$. There are a number of easy variations on the same theme. If $S$ 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 $S$ 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 $R$ and forget its additive structure we’re left with a semigroup. If we take any semigroup homomorphism from $S$ to this “underlying semigroup” of $R$ we can uniquely extend it to a ring homomorphism from $\mathbb{Z}[S]$ to $R$. 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 $xy=s$ only if $y=x^{-1}s$ we can rewrite the product formula in the function case $\left[fg\right](s)=\sum\limits_{x\in S}f(x)g(x^{-1}s)$. 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 | Ring theory

1. […] Now we come to a really nice example of a semigroup ring. Start with the free commutative monoid on generators. This is just the product of copies of the […]

Pingback by Polynomials « The Unapologetic Mathematician | April 16, 2007 | Reply

2. […] Free Ring on an Abelian Group Last week I talked about how to make a ring out of a semigroup by adding an additive structure. Now I want to do the other side. Starting with an abelian group […]

Pingback by The Free Ring on an Abelian Group « The Unapologetic Mathematician | April 19, 2007 | Reply

3. […] can also start with any semigroup and build the semigroup algebra just like we did for the semigroup ring . As a special case, we can take to be the free commutative monoid on generators and get the […]

Pingback by Algebras « The Unapologetic Mathematician | May 8, 2007 | Reply

4. […] representation properly extends that of a group representation. Given any group we can build the group algebra . As a vector space, this has a basis vector for each group element . We then define a […]

Pingback by Algebra Representations « The Unapologetic Mathematician | October 24, 2008 | Reply

5. […] Group Algebra A useful construction for our purposes is the group algebra . We’ve said a lot about this before, and showed a number of things about it, but most of […]

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6. […] it’s just a formal linear combination of singular -cubes. That is, for each we build the free abelian group generated by the singular -cubes in […]

Pingback by Chains « The Unapologetic Mathematician | August 5, 2011 | Reply

7. 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.

Comment by Alexey Muranov | July 9, 2014 | Reply

8. I don’t know that there is such a standard term.

Comment by John Armstrong | July 9, 2014 | Reply

• 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 $R$.

Comment by Alexey Muranov | July 10, 2014 | Reply