## 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 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 . 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 . This way of multiplying two functions on a group is called “convolution”, and it shows up all over the place.

[…] 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 […]

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[…] 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 […]

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[…] 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 […]

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[…] 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 […]

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[…] 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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[…] 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 […]

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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.

Comment by Alexey Muranov | July 9, 2014 |

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

Comment by John Armstrong | July 9, 2014 |

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 |