# 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 | 9 Comments

## I’m going to Faro

The organizers want me to give my 30-minute talk on bracket extensions. Nice way to go out, if it turns out I’m going out of the tower.

## God Bless You, Mr. Vonnegut

This is a little to the side of my usual topics, but I wanted to mention the passing of Kurt Vonnegut. So it goes.

## Direct sums of Abelian groups

Let’s go back to direct products and free products of groups and consider them just in the context of abelian groups.

The direct product $G\times H$ of abelian groups $G$ and $H$ works as we expect it to, because the elements coming from $G$ and $H$ already commute inside $G\times H$. The free product of $G$ and $H$ as groups gives $G*H$ as before, but now this is not an abelian group. Let’s consider the property that defined free products a little more closely. Here’s the diagram.

We want to read it slightly differently now. The new condition is that for any abelian group $X$ and homomorphisms $f_G:G\to X$ and $f_H:H\to X$ there is a unique homomorphism of abelian groups from the free product to $X$ making the diagram commute. We know that there’s a unique homomorphism from $G*H$ already, but we need to “abelianize” this group. How do we do that?

We just move to the quotient of $G*H$ by its commutator subgroup of course! Recall that any homomorphism $f:G\to A$ to an abelian group $A$ factors uniquely through this quotient: $G\to G/[G,G]\to A$. So now $(G*H)/[(G*H),(G*H)]$ is an abelian group with a unique homomorphism to $X$ making the diagram commute, it works for a free product in the context of abelian groups. This sort of thing feels odd at first, but you get used to it: when you change the context of a property (here from all groups to abelian groups) the implications change too.

Okay, so $(G*H)/[(G*H),(G*H)]$ is like the free product $G*H$, but we’ve thrown in relations making everything commute. We started with abelian groups $G$ and $H$, so all we’ve really added is that elements coming from the two different groups commute with each other. And that gives us back (wait for it..) the direct product! When we restrict our attention to abelian groups, direct products and free products are the same thing. Since this is such a nice thing to happen and because we change all our notation when we look at abelian groups anyhow, we call this group the “direct sum” of the abelian groups $G$ and $H$, and write it $G\oplus H$.

Now I didn’t really talk about this much before in the context of groups, but I’m going to need it shortly. We can take the direct sum of more than two groups at a time. I’ll leave it to you to verify that the groups $(G_1\oplus G_2)\oplus G_3$ and $G_1\oplus(G_2\oplus G_3)$ are isomorphic (use the universal property), so we can more or less unambiguously talk about the direct sum of any finite collection of groups. Infinite collections (which we’ll need soon) are a bit weirder.

Let’s say we have an infinite set $S$ and for each of its elements $s$ an abelian group $G_s$. We can define the infinite direct sum $\bigoplus\limits_{s\in S}G_s$ as the collection of all “$S$-tuples” $(g_s)$ where $g_s\in G_s$ for all $s\in S$, and where all but a finite number of the $g_s$ are the zero element in their respective groups. This satisfies something like the free product’s universal property — each $G_s$ has a homomorphism $\iota_s:G_s\rightarrow\bigoplus\limits_{s\in S}G_s$, and so on — but with an infinite number of groups on the top of the diagram: one for every element of $S$.

The direct product $\prod\limits_{s\in S}G_s$, on the other hand, satisfies something like the product condition but with an infinite number of groups down on the bottom of the diagram. Each of the $G_s$ comes with a homomorphism $\pi_s:\prod\limits_{s\in S}G_s\to G_s$, and so on. We can realize this property with the collection of all $S$-tuples, whether there are a finite number of nonzero entries or not.

What’s really interesting here is that for finite collections of groups the free product comes with an epimorphism onto the direct product. Now for infinite collections of abelian groups, the free product (direct sum) comes with a monomorphism into the direct product. The free product was much bigger before, but now it’s much smaller. When all these weird little effects begin to confuse me, I find it’s best just to plug my ears and go back to the universal properties. They will never steer you wrong.

April 12, 2007