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Let $S$ be an abelian group under an operation denoted by $+$. Suppose further that $S$ is closed under a commutative, associative law of multiplication denoted by $\cdot$. Say that $\cdot$ distributes over $+$ in the usual way. Finally, for every $s\in S$, suppose there exists some element $t$, not necessarily unique, such that $s\cdot t=s$.

Essentially, $S$ is one step removed from being a ring; the only problem is that the multiplicative identity is not unique. Here is an example.

Let $S=\{\text{Continuous functions} f: \mathbb{R}\rightarrow \mathbb{R} \ \text{with compact support}\}$ with addition and multiplication defined pointwise. It is clear that this is an abelian group with the necessary law of multiplication. Now, let $f\in S$ be supported on $[a,b]$. Let $S'\subset S$ be the set of continuous functions compactly supported on intervals containing $[a,b]$ that are identically 1 on $[a,b]$. Clearly, if $g\in S'$, then $f\cdot g=f$ for all $x$. Also, there is no unique multiplicative identity in this collection since the constant function 1 is not compactly supported.

I've observed that this example is an increasing union of rings, but I don't know if this holds for every set with the property I've defined.

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No distributive law? – Thomas Andrews Nov 2 '12 at 19:52
Oops, forgot about that. Thanks. I guess we can just say that $\cdot$ distributes over $+$ in the usual way? I'll edit the question. – Alexander Sibelius Nov 2 '12 at 19:55
Your algebra has the added feature that if $f^n=0$ then $f=0$, for example. – Thomas Andrews Nov 2 '12 at 19:59
Note that for any $s\in S$, $S_s=\{r\in S: rs=r\}$ is either $\{0\}$ or a ring with $s$ the identity. You can define a pre-order on $S$ by $s\leq s' \iff \forall r\in S: rs=r \implies rs'=r$ – Thomas Andrews Nov 2 '12 at 20:02
up vote 3 down vote accepted

Your example is a rng with an approximate identity.

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I just wanted to throw out the most obvious and broad class of examples which has gone unmentioned so far: von Neumann regular rngs.

A ring (possibly without identity) is called von Neumann regular if for every $a$ in the ring there exists $x$ in the ring such that $axa=a$. These rings have the "local identity" property you described (on both sides in fact, although without commutativity the local identity might not be the same on both sides :) ).

One thing I find particularly interesting about VNR rings is that it seems like functional analysts have particularly natural uses for VNR rings without identity...

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This is a pseudo-ring, or rng, or ring-without-unit. The article linked in fact actually mentions the example of functions with compact support. The fact that you have a per-element neutral element is probably not sufficiently useful to give a special name to pseudo-rings with this additional property.

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Except that it is a rng with the additional property of having "local" identies. There are rngs which do not have this property. – Thomas Andrews Nov 2 '12 at 20:03
Yes, I've come across pseudo-rings before. I was wondering whether the "local" identities, in Thomas' words, gave this thing a special title. I suppose it would deserve a special title if pseudo-rings of this sort had something especially useful about them, which is another question. – Alexander Sibelius Nov 2 '12 at 20:07
@ThomasAndrews: That's what I said (calling it a per-element neutral element instead). However the fact that some rngs don't have them does not imply that having them makes a great difference, algebraically speaking. I guess if it really made a big difference, the notion would have had a special name, but I'm really quite ignorant about this. – Marc van Leeuwen Nov 2 '12 at 20:09
Just thought I'd add that there's a nearby example of a rng without the "local" identity property. Alexander's ring $S$ sits in the larger ring of continuous functions $f : \mathbb{R} \to \mathbb{R}$ which vanish at infinity. Any nowhere-vanishing function in the latter ring has no "local" identity. – Mike F Jul 7 '13 at 4:29

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