Tricki

## Decompose your ring using idempotents

### Quick description

If is an idempotent in a ring (not necessarily with identity), then we can decompose as the direct sum of four components (subrings), each of them related to . Concretely,

note that here is just notation that means , and similarly for .

This decomposition is known as the Peirce decomposition of with respect to .

An important feature of this decomposition is that each of and is a subring of , and the former even has an identity, namely, the element , which does indeed lie in , since ; also, for all .

If does contain an identity , then is defined, and is again an idempotent, and one has (obviously), and since as is idempotent. One says that and form a pair Furthermore, in this case, the subring as defined above, coincides with the set of products . (A similar remark applies to each of and .)

More generally, continuing to assume that contains an identity, if is a set of orthogonal idempotents for with the property that then

### Prerequisites

The basic idea of decomposing a ring via idempotents should be accessible to anyone knowing the basics of ring theory. Several of the examples below refer to contexts which require a more specialized degree of knowledge; in such cases, this is indicated at the beginning of the example.

### Example 1

If lies in the centre of (in particular, if is commutative), then , and Thus we obtain the simpler decomposition

### Example 2

This example requires a basic knowledge of the costruction of the spectrum of a commutative ring with identity.

If is commutative with and is an idempotent, then , as noted in Example 1. As remarked above, we may form the idempotent , and so in particular, both and are commutative rings with identity (namely and respectively). Thus , and are all defined, and the factorization induces a decomposition

of into a union of two open subsets. Conversely, any such decomposition of arises from an idempotent in this way.

In short, if we think of as being the ring of regular functions on , then idempotents in serve precisely as the indicator (or characteristic) functions of simultaneously open and closed subsets of . (In particular, is the indicator function of itself, while is the indicator function of the empty subset.)

Note that in the case that is a Boolean ring, so that every element is idempotent (by definition), the space is totally disconnected, and the above discussion specializes to the Stone representation theorem.

### Example 3

This example requires some familiarity with topological group theory, including the notion of Haar measure on a locally compact topological group, and various related notions.

Suppose that is a topological group which admits a the identity consisting of compact open subgroups. (A basic example of such a group is , the general linear group of invertible -matrices over the field of -adic numbers.) One consequence of this assumption is that is locally compact, and so we can choose a Haar measure on .

Let denote the -vector space of all compactly supported locally constant -valued functions on . We can make into a -algebra by defining a convolution product on its elements as follows:

for any two functions . The -algebra is then referred to as the Hecke algebraof the group .

If is any compact open subgroup of , then we may define an idempotent via the formula where is the Since is open, the function is locally constant and compactly supported, and thus lies in , and hence so does .

In this situation, for any , the convolution is the function obtained by averaging on the left via the action of the compact open subgroup , and simililarly is the is the function obtained by averaging via the action of on the right. In particular if and only if for all and , and simililarly, if and only if for all K and . Thus the subalgebra of consists of those elements of that are bi-invariant under the action of , i.e. such that for all and .

Since any is compactly supported and locally constant, by virtue of our assumption on we may find some sufficiently small compact open subgroup such that is bi--invariant. Thus we find that

where the union is indexed by the collection of all compact open subgroups of .

### Change of title, and other matters

I propose changing the title of this article to How to exploit idempotents in a ring''.

I have used this idea many times in my research, but never knew it was called the Peirce decomposition until reading this article. Indeed, it is a standard example in the theory of schemes, and in the representation theory of p-adic groups, which motivate two of the above examples, but I have never seen it referred to by this name in any books on those subjects. (I say this just as a defense in advance of my ignorance of this name; I don't think I will be unique in having an interest in this article, while not knowing the name Peirce decomposition".)

If there are no objections within a day or so, I will make this change. I will then add a line to the quick description along the lines of this decomposition is sometimes refereed to as the Peirce decomposition of with respect to ''.

Also, the More generally'' remark about a complete system of orthogonal idempotents should be elaborated on. (Because if does not have an identity, then is not actually defined as an element of , and so the Peirce decomposition is itself more general than decomposing with respect to orthogonal idempotents. So while the decomposition by a complete system of orthogonal idempotents is very closely related to the Peirce decomposition, it does not always contains the latter as a special case. Indeed, the notion of a complete system of orthogonal idempotents probalby only makes sense if has an identity, at least as far as I can tell.)

Finally, I noticed that the earlier version of this article was to some extent lifted from PlanetMath. Maybe we should be careful doing this in general, just for copyright reasons.

### I've discussed titles quite a

I've discussed titles quite a bit in this forum post. If you buy my view about imperative titles, then you could consider "Exploit idempotents in a ring." But it depends whether the point of the article is more "You might not have thought of it, but idempotents are actually very useful," or "You have seen or heard that idempotents can be useful, but here is how to produce such arguments for yourself." If the former, then I would be very much in favour of the imperative title. Or one could go for something more specific, such as "Decompose your ring using idempotents." The ideal, in my view is a title that serves as a super-quick summary of the message of the entire article. It's not always possible to achieve that, but often it is.

### Title explanation

I agree mostly with you in this issue - Imperative and how-to titles are desirable. That's why I created stubs for "How to work without an identity element when you clearly need it", etc.

But my main purpose creating this article was not to explain how to use idempotents (for that, I created the other articles). My main purpose was to help those people that has heard about Peirce decomposition but does not see anything special about it.

The main example I have in mind (I'm new to the field, so there's not that much I can think of!)is when you want to prove a categorical statement about a ring without identity (mostly about R-Mod), and typical "universal" or homological techniques with pretty diagrams do not seem to work; then your "other option" is to switch to ring decompositions like Peirce's (or, in a sort of generalization of this, to exploit some Morita context if you have it).

### Based on Tim's suggestion,

Based on Tim's suggestion, let me second his proposal of Decompose your ring using idempotents''.

Peirce decomposition'' may be standard in some ring theory circles, but it is not at all standard among algebraic geometers/commutative algebraists, nor among representation-theorists, both of whom would be interested in this article. As I said, I had never before heard of it (the name, that is, not the concept), and I have been studying these fields for close to fifteen years. I'm not saying this to pull rank, but rather to point out that this name is not nearly as well-known as the concept, and so there will be many people interested in the contents of this page who wouldn't have any idea about its contents under the present name.

### As a possible compromise,

As a possible compromise, what about a link to this article on the "How to use" page, called "How to use the Peirce decomposition"? The title of the article would be different, but there's no problem with that. There could be a hidden-text quick description that said that the Peirce decomposition was a decomposition of rings using idempotents.

### That sounds sensible. I

That sounds sensible. I don't object at all to using the phrase Peirce decomposition to describe this procedure, as long as we don't rely on it as the only means of navigating to this page. I think in general that it is better to avoid using names (especially people's names) for things if one can avoid it, unless the name is absolutely universally known or absolutely unavoidable.

Thus Prove that your holomorphic function is bounded" (or some variant of this) would be better than Apply Liouville's theorem" as an instruction related to trying to show that a holomorphic functions is constant. After all, there is always the possibility that someone doesn't know or has forgotten what Liouville's theorem says, and then the name carries absolutely no hint about the method. (As a contrast, Apply the maximum modulus theorem'' is probably not quite as bad in this respect, because it contains in its name at least some semblance of what it is about.)

### Not sure about title change

Thanks a lot for the editions!

As far as I can tell (that is, very little!), the name "Peirce decomposition" is standard in ring theory (which is my area of expertise, but I've been on it just for six months now!). I can't be sure about changing the title to something more general about idempotents, because there are other, related but different (if I'm not mistaken!), ways of using idempotents. For example, we say that a ring (not necessarily with unity)has enough idempotents if there exists a subset of orthogonal idempotents (, ), so that we can decompose as a direct sum of the principal right ideals (equivalently ). Note that must be infinite in order to be interesting. I wanted to write different articles for these two concepts (and one more general, titled "How to work without an identity element when you clearly need it" introducing also sets of local units, firm rings, etc).

I agree that the "more general" statement is not right just now: indeed, we must ask for to have identity and also the set should be finite (I wrote it late and I somewhat messed the idea up with the enough idempotents thing!).

Actually, I didn't lift anything of PlanetMath! I just made a quick summary on my own. I checked now with PlanetMath, there is an obvious resemblance because it is a definition but I don't see that much of a coincidence!

### Dear Jose, Sorry for the

Dear Jose,

Sorry for the unfair PlanetMath comment; indeed, it must just be that there are not all that many ways to describe the basics of this concept.

Best wishes,

Matthew