What are the reasons for considering rings without identity?

Question

I think a major reason is because Lie algebras don't have an identity, but I'm not really sure.

Answer 1

The reason is simple: There are many non-unital rings which appear quite naturally.

If $X$ is a locally compact space (in the following every space is assumed to be Hausdorff), then $C_0(X)$, the ring of continuous complex-valued functions on $X$ vanishing at infinity, is a $C^\ast$-algebra which is unital if and only if $X$ is compact. If $X = \mathbb{N}$, this is just the ring of sequences converging to $0$. Gelfand duality yields an anti-equivalence between unital commutative $C^\ast$-algebras and compact spaces, and also between (possibly non-unital) commutative $C^*$-algebras (with "proper" homomorphisms) and locally compact spaces (with proper maps). In a very similar spirit ($\mathbb{C}$ is replaced by $\mathbb{F}_2$), there is an anti-equivalence between unital boolean rings and compact totally disconnected spaces, and also between Boolean rings and locally compact totally disconnected spaces. One-point-Compactification on the topological side corresponds here to the unitalization on the algebraic side. Perhaps we have the following conclusion: As locally compact spaces appear very naturally in mathematics (e.g. manifolds), the same is true for non-unital rings.

If $A$ is a ring (possibly non-unital), its unitalization is defined to be the universal arrow from $A$ to the forgetful functor from unital rings to rings. An explicit construction is given by $\tilde{A} = A \oplus \mathbb{Z}$ as abelian group with the obvious multiplication so that $A \subseteq \tilde{A}$ is an ideal and $1 \in \mathbb{Z}$ is the identity. Because of the universal property, the module categories of $A$ and $\tilde{A}$ are isomorphic. Thus many results for unital rings take over to non-unital rings.

Every ideal of a ring can be considered as a ring. Important examples also come from functional analysis, such as the ideal of compact operators on a Hilbert space.

Answer 2

What is the reason for considering any algebraic structure? Because it comes up naturally when trying to do other things!

Here's a concrete example. In the Langlands programme one of the main local conjectures is relating representations of a (connected reductive) $p$-adic group to representations of a (group related to a) Galois group. Now most of the interesting representations of the $p$-adic group are infinite-dimensional, so this precludes one of the most powerful things that a representation theorist has in his arsenal---namely the possibility of taking traces. But in fact this can be fixed up very nicely! There is an analogue of the "group ring" of our $p$-adic group, namely the space of locally-constant complex-valued functions on the group with compact support. This space interits an addition (obvious) and a multiplication (convolution: the group has a natural measure on it, namely the Haar measure). So it's an algebra. Furthermore it is easily checked to have no identity element (the "delta function" isn't a locally-constant function!). However it's also not hard to check that there's an equivalence of categories between (certain) representations of the $p$-adic group that one is interested in, and (certain) representations of this algebra---the so-called Hecke algebra. Furthermore elements of the Hecke algebra act via maps with finite image, and so have traces! This is a big win. One can prove linear independence of characters etc etc, and get the powerful techniques back. But no way can the identity map be in this Hecke algebra---it certainly doesn't have finite image in general, and hence no trace.

Representations of the Hecke algebra are absolutely crucial in many works on this part of the Langlands correspondence, but they have no identity element. So there is one reason, in my area, at least.

Answer 3

Here is a favorite example. (See also Martin's answer.) Consider $C[0,\infty)$, the continuous complex-valued functions on $[0,\infty)$ with the "multiplication" operation of convolution... $$ f * g (x) = \int_0^x f(t) g(x-t)\,dt $$ It is a ring. Without unit. Even an integral domain. Mikusinski[*] said, take the field of fractions. Great. A simple introduction to generalized functions. Now if the student had studied algebra from some perverse textbook that constructed the field of fractions only in the unital case, what is the student to do? Go back to the textbook and check that it works without unit? A good exercise for that student, I guess.

[*] Jan Mikusinski, OPERATIONAL CALCULUS, 1959

Answer 4

Perhaps you will find the following remarks of interest, excerpted from the preface of Gardner and Wiegandt: Radical Theory of Rings, 2004.

Some authors deal exclusively with rings with unity element. This assumption is all right and not restrictive, if the ring is fixed, as in module theory or group ring theory or sometimes investigating polynomial rings and power series rings (if the ring of coefficients does not possess a unity element. the indeterminate x is not a member of the polynomial ring). Dealing, however, simultaneously with several objects in a category of rings, demanding the existence of a unity element leads to a bizarre situation. Rings with unity element include among their fundamental operations the nullary operation $\mapsto$ 1 assigning the unity element. Thus in the category of rings with unity element the morphisms, in particular the monomorphisms, have to preserve also this nullary operation: subrings (i.e. subobjects) have to contain the same unity element, and so a proper ideal with unity element is not a subring, although a ring and a direct summand; there are no infinite direct sums, no nil rings, no Jacobson radical rings, the finite valued linear transformations of an infinite dimensional vector space do not form a ring, etc. Thus, in many, maybe most, branches of ring theory the requirement of the existence of a unity element is not sensible, and therefore unacceptable. This applies also to radical theory. and so in this book rings need not have a unity element.

Answer 5

A low-level answer, but I found it pretty surprising: Dimension shifting for Hochschild cohomology is easier to prove for non-unital rings than for unital rings. Let me explain these notions:

Let $A$ be a (not necessarily unital) $k$-algebra (with $k$ a commutative ring), and $P$ an $\left(A,A\right)$-bimodule. We denote by $C^n\left(A,P\right)$ the (additive) $k$-module of all $k$-linear homomorphisms $A^{\otimes n}\to P$. We define the differential $\delta:C^n\left(A,P\right)\to C^{n+1}\left(A,P\right)$ by

$\left(\delta f\right)\left(a_1\otimes a_2\otimes ...\otimes a_{n+1}\right)$

$= a_1 f\left(a_2\otimes a_3\otimes ...\otimes a_{n+1}\right) + \sum\limits_{i=1}^n \left(-1\right)^i f\left(a_1\otimes a_2\otimes ...\otimes a_{i-1} \otimes a_i a_{i+1} \otimes a_{i+2} \otimes a_{i+3} \otimes ... \otimes a_{n+1}\right)$

$ + \left(-1\right)^{n+1} f\left(a_1\otimes a_2\otimes ...\otimes a_n\right) a_{n+1}$.

This satisfies $\delta^2 =0$, so we get a cohomology $k$-module $H^n\left(A,P\right)$, which is called the $k$-th Hochschild cohomology of $A$ and $P$.

Dimension-shifting now states that $H^{m+1}\left(A,P\right) = H^m\left(A,Q\right)$ for any $m\geq 1$, where the $\left(A,A\right)$-bimodule $Q$ is the $k$-vector space $C^1\left(A,P\right)=\mathrm{Hom}_k\left(A,P\right)$ with $\left(A,A\right)$-bimodule structure defined by

$\left(a*f\right)\left(b\right)=a\cdot f\left(b\right)$ for any $a\in A$, $f\in Q$, $b\in A$;

$\left(f*a\right)\left(b\right)=f\left(ab\right)-f\left(a\right)b$ for any $a\in A$, $f\in Q$, $b\in A$.

Now, if you try to do this all for rings $A$ with unity and for unital $\left(A,A\right)$-bimodules $P$ (id est, the unity of $A$ acts as identity from both sides on $P$), you are in for a bad surprise: Even if $P$ is a unital $A$-module, $Q$ isn't necessarily. It's the right $A$-action which causes the troubles. What you can do instead is replacing $Q$ by the subset of $Q$ formed by all those $f\in Q$ which satisfy $f\left(1\right)=0$. But now proving $H^{m+1}\left(A,P\right) = H^m\left(A,Q\right)$ isn't as easy anymore, as we have to show that cohomology of normalized cochains is the same as cohomology of cochains (this amounts to finding a chain homotopy, something which is implicit in Hochschild's Annals 1946 paper).

Answer 6

To answer a slightly generalized question, there are nonunital ring maps between unital rings that come up naturally. If $e \in A$ is an idempotent, then the ring of elements of the form $eAe$ inherits its additive and multiplicative structure from $A$, but its identity element is $e$ and not $1_A$. For example, if $k$ is a commutative ring and $m < n$ then the map $M_m(k) \to M_n(k)$ given by "padding by 0's" is a natural nonunital map of unital algebras. Under certain circumstances the rings $A$ and $eAe$ are Morita equivalent, so this type of situation can be useful in representation theory.

Answer 7

For work related to radicals of rings, the Köthe Conjecture, etc., it's very useful to consider "rngs" (Louis Rowen's term for rings without identity).

Answer 8

A consequence of using non-unital rings is that you have to postulate that its additive structure is commutative, so that the group additive structure is abelian. This is not necessary for unital rings, since the commutativity of addition is a consequence of the associativity of addition, of the definition of the multiplicative neutral element and of the left and right distributivity of addition for multiplication. I do not know if this fact can be considered as a reason for considering rings without identity. Gérard Lang

Answer 9

One can imagine a point of view that rings without identity appeared at a certain stage of the development of mathematics as a result of the lack of convenient categories. And this concept will die in the future, when new convenient categories become popular enough (and mathematicians will accept the view that non-unital algebras are just ideals of "normal", unital algebras).

For example it's clear to me that the algebra $C_0(X)$, which Martin Brandenburg mentions, appear from the idea that algebras in functional analysis must be Banach algebras (a special case: $C^*$-algebras), otherwise they become "too complicated". This idea comes from the fact that until recently in functional analysis the only monoidal closed category of topological vector spaces was the category of Banach spaces.

That's why many functional algebras were constructed as Banach algebras. The same is true for the algebras playing the role of group algebras in functional analysis (Kevin Buzzard mentiones one of them, this is the diversity of convolution algebras like $L_1(G)$, or the Banach algebra of measures on $G$, etc.) -- they are mostly also Banach algebras and that is the reason why they often have no identity (the Hecke algebra however has no identity for another reason). And why they are not Hopf algebras (in contrast to the original models in pure algebra). They resemble Hopf algebras, and a mathematician with algebraic training will recognize them as strange substitutes for the usual Hopf algebra of functions on a finite (or algebraic) group. But they are not Hopf algebras. A cynical opinion would be that they are "too ugly for evolution".

As an alternative one can look at the picture from the point of view of another monoidal closed category in functional analysis, the category of stereotype spaces (I am sorry, this is my usual refrain), and everything becomes amazingly simple. The group algebras become "true Hopf algebras" (and "true group algebras"), the necessity to consider algebras without identity disappears, the properties of finiteness like the approximation property are inherited by tensor products, functional analysis becomes much closer to algebra and geometry, unexpected duality theories appear etc., etc., etc.

That would be a point of view of an outside observer.