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

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I don't think this is a reason. The natural ring associated with a Lie algebra is its universal enveloping algebra, which does have an identity. –  José Figueroa-O'Farrill Apr 26 '10 at 10:29
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For the same reason you think ideals are interesting. :) –  Gjergji Zaimi Apr 26 '10 at 11:10
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Lie algebras are not associative, so I don't think they should be considered a major answer to the question. The right answer, which is consistent with the examples illustrated in the answers below, is that there are important rings without identity in analysis. This is a nice example of why mathematicians shouldn't closet themselves into their own little area but have some awareness of what goes on in the rest of math: basic examples you seek might not be in your area but could be mother's milk for other areas. –  KConrad Apr 26 '10 at 17:18
    
Added the ra tag because I had trouble refinding this question. –  Mark Grant Feb 11 '11 at 8:26
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Another argument in favor of requiring rings to have identity: www-math.mit.edu/~poonen/papers/ring.pdf –  Sam Hopkins Aug 13 at 2:21

7 Answers 7

up vote 46 down vote accepted

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 continous 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.

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What is the reason for considering any algebraic structure? Becuase 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.

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I'm just curious, Kevin, if you're looking at the space of locally constant complex valued functions on a locally compact group G, which is a subspace of L^1(G), can't you just view it as a subspace of the Banach algebra (with identity) M(G) of complex Borel measures on G with convolution (where the "delta function" does exist as a measure)? Perhaps you don't want to leave the space you're working in? I don't really know anything about the applications you're talking about. –  Keenan Kidwell Apr 26 '10 at 11:25
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I'm not so sure that this bigger ring would act on the vector space underlying a smooth representation of $G$. Let me try you on a trivial example. Let $G$ be a locally compact $p$-adic group. Fix once and for all a Haar measure on $G$. Let $V$ be the trivial 1-dimensional representation of $G$. Then the Hecke algebra of locally constant functions with compact support also acts on $V$: an element $f$ in this Hecke algebra acts by the constant equal to the integral of $f$ over $G$. Would your bigger ring also act naturally on $V$? More worryingly, what if $V$ is countably infinite-diml? –  Kevin Buzzard Apr 26 '10 at 12:15
    
That's a good point. I hadn't even thought of that and it's not clear to me that the bigger ring does have a natural action on V. –  Keenan Kidwell Apr 26 '10 at 12:24
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I was going was going to give the same answer, but you beat me to it! +1! By the way, isn't there a way to construct the Hecke algebra without a choice of a Haar measure by looking at an operation on the measures themselves (the reason, if I remember correctly (though I may be wrong), that we have to fix a Haar measure is because we're applying a duality from integration theory)? –  Harry Gindi Apr 26 '10 at 12:57
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@Harry Gindi: there is, but if you do it like that then you can't interpret the elements as locally-constant functions on the group, it's slightly more elaborate. It's all explained in Cartier Corvallis. If you're going to go for locally-constant functions on the group as elements then you clearly need a Haar measure because if K is a compact subgroup with characteristic function f then f*f tells you mu(K). Different choices of Haar measure give you canonically isomorphic algebras though, so another way of doing it would be taking all Haar measures at once and then taking the projective limit. –  Kevin Buzzard Apr 26 '10 at 17:59

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 $\delta²=0$, 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).

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I agree, this is surprising. –  Martin Brandenburg Apr 26 '10 at 13:04
    
I remember this fact annoying/confusing me greatly when I first started to learn the rudiments of Hochschild cohomology. The dimension shift is also annoying in the Banach versions of this, because not only is your shifted module not "unital" or "unit-linked", but the action of the algebra might not be contractive. –  Yemon Choi Apr 26 '10 at 19:57
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This is surely not the reason (nor even a reason) for considering non-unital rings, though! –  Mariano Suárez-Alvarez Feb 11 '11 at 17:23

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

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I guess MOST textbook only introduce the field of fractions in the unital case. Actually I don't think I'd able to cite one who does it the other way. By the way, what has to do this construction with usual generalized functions (distributions, hyperfunctions...)? –  Andrea Ferretti Apr 26 '10 at 12:26
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@Andrea Ferretti: I think that Hungerford is one such textbook (see the first paragraph above Theorem 4.2 in books.google.com/…). –  user2734 Apr 26 '10 at 12:40
    
(But I confess that it is the only such textbook that I have found :) ) –  user2734 Apr 26 '10 at 13:09
    
@Andrea: Mikusinski's construction gives you the same generalized functions as other methods. Advantage over the other constructions: much simpler, can be understood by students much earlier in their education. Disadvantage: Only for the underlying space $[0,\infty)$. Generalization to $\mathbb{R}^n$ or to manifolds still has to wait until after some functional analysis. –  Gerald Edgar Apr 26 '10 at 13:27
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@Gerald: with respect, your argument about as to why algebra texts should discuss the non-unital case seems exactly wrong to me. Many aspects of the general theory of rings (especially commutative rings) do indeed require a unit (e.g. the existence of maximal ideals), to an extent that a "general non-unital ring" just doesn't behave in the same way as a "general ring". On the other hand, many individual results do carry over to a nonunital context in a straightforward way. If you want to construct fraction fields in rings without unit: sure, just modify the proof in the unital case. –  Pete L. Clark Feb 3 '11 at 22:40

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

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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.

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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).

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