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## Non-unital rings

(This post was originally intended to go up immediately after the sequence on Gelfand duality.)

A rng (“ring without the i”) or non-unital ring is a semigroup object in $\text{Ab}$. Equivalently, it is an abelian group $A$ together with an associative bilinear map $m : A \otimes A \to A$ (which is not required to have an identity). This is what some authors mean when they say “ring,” but this does not appear to be standard. A morphism between rngs is an abelian group homomorphism which preserves multiplication (and need not preserve a multiplicative identity even if it exists); this defines the category $\text{Rng}$ of rngs (to be distinguished from the category $\text{Ring}$ of rings).

Until recently, I was not comfortable with non-unital rings. If we think of rings either algebraically as endomorphisms of abelian groups or geometrically as rings of functions on spaces, then there does not seem to be any reason to exclude the identity endomorphism resp. the identity function on a space. As for morphisms which don’t preserve identities, if $X \to Y$ is any map between spaces of some kind, then the identity function $Y \to F$ ($F$ is, say, a field) is sent to the identity function $X \to F$, so not preserving identities when they exist seems unnatural.

However, not requiring or preserving identities turns out to be natural in the theory of C*-algebras; in the commutative case, it corresponds roughly to thinking about locally compact Hausdorff spaces rather than just compact Hausdorff spaces. In this post we will discuss rngs generally, including a discussion of the geometric picture of commutative rngs, to get more comfortable with them. It turns out that we can study rngs by formally adjoining multiplicative identities to them. This is an algebraic version of taking the one-point compactification, and it allows us to extend Gelfand duality, in a suitable sense, to locally compact Hausdorff spaces (see this math.SE question for the precise statement, which we will not discuss here).

List o’ examples

First we should have a handy supply of examples. Some of these were already given previously.

Example. If $A$ is an abelian group, the zero multiplication $A \otimes A \to A$ defines a rng structure which has no multiplicative identity unless $A$ is trivial; these are the zero rngs. If $A = 0$, the resulting rng, the trivial rng, is both initial and terminal in $\text{Rng}$, so it is a zero object in this category.

A morphism between zero rngs is precisely an abelian group homomorphism; this exhibits $\text{Ab}$ as a full subcategory of $\text{Rng}$. The inclusion functor $\text{Ab} \to \text{Rng}$ has a left adjoint, the zeroification, sending a rng $R$ to the quotient $R/R^2$ of $R$ by the two-sided ideal $R^2$ of elements which occur as a product of two elements of the rng. Hence $\text{Ab}$ is a reflective subcategory of $\text{Rng}$.

Example. The forgetful functor $\text{Rng} \to \text{Ab}$ has a left adjoint sending an abelian group $A$ to the tensor rng

$\displaystyle A \oplus (A \otimes A) \oplus (A \otimes A \otimes A) \oplus ...$

with multiplication the tensor product. (Consequently, the forgetful functor preserves limits.) This rng generally has no multiplicative identity.

Example. Any left (resp. right) ideal $I$ of a rng $R$ is a rng. If $e \in I$ is a multiplicative identity, then in particular $e^2 = e$, so $e$ is idempotent, and more generally $ea = ae = a$ for every $a \in I$.

Given $e$, any element of $R$ of the form $ere, r \in R$ satisfies this property, and conversely if $r$ already satisfies this property then $r = ere$. Thus if $I$ has a multiplicative identity $e$, then $I$ is a subring of $eRe$, the maximal subset of $R$ on which $e$ acts as an identity.

In particular, if $R$ has no nontrivial idempotents, then its proper left (resp. right) ideals $I$ are never unital.

Example. Let $X$ be a locally compact Hausdorff space and let $C_0(X)$ be the space of continuous functions $X \to \mathbb{C}$ vanishing at infinity as before. $C_0(X)$ is unital if and only if $X$ is compact. In the non-compact case, $C_0(X)$ may be regarded as the maximal ideal of $C(X \cup \{ \infty\})$ (where $X \cup \{ \infty \}$ is the one-point compactification) of functions which vanish at $\infty$.

Example. Consider the algebra of continuous functions $S^1 \to \mathbb{C}$ under convolution

$\displaystyle (f \ast g)(s) = \int_{S^1} f(t) g(t^{-1} s) \, dt$.

A multiplicative identity for this rng must be zero at any point with the possible exception of the identity (and so, less rigorously, must somehow take the value $\infty$ at the identity), so must be identically zero by continuity. (What it ought to be is the Dirac delta function at the identity.)

Example. Let $R$ be a rng and let $\mathcal{M}_{\infty}(R)$ be the colimit of the family of upper-left inclusions

$\displaystyle R \cong \mathcal{M}_1(R) \hookrightarrow \mathcal{M}_2(R) \hookrightarrow \mathcal{M}_3(R) ...$.

Explicitly, $\mathcal{M}_{\infty}(R)$ is the rng of matrices over $R$ with countably many rows and columns but finitely many nonzero entries. An identity for this rng would need to have all diagonal entries $1$, so it does not exist.

Example. Let $C$ be a small category and let $R[C]$ be the category $R$-rng of $C$ ($R$ a rng). Explicitly, this consists of finite sums of the form

$\displaystyle \sum r_i f_i$

where $r_i \in R, f_i \in \text{Mor}(C)$, and the multiplication is defined by $fg = 0$ if $f, g$ cannot be composed in $C$ and their composition otherwise, and by requiring that every element of $R$ commute with every morphism. This rng has a multiplicative identity if and only if $R$ has one and $C$ has finitely many objects.

We recover the above example by taking $C$ to be the groupoid consisting of countably many objects, each of which are isomorphic to each other via a unique isomorphism.

Example. This is an example about morphisms rather than objects. Recall that in $\text{Ring}$ the integers $\mathbb{Z}$ are the initial object, since there is a unique morphism $\mathbb{Z} \to R$ from $\mathbb{Z}$ to any ring. In $\text{Rng}$, the functor $\text{Hom}(\mathbb{Z}, -)$ instead sends a ring $R$ to the set of idempotents in $R$. (This functor is represented by $\mathbb{Z}[e]/(e^2 - e) \cong \mathbb{Z} \times \mathbb{Z}$ in $\text{Ring}$.)

Some categorical properties

$\text{Rng}$ is in some ways a nicer category than $\text{Ring}$ in that it more closely resembles the category of abelian groups. For example, kernels exist. When dealing with rings the kernel $R \to S$ (in the abelian group sense) of a ring homomorphism is a two-sided ideal of $R$ and therefore almost never an object in $\text{Ring}$ (unless it is all of $R$) but in the category of $(R, R)$-bimodules. But in $\text{Rng}$ we can talk about ideals as actual subobjects; in fact they are precisely the subrngs which fit into short exact sequences

$\displaystyle 0 \to I \to R \to R/I \to 0$

(where $0$ is the trivial rng). These include the split exact sequences of the form

$\displaystyle 0 \to R \to R \oplus S \to S \to 0$

which also don’t exist in $\text{Ring}$ because the first map is not generally a ring homomorphism even if $R, S$ are unital. Although for unital rings we can make sense of the above as short exact sequences of bimodules, the same is not true for a longer exact sequence of rngs, say

$\displaystyle 0 \to Q \to R \to S \to T \to 0$

which one can obtain by starting from a rng homomorphism $f : R \to S$ whose image is an ideal of $S$ and taking kernels and cokernels:

$\displaystyle 0 \to \text{Ker}(f) \to R \xrightarrow{f} S \to \text{Coker}(f) \to 0$.

Similarly, arbitrary direct sums $\bigoplus_i R_i$ of rngs exist. For rings we can take finite direct sums and these give the categorical product, but the infinite direct sum (as abelian groups) never has a multiplicative identity. Thus there are at least four functorial ways to put together an arbitrary family of rngs:

1. the categorical product $\prod_i R_i$,
2. the categorical coproduct $\coprod_i R_i$ (the free product),
3. the tensor product $\bigotimes_i R_i$,
4. the direct sum $\bigoplus_i R_i$.

The tensor product of a family of rngs is the universal rng $R$ with inclusions $R_i \to R$ whose images commute with each other. Note that it does not agree with the tensor product of underlying abelian groups; for example, $R \otimes S$ has underlying abelian group $R \oplus S \oplus (R \otimes S)$ because the first two summands receive the inclusions. Similarly, the direct sum of a family of rngs is the universal rng $R$ with inclusions $R_i \to R$ whose images multiply to zero with each other.

A thing that used to worry me

One reason I wasn’t comfortable with rngs is that the correspondence, in the case of commutative rings, between maximal (resp. prime) ideals and surjective maps to fields (resp. integral domains) breaks. If $R$ is a commutative rng and $m$ a maximal ideal of it, then $R/m$ is a commutative simple rng, but there are more of these than just fields. For example, if $R = 2 \mathbb{Z}$ and $m = 4 \mathbb{Z}$ then $R/m$ is $\mathbb{Z}/2\mathbb{Z}$ as an abelian group, but with the zero multiplication. The geometric significance of this (in the sense of understanding $\text{CRng}^{op}$) was unclear to me.

Resolution

One way to resolve the above issue is as follows. The forgetful functor $\text{Ring} \to \text{Rng}$ has a left adjoint, the unitization $R^{+}$. This is the universal unital ring admitting a map $R \to R^{+}$ from a given rng $R$, and it is obtained by formally adjoining a multiplicative identity. As an abelian group it is just $R \oplus \mathbb{Z}$, but with multiplication given by

$\displaystyle (r, n) \times (s, m) = (rs + ns + mr, nm)$.

$R$ sits in $R^{+}$ as an ideal, so we have a short exact sequence of rngs

$\displaystyle 0 \to R \to R^{+} \to \mathbb{Z} \to 0$.

Hence $R^{+}$ is canonically an augmented ring: it is equipped with a morphism $\varepsilon : R^{+} \to \mathbb{Z}$ (the augmentation) of rings. We can then recover $R$ as the kernel of the augmentation (the augmentation ideal). In fact, more is true.

Theorem: The unitization functor is an equivalence of categories from $\text{Rng}$ to the category $\text{Ring}_{\bullet}$ of augmented rings (notation to be explained later).

Proof. It suffices to show that unitization is fully faithful and essentially surjective, all three of which are fairly straightforward verifications. An augmented ring homomorphism $R^+ \to S^+$ determines and is uniquely determined by a rng homomorphism $R \to S$ (in one direction by taking kernels of the augmentation and in the other direction by uniquely extending so that the identity is sent to the identity), which gives fullness and faithfulness. Taking kernels of the augmentation also shows that unitization is essentially surjective. $\Box$

The above internalizes to, for example, the category of $k$-algebras for some commutative ring $k$; there the unitization functor is given by taking the direct sum with a copy of $k$. The above theorem shows that $\text{CRng}^{op}$ is equivalent to the category of pointed affine schemes, namely affine schemes together with a morphism $\text{Spec } \mathbb{Z} \to \text{Spec } R$, and therefore suggests that the geometric object underlying a rng should be thought of as like a pointed space, or more basically like a pointed set.

The category of pointed sets and point-preserving functions can alternately be thought of as the category of sets and partial functions (by forgetting the point), so $\text{CRng}^{op}$ can also be understood as a suitable category of “affine schemes and partial functions.”

1. $A = 0$ … is a zero object in this ring” should surely be “… in this *category*”, right? (I dunno the Markdown flavour here, so I apologise if my markup is grotesque.)