Consequences of not requiring ring homomorphisms to be unital?

Question

As defined in many modern algebra books, a homomorphism of unital rings must preserve the unit elements: $f(1_R)=1_S$. But there has been a minority who do not require this, one prominent example being Herstein in Topics in Algebra.

What are some of the most striking consequences of not requiring ring homomorphisms to be unital? For example, what aspects of algebraic geometry would need to be reworked if we no longer required it? What interesting theorems or techniques arise in the not-necessarily-unital theory which do not apply (or are degenerate) for unital homomorphisms?

Answer 1

Let's suppose that our rings are commutative (which is the case that is immediately relevant to algebraic geometry).

If $\phi:A \to B$ is a (possibly non-unital) homomorphism, then $e := \phi(1_A)$ is an idempotent in $B$, and so we get a decomposition $B = eB \times (1-e)B,$ and the map $\phi$ factors as $A \to eB \to B,$ where the first map is unital, and the second map is simply the inclusion, which is the inclusion of a direct factor.

On Specs, we thus get the composite of the map Spec $eB \to $ Spec $A$, composed with the "map" (this is not necessarily an honest map of schemes, because it corresponds to the possibly non-unital map $e B \to B$) Spec $B \to$ Spec $eB$, which just vaporizes the open and closed subset Spec $(1-e)B$ of Spec $B$, and is the identity on the open and closed subset Spec $eB$.

So the upshot is that nothing much new happens in algebraic geometry, except that we allow maps which are only defined on some open and closed subset of a given scheme. Of course, this is a big except, because these are not honest maps at all (they are simply not defined on some part of their "domain"). There doesn't seem to be any reason to add them into the mix, which is surely one reason why this generalized notion of homomorphism is not used much in practice.

P.S. One could argue another way, beginning with geometry, and passing to algebra by remembering that rings are rings of functions. If we have a map $\phi:X \to Y$ of spaces (of some type, e.g. affine schemes, or anything else), then surely the constant function 1 on $Y$ will pull-back to the constant function 1 on $X$. Thus the induced homomorphism on rings of functions will have to be unital, and so one simply has no cause to consider non-unital homomorphisms in the geometric setting.

P.P.S. The argument in the first paragraph shows that allowing non-unital homomorphisms in the category of commutative rings is the same as adding, in addition to unital homomorphisms, homomorphisms of the form $B_1 \to B_1\times B_2,$ given by $b_1\mapsto (b_1,0),$ for any pair of commutative rings $B_1$ and $B_2$. So it's not really a very exciting change from the purely algebraic point of view either.

Answer 2

This is just an amplification of Emerton and Pietro Majer's answers, but got a bit too big.

You can give a geometric interpretation of all not necessarily unital (NNU) $k$-algebras, for any ring k. The category of NNU $k$-algebras is equivalent to the category of augmented (unital) $k$-algebras, by which I mean $k$-algebras $A$ with a retraction $A\to k$ of the structure map, where the morphisms are the $k$-algebra maps that commute with the retractions. An augmented $k$-algebra $A$ corresponds to the NNU $k$-algebra which is the kernel of its augmentation map $A\to k$. Thus the category of NNU $k$-algebras is anti-equivalent to the category of pointed affine $k$-schemes, where the maps preserve the pointed structure. The forgetful functor from $k$-algebras to NNU $k$-algebras then has a geometric interpretation. It corresponds to the functor from affine $k$-schemes to pointed affine $k$-schemes that sends $X$ to the disjoint union of $X$ and $\mathrm{Spec}(k)$, where the second component is the distinguished point.

In particular, an NNU map $A\to B$ of $k$-algebras corresponds to a scheme map $$\mathrm{Spec}(B)\coprod \mathrm{Spec}(k) \to \mathrm{Spec}(A)\coprod \mathrm{Spec}(k)$$ which is the identity on the second component. The original map is unital if and only if the map takes the $\mathrm{Spec}(B)$ component to the $\mathrm{Spec}(A)$ component. But in general there could be a connected component of $\mathrm{Spec}(B)$ taken to the $\mathrm{Spec}(k)$ component. These are exactly the vaporized components in Emerton's answer.

Here's a little exercise. What does the NNU subring $n\mathbf{Z}$ of $\mathbf{Z}$ look like geometrically?

So the question about whether it's better to look at unital or NNU rings is (at least in the commutative case) the same as the question of whether it's better to looked a pointed or unpointed spaces, which also comes up in homotopy theory. I prefer the unital/unpointed approach (no doubt because of my education), but it's easy to translate back and forth between the two.

Answer 3

Of course, if we assume that the ring have a unit, then there is absolutely no reason not to assume that homomorphisms preserve it (or are there books that do that?)

My impression is that there has been a change in the last 40 years or so; algebra texts written in the '60's or earlier mostly did not require ring to be unital, while, say, from the '70's on most of them did. Of course, rings without unit are still very much present in functional analysis (for example, Banach algebras are not assumed to be unital, as many of the standard examples are not). The first edition of Herstein's book is from 1964.

I don't know how much of this is due to Grothendieck's influence. Certainly algebraists now are less conversant with functional analysis than they used to be, due to an inevitable increase in specialization. In the kind of mathematics I practice, units are a fact of life. The only advantage I can see in not assuming the existence of a unit is that ideals become rings, and one can apply theorems about rings; but this is minor, and is more than offset by the conceptual disadvantages of working with the "wrong" category.

Answer 4

There are natural situations where non-unit preserving homomorphisms are needed. For instance if $R$ is a unital ring and $e$ an idempotent, then $eRe$ is a unital ring with identity $e$ and the inclusion $eRe\to R$ is not unital. Nonetheless it plays a role in Morita theory and in Chapter 6 of Green's lecture notes on polynomial representations of $Gl_n$. It also plays a role in the construction of irreducible representations of finite monoids.

Answer 5

Sometimes you work with some non-unital rings and homomorphisms between them, but the rings happen to be unital. This happens in the context of AF-algebras: This is a $C^{\ast}$-algebra, which is a directed colimit of finite-dimensional $C^{\ast}$-algebras. It turns out that every finite-dimensional $C^{\ast}$-algebra is a direct product of matrix algebras, thus unital. In order to classify AF-algebras, you have to study the transition maps, which are non-unital ring homomorphisms between unital rings. So this happens quite naturally. It turns out that there is a way of visualizing these direct systems via so called Bratteli diagrams. The simplest example is the colimit of the imbeddings $M_n \to M_{n+1}, A \mapsto diag(A,0)$, which is the stabilized matrix algebra $M_\infty$. If you are interested in this, check out Davidson's "$C^{\ast}$-algebras by example". The commutative case is equivalent to the study of non-unital boolean rings, which may also be seen as colimits of finite, thus unital boolean rings.

Now you asked about algebraic geometry. I think one main issue is that if $\phi : A \to B$ is a non-unital homomorphism between unital rings, then there is no reason why prime ideals of $B$ pull back to prime ideals of $A$. Indeed, the inclusion $A \to A \times B$ pulls back $A \times \mathfrak{p}$ to $A$. So you cannot define an associated map of spectra $Spec(B) \to Spec(A)$. As Emerton already pointed out, anyway you get a map $Spec(Be) \to Spec(A)$, where $e=\phi(1)$.

This can be seen functorial in the following sense: Let $C$ be the category of (perhaps non-unital) rings together with a choosen idempotent element; homomorphisms should respect this idempotent element. Then there is a functor $C \to Ring$, which sends $(A,e) \mapsto Ae$, which can be composed with $Spec : Ring \to LRS$. If you plug in $(A,1)$ and $(B,e)$, you get the map above.

Answer 6

I'd like to add some obvious remarks, as a non-specialist (please in case correct or rectify me). My impression is that the matter is important, but not substantial.

Non-unital commutative rings and non-unital homomorphisms do exist in the mathematical world, nevertheless a non-unital ring $A$ may always be seen as an ideal of a unital ring, via the usual construction $\eta_A:A\to\mathbb{Z}\times A$, and a non-unital homomorphism may be seen as the restriction of a unital homomorphism. In some case, there is also a "concrete" embedding : for instance one can embed the non-unital convolution algebra of $L^1(\mathbb{R}^n)$ in the distributions. However, I do have in mind convolution equations in $L^1(\mathbb{R}^n)$ were the computations are greatly simplified just by introducing an even formal unity.

In other words, one loses nothing in generality by stating : commutative rings have a unity, and homomorphism preserve it, as books of commutative algebra usually do. So the situation is similar to the assumption of completeness for metric spaces: we may, and often do consider, a metric space as a dense subset of a complete metric space. The analogy is made precise by the category language: the category of unital commutative rings is a non-full, reflective subcategory of the category of non-unital ring with non-unital homomorphisms (non=not necessarily); the unity of the corresponding adjunction is of course given by the above homomorphisms $\eta_A$ (the co-unity being $\epsilon_R:\mathbb{Z}\times R\ni (n,x)\mapsto n\mathbf{1}_R+x\in R$ for unital rings $R$).

There are of course other situations in mathematics were one prefers not to profit of a canonical embedding; certainly (almost) nobody would see $\mathbb {N}$ as a sub-monoid of $\mathbb{R}$. Clearly these choices depend on practicality, and partially on fashion.

Answer 7

If we permit rings to have 1 = 0, then the trivial ring {0} becomes the zero object of the category.

For comparison, if we required morphisms to keep 1 (but still permitted for 1 = 0), then {0} would be the terminal object only – any object has (unique) morphism to {0}, and 1 maps to 1 because 1 = 0 in the trivial ring. But there would be no morphisms from {0} to anything else since 1 ≠ 0 in any non-trivial ring.