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## Structures on hom-sets

Suppose I hand you a commutative ring $R$. I stipulate that you are only allowed to work in the language of the category of commutative rings; you can only refer to objects and morphisms. (That means you can’t refer directly to elements of $R$, and you also can’t refer directly to the multiplication or addition maps $R \times R \to R$, since these aren’t morphisms.) Geometrically, I might equivalently say that you are only allowed to work in the language of the category of affine schemes, since the two are dual. Can you recover $R$ as a set, and can you recover the ring operations on $R$?

The answer turns out to be yes. Today we’ll discuss how this works, and along the way we’ll run into some interesting ideas.

Geometry

The following geometric special case of the more general construction might be a useful introduction. If we restrict ourselves to, say, finitely-generated integral domains over $\mathbb{C}$ (dual to (irreducible) affine varieties), then a ring $R$ can be completely recovered from the ring of regular functions $\text{Spec } R \to \mathbb{C} \simeq \text{Spec } \mathbb{C}[x]$. On the algebraic side we are looking at morphisms $\mathbb{C}[x] \to R$. Certainly it’s clear that $\mathbb{C}[x]$ represents the forgetful functor to $\text{Set}$ since such a morphism is determined freely by the image of $x$, which is an arbitrary element of $R$. But it isn’t obvious where the ring structure on what is a priori just a hom-set

$\displaystyle X_R = \text{Hom}_{\text{CRing}}(\mathbb{C}[x], R)$

is coming from. That is, it’s not obvious how to define the structure maps in this setting: the addition map $+ : X_R \times X_R \to X_R$, the multiplication map $\times : X_R \times X_R \to X_R$, the additive identity $0 : 1 \to X_R$ (where $1$ is the one-element set), and the multiplicative identity $1 : 1 \to X_R$.

In the setting of affine varieties, we see that $X_R$ gets its structure from the fact that $\mathbb{C} \simeq \text{Spec } \mathbb{C}[x]$ is not just a ring but a ring object in the category of affine varieties. This is worth explaining in some detail, so we’ll explain an analogous concept in a situation where there are less structure maps.

Group objects

Let $C$ be a category with finite products; in particular, it has a terminal object $1$. A group object in $C$ is an object $G \in C$ equipped with maps

$m : G \times G \to G, e : 1 \to G, i : G \to G$

such that the obvious diagrams commute. These three structure maps are the multiplication, identity, and inverse maps (we also call the second map the unit map), and this is a straightforward generalization of the ordinary definition of a group in $\text{Set}$. We obtain the definition of a monoid object by ignoring the inverse map.

Example. A group object in $\text{Top}$ is a topological group. Similarly, a group object in the category $\text{Diff}$ of smooth manifolds is a Lie group.

Example. A monoid object in the category $\text{Mon}$ of monoids is a commutative monoid. This is a form of the Eckmann-Hilton argument. Similarly, a group object in the category $\text{Grp}$ of groups is a commutative group.

Example. A monoid object in the category $\text{Cat}$ of (small) categories is a monoidal category.

Given any group object $G$ in a category, the set $\text{Hom}(X, G)$ acquires the structure of a group as follows. The multiplication map gives rise to a map

$m : \text{Hom}(X, G \times G) \simeq \text{Hom}(X, G) \times \text{Hom}(X, G) \to \text{Hom}(X, G)$

via composition, thanks to the universal property of products. Similarly, the identity map gives rise to a map

$e : \text{Hom}(X, 1) \simeq 1 \to \text{Hom}(X, G)$

thanks to the universal property of the terminal object, and the inverse gives rise to a map

$i : \text{Hom}(X, G) \to \text{Hom}(X, G)$.

Since all of these maps are defined via composition from an element of $\text{Hom}(X, G)$ it is fairly clear that all of the relevant diagrams commute. This is the abstract form of “pointwise” multiplication, but it is worth noting that we don’t have to refer to points to give this definition.

It is worth giving a nontrivial example of how this works. Fix a field $F$, not necessarily algebraically closed, and let $C$ be the opposite of the category of finitely-generated integral domains over $F$ (essentially, affine varieties over $F$). Let $G$ be a group object in $C$ (an algebraic group), such as the general linear group $\text{GL}_n(F)$ (regarded as an affine subvariety of $\mathbb{A}^{n^2+1}(F)$). Then $G$ gives rise to a group

$\displaystyle G_K : \text{Hom}(\text{Spec } K, G)$

for every finite extension $K$ of $F$, the group of $K$-points of $G$. (Any $F$-variety is determined by its $K$-points over all finite extensions $K$: this is a form of the Nullstellensatz. Hence, for example, any real algebraic group is determined by its $\mathbb{R}$-points and its $\mathbb{C}$-points.)

Cogroup objects

By the Yoneda lemma, giving $\text{Hom}(X, G)$ the structure of a group in a way that is natural in $X$ is equivalent to giving $G$ the structure of a group object, at least for categories with finite products. It is natural to ask the dual question: what structure does an object $H$ need to have in order for $\text{Hom}(H, X)$ to have the structure of a group in a way that is natural in $X$?

This is the same as asking the original question in the opposite category $C^{op}$, so let us assume that $C^{op}$ has finite products (that is, $C$ has finite coproducts). Then it is necessary and sufficient that $H$ (as an object in $C^{op}$) is a group object. Dualizing all of the maps, it follows that $H$ (as an object in $C$) is a cogroup object. This is an object equipped with structure maps

$m : H \to H + H, e : H \to 0, i : H \to H$

where $m$ is called the comultiplication (because it is dual to multiplication), $e$ is called the counit (where $0$ is the initial object), and I’m not sure what people call $i$. In some circumstances it’s called the antipode; perhaps we should go with coinversion.

Many natural categories have no nontrivial cogroup objects. For example, the only cogroup object in $\text{Set}$ is the empty set because nonempty sets have no maps to the initial object (which is the empty set itself). More generally, any category $C$ with an initial object and a faithful functor to $\text{Set}$ preserving it has the same property. This includes categories such as $\text{Top}$.

One of the most important examples of a cogroup object in mathematics is the circle $S^1$ in the homotopy category $\text{hTop}_{\ast}$ of pointed topological spaces. The coproduct in this category is the wedge sum $\vee$, and the initial object (the empty coproduct) is the one-point space, which we’ll confusingly continue to denote by $0$. The comultiplication is the map $m : S^1 \to S^1 \vee S^1$ given by tracing the path of both circles, the counit $e : S^1 \to 0$ is unique, and the coinverse $i : S^1 \to S^1$ is given by tracing the circle in the opposite order. For any pointed topological space $X \in \text{hTop}_{\ast}$ with basepoint $x_0$, $\text{Hom}(S^1, X)$ is precisely the fundamental group $\pi_1(X, x_0)$. More generally, the sphere $S^n$ is a cogroup object in $\text{hTop}_{\ast}$, and $\text{Hom}(S^n, X)$ is the homotopy group $\pi_n(X, x_0)$. So cogroup objects are not as weird-sounding as they may first appear.

Recovering groups

Consider our original question, but in the simpler situation of the category $\text{Grp}$ of groups. Can we recover the underlying set and group operations of a group purely by working in the language of $\text{Grp}$? The answer is yes. First of all, the forgetful functor $U : \text{Grp} \to \text{Set}$ is representable: it is given by $\text{Hom}(\mathbb{Z}, -)$, since a morphism $\mathbb{Z} \to G$ is freely determined by the image of $1$, which is an arbitrary element of $G$. In this context it makes sense to write $\mathbb{Z}$ as the free group $F(1)$ on one generator because this is part of a more general statement: $U$ has a left adjoint, the free set functor $F : \text{Set} \to \text{Grp}$, hence

$\text{Hom}_{\text{Grp}}(F(X), G) \simeq \text{Hom}_{\text{Set}}(X, U(G))$

and any functor from a category to $\text{Set}$ with a left adjoint $F$ is representable by $F(1)$, since

$\text{Hom}_{\text{Grp}}(F(1), G) \simeq \text{Hom}_{\text{Set}}(1, U(G)) \simeq U(G)$.

Now, giving $\text{Hom}(F(1), G)$ the structure of a group in a way that is natural in $G$ is equivalent to giving $F(1)$ a cogroup structure. The coproduct in $\text{Grp}$ is the free product, so the comultiplication is given by a map $F(1) \to F(2)$; it suffices to send the generator of $F(1)$ to the product of the generators of $F(2)$, and then it’s not hard to verify that this comultiplication induces the multiplication on $G$. Similarly the counit is the unique map $F(1) \to F(0)$, and the coinverse is the inverse $F(1) \to F(1)$.

More generally, for any set $X$, the hom-set $\text{Hom}(F(X), G) \simeq G^X$ has a natural group structure (the $|X|$-fold product) giving a cogroup structure on the free group $F(X)$. It is a theorem of Kan that these are the only cogroups in $\text{Grp}$.

Hopf algebras

Before we recover the full ring structure on a commutative ring, let’s try two easier things: recovering the abelian group structure, and recovering the group of units. The answer to the full question will essentially come from combining these, but they are already independently interesting. The forgetful functor $A : \text{CRing} \to \text{Ab}$ sending a ring to its underlying abelian group has the property that its composition $UA : \text{CRing} \to \text{Set}$ is just the usual forgetful functor. $UA$ has a left adjoint $F : \text{Set} \to \text{CRing}$ given by sending a set to the free commutative ring on it (a polynomial ring over $\mathbb{Z}$), so $UA$ is representable by $F(1) \simeq \mathbb{Z}[x]$. We want to put an abelian group structure on $\text{Hom}(\mathbb{Z}[x], R)$ in a way that is natural in $R$ to recover $A$, and now we know this is equivalent to putting a cocommutative cogroup structure on $\mathbb{Z}[x]$.

The coproduct in $\text{CRing}$ is the tensor product over $\mathbb{Z}$, so we want to find a comultiplication $m : \mathbb{Z}[x] \to \mathbb{Z}[x] \otimes_{\mathbb{Z}} \mathbb{Z}[x]$. Such a comultiplication is freely determined by $x$, and it’s not hard to see that the correct choice is $x \mapsto x \otimes 1 + 1 \otimes x$ (dualize this to see what happens). The initial object is $\mathbb{Z}$, so the counit $e : \mathbb{Z}[x] \to \mathbb{Z}$ is $x \mapsto 0$, and the coinverse $i : \mathbb{Z}[x] \to \mathbb{Z}[x]$ is $x \mapsto -x$.

This defines the structure of an affine group scheme on $\text{Spec } \mathbb{Z}[x]$. Since it represents the “additive group” functor, it is called the additive group scheme $\mathbb{G}_a$. An affine group scheme, when viewed in $\text{CRing}$ as a cogroup object, is also known as a Hopf algebra (this is a context in which people call coinversion the antipode), and this is one of the many reasons why Hopf algebras are important in mathematics.

The story for the group of units is similar. The composition of the forgetful functor $\text{CRing} \to \text{Ab}$ sending a ring to its group of units with $U$ is representable by $\mathbb{Z}[x, x^{-1}]$, so we want to make this into a Hopf algebra. This is straightforward: the comultiplication is the diagonal $x \mapsto x \otimes x$, the counit is $x \mapsto 1$, and the coinverse (or antipode) is $x \mapsto x^{-1}$. The corresponding affine group scheme $\text{Spec } \mathbb{Z}[x, x^{-1}]$ is the multiplicative group scheme $\mathbb{G}_m$.

Warning: the group operation on a group scheme (as opposed to the group operation on its set of points over a field, say) is more subtle than a set-theoretic map $S \times S \to S$ where $S$ is the set of prime ideals. While direct product of rings corresponds to disjoint union of prime ideals, tensor product of rings does not correspond to Cartesian product of prime ideals. For example, the tensor product of two fields of different characteristic is the trivial ring, which has empty spectrum.

In other words, the forgetful functor $\text{CRing}^{op} \simeq \text{Aff} \to \text{Set}$ sending an affine scheme to its set of prime ideals preserves coproducts, but does not preserve products. (This implies that it is not representable, and is one limitation of the prime ideal picture of algebraic geometry.) On the other hand, the functor $\text{Hom}(\text{Spec } S, -)$ sending a scheme to its set of $S$-points (for any $S$, although the picture is nicest when $S$ is a field) is representable, so it does more than preserve products: it preserves all limits. This is a basic reason thinking about $S$-points is nicer than thinking about all the prime ideals at once (Grothendieck’s relative point of view), and why we can get away with thinking geometrically when we work relative to $S$.

Addition and multiplication on the arithmetic plane

The full answer to our original question is now clear. We want to define the structure of a ring object on $\text{Spec } \mathbb{Z}[x]$, which has a definition completely analogous to that of a group object: it’s precisely a group object structure and a monoid object structure, the latter distributing over the former. In $\text{CRing}$ this translates into wanting the structure of a coring object on $\mathbb{Z}[x]$, and thanks to our work above we already know what the structure maps ought to be (just combine the additive and multiplicative group schemes and ignore multiplicative inversion). This generalizes: in the category of affine schemes over an affine scheme $\text{Spec } S$ (the opposite of the category of $S$-algebras), taking the tensor product with $\mathbb{Z}[x]$ gives a coring structure on $S[x]$ with a compatible coaction of $S$ (that is, $\text{Spec } S[x]$ acquires an action of $S$), and this gives back the $S$-algebra structure on any $S$-algebra. (This is the abstract explanation for the geometric picture we started with, when $S = \mathbb{C}$.)

So we have accomplished our goal: a specific set of structure maps on $\text{Spec } \mathbb{Z}[x]$ allows us to completely recover the commutative ring associated to an affine scheme using only the language of the category of affine schemes.

I was mildly surprised when I realized this. $\text{Spec } S[x]$ is exactly what you expect it to be if $S$ is an algebraically closed field, and it’s similar if $S$ is a field, but for $S = \mathbb{Z}$ it’s the arithmetic plane. This is a somewhat more complicated object, and it’s not immediately clear exactly what the function $\text{Spec } R \to \text{Spec } \mathbb{Z}[x]$ associated to an element $r \in R$ looks like in terms of the arithmetic picture.

So let’s work it out. First we’ll deal with a simpler universal object, $\mathbb{Z}$. It is the initial object in $\text{CRing}$, hence the terminal object in $\text{Aff}$, so there is always a unique morphism $\text{Spec } R \to \text{Spec } \mathbb{Z}$. A prime ideal $P$ in $R$ gives rise to a quotient map $R \to R/P$ into an integral domain, giving a composite $\mathbb{Z} \to R \to R/P$ (which is already unique). If $R/P$ has characteristic $p$, then $P$ is sent to $(p) \in \text{Spec } \mathbb{Z}$; otherwise, it’s sent to the generic point $(0)$. So the unique morphism $\text{Spec } R \to \text{Spec } \mathbb{Z}$ sorts points in $\text{Spec } R$ by the characteristic of their residue fields. (Note that sending a point to another point on the geometric side induces a morphism between residue fields in the other direction on the algebraic side.)

Next we’ll talk about $\mathbb{Z}[x]$ in isolation. Recall that $\text{Spec } \mathbb{Z}[x]$ consists of the generic point $(0)$, the prime ideals $(f(x))$ where $f$ is irreducible (not necessarily monic), and the prime ideals $(p, f(x))$ where $p$ is a prime and $f$ is irreducible over $\mathbb{F}_p$. One should think of these as organized by a “prime axis” where projection to the prime axis corresponds to the morphism $\text{Spec } \mathbb{Z}[x] \to \text{Spec } \mathbb{Z}$ above. The fiber of this map over $(p)$ is $\text{Spec } \mathbb{F}_p[x]$, and the fiber of this map over the generic point $(0)$ is $\text{Spec } \mathbb{Q}[x]$. Motivated by this uniformity, we will refer to $0$ as the zero prime and $\mathbb{Q}$ as the characteristic zero prime field $\mathbb{F}_0$; then we can say uniformly that the fiber over $(p)$ is $\text{Spec } \mathbb{F}_p[x]$.

Now let’s turn to the full picture. Given an element $r \in R$ of an arbitrary commutative ring, consider the induced morphism $r_{\ast} : \text{Spec } R \to \text{Spec } \mathbb{Z}[x]$. Given a prime ideal $P$ of $R$, we get a morphism $\mathbb{Z}[x] \xrightarrow{r} R \to R/P$, and the image of $P$ under $r_{\ast}$ is as follows. If $R/P$ has characteristic $p$, then $P$ lands in $\text{Spec } \mathbb{F}_p[x]$. Next, if the image of $r$ in $R/P$ satisfies a polynomial relation $f(r) = 0$ with coefficients in $\mathbb{F}_p$ (which always happens if, for example, $R/P$ is a finite field), then $P$ is sent to $(p, f(x)) \in \text{Spec } \mathbb{F}_p[x]$. (Remember that $p$ can be zero here!) Otherwise, it is sent to the generic point of $\text{Spec } \mathbb{F}_p[x]$.

So $r_{\ast}$ is actually something quite nice: it records all the possible arithmetic behavior of $r$. This is nice if $R$ is some ring closely related to $\mathbb{Z}[x]$ such as an “arithmetic curve” $\mathbb{Z}[x]/(f(x))$ where $f$ is an irreducible, ideally monic, polynomial. But if $R$ is something like the coordinate ring of an affine variety over $\mathbb{C}$ then there is no arithmetic data and this picture is less useful. (Again, if you are in this situation you should be working relative to $\mathbb{C}$ anyway.)

### 4 Responses

1. Quibble: “a monoid object in the category of abelian groups is a ring” — I think you mean monoid object where the monoidal product is taken to be tensor product. (This wasn’t clear from the discussion which precedes it — in fact the discussion only concerned the case where the monoidal product is cartesian product.)

• Right. I remember worrying about this and then forgetting that I worried about it. Since the whole point of this exercise is to impose structure on Hom-sets I guess I should just get rid of that example.

2. Reading your post, I was reminded that in any symmetric monoidal category, the cocommutative comonoid objects form a cartesian category, where the cartesian product is the tensor product of the underlying monoidal category. So that, for example, a cocommutative Hopf algebra is literally a group object in this cartesian category if the monoidal category one starts with is Vect with the usual tensor product.

(A lot of rich mathematics can be developed by taking, in place of Set, cocommutative coalgebras as a base of hom-enrichment. It’s a cartesian closed category, for one thing.)

3. […] Example. In the category of schemes over a base ring , the endomorphism Lawvere theory of the affine line is the Lawvere theory of polynomials over , or equivalently the Lawvere theory of commutative -algebras. Hence naturally acquires the structure of a commutative -algebra. (We previously discussed the case in this blog post.) […]