### What's a point of this?

Monday, November 19th, 2012 | Author:

I recently came across a paper using a "universal domain" to discuss "generic points" of a variety, using Weil's foundations of algebraic geometry instead of Grothendieck's. First I had to learn that stuff, then I wanted to translate it. This lead to a more systematic study of what it means to be a point of a variety or scheme, in the various different definitions.

So in this post I will explain closed points, generic points, points in general position, schematic points, generalized points, rational points, geometric points, and in particular, which of these notions can be considered a particular case of another of these. I will try to give you a hint why one wants to generalize the ordinary (closed) points of a variety that much, to answer the question in the title: "What's the point of this?".

To make it very short: closed points are generic points; "the" generic point of a scheme which isn't a point itself is not a closed point, but a generic point; schematic point is a synonym for generic points, probably used to not confuse "the" generic point of a scheme with more general generic points; general position refers to not lying in some closed subset and is related to the concept of "the" generic point; generalized points are morphisms, in the Yoneda-embedding/functor-of-points sense; rational points are generalized points which are morphisms from honest points, i.e. topologically they're just points, not something more/else.

Now for the longer version. I will assume that you know the definitions of $MaxSpec(R) = \{\text{maximal ideals of }R\}, Spec(R) = \{\text{prime ideals of }R\}$ with the corresponding Zariski topologies and $Hom_k(R,A)$ for $k$-algebras $R,A$. I will only talk about affine varieties/schemes $Spec(R)$, since it should be obvious how one generalizes. I guess Hartshorne's Chapter II.3 will be a sufficient background.

#### Generalized points: closed points as rational points

Definition: A closed point $p$ of a scheme $Spec(R)$ is an element $p \in Spec(R)$ which is its own closure in the Zariski topology: $\{p\} = \overline{\{p\}}$. This amounts to the same as saying that a closed point corresponds to a maximal ideal. So, $MaxSpec(R)$ consists of precisely the closed points of $Spec(R)$.
Example: $(x,y) \subset k[x,y]$ is the maximal ideal corresponding to the point $0 \in k^2 \simeq Spec(k[x,y])$.

Definition: A generalized $A$-point $P$ of a scheme $Spec(R)$ is a scheme morphism $Spec(A) \to Spec(R)$ (in the affine context, we could just say a ring morphism $R \to A$). Therefore, the set of all (generalized) $A$-points is $Mor(Spec(A),Spec(R))$. Consequently, $Mor(-,Spec(R))$ is a (contravariant) functor from the category of commutative rings to the category of sets, and we call this functor "functor of points". We can also consider the map $R \mapsto Mor(-,Spec(R))$ that yields a (covariant) functor from the category of rings to the category of contravariant set-valued functors on rings, called "Yoneda embedding".

Comparison: So, to understand a closed point $p \in Spec(R)$ as a generalized $A$-point, we need to specify a ring $A$ and a morphism $P : Spec(A) \to Spec(R)$.
We start with the case of schemes defined over a field $k$, which is algebraically closed ($k = \overline{k}$), where we take $A = k$ and define $P : Spec(k) \to Spec(R)$ by the morphism of topological spaces which sends the unique point of $Spec(k)$ to the point $p \in Spec(R)$ and the morphism of sheaves $P^{-1} \mathcal{O}_{Spec(R)} \to \mathcal{O}_{Spec(k)}$ (which is just $R \to k$ on global sections, and that's all there is) defined by the surjective homomorphism to the residue field of $p$, i.e. the homomorphism $R \to R/p = k$, when we consider $p$ as a maximal ideal in $R$.
If $k$ is not algebraically closed (or the ring $R$ is not even a $k$-algebra), the residue field of a point $\kappa(p) := R/p$ might not be $k$, but some algebraic field extension. In that case, we need to take $A = \kappa(p)$ and then the homomorphism to the residue field $R \to \kappa(p)$ defines a sheaf morphism that, together with the continuous map $pt \mapsto p \in Spec(R)$, constitutes a $k$-morphism $Spec(\kappa(p)) \to Spec(R)$. So, we get a generalized point for each closed point.
For schemes over a field $k$, we could have just used $A = \overline{k}$ to define the generalized point corresponding to a closed point $p$. In fact, every algebraic field extension of $\kappa(p)$ does the job. This means that there are many generalized points $P : Spec(K) \to Spec(R)$, with $K$ some field, whose image in $Spec(R)$ is the same closed point.

Definition: A $K$-rational point $P$ of a scheme $Spec(R)$ is a morphism of schemes $Spec(K) \to Spec(R)$, in other words, a generalized $K$-point, for $K$ a scheme. If $L/K$ is a field extension (so we have $Spec(L) \to Spec(K)$) and $P : Spec(L) \to Spec(R)$ is an $L$-rational point that factors over a $K$-rational point $Spec(K) \to Spec(R)$, we call this point also $P$ and say that $P$ is defined over $K$. If a scheme is defined over some field $k$, one calls a $k$-rational point colloquially just a rational point. If a scheme is defined over some field $k$, one calls a $k^{sep}$-rational point a geometric point (where $k^{sep}$ is the separable closure, which in characteristic $0$ is just the algebraic closure $\overline{k}$).

To summarize, we just understood how a closed point $p$ corresponds to a unique $\kappa(p)$-rational point.

#### Generic points and the general position

Definition: A generic point $\eta$ of a scheme $Spec(R)$ is an element $\eta \in Spec(R)$ (which is not necessarily closed, i.e. it corresponds to a prime ideal which isn't necessarily a maximal ideal). The term "schematic point" is a synonym (which emphasizes looking at topological spaces, not generalized points). The closure of $\{\eta\}$ in the Zariski topology is some closed subspace which contains $\eta$. If the closure $\overline{\{\eta\}}$ is the whole $Spec(R)$, we call $\eta$ the generic point of $Spec(R)$. In general, a scheme might not have a "the" generic point, but each irreducible component will have one. For an affine scheme $Spec(R)$ this corresponds to the prime ideal $Nil(R) = \sqrt{(0)}$, so for varieties this is just the $0$-ideal.

Every generic point $\eta$ is "the" generic point of a subscheme of $Spec(R)$ with underlying topological space $\overline{\{\eta\}}$. The closed points are generic points of single-point subschemes.

Explanation of general position: Suppose you have a space $Spec(R)$ with some curve $C \subset Spec(R)$. Then, we can say "take a point in general position" to really mean "take a point not on the curve". Maybe more common would be "take another curve, in general position" to really mean "take a curve which intersects the other curve only in finitely many points". In each of these examples, one has a parameter space (in the first example just $Spec(R)$, in the second the space of all curves on $Spec(R)$) and some closed subset of the parameter space to avoid. So, in some sense, "in general position" means "avoiding some closed subset of the parameter space". Avoiding a closed set means being in an open set.

Comparison: The generic point of $Spec(R)$ doesn't sit in some small closed subset, since by definition its closure is the whole $Spec(R)$ and not any smaller closed subset. This means that the generic point is contained in the complement of any closed (proper) subset of $Spec(R)$, i.e. the generic point is contained in every open subset. Let's say we have some property of schematic points that depends only on the local ring of that point and is stable under further localization. Then, if such a property holds for a point $\eta$, it automatically holds for all points in $\overline{\{\eta\}}$. In this sense, such a property of the generic point is a generic property of $Spec(R)$.

Comparison: Generic points yield generalized points, too. I suppose you can figure it out yourself, but for clarity I'll do it here: Take $\eta \in Spec(R)$, then define $\kappa(\eta) := Quot((R/\eta)^{red})$, where $A^{red} := A/Nil(A)$ means quotienting out the Nilradical $Nil(A) := \sqrt{(0)}$ of a ring $A$, to get a nilpotent-free ring $A^{red}$, and $Quot(A)$ is the quotient field of $A$. We obtain a morphism as composition $R \to R/\eta \to (R/\eta)^{red} \to \kappa(\eta)$ and this yields $Spec(\kappa(\eta)) \to Spec(R)$, a $\kappa(\eta)$-rational point with image exactly $\eta$.
Sure, we could've just taken $R \to R/\eta$ to define a morphism $Spec(R/\eta) \to Spec(R)$, which would be a generalized $R/\eta$-point, but not a rational point. In particular, the image wouldn't be just the point $\eta$ but in fact the whole $\overline{\{\eta\}}$.

Easy exercise: For $R := k[x_1,...,x_n]$ figure out how residue fields $\kappa(\mathfrak{p})$ of prime ideals $\mathfrak{p} \subset R$ look like and how the global function fields of the subvarieties $V(\mathfrak{p}) := \{p \in k^n : \forall f \in \mathfrak{p} : f(p)=0\}$ look like. In particular, for $\mathfrak{p} := (0)$, what is $\kappa(\mathfrak{p})$, what is the generic point $\eta$ of $Spec(R)$ and what is their relation?

#### Why rational points are better than closed points, and base change

Look at the $\mathbb{R}$-algebra $A := \mathbb{R}[x,y]/(xy-1)$. Its $\mathbb{R}$-rational points are a set isomorphic to $\mathbb{R}^\times$, and we can write the multiplication in $\mathbb{R}^\times$ as a morphism of sets $\mathbb{R}^\times \times \mathbb{R}^\times \to \mathbb{R}^\times$ which comes from a morphism of rings $\Delta : A \to A \otimes A$ which is defined by $x \mapsto (x,x)$ and $y \mapsto (y,y)$ (called comultiplication). There, "comes from" means that $\Delta$ induces a map of sets $Hom_{\mathbb{R}}(A \otimes A,\mathbb{R}) \to Hom_{\mathbb{R}}(A,\mathbb{R})$ and we see that $Hom_{\mathbb{R}}(A \otimes A,\mathbb{R}) = Hom_{\mathbb{R}}(A,\mathbb{R}) \times Hom_{\mathbb{R}}(A,\mathbb{R})$ and $Hom_{\mathbb{R}}(A,\mathbb{R}) \simeq \mathbb{R}^\times$, so $\Delta$ induces the multiplication.
We observe here that $\Delta$ is defined over $\mathbb{Z}$, i.e. we don't need any other coefficients in the definition of $\Delta$. We could as well define $\mathcal{A} := \mathbb{Z}[x,y]/(xy-1)$, have $\Delta : \mathcal{A} \to \mathcal{A} \otimes \mathcal{A}$ and define $A := \mathcal{A} \otimes \mathbb{R}$. We say that $A$ is "really" defined over $\mathbb{Z}$.

Now look instead at the closed points of $Spec(A)$. There we have, for each $\lambda \in \mathbb{R}^\times$ a maximal ideal $(x-\lambda) \in Spec(A)$, so in some sense $\mathbb{R}^\times \subset MaxSpec(A)$. But this is not the whole story, as there is also $(x^2+1) \in Spec(A)$, which doesn't correspond to any point in $\mathbb{R}^\times$. In fact, every point of $Spec(A)$ corresponds to an equivalence class of a point in $\mathbb{C}^\times$ under complex conjugation. If we look at $A_{\mathbb{C}} := A \otimes_{\mathbb{R}} \mathbb{C}$, we're in a better situation, as $x^2+1 = (x+i)(x-i)$ and the ideals $(x+i),(x-i)$ are maximal ideals of $A_{\mathbb{C}}$. As noted earlier, the closed points of $Spec(A_{\mathbb{C}})$ are precisely the $\mathbb{C}$-rational points, so as sets we have $Spec(A_{\mathbb{C}}) = \mathbb{C}^\times$. We have just seen that this fails for non-algebraically closed base fields. And it also fails if the base is not even a field, as for example $\mathbb{Z}$, the base of $Spec(\mathcal{A})$. The $K$-rational points of $Spec(\mathcal{A})$ however, are always $K^\times$, as a set. If we consider $Spec(\mathcal{A})$ together with $\Delta^\ast$, the $K$-rational points even carry the group structure of $K^\times$.

For the sake of completeness: $Spec(\mathcal{A}) =: \mathbb{G}_m$ together with $\Delta^\ast : \mathbb{G}_m \times \mathbb{G}_m \to \mathbb{G}_m$ is called the multiplicative group (scheme). I consider the functor-of-points approach as particularly useful in the theory of algebraic groups.

Fun exercise: How does the multiplication induced by $\Delta : A \to A \otimes A$ look like on $Spec(A)$? In particular, if we picture the topological space $Spec(A)$ as an closed half-plane in $\mathbb{C}$ with removed origin, what is the product of $\pm i$ and $\pm i$? This exercise shows that some things are more complicated in the world of schemes than in the world of topological spaces or sets. On the other hand, this complication doesn't appear if we look at generalized points.

#### Generalized points in Weil's Foundations

Definition: Let $\Omega / k$ be a field extension of infinite transcendence degree, with $\Omega$ algebraically closed. Then $\Omega$ is called a universal domain for $k$.

Comparison: We can embed any field extension $K$ of finite transcendence degree over $k$ into $\Omega$, so every homomorphism $R \to K$ can be seen as homomorphism $R \to \Omega$. In particular, every generalized point corresponding to any generic point $\eta \in Spec(R)$ for $R$ a $k$-algebra can be seen as a generalized $\Omega$-point, which is defined over $\kappa(\eta)$, since $\kappa(\eta) = Quot((R/\eta)^{red})$ is a field extension of $k$ of finite transcendence degree.
It is my impression that, back in the days of Weil's foundations of algebraic geometry, one didn't care so much about the field of definition, but wrote just "let $t \in X(\Omega)$ be a generic point". The problem is, that "the" generic point is not unique, so one had to write "a generic point", where one really didn't care about the choices. These choices are the embeddings $\kappa(\eta) \to \Omega$, and they're completely irrelevant.

Fun: If you are unhappy with schematic/generic points, and want to stick to closed points without using the functor-of-points approach, there is a cute workaround. Take a $k$-scheme $X := Spec(R)$ (or just any $X$, in fact) and look at the base change $X_{k(X)} := X \times_k Spec(k(X))$. For example, if $R=k[x]$ we have $X = \mathbb{A}^1$ and $k(X) = k(x)$, so $X_{k(X)} = Spec(k[x] \otimes k(y))$. The scheme $X_{k(X)}$ is now a $k(X)$-scheme and the generic point $\eta$ of $X$ (which was a $k(X)$-rational point) now gives us (by base change) a $k(X)$-rational point of $X_{k(X)}$. In particular, it is a closed point of $X_{k(X)}$ (which might be clear if you consider the base change to the algebraic closure $\overline{k(X)}$ instead). Using a universal domain we can thus write generic points as closed points of $X_\Omega$.

#### Fun with generalized points

Take $A := k[\epsilon]/(\epsilon^2)$ (bad joke warning: $\epsilon$ is so small, its square is zero), and $R$ a $k$-algebra, then a morphism $Spec(R) \to Spec(A)$ corresponds to a $k$-algebra homomorphism $k[\epsilon]/(\epsilon^2) \to R$, which is given by choosing an image $e \in R$ of $\epsilon$, such that $e^2 = 0$. The set $Hom_k(Spec(R),Spec(A))$ is also called cotangent space of $Spec(R)$. As a topological space $Spec(A)$ consists of two points: the maximal ideal $(\epsilon)$ and the prime ideal (the generic point) $(0)$. A morphism $Spec(A) \to Spec(R)$ is just a point of $Spec(R)$ together with a tangent direction. The same game can be played with higher nilpotents, as $k[\epsilon]/(\epsilon^n)$.
This is useful to talk about Jets, in deformation theory, and I think one of the main reasons one should allow nilpotents in rings of functions in algebraic geometry (i.e. to use schemes and not just varieties).

#### Points I ignored

I could've talked about topos-theoretic points. A topos-theoretic point is something that allows building stalks and skyscraper sheaves (adjoint functions), so each generalized point gives us a topos-theoretic point for the Zariski topology. There are other Grothendieck topologies which are not so lucky, where there are not enough points, in the precise sense that one cannot check whether a sheaf morphism is an isomorphism by checking it on all stalks.

The valuative criteria for separatedness and properness, as in Hartshorne's book, can be seen more geometrically by interpreting a DVR as the spectrum of the local ring of a curve, in some sense a "curve germ". I always wanted to investigate that a bit more, but don't have the time right now.

Did I miss something else?

Category: English, Mathematics

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4 Responses

1. 1
Allen Knutson
2012-12-02 (2. December 2012)

You didn't precisely leave out geometric generic points (the separable closure of the function field), but you could emphasize more how they are more analogous to general points than generic points are.

Example: let s:A^1 -> A^1 be the squaring map. The fiber over a general point is two points. The fiber over the generic point is irreducible (!). The fiber over the geometric generic point is two points.

2. You wrote "To summarize, we just understood how a closed point p corresponds to a unique κ(p)-rational point". I think it's not entirely correct, as Spec(Qbar) (Qbar the algebraic closure of the rational numbers) has a lot of field automorphisms, each of which gives a different Qbar-rational (in fact geometric) point s : Spec(Qbar)-->Spec(Qbar). What we can say, I think, is that there's a bijection between the set of closed points in |X| (|X| is the topological space underlying the scheme X) and the equivalence classes of rational points by the equivalence relation given, say, at page 39 of Laumon--Moret-Bailly "Champs algébriques".

3. ...Or, probably what you said is still true when the "unique κ(p)-rational point corresponding to the closed point" p ∈ |X| is taken to be the one induced by the identity automorphism of κ(p).

4. Thank you, good point!

Now I have to find some time to look it up in "Champs algébriques"...