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The notion of a surjective map of schemes is somewhat confusing. For example, consider an elliptic curve ${E}$ over ${{\mathbb Q}}$ of positive rank. The multiplication-by-${m}$ map is obviously not surjective as a map ${E({\mathbb Q}) \rightarrow E({\mathbb Q})}$, but in fact it is surjective as a map of schemes, because schemes carry baggage associated to the points we haven’t yet added in’ by passing to an algebraic closure. In this post we investigate the condition of a map being surjective on schemes as related to its inducing a surjection on the geometric points. I looked briefly in Hartshorne and did not seem to find anything about this, and I still do not have the whole picture (I have only considered what happens on closed points), so if readers can suggest a proper reference I would be grateful. Also, this post is likely to contain errors which might need fixing, so I urge readers not to assume I know what I’m doing.

Proposition 1 Let ${\phi: Y \rightarrow X}$ be a map of schemes of finite type over ${k}$, and ${k^a}$ an algebraic closure for ${k}$. The induced map ${Hom_k(Spec \ k^a, Y) \rightarrow Hom_k(Spec \ k^a, X)}$ is surjective if and only if ${Y \rightarrow X}$ is surjective on closed points.

I think one direction is reasonably straightforward. Given a closed point ${x \in X}$, take any geometric point ${f:Spec\ k^a \rightarrow X}$ with image ${x}$, for example the point obtained from ${O_U \rightarrow O_{U,x} \rightarrow k(x) \rightarrow k^a}$ for some affine ${U}$ containing ${x}$ (since we are of finite type over ${k}$ and ${x}$ is maximal, ${k(x)}$ is a field of finite type over ${k}$, so finite over ${k}$ by the Nullstellensatz). Then by assumption, there is a ${g:Spec\ k^a \rightarrow Y}$ with ${\phi g = f}$, so in particular any closed point in the closure of ${g(0)}$ is a closed point of ${Y}$ lying above ${x}$.

Let’s try to prove the other direction. So now we are given a geometric point ${f:Spec\ k^a \rightarrow X}$ and we want to lift it to such a point of ${Y}$. Well, such an ${f}$ determines a unique pair ${(x,\phi)}$ with ${x\in X}$ and ${\psi: k(x) \rightarrow k^a}$, and ${x}$ must be closed by a dimension argument, so let us choose a closed point ${y \in Y}$ with ${\phi(y)=x}$. There is an induced map in the wrong direction ${k(x) \rightarrow k(y)}$. However, since ${y}$ is closed, ${k(y)}$ is a finite extension of ${k}$, so in particular we can generate a map ${\psi'k(y) \rightarrow k^a}$. Furthermore, it is possible to do so in such a way as to ensure that ${k(x) \rightarrow k(y) \rightarrow k^a}$ is the map ${\psi}$.

But then the pair ${(y,\psi')}$ determine a geometric point ${g:Spec\ k^a \rightarrow Y}$ in the obvious fashion, and it is clear from our construction that ${\phi g = f}$, so we have proved that the map on geometric points is a surjection.

There are at least two things I still don’t know which would be good to know. Firstly, what conditions do I need to impose to guarantee that ${k^a}$ can be replaced with ${k^s}$, a separable closure? There is some kind of condition needed to say that induced maps are separable (so I guess insisting that ${\phi}$ is unramified) but also that the extensions ${k(x)/k}$ are separable, which is probably some simple condition but I haven’t yet worked out what. Secondly, can I extend the result to surjectivity of the map ${Y \rightarrow X}$ on all points? Again, hopefully this is fairly easy messing around with taking closures, etc. but I still don’t seem to have managed it.

It would also be nice to know if the above argument is actually correct and/or necessary. It’s been assembled in a very ad-hoc way to convince me that things I was doing with étale covers of elliptic curves weren’t nonsense, so I can believe my sketchy knowledge of scheme theory could have allowed a flaw to pass. Note that the result claimed about multiplication by ${m}$ on an elliptic curve follows (on the closed points) immediately, because given any geometric point I can divide it by ${m}$ by solving an appropriate degree ${m^2}$ polynomial, hence I get a geometric point with co-ordinates in ${k^a}$.

In the third post of this series, we shall very briefly introduce the formalism of ideles and restate the Artin reciprocity theorem in our new language, following the ideas of Chevalley. It is a little difficult to get used to the language of adeles and ideles, but they turn out to be a very natural setting in which to study a global field (like a number field) in relation to its completions with respect to absolute values. In fact, one of the reasons for their introduction was to facilitate a proof of global class field theory from local class field theory. However, the main results of global class field theory also become much cleaner to state, as we shall see next time when we finally get to them.

Firstly, we have to define these things! Fix a global field ${K}$ (a number field or a 1-dimensional ${\mathbb{F}_p}$-function field). We want an adele to be an element of the product ${\prod_{v \in M_K} K_v}$, where ${M_K}$ is the set of valuations on ${K}$. In other words, we want to take one element of each completion of ${K}$. However, there is far too much junk in this product! We are after all studying ${K}$ itself, and the diagonal map ${K \rightarrow \prod_{v \in M_K} K_v}$ is ridiculously far from being surjective. For example, a real life element of ${K}$ will have to be integral at all but finitely many places. In fact, once we’ve noted this, we are actually ready to define the adeles.

The ring of adeles ${\mathbb{A}_K}$ is the set of ${(x_v) \in \prod_{v} K_v}$ such that ${x_v \in \mathcal{O}_v}$ for all but finitely many ${v}$ (note that there are only finitely many archimedean places, so we can ignore them and the definition still makes sense).

This is a fairly hefty ring. It had better possess some kind of interesting topology if we’re going to be able to understand it. One would like to take the subspace topology coming from the product topology, but annoyingly a product of locally compact spaces need not be locally compact (think ${{\mathbb R}^{\mathbb N}}$), so if we want local compactness (which would be nice) we need a different topology. It turns out there is a fairly natural topology. We take as a base of open sets the products ${\prod_v U_v}$ such that ${U_v}$ is open for each ${v}$ and for all but finitely many ${v}$, we have ${U_v=\mathcal{O}_v}$. So this looks kinda similar to the product topology, but is more suited to situations where we have imposed a restriction like the one we did: it is called the restricted product topology.

With this shiny new topology, ${\mathbb{A}_K}$ becomes a locally compact topological ring (noting that ${\mathcal{O}_v}$ is in fact always compact and that if you’re “locally locally compact” then you’re locally compact), and that’s absolutely fantastic! However, life just keeps getting better for us. Remember how earlier we were talking about how we’re actually studying ${K}$ and so we want the image of the diagonal map ${K \rightarrow \mathbb{A}_K}$ to not be too insignificant. Well, we’re in unbelievably good luck! Firstly the image of ${K}$ is discrete (which seems like it should be true). However, there are so many elements of ${K}$ discretely scattered in ${\mathbb{A}_K}$ that in fact the quotient ${\mathbb{A}_K/K}$ is compact! The main ingredient of the proof is again the weak approximation theorem (which is very similar in flavour and sophistication, but also ubiquity, to the chinese remainder theorem).

There is some fairly cool measure theory going on here (in fact, if you develop this theory correctly, you can bypass the classical Minkowski proofs of finiteness of the class group and Dirichlet’s unit theorem). Locally compact groups have a unique (up to scaling) Haar measure, and being compact, ${\mathbb{A}_K/K}$ has finite quotient measure. Since the effect of multiplication by ${a \in K}$ on the measure of a subset of ${\mathbb{A}_K}$ can be shown to be precisely the product of the normalised valuations, this gives an instant and satisfying proof of the classical product formula for all norms on ${K}$. We shall not explore these issues here, but some other post may, and they are definitely worth attention.

Anyway, back to the task in hand. Now we have the adeles. How about the ideles? Well, they are just the multiplicative analogue. Indeed, the idele group of ${K}$ is just the group of units ${J_K = \mathbb{A}_K^*}$. However, things aren’t quite as nice as that. We have no guarantee that inversion is continuous in the subspace topology, so we have to take a slightly refined topology: the subspace topology on ${\mathbb{A}_K^* \hookrightarrow \mathbb{A} \times \mathbb{A}}$ (where the embedding is ${x \mapsto (x,x^{-1})}$). Soon we will pass to a subgroup of the ideles which makes this issue go away, so it probably isn’t worth thinking about for too long.

So what do we know? It’s clear that the principal ideles ${K^* \hookrightarrow J_k}$ are discrete in this topology (since it’s stronger than the previous one). It can also be seen that in fact the ideles in this topology are just the elements ${(x_v) \in \prod_v K_v^*}$ such that ${x_v \in \mathcal{O}^*_v}$ for all but finitely many ${v}$, with a restricted product topology like that on the adeles, so they really are exactly the multiplicative analogue (even their topology is!).

But always we should get a nagging feeling that we should be studying real life elements of ${K}$ and maybe that ${J_K}$ is just a bit too big to be studying. What other properties do elements of ${K}$ have? Well, they satisfy a product formula

$\displaystyle \prod_v |x|_v = 1.$

We cannot guarantee this for every idele. We therefore consider the map

$\displaystyle c: (x_v) \mapsto \prod_v |x_v|_v$

which gives a continuous homomorphism ${J_k \rightarrow {\mathbb R}_{>0}^*}$, and let ${J_K^1}$ be its kernel.

This set ${J_K^1}$ turns out to be often a more natural object of study. It contains all of the principal ideles, as we remarked above. It also is a closed subgroup of ${J_K}$ and has the surprising property that its topology is equal to its subspace topology inherited from ${\mathbb{A}_K}$. However, the main reason it is a good thing to study comes from its receiving the same seal of approval as that of the ideles:

Theorem 1 The topological group ${J^1_K/K^*}$ is compact.

This is all great fun, at least for me, but lots of readers have either now probably stopped reading or started wondering how any of this is remotely related to class field theory. Well, there is a natural map ${\alpha}$ from ${J_K}$ to the ideal group ${I_K}$ of ${K}$, obtained by just trying to construct the fractional ideal which feels most likely to be generated by’ our idele. Indeed, we define

$\alpha: \displaystyle (x_v) \mapsto \prod_{v \text{ finite}} \mathfrak{p}_v^{v(x_v)}.$

This makes sense because all but finitely many of the ${x_v}$ are units in their valuation rings.

Notice how by fudging things into the archimedean valuations (which don’t get detected by this map) we get something better. In fact, restricting to ${J_K^1}$ we still get a surjection ${J_K^1 \rightarrow I_K \rightarrow 0}$. This has all kinds of cool consequences. Firstly, including the finiteness of the class group (${I_K/K^*}$ is discrete, and the above surjection implies it is compact, whence finite).

So it’s worth bearing in mind we have this nice ${J_K^1}$ thing lying around. However, for class field theory we will decide to work with ${J_K}$ instead. I am not entirely sure why – I guess it is ultimately a bigger object, and given we are able to prove stuff about it, we might as well use it (since the things proved are presumably in some sense more general’). Having not seen a proof myself it is difficult to say, but I shall try to post back here when I do find out (or you might want to tell me why we use ${J_K}$ and not ${J_K^1}$ if you know).

Now, let us take ${L/K}$ a finite abelian extension. The classical Artin map ${F_{L/K}: I^S \rightarrow G(L/K)}$ can be composed with our canonical map ${\alpha: J_K \rightarrow I_K}$ to obtain a map ${\psi_{L/K}: J_K \rightarrow G(L/K)}$ which from now on is what we mean when we say the phrase Artin map’. The reciprocity law can be re-written in the following form:

Theorem 2 (Artin Reciprocity, idelic form) There exists a continuous map ${\psi_{L/K}: J_K \rightarrow Gal(L/K)}$ with the following properties. The kernel of ${\psi}$ must contain ${K^*}$, and any idele ${x}$ which takes the value ${1}$ at all the ramified and archimedean primes must satisfy ${\psi(x)=F_{L/K}(\alpha(x))}$.

In fact, the condition of being trivial on the principal ideles makes it natural to define the idele class group to be ${C_K = J_K/K^*}$ and regard the Artin map as a continuous homomorphism ${C_K \rightarrow Gal(L/K)}$. It is this language which turns out to seem optimal for stating the main results of global class field theory in their greatest generality, and that is what we will do next time. Indeed, we shall see that the main result of global class field theory is the now simple statement that there is an astonishing correspondence between the finite index open subgroups of ${C_K}$ and the finite abelian extensions of ${K}$.

In this short post we look at the general definition of a zeta function for a scheme of finite type over ${Spec {\mathbb Z}}$ and record its relationship with the zeta function of an algebraic variety ${X}$ over a finite field ${\mathbb{F}_q}$, defined as a kind of generating function for the number of points of ${X}$ over all finite extensions of ${\mathbb{F}_q}$.

Let ${X}$ be a scheme of finite type over ${Spec {\mathbb Z}}$. In fact, for concreteness, we start by looking at a single affine chart: imagining ${X=Spec A}$ as a geometrical space whose ring ${A}$ of functions is a quotient of some ring ${{\mathbb Z}[t_1,...,t_n]}$ of polynomials (in finitely many variables).

Then for every maximal ideal ${m}$ of ${A}$, the quotient ${A/m}$ is in fact a finite field, so we can define the norm ${N_m = |A/m|}$ to get a positive integer ${\geq 2}$ associated with ${m}$. We can use this to define the zeta function associated with ${X}$ to be the following formal Dirichlet series, by analogy with the classical Euler product formula.

$\displaystyle \zeta_X(s) = \prod_{m \text{ a maximal ideal of }A} \frac{1}{1-N_m^{-s}}.$

Notice that taking ${X=Spec {\mathbb Z}}$ we recover the classical Riemann zeta function, and taking the ring of integers of a number field recovers the zeta function of a number field. What happens if we take ${X}$ to be a variety over a finite field ${\mathbb{F}_q}$?

Well, ${log \zeta_X(s) = - \sum_{x} log(1-N_x^{-s})}$ where the sum is over all closed points ${x \in X}$.

What is a closed point of ${X}$? What is an ${\mathbb{F}_{q^r}}$-rational point of ${X}$? This is one of the confusing subtleties of scheme theory: these concepts are related but not exactly the same. Since the more classical definition of a zeta function is in terms of rational points, we will have to make this leap.

In fact, an ${\mathbb{F}_{q^r}}$-rational point (which we will also call a geometric point’, perhaps slightly nonstandardly) is simply a map ${Spec(\mathbb{F}_{q^r}) \rightarrow X}$. This makes sense for two reasons. Firstly, a point in ordinary geometry can be described as just a map from a one point space (or as a zero-simplex, if you like), so our definition seems plausible. Secondly, and more convincingly, in the case where ${X}$ is affine, for example the elliptic curve over ${{\mathbb Z}}$, ${y^2=x^3-x}$, such a map corresponds to a map of rings

$\displaystyle {\mathbb Z}[x,y]/(y^2-x^3+x) \rightarrow \mathbb{F}_{q^r}.$

In other words, it corresponds to a choice of values for ${x}$ and ${y}$ in ${\mathbb{F}_{q^r}}$ satisfying the governing equation – exactly what we should mean by specifying a ${\mathbb{F}_{q^r}}$-rational point.

On the other hand, a closed point is a maximal ideal of the ring of functions of an affine neighbourhood. Whenever we have a map ${Spec\mathbb{F}_{q^r} \rightarrow X}$, it corresponds to a map of rings whose image is a field. In other words, a map whose kernel is a maximal ideal. Therefore, every ${\mathbb{F}_{q^r}}$-rational point is associated with a unique closed point ${x}$. In fact, we get an induced map ${k(x) \rightarrow \mathbb{F}_{q^r}}$, where ${k(x)}$ is the residue field at ${x}$. Conversely, any such map can be extended and determines a geometric point.

So finally we have arrived at the relationship between closed points and geometric points. Namely, each closed point ${x}$ determines the a set of ${\mathbb{F}_{q^r}}$-rational points ${X_x}$ which can be canonically identified with ${Hom_{\mathbb{F}_q}(k(x), \mathbb{F}_{q^r})}$. But by basic Galois theory, this says that, if ${|k(x)|=q^d}$, ${X_x}$ is nonempty iff ${d|r}$, in which case ${|X_x|=d}$.

So let us return to our zeta function. Writing ${N_x=q^d}$, we can expand

$\displaystyle -log(1-N_x^{-s}) = \sum_{m \geq 1} \frac{q^{-sdm}}{m}.$

Now, set ${T=q^{-s}}$ and bearing in mind where we are trying to get (to an expression involving the numbers ${|X_x|}$), we can define ${N(m,x)}$ to be ${d}$ if ${d|m}$ and ${0}$ otherwise, and rewrite this sum as

$\displaystyle \sum_{m \geq 1} N(m,x)\frac{T^m}{m} = \sum_{m \geq 1} |Hom_{\mathbb{F}_q}(k(x), \mathbb{F}_{q^r})|\frac{T^m}{m}.$

Now, take the sum over all closed points ${x}$, and conclude that

$\displaystyle log \zeta_X(s) = \sum_{m \geq 1} (\sum_x |Hom_{\mathbb{F}_q}(k(x), \mathbb{F}_{q^r})|)\frac{T^m}{m} = \sum_{m \geq 1} |X(\mathbb{F}_{q^m})|\frac{T^m}{m}.$

This gives the familiar expression for the zeta function as an object whose logarithmic derivative is a generating function for the number of points of ${X}$ over all the finite field extensions of ${\mathbb{F}_q}$, as appears in the statement of the Weil Conjectures and is well studied. One can check further that given the analysis above, the Riemann Hypothesis’ for varieties of dimension ${d}$ over a finite field asserts that all the zeroes are on the lines ${Re(s) = 1/2,3/2,....,(2d-1)/2}$ and the poles are on the lines ${Re(s) = 0,1,2,...,d}$. Noting that ${{\mathbb Z}}$ is an object of dimension 1, the analogy with the classical Riemann hypothesis could not be clearer.