Archives for category: number theory

I’m somewhat late to the mark on this, but I just read a wonderful short story by Lieven Le Bruyn that combines two things that have been on my brainwaves during the last few months: David Lynch’s Twin Peaks and Shinichi Mochizuki’s Inter-Universal Teichüller Theory.  It’s called The Log Lady and the Frobenioid of $\mathbb{Z}$!

It involves the mysterious development of ideas in dreams, a certain Japanese tourist and compactifying Norma’s famous cherry pies at the Double-R Diner (Edit: I have just realised it’s $\pi$-day, so this is extremely fitting!)

Like many other things I have read on neverendingbooks, it’s smart, funny, philosophical and left me feeling both more and less mystified. I urge you to give it a read!

(P.S. there are no Twin Peaks spoilers in it either, so there’s no danger in reading if you – like myself – have not yet finished watching.)

In a previous post I wrote about and drew pictures of certain special schemes (some of which I would now draw differently after a bit more experience). There was one scheme here that I really wanted to look at again – and which we might be able to say a bit more about now we’ve looked at the functor of points.

That scheme is “affine space over the integers”

$X = \mathbb{A}^1_{\mathbb{Z}} = \text{Spec}\mathbb{Z}[x]$

The inclusion $\mathbb{Z}\hookrightarrow \mathbb{Z}[x]$ induces a morphism

$\phi: X\rightarrow \text{Spec}\mathbb{Z}$

and the “fibres” of this morphism – that is, the preimages of the points in $\text{Spec}\mathbb{Z}$ – in certain subschemes of $X$ are of substantial number-theoretic interest.

A really quick definition of fibre products and inverse images

In general if $X$ is a scheme, $Y\hookrightarrow X$ is a subscheme, and $\phi: Z\rightarrow X$ is a morphism, then the “inverse image” $\phi^{-1}(Y)$ of $Y$ under $\phi$ is the fibre product

$\phi^{-1}(Y) = Y\times_X Z$

Note that in the category of affine schemes, the fibre product

$\text{Spec}(A)\times_{\text{Spec}(C)} \text{Spec}(B)$

is just the spectrum of the tensor product

$\text{Spec}(A\otimes_C B)$

which you can check by checking this has the right universal mapping property for the fibre product. In the general case, since schemes are glued up from affine schemes we can show that fibred products exist and are made by glueing spectra of tensor products together. I won’t say any more about fibre products except that if $x\in X$ is a point then we can describe $x$ as a morphism

$\bar{x}: \text{Spec}\kappa(x) \rightarrow X$

from the spectrum of the residue field at $x$ into $X$. Then the fibre above $x$ is the fibre product

$\phi^{-1}(x) = \phi^{-1}(\left\{x\right\}) = \text{Spec}\kappa(x)\times_X Z$

Geometrically, this is just all the points of $Z$ that map to $x$ under $\phi$, and this constitutes a subscheme of $Z$ since the category of schemes has fibre products.

The fibres of $\mathbb{A}^1_{\mathbb{Z}} \rightarrow \text{Spec}\mathbb{Z}$

Okay, so back to our arithmetic space $X = \mathbb{A}^1_{\mathbb{Z}}$ and its morphism $\phi: X\rightarrow \text{Spec}\mathbb{Z}$. Then the fibre above $[(0)]$ is

$\phi^{-1}([(0)]) = \text{Spec}\mathbb{Q} \times_{\text{Spec}\mathbb{Z}} X = \text{Spec}(\mathbb{Q}\otimes_\mathbb{Z} \mathbb{Z}[x]) = \text{Spec}\mathbb{Q}[x] = \mathbb{A}^1_{\mathbb{Q}}$

So above the generic point $[(0)]$, the points of $X$ look like the spectrum of $\mathbb{Q}[x]$; this looks like a one-dimensional line with points for all the rational numbers $p/q$ corresponding to maximal ideals of the form $(x-p/q)$ , plus a “one-dimensional” generic point corresponding to the zero ideal; it also has extra points owing to the fact that $\mathbb{Q}$ is not algebraically closed, and in fact for every irreducible monic polynomial $f\in\mathbb{Q}[x]$ there is a point $[(f)]$ in $\text{Spec}\mathbb{Q}[x]$. Hence these extra points corresponds to orbits of algebraic numbers under the action of $\text{Gal}(\mathbb{Q})$, the absolute Galois group of the rationals. So already the fibre above the generic point of $\text{Spec}\mathbb{Z}$ carries a huge amount of arithmetic information. We can draw $\text{Spec}\mathbb{Q}[x]$ as looking something like this:

Here the single black line is one single point – the generic point corresponding to the zero ideal. The red lines are meant to denote rational numbers (in lowest terms) spread densely along this line. Then the pair of purple points denotes a single point of this scheme – here it’s a pair of algebraic numbers of degree 2: for example the pair $\pm\sqrt{2}$ under the action of the Galois group which swaps these two points. Similarly the blue quadruple of points and arrows represents a single point in this scheme corresponding to the orbit of an algebraic number of order 4.

Okay, cool. So the fibre above the generic point looks like affine space over the rationals – a line of rational points packaged with extra info about algebraic numbers of higher degrees and their symmetries. What about the fibre over a closed point $[(p)]$ corresponding to a prime number $p$? Here the point $[(p)]$ can be described as a morphism $\text{Spec}\mathbb{F}_p \rightarrow \text{Spec}\mathbb{Z}$, and if you do the easy computation you see that the fibre above this point looks like $\text{Spec}\mathbb{F}_p [x] = \mathbb{A}^1_{\mathbb{F}_p}$ – the affine line over the finite field $\mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$. Again, the points of this scheme are orbits of the action of the absolute Galois group of $\mathbb{F}_p$ on its algebraic closure.

So the first interesting thing about $\mathbb{A}^1_{\mathbb{Z}}$ is that it basically “parameterises” affine spaces over all of the simplest fields as the characteristic varies: it is a collection of fibres, varying over the prime numbers and zero, which “package” affine spaces over different fields together into one whole:

Algebraic numbers as functions on the primes

Here’s one thing that struck me as extremely cool: $\text{Spec}\mathbb{Z}$ was created as a geometric object on which the integers $\mathbb{Z}$ arose as the ring of globally-defined “functions”. The affine space $X=\mathbb{A}^1_{\mathbb{Z}}$ allows us to draw these functions, but they’re not especially interesting in themselves. What’s more interesting is that every single algebraic number can be realised as a function on a subscheme of $\text{Spec}\mathbb{Z}$ and drawn using a subscheme of $X$!

For, starting with the integers: the point $(x)\in X$ (I’ll omit the square brackets around ideals to denote points of the scheme to ease notation) corresponds to the value “zero” of these functions; that’s because when $x = 0$, $(x) = (0)\subseteq \mathbb{Q}$, and when $x = p$, $(x) = (0) \subseteq\mathbb{F}_p = \mathbb{Z}/(p)$. Now for example the point $(x-2)\in X$ – that is, the ideal – only equals $(x)$ in the ring $\mathbb{F}_2 [x]$, and so we see that $(x-2)$ intersects the zero-line when $x=2$. This corresponds to the “function” “2” on the integers just vanishing at the prime 2. Conversely, the ideal $(x-6)$ only equals $(x)$ when $x=2$ or $3$, and so the “function” 6 vanishes on the primes at 2 or 3. We can make sense of the “function” $9 = 3^2$ vanishing “to order 2” at the prime 3, but let’s leave what this means precisely for now (visually, the line corresponding to the graph of this function is tangent to the graph of $(x)$ at 3). We can draw these things like so:

But we can do much more: we can consider rational numbers (in lowest form) $p/q$, which correspond to prime ideals of $\mathbb{Z}[x]$ of the form $(px - q)$. These have a zero at the prime numbers in the factorisation of $p$ and a pole at the primes in the factorisation of $q$, and by putting the fraction in lowest terms we ensure that we don’t get primes on which ther are both zeros and poles. For example, the graph of the function $49/5$ looks something like this:

There is a double-zero at 7 and a pole at 5. The pole appears because modulo 5,

$5x - 49 \equiv -49 \equiv 1 \pmod 5$

Thus with some imagination we see that at the prime 5, the “function” 49/5 corresponding to the ideal $(5x -49)$ doesn’t coincide with any points along the vertical “geometric axis”. Instead we get $5x - 49 \equiv 1 \pmod 5$ (which you can check will always happen when the fraction is in lowest terms), and then the ideal $(1)$ always equals the whole ring (whichever ring we’re in). Now prime ideals are by definition not equal to the whole ring, so the ideal $(1)$ can never intersect one of the fibres $\mathbb{A}_K^1$, because if it did it would intersect at a point, which is a prime ideal. Hence it is reasonable to regard 5 as a pole of the function $49/5$ on the primes.

I’ve also drawn the values of the function at the primes 2 and 3, which are obtained by reducing the polynomial modulo those primes. Hence we can meaningfully say that modulo 2, $49/5 = 1$ and modulo 3, $49/5 = 1/2$ because the ideal $(5x - 49)$ equals the ideals $(x-1)$ and $(2x-1)$ at these respective points.

Another useful fact: the graphs of all of these “functions” emerge as  closed subschemes of $X$. The first 3 on the first picture are all isomorphic to $\text{Spec}\mathbb{Z}$ via e.g. $\mathbb{Z}[2] = \mathbb{Z}[x]/(x-2)\cong\mathbb{Z}$. On the graph we see that indeed these graphs look like squished-up versions of the copy of $\text{Spec}\mathbb{Z}$ below that they map down onto via $\phi$. The second scheme looks different – the pole at 5 means there is a “gap” in the scheme corresponding to (the closure of) $(5x-49)\in X$, and indeed this scheme emerges as

$\text{Spec}\mathbb{Z}[x]/(5x-49) \cong \text{Spec}\mathbb{Z}[49/5] \not\cong\text{Spec}\mathbb{Z}$

However, we see that in the way I’ve drawn it, there is only ever at most one point in the fibre of this scheme over a point of $\text{Spec}\mathbb{Z}$ – this corresponds to the degree of the polynomial $5x - 49$ being 1.

Algebraic numbers of higher degree and $\sqrt{3}$

Here is where algebraic numbers of higher degree pop up – they appear as subschemes of $X$ where the fibres above $\text{Spec}\mathbb{Z}$ contain more than 1 point. For example, consider the closure of the point $(x^2 - 3)\in\text{Spec}\mathbb{Z}[x]$. We can factor it as

$x^2 - 3 \equiv (x-1)^2 \pmod 2, \quad x^2 - 3 \equiv x^2 \pmod 3$

whereas over $\mathbb{F}_{5}$, in contrast, the polynomial remains irreducible, and over $\mathbb{F}_11$, it splits into two distinct factors. We can draw this as follows:

Here there are three different types of points in the subscheme $Y = \text{Spec} \mathbb[Z][\sqrt{3}] \cong \text{Spec}\mathbb{Z}[x]/(x^2 - 3)$ that arises as the closure of this point. There are “ordinary” pairs of points above primes 11 and 13 – and more generally, over any prime $p$ where $3$ is a square modulo $p$. Here the restriction of $\phi$ to $Y$ is “unramified” at these points, to use familiar geometric language from varieties, and the residue fields at these points are just the residue fields $\mathbb{F}_p$. So in this sense, these points are pairs of points on the “most simple part” of the fibre $\mathbb{A}^1_{\mathbb{F}_p}$ – the degree 1 points of $\mathbb{F}_p$ which are actually in this field. These points are called reduced because the local ring at these points is reduced – it has no nilpotents.

The second type of point is one like that above the primes $p= 2, 3$, where the polynomial $x^2 - 3$ is the square of another irreducible polynomial $g\in\mathbb{F}_p [x]$. These points are nonreduced, because they are subschemes $Z = \text{Spec}\mathbb{F}_p [x]/(g^2)$ and thus have the nilpotent element $g$. These are “one-dimensional shreds of points”, and the data of a function vanishing on one of these points means it and its first derivative must also vanish. So they are “double-points” consisting of the point and an “infinitesimal neighbourhood” around the point. However, the residue field at these points is still $\mathbb{F}_p$, so we have a “double point” on the most simple part of the affine line $\mathbb{A}^1_{\mathbb{F}_p}$ – i.e. an orbit consisting of a single point. These points are called ramified points, because the two fibres join together here to create a double point. We see this as the two fibres joining vertically at the point.

The third type of point is like that above the primes $p =5, 7$, and more generally any prime such that $3$ is not a square modulo $p$. Here the polynomial remains irreducible since $3$ is not a square modulo that prime, and hence $(x^2 - 3)$ corresponds to a single point in the fibre. This point is also reduced, having coordinate ring $\mathbb{F}_p [x]/(x^2 - 3)$, which also equals its residue field; however, the difference here is that the residue field is now not $\mathbb{F}_p$ but a quadratic extension $\mathbb{F}_{p^2}$. This means that the point in the fibre above $p$ contained in the closure of $(x^2 - 3)$ is not one of the most simple “degree 1” points, but in fact a point corresponding to an orbit of 2 Galois-conjugate quadratic-extension points. This corresponds to the two lines joining tangentially to the point. It’s not a perfect way to draw it but it works, and I’ve attempted a localised picture of this intersection below:

Here the black line represents the generic point of the fibre $\mathbb{A}^1_{\mathbb{F}_p}$ above $p$, the blue dashes represent the simplest “degree 1” points of the fibre corresponding to ideals of $\mathbb{F}_p [x]$ of degree 1, the two green dots represent a single point of the fibre corresponding to a Galois-orbit of two conjugate quadratic points, and the red lines represent the closure of $(x^2- 3)$ as it moves through $X = \mathbb{A}^1_{\mathbb{Z}}$ – it’s part of the generic point. So this red line intersects the fibre at one or two points; in the first case it’s at two distinct blue points, the nicest “unramified” kind of intersection. In the second case the fibre itself comes together at a blue point, giving us a point of “intersection multiplicity 2” – the ramified case. In the third case the closure of $(x^2 - 3)$ doesn’t really intersect itself at the fibre above $p$ – geometrically, it should really still be pictured as two distinct lines. The problem is these two lines meet the fibre at the pair of Galois-conjugate points, which really count as just one point in $\mathbb{A}^1_{\mathbb{F}_p}$; this is why we just see one point in the scheme. However, if we instead “pulled back” this fibre to $\mathbb{A}^1_{\mathbb{F}_{p^2}}$, so that we could see these two Galois-conjugate points were genuinely distinct, we’d somehow “resolve” these sorts of ramifications. It’s not a genuine ramification as in the type-2 case. I guess this is the sort of thing that can be dealt with using “blow-ups” or some such thing, although I’m not at this level yet.

Finally, I’ll leave you with this relation to algebraic number theory: the primes above which we get really pathological “ramification” of the map $Y\rightarrow \text{Spec}\mathbb{Z}$ above are $p = 2,3$, which are the primes dividing the discriminant of the quadratic number field $\mathbb{Q}(\sqrt{3})$. This geometric picture explains why we call these primes ramified in algebraic number theory; the ring of integers of $\mathbb{Q}(\sqrt{3})$ is $\mathbb{Z}[\sqrt{3}]$ and its spectrum is exactly the subscheme of $X$ that I drew above. We see that the ramification points of this map correspond to the truly ramified primes. The other primes are controlled by the Legendre symbol $(3/p)$ and quadratic reciprocity laws and there are only two possible types of other behaviour here, corresponding to type 1 and type 3 points. I’m particularly interested in learning more about this intersection at a pair of conjugate points and how this can be remedied by pulling back to an algebraic extension field.

Further, it hints at how traditional algebraic number theory – the study of finite extensions of the rationals and their rings of integers – can all be interpreted in a geometric context. The spectrum of the ring of integers of every number field arises as a closed subscheme of $X$, and spectra of many more types of rings – such as $p$-adic rings – occur as localisations here. So we need to develop and refine our geometric tools – with a special focus on this interesting intersection theory – to be able to get arithmetic information back from the geometry.

I found this a fascinating post, and given the context of this blog I hope other readers may do too!

Brian Conrad is a math professor at Stanford and was one of the participants at the Oxford workshop on Mochizuki’s work on the ABC Conjecture. He is an expert in arithmetic geometry, a subfield of number theory which provides geometric formulations of the ABC Conjecture (the viewpoint studied in Mochizuki’s work).

Since he was asked by a variety of people for his thoughts about the workshop, Brian wrote the following summary. He hopes that a non-specialist may also learn something from these notes concerning the present situation. Forthcoming articles in Nature and Quanta on the workshop will be addressed at the general public. This writeup has the following structure:

1. Background
2. What has delayed wider understanding of the ideas?
3. What is Inter-universal Teichmuller Theory (IUTT = IUT)?
4. What happened at the conference?
5. Audience frustration
6. Concluding thoughts
7. Technical appendix

1.  Background

The ABC Conjecture is one of the outstanding conjectures in number…

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I wasn’t entirely satisfied with my closing remarks in my last post; in particular I don’t think I explained the term “functor of points” well in an arithmetic context. Luckily I just read something in Liu’s book which completely cleared this up for me!

We’re going to need to define these $S$-schemes that I alluded to last time; these are basically just schemes over a fixed scheme $S$. Formally, if $S$ is a scheme then an $S$-scheme is a morphism of schemes

$\pi: X\rightarrow S$

(called the “structural morphism”) and a morphism of $S$-schemes

$f: (\pi: X\rightarrow S)\rightarrow (\rho: Y\rightarrow S)$

is just a morphism of schemes $f: X\rightarrow Y$ such that $\rho \circ f = \pi$. Thus if we denote the category of schemes by $\text{Sch}$ then the category of $S$-schemes, or schemes over $S$, is just the slice category $\text{Sch}/S$. So this category encodes all the information about morphisms into a fixed scheme $S$. If $S = \text{Spec}(R)$ for some ring $R$ then we call $\text{Sch}/S$ the category of schemes over $R$ or $R$-schemes.

Note that the category of $\mathbb{Z}$-schemes is equivalent to the general category of schemes; this is because $\mathbb Z$ is the initial object in the category of rings, and by the adjunction isomorphism

$\text{Mor}(X,\text{Spec}(\mathbb{Z})) \cong \text{Hom}(\mathbb{Z}, \mathcal{O}_X (X))$

we know $\text{Spec}(\mathbb{Z})$ is the terminal object in the category of schemes. Taking slice categories over terminal objects yields an equivalent category.

Definition: If $\pi: X\rightarrow S$ is an $S$-scheme then a section of $X$ is a morphism of $S$-schemes $\sigma: S\rightarrow X$; if you unpack this, you get the usual notion of a “section” of $\pi$ i.e. a morphism $\sigma: S\rightarrow X$ such that

$\phi \circ \sigma = \text{id}_S$

The set of sections of $X$ is denoted $X(S)$. If $S = \text{Spec}(A)$ for some ring $A$ then we write $X(A)$ for the set of sections of $X$.

It’s easy to see by the bijections between morphisms of schemes and morphisms of rings (in the opposite directions) that equipping $X$ with the structure of a scheme over a ring $A$ is the same as giving a sheaf of $A$-algebras on $X$. Now let’s see how this helps us understand “rational points” as in the functor of points approach. We have the following nice identification:

Theorem: Let $(X, \pi)$ be a scheme over a field $K$ (i.e. a $\text{Spec}(K)$-scheme). Then there is a bijection

$X(K) \cong \left\{x\in X: k(x) = K\right\}$

where $k(x) = \mathcal{O}_{X,x}/\mathfrak{m}_x$ denotes the residue field at $x$.

Proof: Pick a section $\sigma \in X(K)$, so $\sigma$ is a morphism of schemes $\text{Spec}(K)\rightarrow X$ with $\pi \circ \sigma = \text{id}_\text{Spec}(K)$. Let $\sigma^\# : \mathcal{O}_X \rightarrow K$ denote the morphism of structure sheaves, and let $x$ be the image of the unique point in $\text{Spec}(K)$ under $\sigma$. Then the stalk homomorphism

$\sigma_x^\#: \mathcal{O}_{X,x} \rightarrow K$

induces a field homomorphism

$\mathcal{O}_{X,x}/\mathfrak{m}_x = k(x) \rightarrow K$

because $\sigma_x^\# (\mathfrak{m}_x) = \left\{0\right\}$ since every homomorphism takes units to units. Now $k(x)$ is a $K$-algebra i.e. there is a field homomorphism $K\rightarrow k(x)$; therefore since field homomorphisms are injective it follows that $k(x) = K$.

Conversely, suppose that $x\in X$ satisfies $k(x) = K$. Then the natural map

$\mathcal{O}_{X,x}\rightarrow k(x)$

induces a morphism of schemes

$\text{Spec}(k(x))\rightarrow \text{Spec}(\mathcal{O}_{X,x})$

Now take an open set $U\subseteq X$ containing $x$. The canonical inclusion homomorphism into the colimit $\mathcal{O}_X (U)\rightarrow \mathcal{O}_{X,x}$ induces a morphism of schemes

$\text{Spec}(\mathcal{O}_{X,x})\rightarrow \text{Spec}(\mathcal{O}_X (U)) = U$

Composing this with the open immersion $U\hookrightarrow X$ gives a morphism $\mathcal{O}_{X,x}\rightarrow X$. Then the composition of all these maps:

$\text{Spec}(k(x)) \rightarrow \text{Spec}(\mathcal{O}_{X,x}) \rightarrow X$

sends the unique point of $\text{Spec}(k(x))$ to $x$, and thus since $k(x) = K$ we get a unique section $\sigma: \text{Spec}(K)\rightarrow X$ i.e. an element $\sigma \in X(K)$. Both of these constructions can be seen to be mutually inverse to one another. This concludes the proof.

Nice! So here’s another way to think about rational points on a variety – say $X$ is a variety over a field $K$ and $x\in X$ is a point. Then adjoining the coordinates of $x$ to $K$ produces a new field $K(x)$, and $K(x) = K$ precisely when $x$ is a $K$-rational point. Since adjoining the coordinates of $x$ to $K$ is the same as specifying the residue field at $x$, this links everything together.

So we should think of rational points on a variety over a field $K$ as sections $\text{Spec}(K)\rightarrow X$, or as points whose residue field is equal to $K$. In terms of the functor of points approach, therefore, the functor $h_X$ restricts to the subcategory of fields in the category of rings in several different ways:

$h_X (K) = \left\{ K\text{-points of } X \right\}$

$= \left\{\text{sections } \text{Spec}(K)\rightarrow X \right\}$

$= \text{Mor}(\text{Spec}(K), X)$

Here’s a cool toy arithmetic geometry example which also explains “nilpotents”: the Chinese remainder theorem is a statement about schemes!

Recall that in general ring-theoretical form, the chinese remainder theorem says that if we have a prime factorisation of an integer

$n = \prod_{i=1}^{r} p_i^{a_i}$

then there is a ring isomorphism

$\mathbb{Z}/n\mathbb{Z} \cong \bigoplus_{i=1}^r \mathbb{Z}/{p_i}^{a_i} \mathbb{Z}$

The term on the right is both a product and coproduct (as it is a direct sum of finite indexing set) and so under the antiequivalence functor $\text{Spec}$ this transforms to

$\text{Spec}(\mathbb{Z}/n\mathbb{Z}) \cong \coprod_{i=1}^r \text{Spec}(\mathbb{Z}/{p_i}^{a_i}\mathbb{Z})$

where the coproduct is taken in the category of affine schemes (of course, I haven’t properly defined what this is yet, but in this case it is not too different to what you might expect). Then this, along with the canonical quotient map $\mathbb{Z}\rightarrow\mathbb{Z}/n\mathbb{Z}$ imply that there is closed immersion

$\coprod_{i=1}^r \text{Spec}(\mathbb{Z}/{p_i}^{a_i} \mathbb{Z})\hookrightarrow \text{Spec}(\mathbb{Z})$

But what does the scheme on the left look like?

Well, the rings $\mathbb{Z}/{p_i}^{a_i}\mathbb{Z}$ are definitely not integral domains because they contain nilpotent elements; in particular the image of $p_i$ is nilpotent since by definition $p_i^{a_i} \equiv 0 \pmod{p_i^{a_i}}$. So the ideal $(0)$ is not a point in any $\text{Spec}(\mathbb{Z}/{p_i}^{a_i}\mathbb{Z})$. Recall that the prime ideals of $\mathbb{Z}/{p_i}^{a_i}\mathbb{Z}$ are precisely the prime ideals of $\mathbb{Z}$ containing $(p_i^{a_i})$. But the only prime ideal of $\mathbb{Z}$ containing $(p_i^{a_i})$ is $(p_i)$.

Hence $\text{Spec}(\mathbb{Z}/{p_i}^{a_i}\mathbb{Z})$ only has one point, namely $(p_i)$. So the underlying topological space is exactly the same as for $\text{Spec}(\mathbb{F}_{p_i})$ and in fact for any other field. But the difference lies in the structure sheaf – in particular, the ring of global sections is $\mathbb{Z}/{p_i}^{a_i}\mathbb{Z}$ which contains nilpotent elements.

People tend to think of these nilpotent elements as “fattening” the single point of these topological spaces by an infinitesimal neighbourhood because they correspond to “functions” on $\text{Spec}(\mathbb{Z}/{p_i}^{a_i}\mathbb{Z})$ which are not quite zero but can be thought of as “close” to zero, in that application of them several times reduces to zero. This is a sort of formal version of the situation in analysis – if a number $\epsilon$ is very close to zero in absolute value, we may ignore terms containing $\epsilon^2$ and higher powers which will vanish much more quickly as $\epsilon \rightarrow 0$.

So this means in the situation above that we have a scheme with underlying topological space containing $r$ points – the $r$ prime ideals $(p_1), \dots, (p_r)$ of the corresponding rings, and each point carrying with it an infinitesimal neighbourhood of “radius” $a_i$. Here’s the picture:

Here the blue bits correspond to the points of the underlying topological spaces, and the red bits denote structure sheaves. In particular the more red loops on a point, the greater the “degree” of nilpotents in the ring of global sections. Also sneakily shown is the ring $Q$ arising as the ring of global sections of the generic point $(0)$ which is squished out along the line of $\text{Spec}(\mathbb Z)$.

This allowance of nilpotents in the structure sheaf underlies many subtleties in the geometry of schemes, but also means we can use them for very general things in the far-off realms of number theory.

But wait, there’s more! The example above illustrates an important “fibering” property of scheme morphisms. Recall that for all rings $R$ and all schemes $X$ there is a natural bijection

$\text{Mor}(X, \text{Spec}(R)) \cong \text{Hom}(R, \mathcal{O}_X (X))$

Now the ring $\mathbb Z$ is the initial object in the category of commutative rings, so by the above bijection it follows that $\text{Spec}(\mathbb Z)$ is the terminal object in the category of schemes, so every scheme has a unique morphism to $\text{Spec}(\mathbb Z)$. In the example above, each point $x= [(p_i^{a_i})]$ in $\text{Spec}(\mathbb{Z}/n\mathbb{Z})$ maps to the point $y=[(p)]$ in $\text{Spec}(\mathbb Z)$ such that $p$ is the characteristic of the residue field

$k(x) = \mathcal{O}_{X,x}/\mathfrak{m}_x$

where in this case $\mathcal{O}_{X,x} = \mathbb{Z}/{p_i}^{a_i}\mathbb{Z}$ and $\mathfrak{m}_x$ is its unique maximal ideal (which in this case is the ideal generated by $(p_i)$). Hence every scheme $X$ can be considered as a sort of “topologically fibered object”  where each fiber consists of points whose residue fields have a fixed characteristic, either 0 or a prime number.

Given how technical my last few posts have been, I decided that I’d ditch any more technicalities for a couple of posts and instead show you some pictures of schemes. The ones today will all be affine (which already become pretty complex to draw!) but perhaps I’ll do some general pictures too some other time.

Spectrum of a field

As with all things, let’s start on pretty much the simplest example possible. The simplest rings – at least in the context of ideals – are fields. Let $K$ be a field. We want to draw $\text{Spec}(K)$, so we’ll need to know the prime ideals of $K$ to do this. The only prime ideal of any field is the zero ideal $(0)$. Hence $\text{Spec}(K) = \left\{\bullet\right\}$ is just one point. The stalk of the structure sheaf at this point – which, as a one-point space, coincides with the ring of global sections – is just $K$:

Note though that if $K, L$ are two different fields then the schemes $\text{Spec}(K)$ and $\text{Spec}(L)$ are NOT THE SAME! Even though they are homeomorphic as topological spaces, the schemes carry different information in their structure sheaves. It is important to remember that schemes have the structure of their “functions” built in as the structure sheaf.

In all my pictures, I’ve drawn the points of the scheme in blue, and the various sections of the structure sheaf in red.

Spectrum of a DVR

So that was pretty boring. What about another slightly more interesting example? Let $R$ be a discrete valuation ring (DVR) and write $\text{Frac}(R)$ denote its fraction field. What are the prime ideals of $R$? Firstly, since DVRs are integral domains, the zero ideal $(0)$ is prime. I’ll write $t_0$ for the point in $X=\text{Spec}(R)$ corresponding to the zero ideal. The only other prime ideal is the unique maximal ideal $M$ of the DVR, and I’ll write $t_1$ for this point in the affine scheme $\text{Spec}(R)$. So $\text{Spec}(R) = \left\{t_0, t_1\right\}$ has two points. But these two points are very different!

The point $t_1$ corresponds to the maximal ideal $M$; recall from this post that the closed points of an affine scheme correspond exactly to the maximal ideals of the ring. Therefore, $t_1$ is a closed point, meaning the subset $\left\{t_1\right\}$ is closed in the Zariski topology.

That’s not the case with $t_0$; the ideal $(0)$ is not maximal, so $\left\{t_0\right\}$ is not closed in the Zariski topology – in fact, the closure of $\left\{t_0\right\}$ is the whole of $\text{Spec}(R)$! So you can picture $t_0$ as a point “smeared out” over the whole scheme:

Before we look at the sheaf structure, recall there’s a natural ring homomorphism $R\rightarrow \text{Frac}(R)$; therefore we know from this post that there’s a morphism of schemes $\text{Spec}(\text{Frac}(R))\rightarrow \text{Spec}(R)$. Since $\text{Frac}(R)$ occurs as a localisation of $R$, this must be an open immersion and hence must send the unique point of $\text{Spec}(\text{Frac}(R))$ to an open point in $\text{Spec}(R)$. Therefore this morphism has to send the unique point $\text{Spec}(\text{Frac}(R))$ to $t_0$. If instead it was sent to $t_1$, this would still be a morphism of ringed spaces, but not of locally ringed spaces and hence not a morphism of schemes. So this is a first example of subtleties arising from schemes and their morphisms.

Now we know that the ring of global sections is equal to $R$; we can use this to compute the stalk $\mathcal{O}_{X,t_1}$. The stalk is the colimit

$\mathcal{O}_{X,t_1} = \text{colim}_{U\ni t_1} \mathcal{O}_X (U)$

where the limit ranges over the open sets $U\subseteq X$ containing $t_1$. The only such open set containing $t_1$ is $X$ itself, and hence the stalk $\mathcal{O}_{X,t_1}$ is equal to $\mathcal{O}_X (X) = R$.

We can use the fact that the inclusion $\text{Spec}(\text{Frac}(R)) \rightarrow X$ is an open immersion to compute the stalk at $t_0$; since this open immersion of ringed spaces has to induce isomorphisms on the structure sheaves, we see that the stalk $\mathcal{O}_{X,t_0} = \text{Frac}(R)$.

The affine line

Now for a more “geometric example”, which really shows how schemes generalise varieties. Let $K$ be an algebraically closed field and set $X = \text{Spec}(K[x])$. To understand $X$, we need to know what the prime ideals of $K[x]$ are. In fact, there are only two distinct kinds – the zero ideal $(0)$ since $K[x]$ is an integral domain, and maximal ideals of the form $(x-a)$ where $a\in K$. So we can picture $X$ as a line, which we’ll denote $\mathbb A^1_K$, called the affine line. The maximal ideals $(x-a)$ correspond to the closed point $a$ on the line, and the line itself corresponds to the “generic point” (i.e. nonclosed point) $(0)$:

In a sense, the nonmaximal ideal $(0)$ is a point that isn’t anywhere in particular on the line, but its closure is the whole line, so it’s helpful to think of it almost as a “subvariety” of $X$. Thinking of things this way (which we will do lots later) is nice, because it forces us to make the notion of “point” more general, allowing points of schemes to “parameterise” subschemes in a vague hand-wavey sense. We can think of each maximal ideal $(x-a)$ as being a zero-dimensional point of $X$, while the generic point $(0)$ is a one-dimensional point. In fact, just as divisors are formal linear combinations of traditional points in a variety, the notion of algebraic cycles can be thought of as formal linear combinations of subvarieties of higher dimension, or in this sense points of higher dimension. Anyway, I digress – let’s get back to looking at our picture above:

For each $a\in K$ the quotient ring $K[x]/(x-a)$ is isomorphic to $K$, and so we have closed immersions $\text{Spec}(K)\rightarrow \mathbb A^1_K$ where the unique point is sent to the ideal $(x-a)$. We also know that the stalk at $(x-a)$ is equal to the localisation

$K[x]_{(x-a)} = \left\{f/g: f,g\in K[x], g(a)\neq 0\right\}$

The stalk at the generic point $(0)$ is $\mathcal{O}_{X,(0)} = K(x)$, the field of fractions of $K[x]$.

The affine plane

Now for an even more geometric example, to see how these things generalise to higher dimensions:

Let $K$ be an algebraically closed field and let $X = \text{Spec}(K[x,y])$; we call $X = \mathbb A^2_K$ the affine plane. It has prime ideals $(0)$ (now a two-dimensional generic point, delocalised over the whole plane), maximal ideals $(x-a, y-b)$ corresponding to the closed point $(a,b)$ in the plane, and a third kind of prime ideal corresponding to one-dimensional curves in the plane; these are the ideals $(f)$ where $f\in K[x,y]$ is an irreducible polynomial. These ideals are contained in many of the maximal ideals $(x-a,y-b)$, which is reflected by the fact that the “curve” $(f)$ in $\mathbb A^2_K$ passes through the points $(a,b)$ in the plane (as inclusion is reversed when we go between ideals in the ring and subsets of the scheme). So the curve $(f)$ is a point of the scheme, but it can be thought of as living everywhere along a curve of points $(a,b)$:

The stalks pictured tell us the functions we’re allowed to invert on each part of $\mathbb A^2_K$; for example, let $(a,b)$ be the point in $\mathbb A^2_K$ corresponding to the maximal ideal $(x-a,y-b)$; then the “function” $(x-a,y-b)$ is zero on $(a,b)$, so $(x-a, y-b)$ is the unique maximal ideal in the stalk $\mathcal{O}_{X,(a,b)}$. Any “function” which isn’t in $(x-a,y-b)$ is nonzero near $(a,b)$, so in $\mathcal{O}_{X,(a,b)}$ it becomes a unit, meaning we can divide by it.

Spectrum of the integers

Now for a more arithmetic example: $\text{Spec}(\mathbb Z)$. The prime ideals of $\mathbb Z$ are $(0)$ and $(p)$ for all prime numbers $p$. The second type of prime ideal is maximal, corresponding to closed points in $\text{Spec}(\mathbb Z)$, while $(0)$ is not, and so it corresponds to a “one-dimensional” point. In fact, $\text{Spec}(\mathbb Z)$ looks suspiciously similar to $\text{Spec}(K[x])$ when $K$ is an algebraically closed field!

Each closed point $(p)$ is isomorphic to $\text{Spec}(\mathbb F_p)$, and the stalks at $(p)$ are the local rings $\mathbb Z_(p)$ consisting of all fractions whose denominators are not divisible by $p$. The stalk at $(0)$ is $\mathbb Q$, the field of fractions of $\mathbb Z$; in general, when $R$ is an integral domain, the stalk at the generic point of $\text{Spec}(R)$ will be the field of fractions of $R$.

An “arithmetic surface”

The last example for today, with some real arithmetic content, is the affine line over $\mathbb Z$, namely $\text{Spec}(\mathbb{Z} [x])$. It’s often written$\mathbb A^1_{\mathbb Z}$. Here’s a picture of it taken from Mumford’s Red Book:

It’s called a “line” because it’s the spectrum of a polynomial ring in one variable, but it’s really a two-dimensional object, with two different axes – an “arithmetic axis” along the bottom, and a “geometric axis” going upwards. Even though $\mathbb Z$ is a very simple ring – it’s a Euclidean domain, and thus “nearly” a field in terms of nice structure – it’s startling how much more complicated the affine line $\mathbb A^1_{\mathbb Z}$ is compared to $\mathbb A^1_K$ when $K$ is an algebraically closed field. This “arithmetic surface” provides the first glimpse that schemes encode arithmetic information well, and that they’re the right objects to use to get number-theoretical data out of geometry.

In fact, I’m not going to talk about this scheme today – I will probably devote an entire post to it in the future when we’ve got some more tools – because I can’t do a better job than Lieven le Bruyn did on his own blog in his analysis of “Mumford’s Treasure Map“. Please have a look at it!

This post is entirely unrelated to what I’ve been talking about involving sheaves and ringed spaces. But I saw it today in class and thought it was so fitting with the name and purpose of this blog that I’d present it anyway, because I love it so much. It’s a proof that there are infinitely many prime numbers (due to Furstenberg) – but instead of Euclid’s more direct approach, it uses topology!

Let’s create a basis for a topology on $\mathbb{Z}$; for integers $m, d$ with $d > 0$, denote by $A(m,d) = \left\{ m + d n : n\in\mathbb{Z}\right\}$ the infinite arithmetic progression of all integers congruent to $m$ modulo $d$. I claim that $\mathcal{B} = \left\{ A(m,d) : m\in\mathbb{Z}, d\in\mathbb{N}\right\}$ is a basis for a topology on $\mathbb{Z}$. Firstly, note that we can write $\mathbb{Z} = A(0,1)$, and we can agree that the empty set is equal to an empty union of arithmetic progressions. As we’re defining a basis our open sets are just unions of the $A(m,d)$ so we don’t need to check anything here. But we do need to check whether intersections of two basic open sets $A(m_1,d_1), A(m_2, d_2)$ can be expressed as unions of other basic opens of the form $A(m,d)$.

Firstly, if $A(m_1, d_1)\cap A(m_2, d_2) = \varnothing$ (i.e. $d_1 = d_2$ and $m_1 \neq m_2$) then we can vacuously write the intersection as an empty union, as agreed. So instead suppose $A(m_1, d_1)$ and $A(m_2, d_2)$ have nonempty intersection. Then without loss of generality we may assume they intersect at some point $m$. “Re-centering” each arithmetic progression on $m$ (as they are only defined modulo their respective differences $d_1$ and $d_2$ anyway), we now need to compute $A(m,d_1)\cap A(m,d_2)$. But it is not hard to see that this is just $A(m,\text{lcm}(d_1, d_2))$. But this is just another basic open set in $\mathcal{B}$ – it’s not even a union of more than one open set! So the whole construction works very well – we can take arbitrary unions of these sets, take finite intersections, and we know that we can express the whole space $\mathbb{Z}$ and the empty set in terms of the basic opens $A(m,d)$. So we really have a basis for a topology on $\mathbb{Z}$, whose open sets consist of unions of infinite arithmetic progressions.

Now the sets $A(m,d)$ are open, but they are also closed! To see this, note that $A(m,d)^c = \bigcup_{m'\not\equiv m \pmod d} A(m',d)$ is just a union of all those $d-1$ arithmetic progressions of the same step-size $d$ but shifted throughout the whole interval $\left\{m+1, \dots, m+d-1\right\}$. So the complement of a basic open set is a (finite, even) union of basic open sets, and thus is open too. Hence the basic open sets $A(m,d)$ are both closed and open.

Now consider the union $\bigcup_{p \text{ prime}} A(0,p)$. Since every integer except $1$ and $-1$ is divisible by at least one prime (by the fundamental theorem of arithmetic), the above union gives $\bigcup_{p \text{ prime}} A(0,p) = \mathbb{Z} \backslash \left\{1,-1\right\}$. The complement of this set, namely $\left\{1,-1\right\}$, is finite – hence there is no way that it can be expressed as a union of arithmetic progressions of the form $A(m,d)$, which are all infinite sets. But therefore as these are the basic open sets in the topology, the set $\left\{1,-1\right\}$ is not open. Hence its complement, $\mathbb{Z}\backslash \left\{1,-1\right\} =\bigcup_{p \text{ prime}} A(0,p)$, is not closed.

But $\bigcup_{p \text{ prime}} A(0,p)$ is a union of closed sets (as I showed, each arithmetic progression is both open and closed). If this was a union of a finite number of closed sets – and therefore if it were indexed by finitely many primes $p$ – then it would itself be closed, because in a topology finite unions of closed sets are always closed. And we definitely have a base for a topology, as shown above. Therefore, the union must involve an infinite number of closed sets $A(0,p)$, indexed by infinitely many primes $p$. And therefore, there are infinitely many primes!!

Do you prefer Furstenberg’s topological proof or Euclid’s streamlined proof? The latter has the advantage that it’s a lot more direct, quicker, and probably seems like it’s more related to prime numbers. Furstenberg’s proof, on the other hand, takes a long time to set up and it feels very removed from anything to do with primes, using a lot more abstract machinery and a proof by contradiction, while Euclid’s proof is direct. However, the contradiction is very easy to understand, while many people (myself included, until I was wisely advised to the contrary) believe that Euclid’s proof uses a contradiction. I guess this subtlety in Euclid’s proof – depending on how it’s presented – can make it slightly confusing. The topological proof seems more simple in each individual step it takes, and for this reason I prefer it. But both proofs rest on the fundamental fact that every one of the infinitely many integers (except the units) are divisible by at least one prime.

Edit: This post has generated a lot of comments with great advice and helpful pointers, for which I’m really grateful! I have made several changes to the post accordingly.