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.