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## The arithmetic plane

If you haven’t seen them already, you might want to read John Baez’s week205 and Lieven le Bruyn’s series of posts on the subject of spectra. I especially recommend that you take a look at the picture of $\text{Spec } \mathbb{Z}[x]$ to which Lieven le Bruyn links before reading this post. John Baez’s introduction to week205 would probably also have served as a great introduction to this series before I started it:

There’s a widespread impression that number theory is about numbers, but I’d like to correct this, or at least supplement it. A large part of number theory – and by the far the coolest part, in my opinion – is about a strange sort of geometry. I don’t understand it very well, but that won’t prevent me from taking a crack at trying to explain it….

Before we talk about localization again, we need some examples of rings to localize. Recall that our proof of the description of $\text{Spec } \mathbb{C}[x, y]$ also gives us a description of $\text{Spec } \mathbb{Z}[x]$:

Theorem: $\text{Spec } \mathbb{Z}[x]$ consists of the ideals $(0), (f(x))$ where $f$ is irreducible, and the maximal ideals $(p, f(x))$ where $p \in \mathbb{Z}$ is prime and $f(x)$ is irreducible in $\mathbb{F}_p[x]$.

The upshot is that we can think of the set of primes of a ring of integers $\mathbb{Z}[\alpha] \simeq \mathbb{Z}[x]/(f(x))$, where $f(x)$ is a monic irreducible polynomial with integer coefficients, as an “algebraic curve” living in the “plane” $\text{Spec } \mathbb{Z}[x]$, which is exactly what we’ll be doing today. (When $f$ isn’t monic, unfortunate things happen which we’ll discuss later.) We’ll then cover the case of actual algebraic curves next.

Finite fields

As a digression, we should take the time to describe $\text{MaxSpec } \mathbb{F}_q[x]$ geometrically. In the analogy between number fields and function fields, $\mathbb{F}_q[x]$ sits halfway between $\mathbb{Z}$ and $\mathbb{C}[x]$ as it shares properties of both. For example, recall that there are

$\displaystyle M(q, n) = \frac{1}{n} \sum_{d | n} \mu \left( \frac{n}{d} \right) q^d$

irreducible polynomials of degree $n$, hence $\mathbb{F}_q[x]$ satisfies a simple analogue of the prime number theorem. Geometrically, irreducible polynomials of degree $n$ correspond to $n$-element orbits of elements of $\overline{ \mathbb{F}_q }$ under the action of the Galois group $\text{Gal}(\overline{\mathbb{F}_q}/\mathbb{F}_q) \simeq \hat{ \mathbb{Z} }$. Algebraically, this is the map of spectra obtained from the homomorphism $\mathbb{F}_q[x] \to \overline{ \mathbb{F}_q }[x]$; the maximal ideal $(x - a), a \in \overline{ \mathbb{F}_q }$ is sent to the maximal ideal in $\mathbb{F}_q[x]$ generated by the minimal polynomial of $a$. This is for the fairly simple reason that if an element of $\mathbb{F}_q[x]$ vanishes at a particular element of $\overline{ \mathbb{F}_q }$, it must also vanish at every element of the Galois orbit.

The geometry of the plane

One way to understand $\text{MaxSpec } \mathbb{Z}[x]$ via “coordinates” is to consider the quotient homomorphisms $\mathbb{Z}[x] \to \mathbb{F}_p[x]$ for every prime $p$; these correspond to embeddings of spectra $\text{MaxSpec } \mathbb{F}_p[x] \to \text{MaxSpec } \mathbb{Z}[x]$, where the image of the ideal $(f(x))$ is the ideal $(p, f(x))$. In this way, $\text{MaxSpec } \mathbb{Z}[x]$ can be described as the union of these spectra in the same way that $\mathbb{A}^2(\mathbb{C})$ can be described as the union of its vertical lines or its horizontal lines. Alternately, the embedding $\mathbb{Z} \to \mathbb{Z}[x]$ gives a quotient $\text{MaxSpec } \mathbb{Z}[x] \to \text{MaxSpec } \mathbb{Z}$ sending a maximal ideal $(p, f(x))$ to $(p)$, and the fibers of this map are precisely the subvarieties $\text{MaxSpec } \mathbb{F}_p[x]$ we just described; this map can be thought of as “projection” onto the “arithmetic axis” (or $\mathbb{Z}$-axis if you want to be cheeky).

Now that we know what the “plane” looks like, let’s describe some of its subvarieties. The simplest number rings which are not $\mathbb{Z}$ occur as quotients $\mathbb{Z}[x]/(x^2 - d)$, so let’s look at these first. By the correspondence theorem, the prime ideals of $\mathbb{Z}[\sqrt{d}]$ are precisely the prime ideals of $\mathbb{Z}[x]$ containing $x^2 - d$. Since $x^2 - d$ is irreducible, the only principal ideal with this property is $(x^2 - d)$, which is sent to the zero ideal of $\mathbb{Z}[ \sqrt{d} ]$. Any other prime ideals must come from the maximal ideals $(p, f(x))$, and this ideal contains $x^2 - d$ if and only if the ideal $(f(x))$ contains $x^2 - d$ in $\mathbb{F}_p[x]$. So we need to understand how $x^2 - d$ splits modulo a prime. There are three cases:

1. $p | 4d$, hence $x^2 - d \equiv x^2 \bmod p$ has a double root and $p$ is said to ramify;
2. $d$ is a quadratic residue $\bmod p$, hence $x^2 - d \equiv (x - a)(x + a)$ has two distinct linear factors and $p$ is said to split; or
3. $d$ is a quadratic non-residue $\bmod p$, hence $x^2 - d$ remains irreducible and $p$ is said to be inert.

Except for the strange case $p = 2$, these three cases correspond to the three possible values of the Legendre symbol $\left( \frac{d}{p} \right)$; the first case occurs finitely often and quadratic reciprocity is the tool by which we distinguish the last two. Now, when an ideal $(p, f(x))$ contains $x^2 - d$, its image in $\mathbb{Z}[ \sqrt{d} ]$ is said to be an ideal lying over $p$ because it is in the fiber over $p$ in the map $\text{Spec } \mathbb{Z}[ \sqrt{d} ] \to \text{Spec } \mathbb{Z}$ coming from the obvious injection $\mathbb{Z} \to \mathbb{Z}[ \sqrt{d} ]$. Geometrically, this is analogous to a branched covering of Riemann surfaces, and in fact for the function field case that’s exactly what it comes out to; this is the origin of the term ramification. The function field analogue you should be thinking of is that the square root function $\sqrt{z}$ branches at the origin, and this is just the statement that the prime ideal $(z)$ in $\mathbb{C}[z]$ ramifies in $\mathbb{C}[z, t]/(t^2 - z)$. (We’ll discuss this more later.)

In the ramified case, the only prime ideal lying over $p$ is $(p, x)$. In the split case, the only prime ideals are $(p, x + a), (p, x - a)$. And in the inert case, the only prime ideal is $(p, x^2 - d)$. We can summarize this as follows.

Proposition: The prime ideals of $\mathbb{Z}[ \sqrt{d} ]$ are $(0)$ and the maximal ideals $(p, \sqrt{d})$ for all primes $p | d$, the maximal ideals $(p, \sqrt{d} \pm a)$ for all primes $p$ for which $d$ is a quadratic residue and $a$ is an integer satisfying $a^2 \equiv d \bmod p$, and the maximal ideals $(p)$ for all primes $p$ for which $d$ is a quadratic non-residue.

When $d = -1$, for example, the resulting ring $\mathbb{Z}[i]$ (the Gaussian integers) is a Euclidean domain because the norm $N(a + bi) = a^2 + b^2$ is a Euclidean function. There are no ramified primes, and the split primes are exactly the primes representable as a sum of two squares, i.e. the primes $2$ and the primes congruent to $1 \bmod 4$. In the split case the ideals $(p, a \pm i)$ are principal; they are generated by two irreducible elements $c \pm di$ such that $c^2 + d^2 = p$. There is a nice story here involving a different kind of plane geometry; see, for example, this old blog entry.

However, for generic $d$ there will be many non-principal ideals given by the split primes. Algebraically this means that a split prime $p$ is irreducible but not prime in $\mathbb{Z}[ \sqrt{d} ]$; instead, it generates an ideal $(p)$ contained in the two maximal ideals $(p, \sqrt{d} + a), (p, \sqrt{d} - a)$. Geometrically this means that the function $p$ vanishes at two points simultaneously. But a stronger statement can be made: if the ideal $(p, \sqrt{d} + a)$ isn’t principal, there doesn’t exist any element of $\mathbb{Z}[ \sqrt{d} ]$ which vanishes only at this point and at no other point. This is an indication that there is some nontrivial “geometry” going on because we can’t construct functions with arbitrary vanishing behavior.

For general monic irreducible polynomials $f(x)$, the splitting behavior of $f(x)$ is again determined by the factorization $f(x) \equiv \prod_{i=1}^{k} f_i(x)^{e_i} \bmod p$, and the primes lying over $p$ are precisely the maximal ideals of the form $(p, f_i(x))$. When $p$ divides the discriminant of $f$, $f(x) \bmod p$ has repeated roots and $p$ is again said to ramify, and the $e_i$ are the ramification indices. Otherwise, the roots of $f(x)$ are distinct over $\overline{ \mathbb{F}_p }$ and $e_i = 1$. If $f(x) \bmod p$ has multiple irreducible factors, $p$ is again said to split; if $f(x) \bmod p$ remains irreducible, $p$ is again said to remain inert. When the Galois group of the splitting field of $f(x)$ is abelian, class field theory implies that the splitting behavior of a given prime can be predicted by its membership in certain residue classes just as in the case of quadratic extensions.

The nicest case beyond quadratic extensions here is that $f(x) = \Phi_n(x)$ is a cyclotomic polynomial, which corresponds to the ring $\mathbb{Z}[\zeta_n], \zeta_n = e^{ \frac{2\pi i}{n} }$, and which corresponds to the abelian Galois group $(\mathbb{Z}/n\mathbb{Z})^{\times}$. Here it is easy to see explicitly that the splitting behavior of a rational prime $p$ depends only on the residue of $p \bmod n$, as follows: if $(p, n) = 1$ (the non-ramified case) then $x^n - 1$ has distinct roots $\bmod p$, hence the roots of $\Phi_n(x)$ in $\overline{ \mathbb{F}_p }$ are precisely the elements of multiplicative order $n$. The Frobenius map $x \mapsto x^p$ then preserves order, which is another way to say that it acts on the roots of $\Phi_n(x)$. Each orbit has the same size, which is the smallest positive integer $f$ such that $p^f \equiv 1 \bmod n$ (the order of the Frobenius map), so $\Phi_n(x)$ splits up into the product of $\frac{\varphi(n)}{f}$ distinct irreducible factors of degree $f$, and $f$ depends only on the residue of $p \bmod n$. Note that $p$ remains inert if and only if $f = \varphi(n)$, which is the case if and only if $p$ is congruent to a primitive root $\bmod n$. But these exist if and only if $n = 1, 2, 4, q^k, 2q^k$ for $q$ an odd prime and $k \ge 1$, and if $n$ is not of this form then no rational primes remain inert.

When $p | n$ (the ramified case), write $n = p^k m$ where $(p, m) = 1$. Recall that

$\displaystyle \Phi_n(x) = \prod_{d | n} (x^{ \frac{n}{d} } - 1)^{\mu(d)}$,

which then gives

$\displaystyle \begin{array}{ccc} \Phi_{p^k m}(x) &=& \prod_{d | m} (x^{ \frac{n}{d} } - 1 )^{\mu(d)} (x^{ \frac{n}{pd} } - 1)^{\mu(pd)} \\ &=& \Phi_m(x^{p^k}) \Phi_m(x^{p^{k-1}})^{-1} \\ & \equiv & \Phi_m(x)^{p^k - p^{k-1}} \bmod p \\ \end{array}$

and we can factor $\Phi_m(x)$ by the same method as above. (Note that it is possible to determine the ramification behavior of primes without explicitly computing the discriminant of $\Phi_n(x)$.)

A historical note: if $n > 2$ is prime and $\mathbb{Z}[\zeta_n]$ has unique factorization, then it follows that the Fermat equation $x^n + y^n = z^n$ has no nontrivial integer solutions by factoring the LHS over $\mathbb{Z}[\zeta_n]$. Although it was not appreciated for quite a long time, unique factorization doesn’t always hold, but the first counterexample doesn’t appear until $n = 23$: Kummer showed that

$(1 + \zeta_{23}^2 + \zeta_{23}^4 + \zeta_{23}^5 + \zeta_{23}^6 + \zeta_{23}^{10} + \zeta_{23}^{11})(1 + \zeta_{23} + \zeta_{23}^5 + \zeta_{23}^6 + \zeta_{23}^7 + \zeta_{23}^9 + \zeta_{23}^{11})$

is divisible by $2$! Some people think Fermat incorrectly assumed that $\mathbb{Z}[\zeta_n]$ always has unique factorization in claiming a proof of his celebrated theorem. For what it’s worth, Kummer was able to successfully generalize from the case that $\mathbb{Z}[\zeta_n]$ is a UFD to the case that $n$ is a regular prime.

Smoothness

The ring $\mathbb{Z}[ \sqrt{-3} ]$ does not have unique factorization of elements, since $(1 + \sqrt{-3})(1 - \sqrt{-3}) = 2 \cdot 2$ and it can be shown that all of the elements involved are irreducible. Actually, the situation is even worse: this ring doesn’t have unique factorization of ideals, either. The ideal $(2, 1 + \sqrt{-3})$ is the only prime lying over $(2)$, which ramifies, but $(2)$ is not the product of two copies of this ideal. Instead,

$(2, 1 + \sqrt{-3})^2 = 2(2, 1 + \sqrt{-3})$

as you can check yourself. So ideals don’t have unique factorization either. However, $\mathbb{Z}[ \sqrt{-3} ]$ happens to be contained in the larger ring $\mathbb{Z}[\omega]$ where $\omega = \frac{1 + \sqrt{-3}}{2}$ is a primitive sixth root of unity, and adding $\omega$ turns out to repair unique factorization of ideals. In particular, $(2)$ no longer ramifies in $\mathbb{Z}[\omega]$; it is in fact inert. $\mathbb{Z}[\omega]$ even has a division algorithm, so like $\mathbb{Z}[i]$ it is even a UFD. These two cases are rather special because they are precisely the intersection of the quadratic case and the cyclotomic case ($n = 6, 4$ respectively) and we can read off the description of their prime ideals using either description.

The way to think about this situation geometrically is in terms of the inclusion $\mathbb{Z}[\sqrt{-3}] \to \mathbb{Z}[\omega]$, which gives a continuous map of spectra $\text{Spec } \mathbb{Z}[\omega] \to \text{Spec } \mathbb{Z}[ \sqrt{-3} ]$ in the other direction. It turns out that in the geometric analogy we should think of $\text{Spec } \mathbb{Z}[\omega]$ as a “smooth” curve and $\text{Spec } \mathbb{Z}[ \sqrt{-3} ]$ as a continuous image of this curve which is “not smooth” at the image of the prime $2$. This will be made precise in later posts.

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1. […] is an algebraically closed field, and it’s similar if is a field, but for it’s the arithmetic plane. This is a somewhat more complicated object, and it’s not immediately clear exactly what the […]