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We begin with a few definitions.

Definition 1: A integral domain $R$ is called a Dedekind domain if it is noetherian, every nonzero prime ideal is maximal, and it is integrally closed in its field of fractions.

Definition 2: A ring $R$ is called a discrete valuation ring (DVR) if it is a principal ideal domain (PID) with exactly one nonzero prime ideal. (In other language, a DVR is a local PID of Krull dimension 0 or 1.)

One very important property of Dedekind domains is that ideals have unique factorizations as products of prime ideals. I used this property in the case of rings of integers in my last post to say that if $L/K$ is an extension of number fields with rings of integers $B/A$, so if $\mathfrak{p}\subset A$ is a prime ideal, then we can write $\mathfrak{p}B=\prod_{i=1}^g \mathfrak{P}_i^{e_i}$. But this result holds in more generality, for any Dedekind domain.

Also, it is very easy to check that a DVR is a Dedekind domain. But one very common occurrence of DVRs is as localizations of rings of integers. In particular, if $A$ is a Dedekind domain and $\mathfrak{p}$ is a prime ideal of $A$, then $A_{\mathfrak{p}}$ is a DVR.

One way to interpret a DVR is through the following filtration of ideals. Let $R$ be a DVR, and let $\mathfrak{p}$ be the unique nonzero prime ideal of $R$. Then every nonzero ideal of $R$ is of the form $\mathfrak{p}^n$ for some $n\ge 0$ (where by $\mathfrak{p}^0$ I mean $R$). Now, for any $x\in R\setminus\{0\}$, there is an integer $n$ so that $x\in\mathfrak{p}^n\setminus\mathfrak{p}^{n+1}$. We can now define a function $v:R\setminus\{0\}\to\mathbb{N}$ (where $\mathbb{N}$ includes 0 in this case) by $v(r)=n$ as above. We can extend our definition of $v$ to all of $R$ by setting $v(0)=+\infty$.

It is also possible to extend $v$ to the quotient field $K$ of $R$ by setting $v(x/y)=v(x)-v(y)$; it is easy to check that this is well-defined. Now, $v$ satisfies the following properties:

1) $v:K\setminus\{0\}\to\mathbb{Z}$ is a surjective homomorphism.

2) $v(x+y)\ge\min(v(x),v(y)$.

We call such a function $v$ a valuation of the field $K$.

Knowing $v$ is enough to reconstruct $R$, since $R=\{x\in K:v(x)\ge 0\}$. Furthermore, $\mathfrak{p}=\{x\in K:v(x)\ge 1\}$. We call $R$ the valuation ring of $K$.

Let’s look at a few places where DVRs arise naturally.

1) As we mentioned earlier, the localization of a Dedekind domain at a prime ideal is a DVR. So, for example, $\mathbb{Z}_{(p)}=\{x/y\in\mathbb{Q}:p\nmid y\}$ is a DVR if $p$ is a prime. The unique prime ideal is $p\mathbb{Z}_{(p)}$.

2) The ring $\mathbb{Z}_p$ of $p$-adic integers is a DVR with unique prime ideal $p\mathbb{Z}_p$. Also, finite extensions of the field $\mathbb{Q}_p$ of $p$-adic numbers inherit valuations from $\mathbb{Q}_p$, and so they contain DVRs as described above. In particular, if $K/\mathbb{Q}_p$ is a finite field extension, then the integral closure of $\mathbb{Z}_p$ in $K$ is a DVR.

Now, if $R$ is a DVR and $\mathfrak{p}$ is its prime ideal, then $R/\mathfrak{p}$ is a field. In the cases described above, this will always be a finite field; in what follows, we always assume that this field is finite. We call $R/\mathfrak{p}$ the residue field.

We can also put a topology on a valued field $K$ by letting the following sets be a basis for the topology: if $x\in K$ and $n\ge 0$ is an integer, then $\{y\in K:v(x-y)\ge n\}$ is an open set. These sets generate the topology. In what follows, we will assume that $K$ is complete as a topological space with this topology. Finite extensions of $\mathbb{Q}_p$ are complete with respect to this topology, so this will be our motivating example. The residue fields will also be finite.

Last post, I pointed out that if $L/K$ is a Galois extension of number fields, then $efg=[L:K]$. This holds more generally, however. If $L/K$ is a finite Galois field extension, and $A\subset K$ is a Dedekind domain so that $K$ is the quotient field of $A$, and $B$ is the integral closure of $A$ in $L$, then we still have $efg=[L:K]$.

We now interpret this in the case of $K$ a field complete with respect to a discrete valuation $v$, and $A$ the valuation ring of $K$. Let $L/K$ be a finite Galois extension, and let $B$ be the integral closure of $A$ in $L$, or, equivalently, the valuation ring of $L$. Then $L$ is also complete with respect to a discrete valuation $w$ that is very closely related to $v$, as we will see soon.

Let $\mathfrak{p}$ be the prime of $A$, and let $\mathfrak{P}$ be the prime of $B$. Since there is only one prime, $g=1$. Hence $\mathfrak{p}B=\mathfrak{P}^e$. Now, if $x\in K$, then $w(x)=ev(x)$, and if $x\in L$, then $fw(x)=v(N_{L/K}x)$. (But we won’t need these results in what follows, at least today.) The implication is the decomposition group of the extension $L/K$ is the entire Galois group.

We can put a filtration on the Galois group as follows: For $i\ge -1$, let $G_i=\{\sigma\in Gal(L/K):w(\sigma(x)-x)\ge i+1 \hbox{ for all } x\in L\}$. We call $G_i$ the $i^{\hbox{th}}$ ramification group of $L/K$. $G_{-1}=G$ is the entire Galois group (or the decomposition group; $G_0$ is the inertia group. Also, each $G_i$ is normal in $G$.

Now, I won’t prove it here, but it can be shown that if the residue field is finite of characteristic $p>0$ and $K$ is complete, then for each $i\ge 1$, $G_i/G_{i+1}$ is a direct product of copies of $\mathbb{Z}/p\mathbb{Z}$, and $G_0/G_1$ is a subgroup of the roots of unity of $B/\mathfrak{P}$ (and hence finite and cyclic of order prime to $p$). Hence, by basic group theory or otherwise, $G_0$ is a semidirect product of a normal Sylow $p$-subgroup and a cyclic group of order prime to $p$. In particular, $G_0$ is solvable. However, as shown in the last post, $G/G_0\cong Gal((B/\mathfrak{P})/(A/\mathfrak{p}))$ is cyclic since it is the Galois group of an extension of finite fields. Hence:

Theorem: $G$ is solvable.

I have been reading Joseph Silverman’s new book on arithmetic dynamics lately. There’s a lot of really fascinating stuff in there, including a large number of potential research problems that are currently way beyond me, but I’ll continue thinking about them! Most interesting so far is the Uniform Boundedness Conjecture:

Let $d\ge 2$, $N\ge 1$, and $D\ge 1$ be integers. Then there exists a constant $C=C(d,N,D)$ such that for any number field $K$ with $D \ge [ K : \mathbb{Q} ]$ and any morphism $\phi:\mathbb{P}^N(K)\to\mathbb{P}^N(K)$ of degree $d$, the number of preperiodic points of $\phi$ in $\mathbb{P}^N(K)$ is at most $C$.

Not much is known about this conjectures; even the case $d=2$, $N=1$, and $D=1$ is open. It’s even open if we restrict to morphisms of the form $\phi_c(z)=z^2+c$. Bjorn Poonen has shown, however, that these maps have no rational periodic points of exact period 4 or 5; it is conjectured that they have no rational periodic points of exact period greater than 3.

However, there is a positive result of the above type that doesn’t depend that much on some of the above quantities:

Let $K$ be a number field and $\phi:\mathbb{P}^1\to\mathbb{P}^1$ be a rational map over $K$. Let $\mathfrak{p}$ and $\mathfrak{q}$ be prime ideals of $\mathfrak{o}_K$ so that $\phi$ has good reduction at $\mathfrak{p}$ and $\mathfrak{q}$ (meaning that when we reduce $\phi$ modulo $\mathfrak{p}$ and $\mathfrak{q}$, we end up with a map $\tilde\phi$ of the same degree as $\phi$) and whose residue characteristics are distinct. Then the period $n$ of any periodic point of $\phi$ in $\mathbb{P}^1(K)$ satisfies $n\le (N\mathfrak{p}^2-1)(N\mathfrak{q}^2-1)$, where $N$ denotes the (absolute) norm.

(See, for instance, my algebraic number theory notes for definitions of some of these terms.)

Anyway, that wasn’t really the point of this post, as you may have guessed from the title. I meant to talk about theorems that pretend not to be related to dynamical systems but actually are. First we need to discuss height functions a bit; there’s a lot more about them in Silverman’s book and in my elliptic curve notes.

We let $K$ be a number field and $M_K$ the set of standard absolute values on $K$ (These are the absolute values on $K$ whose restriction to $\mathbb{Q}$ is either the standard absolute value or one of the $p$-adic absolute values.) We write $n_v=[K_v:\mathbb{Q}_v]$ (where $F_v$ denotes the completion of $F$ with respect to the absolute value $v$). Suppose $P\in\mathbb{P}^N(K)$; we can then write $P=[x_0,\ldots,x_N]$ for some $x_0,\ldots,x_N\in K$. We then define the height of $P$ with respect to $K$ to be $H_K(P)=\prod_{v\in M_K} \max\{|x_0|_v,\ldots,|x_N|_v\}^{n_v}$. One can check that this is well-defined, and that if $L/K$ is a finite extension of number fields and $P\in K$, then $H_L(P)=H_K(P)^{[L:K]}$. Hence it is possible to define the absolute height of $P$ by $H(P)=H_K(P)^{1/[K:\mathbb{Q}]}$.

One of the key facts about heights is the following: If $B$ and $D$ are constants, then $\{P\in\mathbb{P}^N(\overline{\mathbb{Q}}):H(P)\le B \text{ and } [\mathbb{Q}(P):\mathbb{Q}]\le D\}$ is finite. A corollary is the following well-known result of Kronecker:

Let $\alpha\in\overline{\mathbb{Q}}$ be nonzero. Then $H(\alpha)=1$ if and only if $\alpha$ is a root of unity.

Proof: If $\alpha$ is a root of unity, then $H(\alpha)=1$ is clear. Now suppose that $H(\alpha)=1$. For any $\beta$ and $n$, we have $H(\beta^n)=H(\beta)^n$, so $H(\alpha^n)=H(\alpha)^n=1$, so $\{\alpha,\alpha^2,\alpha^3,\ldots\}$ is a set of bounded height and is therefore finite. Therefore there are integers $i>j>0$ such that $\alpha^i=\alpha^j$, so $\alpha$ is a root of unity.

And now for linear algebra. Sheldon Axler has a well-known book on linear algebra without determinants. He therefore uses dynamical systems to show the following familiar result:

Theorem: Every operator on a finite-dimensional nonzero complex vector space has an eigenvalue.

Proof: Let $V$ be such a vector space of dimension $n$. Let $T$ be an operator on $V$, and let $v\in V$ be nonzero. Then $\{v,Tv,T^2v,\ldots,T^nv\}$ cannot be a linearly independent set. Hence there exist $a_0,\ldots,a_n\in\mathbb{C}$, not all zero, so that $a_0v+a_1Tv+\cdots+a_nT^nv=0$. Suppose $m$ is maximal with respect to $a_m\neq 0$. Then $a_0v+a_1Tv+\cdots+a_mT^mv=0$. Since we’re working over $\mathbb{C}$, the polynomial $a_0+a_1z+\cdots+a_mz^m$ factors as $a_m(z-\lambda_1)\cdots(z-\lambda_m)$. We then have $0=a_0v+a_1Tv+\cdots+a_mT^mv=a_m(T-\lambda_1I)\cdots(T-\lambda_mI)v$. Hence some $T-\lambda_jI$ is not injective. This $\lambda_j$ is an eigenvalue for $T$.