The Algebra and Geometry of $\mathbb Q_p$

In Absolute Values and Completions of $\mathbb Q$ we looked at the completions of $\mathbb Q$, and in particular the non-archimedean completions $\mathbb Q_p$, from the viewpoint of analysis and topology. Here we investigate the algebraic and geometric properties of the $p$-adic numbers, though we will not ignore the topology completely.

The archimedean property of the real numbers asserts that for any real number $x$ there is an integer $n$ such that $| x – n |_{\infty} \leq 1/2$. That is, every real number is at most $1/2$ unit away from its closest integer. We contrast this with $\mathbb Q_p$ where $| n |_p \leq 1$ for all $n \in \mathbb Z$, and where we can find $x \in \mathbb Q_p$ with $|x|_p$ as large (or small) as we want. Suffice it to say $\mathbb Q_p$ does not have the archimedean property, hence the adjective non-archimedean for such fields.

The completion of $\mathbb Z$ in $\mathbb Q_p$ is denoted $\mathbb Z_p$ and explicitly given by $$\mathbb Z_p = \{ x \in \mathbb Q_p : |x|_p \leq 1 \}.$$ Put another way, $p$-adic integers are equal to the closed unit ball in $\mathbb Q_p$.

It is easy to verify that $\mathbb Z_p$ is a ring; the only property in doubt is closure under addition, but this comes from the strong triangle inequality. If $x, y \in \mathbb Z_p$ then $$|x+y|_p \leq \max\{ |x|_p, |y|_p\} = 1.$$ This same argument shows that $\mathfrak m_p = \{ x : |x|_{p} < 1 \}$ is itself a ring—in fact a maximal ideal—of $\mathbb Z_p$. We also define $U_p = \{ x : |x|_p = 1\}$. Note that $U_p$ is not a ring, but it is a group under multiplication: the group of units in $\mathbb Z_p$. We note that the definition of $\mathbb Q_p, \mathbb Z_p, \mathfrak m_p$ and $U_p$ are invariant under substitution of an equivalent absolute value. That is, the $p$ that indexes these sets is associated to the place indexed by $p$ not the specific choice of absolute value $| \cdot |_p$.

Commutative rings and maximal ideals quotient to make fields, and we define the residue field of $\mathbb Q_p$ to be $\mathbb F_p \cong \mathbb Z_p/\mathfrak m_p$, which as the notation suggests is the finite field with $p$ elements. We will prove this in the next section.

An explicit construction of $\mathbb Q_p$

We have defined the $p$-adic numbers as equivalence classes of Cauchy sequences. It is useful to have a proscribed choice of representative for each equivalence class. This is done using series. Consider the formal series $$ \sum_{m=v}^{\infty} a_m p^m$$ where each $a_m \in \{0,1,\ldots,p-1\}$. The $M$th partial sum is a rational number $$n_M = \sum_{m=v}^M a_m p^m,$$ and $|n_M – n_{M-1}|_p \leq p^{-M}$. In fact, $$|n_{M + n} – n_M|_p \leq \max_{\ell=1,\ldots n}\{|n_{M+\ell} – n_{M+\ell-1}|_p\} \leq p^{-M}$$ and hence $(n_M)$ is Cauchy with respect to $| \cdot |_p$. It follows that $$ x = \lim n_M = \sum_{m=v}^{\infty} a_m p^m$$ defines a $p$-adic number with $|x|_p \leq p^{-v}$. Note that when $x$ is a positive rational integer, its series representation has finitely many terms (indexed starting at 0) and is simply its base-$p$ expansion.

So every power series of this form produces a $p$-adic number. What about the converse? Given a $p$-adic number, can we find a representative given as a “base $p$ series”?

Theorem

Suppose $x \in \mathbb Q_p$ then there exists an integer $v$ and a sequence of integers $(a_n)_{n=v}^{\infty}$ with each $0 \leq a_n < p$ such that $x$ is represented by the sequence of partial sums of $$ \sum_{n=v}^{\infty} a_n p^n.$$ Moreover each such series defines an element of $\mathbb Q_p$.

Proof

We first do this for sequences of integers. Let $(\ell_m)$ be a sequence of integers Cauchy with respect to $| \cdot |_p$. We may assume, by taking a subsequence if necessary that $|\ell_m – \ell_{m-1}|_p \leq p^{-m}$. We make a new sequence of integers $(n_m)$ by taking the base-$p$ expansion of $\ell_m$ and truncating it at the $m$th term. That is, if $\ell_m = \sum_{j=0}^J a_j p^j$ then $n_m = \sum_{j=0}^m a_j p^j$. We note that $|\ell_m – n_m| \leq p^{-m-1}$, and that \begin{align}|n_m – n_{m-1}|_p &= |n_m – \ell_m + \ell_m – \ell_{m-1} + \ell_{m-1} – n_{m-1}|_p \\ &\leq \max\{|n_m – \ell_m|_p, |\ell_m – \ell_{m-1}|_p, |\ell_{m-1} – n_{m-1}| \} \leq p^{-m}.\end{align} All this is to say that $(\ell_m)$ and $(n_m)$ are equivalent Cauchy sequences. It remains to show that $(n_m)$ is the sequence of partial sums of an infinite base-$p$ expansion. Currently we know that $n_m$ is a polynomial in $p$ of degree $m$ with coefficients in $\{0, 1, \ldots, p-1\}$, but we don’t know if these coefficients agree with those (of degree $<m$) of $n_{m-1}$ for all $m$. That is we need, to verify that for each $m$, $n_m = n_{m-1} + a_{m} p^m$ for some $a_m \in \{0, 1, \ldots, p-1\}$. Because $|n_m – n_{m-1}|_p \leq p^{-m}$ we know that $n_m = n_{m-1} + A_m p^m$ for some integer $A_m$. We may replace $A_m$ with $a_m \in \{0, 1, \ldots, p-1\}$ congruent modulo $p$, and by replacing $n_m$ with $n_{m-1} + a_m p^m$ (if necessary), we find that the series $(n_m)$ is the sequence of partial sums of $$\sum_{m=0}^{\infty} a_m p^m$$ as desired.

We are almost done, we now simply need to show that if we have any rational sequence $(r_m)$ Cauchy with respect to $| \cdot |_p$ that it can be represented by the partial sums of an infinite base-$p$ expansion of the form $$\sum_{m=v}^{\infty} a_m p^m$$ for some $v < 0$. We could bust out the previous analysis, but here we remark that there is some largest (least negative) integer $v$ such that $p^{v} \mathbb Z_p$ contains $(r_n)$. That is $x_m = p^{-v} r_m$ defines a sequence such that $|x_m|_p \leq 1$. Scaling by $p^{-v}$ is continuous, and so $(x_m)$ is a Cauchy sequence. We define $y_m$ to be the degree $m$ truncation of the base-$p$ expansion of $x_n$. This is exactly what we did before by defining $n_m$ in terms of the $\ell_m$ except here we do not know that the $x_m$ are integers—however we just proved that they have (possibly infinite) base-$p$ expansions because they are all in the closure of the integers—this can be truncated to produce the $y_m$. Regardless all the analysis works and we find that $(y_m)$ is equivalent to $(x_m)$ and is the sequence of partial sums of some $$\sum_{m=0}^{\infty} b_m p^m.$$ We then define $$s_m = p^v \sum_{m=0}^{\infty} b_m p^m.$$ Scaling is still continuous, and so $(s_n)$ is Cauchy and equivalent to $(r_n)$ and is the sequence of partial sums of an infinite base-$p$ expansion (allowing for finitely many negative powers of $p$).

So far we have only constructed positive numbers. Negative numbers can be represented in base-$p$ expansion. In particular, $$ -1 = \sum_{n=0}^{\infty} (p-1) p^n.$$ To compute the negative of a generic number $x$ we simply compute $0-x$ base-$p$.

Example

The base-5 expansion of 429 is $429 = 4 \cdot 5^0 + 0 \cdot 5^1 + 2 \cdot 5^2 + 3 \cdot 5^3$. To compute $-429$ we wish to add powers of 5 that always cause us to “carry the one”.

We find $-429 = (1 \cdot 5^0 + 4 \cdot 5^1 + 2 \cdot 5^2 + 1 \cdot 5^3) + 4 \cdot 5^4 + 4 \cdot 5^5 + \cdots$.

We return to our claim about $\mathbb Z_p/ \mathfrak m_p$.

Theorem

$\mathbb Z_p/ \mathfrak m_p$ is isomorphic to the field with $p$ elements.

The proof is now obvious because $\mathfrak m_p = p \mathbb Z_p$ and thus two base-$p$ expansions are the same modulo $\mathfrak m_p$ if and only if they have the same constant coefficient, and hence $\mathbb Z_p/\mathfrak m_p$ is a field with $p$ elements. It is easy to see that, in fact $p^n \mathbb Z_p/p^{n+1} \mathbb Z_p \cong \mathbb F_p$ (for all $n \in \mathbb Z$).

The Geometric Picture of $\mathbb Z_p$

Here we want to think of the coefficients of $x = \sum_{n=0}^{\infty} a_n p^n \in \mathbb Z_p$ not as coefficients of a power series, but as an address. Imagine driving in a strange town of one-way roads, where at each intersection you have $p$ choices of roads ahead of you (numbered in some consistent way using $0,1, \ldots, p-1$). Then by telling you a sequence of numbers $(a_n)$ with $a_n \in \{0, 1, \ldots, p-1\}$ I am giving you instructions to an address at the end of an infinite sequence of roads.

Six Corners in Chicago. Coming into this intersection along the red arrow, there are five choices, labelled $0, 1,2,3,4$ leaving the intersection. This choice of labelling will be convenient for visualizing elements in $\mathbb Z_5$.

This analogy is not very apt, because we allow no loops in our strange city, but the point remains: we may think of the $(a_n)$ as an “address” for $x = \sum a_n p^n \in \mathbb Z_p$. Each $x \in \mathbb Z_p$ has a unique address, and we may visualize the network of roads as a complete, infinite $p$-nary tree.

Schematic of the “roads” in $\mathbb Z_5$ and $\mathbb Z_3$. These are different embeddings of the complete $5$-nary and $3$-nary trees. The points in $\mathbb Z_p$ are the “boundary” of these trees.

What if we drive only part way to an address? Suppose we start down the roads labelled $(3, 4, 2)$ in $\mathbb Z_5$. This finite tuple then gives us the address of a neighborhood—that consisting of all infinite addresses that start $(3, 4, 2, \ldots )$. Note that $3 + 4 \cdot 5 + 2 \cdot 25 = 73$, and so we can think of this neighborhood as the ball of radius $1/125$ around $73$.

The “roads” to $0, 1$ and $-1$ in $\mathbb Z_5$. Slide left to see the road to the neighborhood defined by the finite tuple $(3,4,2)$. Where is $73$ in that neighborhood?

The positive rational integers can be seen inside $\mathbb Z_p$ as the destination of itineraries which eventually have no turns to the left or right. Negative integers follow itineraries that eventually have a clockwise spiral like that of $-1$. In either event we see visually how $\mathbb Z$ (and indeed $\mathbb N$ and $-\mathbb N$ individually) are dense in $\mathbb Z_p$.

A Bijection Between Balls and Cosets

There is a bijection between the balls in $\mathbb Z_p$ of radius $p^{-n}$ and the cosets of $\mathbb Z_p / p^n \mathbb Z_p$. In the schematic for $\mathbb Z_5$ we may think of a ball as one of the naturally appearing pentagons (of any size). This is not quite right, the ball is actually the fractal bits of the boundary of the tree contained in such a pentagon. In general for $\mathbb Z_p$ there would be a similar schematic with the pentagons (and their fractal tree boundaries) replaced with $p$-gons.

The correspondence between balls and cosets when $p = 5$ and $n=1,2$.

This allows us to index balls of radius $p^{-n}$ by the integers $\{0, 1, \ldots, p^n-1\}$. For instance, the neighborhood in $\mathbb Z_5$ indexed by $(3,4,2)$ is exactly the coset $73 + 125 \mathbb Z_5$.

$\mathbb Z_p$ as a Pro-finite Completion

Another way of specifying the directions to a point $x \in \mathbb Z_p$ is to record the neighborhoods one passes through on the way to $x$. By the correlation between neighborhoods and cosets we can identify that point in $\mathbb Z_p$ with a sequence of cosets $(c_n)$ with $c_n \in \mathbb Z_p/p^n \mathbb Z_p$ represented as integers $0 \leq c_n < p^n$. The relationship between $(c_n)$ and the coefficients $(a_m)$ of the base-$p$ expansion of $x$ is $$c_n = \sum_{m=0}^n a_m p^m.$$ Notice the congruence relations $$c_n \equiv c_{n-1} \bmod p^{\ell}, \qquad \ell=1,\ldots,n.$$ More generally, there is an inverse system of projections $\pi_{\ell \leftarrow n}: \mathbb Z/p^n \mathbb Z_p \rightarrow \mathbb Z/p^{\ell} \mathbb Z_p$, for $0 \leq \ell \leq n$, such that for any $\ell \leq m \leq n$, $\pi_{\ell \leftarrow n} = \pi_{\ell \leftarrow m} \circ \pi_{m \leftarrow n}$. We may thus identify $\mathbb Z_p$ with $$\lim_{\leftarrow n} \mathbb Z/p^n \mathbb Z := \left\{ (c_n) \in \prod_{n=1}^{\infty} \mathbb Z/p^n \mathbb Z : c_{\ell} = \pi_{\ell \leftarrow n}(c_n) \mbox{ for all } 1 \leq \ell \leq n \right\}.$$ This set is exactly the pro-finite completion of the inverse system.

 

A geometric representation of the profinite completion view of the $3$-adic integers. Elements in the $\mathbb Z_3$ are not indexed by the “roads” that you take to get there, but by the cosets aka neighborhoods (here represented as disks) you must pass through to get to your destination. The destination is the same, but the data you use to get there is (only slightly) different.

Addition and multiplication in the pro-finite completion are done component wise, and sums and product remain in $\lim_{\leftarrow } \mathbb Z/p^n \mathbb Z$ because the $\pi_{\ell \leftarrow n}$ are ring homomorphisms. These operations are the same addition and multiplication that comes from the base-$p$ expansion representation of $\mathbb Z_p$.

0

Field Extensions and Number Fields

Here I am storing various basic facts about Number Fields that are useful in other notes. I hope this becomes more complete as time goes on.

Number Fields

Recall that a number field $K$ is a finite extension of $\mathbb Q$. While we often think of number fields as $\mathbb Q(\alpha)$ for some algebraic number embedded in $\mathbb C$ it is useful to recall the general (unembedded) construction. $\mathbb Q[x]$ is the ring of polynomials with rational coefficients in the indeterminant $x$. If $f(x) \in \mathbb Q[x]$ is irreducible, then $f(x) \mathbb Q[x]$, the ideal formed from all rational polynomials divisible by $f(x)$, is a maximal ideal in $\mathbb Q[x]$. It follows that $K = \mathbb Q[x]/f(x) \mathbb Q[x]$ is a commutative ring with all non-zero elements invertible—that is a field.

In this construction, the elements of $K$ are cosets of the form $g(x) + f(x) \mathbb Q[x]$. If $g(x)$ and $h(x)$ generate the same coset, then we will write $g(x) \equiv h(x)$ (or $g(x) \equiv h(x) \bmod f(x)$ if more clarity is necessary). In this situation $f(x) | (g(x) – h(x))$.

Given the coefficients of $f(x)$, the arithmetic in $K$ is easy to perform. Suppose for $a_0, \ldots, a_{d-1}$ are the rational coefficients to $$f(x) = x^d + \sum_{n=0}^{d-1} a_n x^n,$$ then, $$ x^d \equiv -a_0 – a_1 x – \cdots – a_{d-1} x^{d-1}.$$ Now suppose $g(x) + f(x) \mathbb Q[x]$ is an arbitrary coset. By replacing monomials $x^n$ in $g(x)$ when $n > d$ (serially, if necessary) using this congruence, we see that $g(x) \equiv h(x)$ for some $h(x) \in \mathbb Q[x]$ with $\deg(g) < d$. The polynomial $h(x)$ is equivalent to the result of the Division Algorithm in $Q[x]$ for the remainder of $g(x)$ when divided by $f(x)$.

That is, as a group (in fact, as a vector space) $K$ is isomorphic to $\mathbb Q^d$ where the isomorphism is given by $$(b_0, \ldots, b_{d-1}) \mapsto x^d + \sum_{m=0}^{d-1} b_m x^m + f(x) \mathbb Q[x].$$ The only thing missing in this description is the multiplication. If we want to multiply two vectors $\mathbf b, \mathbf c \in \mathbb Q^d$, we set $g(x)$ to be the monic polynomial with coefficient vector $\mathbf b$ and $h(x)$ to be the polynomial with coefficient vector $\mathbf c$. We first multiply $g(x)$ and $h(x)$ as usual in $\mathbb Q[x]$, and then we use the equivalence $ x^d \equiv -a_0 – a_1 x – \cdots – a_{d-1} x^{d-1}$ to replace monomials in $g(x) h(x)$ (repeatedly if necessary) until we arrive at a polynomial $p(x)$ of degree $< d$. The coefficient vector of this polynomial in $\mathbb Q^d$ is the product of $\mathbf b$ and $\mathbf c$.

$K$, What is it Good for?

First, note that $\mathbb Q \hookrightarrow K$ by the map $r \mapsto r + f(x) \mathbb Q[x]$, and by definition (the fact that $K$ is a vector space of dimension $d$ over $\mathbb Q$) it is a number field of degree $d$ over $\mathbb Q$. This implies $\mathbb Q[x] \hookrightarrow K[x]$, and in particular, $f(x)$ has a life in $K[x]$. Because $f(x)$ is irreducible in $\mathbb Q[x]$ it has no zeroes in $\mathbb Q$. However, we will show that this is no longer the case in $K[x]$. And that is what $K$ is good for—producing a number field where $f(x)$ has a zero.

The element $x + f(x) \mathbb Q[x]$ is the root of $f(x)$ in $K$. To see this, we need only calculate $$f(x + f(x) \mathbb Q[x]) = f(x) + f(x) \mathbb Q[x] = 0 + f(x) \mathbb Q[x].$$

The element $x + f(x) \mathbb Q[x]$ is important as well because if we know how to multiply by this element, then we know how to multiply by arbitrary elements (which are, after all, simply linear combinations of its powers).

Multiplication by $x$ is a linear operator on $\mathbb Q[x]$, indeed $x( a g(x) + h(x) ) = a x g(x) + x h(x)$, and multiplication by $x + f(x) \mathbb Q[x]$ is a linear operator on $K$. We know $K$ is a vector space with basis $( x^n + f(x) \mathbb Q[x] : n=0,\ldots, d-1)$, so it makes sense to talk of the matrix of the multiplication operator, call it $T$, with respect to this basis. Note that, if we denote the standard basis of $\mathbb Q^d$ (with coordinates indexed from 0 to $d-1$ for consistency) by $\mathbf e_0, \ldots, \mathbf e_{d-1}$, then for $n < d-1$, $T \mathbf e_{n} = \mathbf e_{n+1}$. This corresponds to the multiplication $x x^{n} = x^{n+1}$ which remains true in $K$ if $n < d-1$. The final calculation, using the same equivalence that has gotten us so far $ x^d \equiv -a_0 – a_1 x – \cdots – a_{d-1} x^{d-1}$, shows that $T \mathbf e_{d-1} = -a_0 \mathbf e_0 – a_1 \mathbf e_1 – \cdots – a_{d-1} \mathbf e_{d-1}$. It follows that the matrix of $T$ with respect to the basis $(\mathbf e_n)$ is $$ \begin{pmatrix} 0 & 0 & \cdots & 0 & -a_0 \\ 1 & 0 & \cdots & 0 & – a_1 \\ 0 & 1 & \cdots & 0 & – a_2 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & -a_{d-1}\end{pmatrix}.$$

If this matrix looks familiar it is because it is the (Frobenius) companion matrix to $f(x)$ and the characteristic polynomial of this matrix (and hence the operator $T$) is $f(x)$. Indeed, the irreducibility of $f(x)$ implies that the minimal polynomial of $T$ is $f(x)$ as well.

But $f(x)$ has roots in $\mathbb C$. What about them?

The Fundamental Theorem of Algebra (ironically a theorem in analysis) guarantees that $f(x)$ has $d$ roots (counting multiplicity) in $\mathbb C$. How are they related to the root of $f(x)$ is $K$?

Let’s start with our favorite $\alpha \in \mathbb C$ such that $f(\alpha) = 0$. We know that $\alpha$ is either in $\mathbb R$ or it has a complex conjugate—more about that later. We can embed $K$ into $\mathbb C$ by sending $x + f(x) \mathbb Q[x] \rightarrow \alpha$. That is, $$ a_{d-1} x^{d-1} + \cdots + a_1 x + a_0 + f(x) \mathbb Q[x] \quad \mapsto \quad a^{d-1} \alpha^{d-1} + \cdots a_1 \alpha + a_0.$$ We denote this embedding by $\mathbb Q(\alpha) \subset \mathbb C$. Notice that if $\alpha \in \mathbb R$, then $\mathbb Q(\alpha) \subset \mathbb R$ and we call it a real embedding of $K$.

A count of the real and complex embeddings products the first of the classical invariants of a number field.

Invariants: Number of real and complex embeddings $r_1$ and $r_2$

Let us distinguish the real and complex roots of $f(x)$ by setting $\alpha_1, \ldots, \alpha_{r_1}$ to be the real roots and $\beta_1, \overline{\beta_1}, \ldots, \beta_{r_2}, \overline{\beta_{r_2}}$ be the non-real complex roots. Clearly $r_1 + r_2 = d$. Then the embeddings $\mathbb Q(\alpha_1), \ldots, \mathbb Q(\alpha_{r_1}) , \mathbb Q(\beta_1) , \ldots, \mathbb Q(\beta_{r_2})$ are called the archimedean embeddings of $K$.

The Norm and Trace

Here we wish to work in some generality and consider field extension $K | k$ where both are number fields. Little generality is lost by keeping the example $k = \mathbb Q$ at the front of your mind. However, as many properties of number fields ‘factor through’ intermediate fields (for instance $[ K : \mathbb Q] = [K : k] [k : \mathbb Q]$) it is useful to maintain some generality in notation etc.

We will also abandon our attempt to denote elements of $K | k$ as cosets in $k[x] / f(x) k[x]$, writing for instance $\alpha, \beta, \gamma, \ldots $ for generic field elements. Often we will implicitly identify $K$ with $k(\alpha)$ for some algebraic number $\alpha$ of degree $d$ over $k$. In this situation $\{1, \alpha, \ldots, \alpha^{d-1} \}$ is a basis for $K$, and the matrix of multiplication by $\alpha$ with respect to this basis is exactly the matrix of $T$ as before (the Frobenius companion matrix of the minimal polynomial of $K | k$).

More generally, given any $\gamma \in K$, we can make the linear operator “multiplication by $\gamma$” $T_{\gamma}$. If $\gamma$ is given as a $k$-linear combination of $\{1, \alpha, \ldots, \alpha^{d-1}\}$ then it is relatively easy to compute the matrix of $T_{\gamma}$ with respect to this basis. Note that this matrix has entries in $k$.

Norm and trace of $K|k$

The norm $N_{K|k} : K \rightarrow k$ and trace $\mathrm{Tr}_{K|k} : K \rightarrow k$ are respectively the determinant and trace of $T_{\gamma}$.

This definition is independent of basis, but can be computed explicitly in the basis $\{1, \alpha, \ldots, \alpha^{d-1}\}$.

If $\beta, \gamma \in K$, then $T_{\beta \gamma} =T_{\beta} \circ T_{\gamma}$ and $T_{\beta + \gamma} = T_{\beta} + T_{\gamma}$. The multiplicativity of the determinant and the additivity of the trace imply that $$N_{K|k}(\beta \gamma) = N_{K|k}(\beta) N_{K|k}(\gamma)$$ and $$\mathrm{Tr}_{K|k}(\beta + \gamma) = \mathrm{Tr}_{K|k}(\beta) + \mathrm{Tr}_{K|k}(\gamma).$$

The norm is a natural homomorphism from $K^{\times}$ onto $k^{\times}$ and the trace is a natural homomorphism from the additive group $(K, +)$ onto $(k,+)$.

Notation

Number fields
$f(x), g(x), h(x),$ etc.Polynomials, often in $\mathbb Q[x]$ or $k[x]$
$k, K, L$Number fields
$\alpha, \beta, \gamma,$ etcGeneric field elements. The field depends on context.
$[K : k]$, $d$The degree of a field extension. The fields depend on context.
$T_{\alpha}$The linear transformation on $K|k$ (fields context dependent) given by multiplication by $\alpha$.
$r_1, r_2$The number of real and complex embeddings (respectively) of $K$ (context dependent).
$N_{K|k}$, $\mathrm{Tr}_{K|k}$The Norm and Trace maps $K \rightarrow k$ given by $\alpha \mapsto \det( T_{\alpha})$ and $\alpha \mapsto \mathrm{Tr}( T_{\alpha})$.
$\mf o$, $\mf O$Rings of integers in $k$ and $K$.
$\mf a, \mf b, \mf A, \mf B,$ etc.Ideals in rings of integers. We often use lower case fraktur letters for ideals in $\mf o$ and capital fraktur letters for ideals in $\mf O$.
$\mf p, \mf q, \mf P, \mf Q$ Prime ideals in $\mf o$ and $\mf O$.
$\mathbb N \mf a, etc$ etcThe ideal norm $\mathbb N \mf a = [\mf o : \mf a]$.
0

Recalling Galois Theory

This is a brief reminder of the main ideas of Galois theory. Any proofs purported here are meant to be suggestive. I learned Galois theory out of Dummit and Foote, which I thought was pretty good. I also have Classical Galois Theory by Gaal on my shelf. This book is in essence one giant worksheet. I have not completed many of the exercises, but I suspect anyone who did would gain a remarkable intuition as to how the theory hangs together.

At any rate, this is mostly for me, since I seem to need to be reminded of the basics of Galois theory every few years.

It may be useful to review Field Extensions and Number Fields before continuing.

Automorphisms of Field Extensions

We want to work in a bit of generality here, so we assume $K | k$ is an extension of number fields. Little generality is lost at this point if you take $k = \mathbb Q$.

Definition

An isomorphism of K is called an automorphism. The set of automorphisms of $K,$ $\mathrm{Aut}(K)$, forms a group under composition. An automorphism σ is said to fix $k$ if $\sigma \gamma=\gamma$ for all $\gamma \in k$. The set of automorphisms of $K$ which fix $k$ is denoted $\mathrm{Aut}(K|k)$ and is a subgroup of $\mathrm{Aut}(K)$.

$\mathrm{Aut}(K|k)$ is a finite group of degree at most $d = [K:k]$. We will use this fact, though the proof would take us too far afield.

Automorphisms of $K$ which preserve $k$ permute roots of polynomials with coefficients in $k$ and roots in $K$.

Proposition

Suppose $g(x) \in k[x]$ is irreducible and there exists $\beta \in K$ such that $g(\beta) = 0$. Then $g( \sigma \beta) = 0 \quad \mbox{for all} \quad \sigma \in \mathrm{Aut}(K|k)$.

Proof

Suppose $g(x) = \sum_m b_m x^m$, then $\sigma b_m = b_m$, and hence $$0= \sigma g(\beta) = \sum_m \sigma b_m (\sigma \beta)^m = \sum_m b_m (\sigma \beta)^m = g(\sigma \beta). \qquad \square$$

One particularly important automorphism is complex conjugation. Suppose that $K | \mathbb Q$ is a number field, and $K \cong \mathbb Q(\alpha)$ for some non-real $\alpha$ (that is the minimal polynomial of $K$ has a non-real root in $\mathbb C$). Then, since complex conjugation is an automorphism of $\mathbb C | \mathbb R$, we have it is also an isomorphism on $K(\alpha) | \mathbb Q$. It follows that if $\beta \in \mathbb Q(\alpha)$ then $\overline \beta \in \mathbb Q(\alpha)$ as well and hence, as sets, $\mathbb Q(\alpha)=\mathbb Q(\overline \alpha)$, and algebraic operations in one of these embeddings can be found from the other by complex conjugation.

Let us distinguish the real and complex roots of $f(x)$ by setting $\alpha_1, \ldots, \alpha_{r_1}$ to be the real roots and $\beta_1, \overline{\beta_1}, \ldots, \beta_{r_2}, \overline{\beta_{r_2}}$ be the non-real complex roots. Clearly $r_1 + r_2 = d$. Then the embeddings $\mathbb Q(\alpha_1), \ldots, \mathbb Q(\alpha_{r_1}) , \mathbb Q(\beta_1) , \ldots, \mathbb Q(\beta_{r_2})$ are called the archimedean embeddings of $K$.

Splitting Fields

Galois theory is concerned about the zeros of rational polynomials and how their zeroes are permuted by the automorphisms of certain extensions of $\mathbb Q$ (which will come to be called Galois extensions). We already noted, that the automorphisms in $\mathrm{Aut}(K|k)$ preserve the set of zeroes in $K$ of any given polynomial $g(x) \in k[x]$. However, the construction $K$ makes no guarantee that a generic polynomial $g(x)$ will have a zero in $K$, and even for the minimal polynomial $f(x)$, the construction of $K$ only guarantees the existence of a single zero of $f(x)$.

In general, if we wanted an extension of $k$ that contains all the zeros of $f(x)$, we would first compute $K = k[x]/ f(x) \mathbb Q[x]$. $K$ contains at least one zero of $f(x)$, and if we factor it in $K[x]$ there will be a linear factor for each of those zeroes. We can then sequentially extend $K$ by constructing field extensions from the remaining irreducible factors of $f(x)$. Each time we extend fields by another irreducible factor, we add another zero of $f(x)$ to the resulting field extension. The process terminates after at most $d$ steps to produce the splitting field of $f(x)$. The splitting field of $f(x)$ has degree bounded by $d!$.

It is possible that the degree of the splitting field is as small as $d$, since it is possible, depending on the nature of $f(x)$, that $K[x]/f(x)\mathbb Q(x)$ itself contains $d$ zeros of $f(x)$.

Example

Suppose $p$ is prime and consider the $p$th cyclotomic polynomial $$\Phi_p(x) = x^{p-1} + x^{p-2} + \cdots + x + 1.$$ Suppose $\zeta$ is a zero of $\Phi_p(x)$ in $\mathbb Q[x]/\Phi_p(x) \mathbb Q[x]$, then it is easily verified that $\zeta^p = 1$.It follows that, if $\ell = 1, \ldots, p-1$, \begin{eqnarray}\Phi_p(\zeta^{\ell}) &=& \zeta^{\ell(p-1)} + \zeta^{\ell(p-2)} + \cdots + \zeta^{\ell} + 1 \\ &=& \zeta^{p-1} + \zeta^{p-2} + \cdots + \zeta + 1 \\ &=& 0.\end{eqnarray} It follows that $\zeta, \zeta^2, \ldots, \zeta^{p-1}$ are all $p-1$ roots of $\Phi_p(x)$ and hence $K = \mathbb Q[x]/\Phi_p(x)\mathbb Q[x]$ is the splitting field of $\Phi_p(x)$.

Galois Theory

Definition

If $K | k$ is a splitting field for a polynomial $g(x) \in k[x]$, then $K$ is said to be Galois over $k,$ and the group of automorphisms of $K$ which fix $k$ is called the Galois group and denoted $\mathrm{Gal}(K|k)$.

Claim

$K | k$ is Galois if and only if $\# \mathrm{Aut}(K|k) = [K : k]$.

We won’t prove this claim (though the only if direction is easy) because it is a bit fiddly with separability and involves a diversion into character theory. Some (most?) authors give this as the definition of Galois and prove that it implies the splitting field definition.

The main result in Galois Theory is a correspondence between intermediate fields of $K | k$ and subgroups of $\mathrm{Gal}(K|k)$. Let us write $G = \mathrm{Gal}(K|k)$ and suppose $H < G$ is a subgroup. Define $$K_H = \{ \gamma \in K : \sigma(\gamma) = \gamma \mbox{ for all } \sigma \in H \}.$$ It is easily verified that $K_H$ is a field, and $k \subset K_H \subset K$ (which we might abbreviate $K | K_H | k$). It will turn out that $H \leftrightarrow K_H$ will be a bijection (called the Galois correspondence) between subgroups of $G$ and intermediate fields of $K | k$.

This correspondence goes beyond a bijection, because there is an interpretation for $H$ and $G/H$ (as a subgroup in the case where $H$ is normal, but to some extent even as a set of cosets in the non-normal case) in terms of the groups of automorphisms $\mathrm{Gal}(K|K_H)$ and $\mathrm{Aut}(K_H|k)$. I hope you objected to the notational switch between $\mathrm{Gal}$ and $\mathrm{Aut}$ in the previous sentence, but it is correct. The fact that $K$ is a splitting field for a polynomial in $k[x]$ means that it is also the splitting field for a polynomial in $K_H[x]$ (namely any one of the irreducible factors of the original polynomial in $k[x]$) and hence $K | K_H$ is Galois and we use the notation $\mathrm{Gal}(K | K_H)$ for the group of automorphisms of $K$ preserving $K_H$. This is, unsurprisingly, equal to $H$. On the other hand, just because $K$ is the splitting field of a polynomial in $k[x]$ doesn’t imply that an intermediate field, such as $K_H$, must be a splitting field for that or any other polynomial in $k[x]$. Thus, in general we need to refer to the automorphism group of $K_H | k$ by $\mathrm{Aut}(K_H | k)$. It will turn out that when $H$ is normal in $G$ then $K_H | k$ is Galois, and $\mathrm{Gal}(K_H | k) \cong G/H$. This will all be enumerated in the Fundamental Theorem of Galois Theory, but we need to develop a few results first.

Given $\gamma \in K$, we call $\sigma \gamma; \sigma \in G$ the Galois conjugates of $\gamma$. Moreover, if $L$ is any intermediate field extension, $K | L | k$, then $\sigma$ gives an isomorphism from $L$ onto $\sigma(L)$ (which fixes $k$). In particular $K_H$ is isomorphic to its image $\sigma K_H$. Notice then that if $\psi \in \mathrm{Aut}(K_H | k)$, then $\sigma \psi \sigma^{-1}$ is an element of $\mathrm{Aut}(\sigma K_H | k)$.

Indeed, $\sigma H \sigma^{-1} = \mathrm{Aut}(\sigma K_H | k)$. We can make this more evocative by denoting the action by conjugation of $G$ on $H$ as $\sigma \cdot \psi = \sigma \psi \sigma^{-1}$, in which case, $$\sigma \cdot \mathrm{Aut}(K_H | k) = \mathrm{Aut}(\sigma K_H | k).$$ If $\sigma K_H = K_H$ for all $\sigma \in G$, then $GHG^{-1} = H$, that is $H$ is normal in $G$. On the other hand, if $H$ is normal in $G$, then $\sigma \psi \sigma^{-1} \in H$ and $\sigma K_H = K_H$ for all $\sigma \in G$.

Now, suppose $H$ is normal and $g(x) \in k[x]$ is a polynomial so that $K_H = k[x]/g(x)k[x]$. From the previous discussion, $x + g(x) k[x]$ is a zero of $g(x)$ in $K_H$, as are $\sigma (x + g(x) k[x])$ for all $\sigma \in G$. To establish $K_H | k$ is Galois, we need to show that the orbit of $x + g(x) k[x]$ under $G$ is equal to $[K_H : k]$. We know for $\sigma \in H$, $\sigma(x + g(x) k[x]) = x + g(x)k[x]$. On the other hand, if $\sigma(x + g(x) k[x]) = x + g(x)k[x]$ then $\sigma \in H$ because $\sigma$ is completely determined by its action on $x + g(x) k[x]$. Thus, the automorphisms of $\mathrm{Aut}(K_H|k)$ are in correspondence with $G/H$. We thus have $[K : K_H] = \#H$, $[K : k] = \#G$ and $[K_H : k] = \#G/H$. It follows that $\# \mathrm{Aut}(K_H|k) = [K_H : k]$ and $K_H | k$ is thus Galois.

To be sure, we have glossed over many details. However, many important observations are captured in the Fundamental Theorem of Galois Theory.

Fundamental Theorem of Galois Theory

Suppose $K | k$ is Galois and $G = \mathrm{Gal}(K|k)$.

CORRESPONDENCE

There is an inclusion reversing correspondence between intermediate fields of $K|k$ and subgroups of $H$.

Normality $\leftrightarrow$ Galois

$H$ is normal in $G$ if and only if $L|k$ is Galois. In this situation $\mathrm{Gal}(L|k) \cong G/H$.

The Correspondence Preserves Lattices

Suppose $H_1 \leftrightarrow L_1$ and $H_2 \leftrightarrow L_2$ for $H_1, H_2 \leq G$ and $L_1, L_2$ intermediate fields of $K|k$. Then $\langle H_1, H_2 \rangle \leftrightarrow L_1 \cap L_2$ and $H_1 \cap H_2 \leftrightarrow L_1 L_2$. (Here $\langle H_1, H_2 \rangle$ is the smallest subgroup of $G$ containing both $H_1$ and $H_2$ and $L_1 L_2$ is the smallest field containing both $L_1$ and $L_2$). Moreover the inclusions (e.g. $L_1 \cap L_2 \subset L_1 \subset L_1 L_2$) are reversed under the correspondence.

Here we write arrows for the inclusion map. The correspondence reverses inclusion.

The correspondence between subgroups of $\mathrm{Gal}(K|k)$ and subfields of $K|k$ is complete as the subfields of $\mathbb{Q}(i, \sqrt[8]{2})$ and subgroups of $G = \langle \sigma, \tau : \sigma^8 = \tau^2 = 1, \sigma \tau = \tau \sigma^3 \rangle$. This example was cribbed from Abstract Algebra, second edition by Dummit and Foote.

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From Measures to Metrics on Pro-finite Completions

The complete 3-nary tree represents the family tree where each individual in a generation spawns (asexually) exactly three progeny in the subsequent generation. The image to the left represents 7 generations beginning from a single ancestor (the root) at the center of the image.

If we imagine the generations continuing ad infinitum, then we arrive at an object called the pro-finite completion of the tree. Loosely speaking this is the topological space which consists of all infinite paths from the root down through the (infinitely many) generations.

The pro-finite completion of the complete 3-nary tree can be put in correspondence with sequences of the form $(m_n)$ where each $m_n \in \{0,1,2\}$ as follows: Each descendent of an individual is labelled either 0, 1 or 2 (this can be done consistently by ordering, say, counterclockwise in the embedding above, but it doesn’t really matter so long as the labels are fixed for all time). A sequence starting, say, $(1,0,2,1,…)$ represents a child of (reading right to left) the first child of the second child of the zeroth child of the first child of the root. Admittedly ‘zeroth child’ sounds awkward, but we think of these as labels and not ordinals.

Visually, we may think of the pro-finite completion to be the boundary of the infinite graph, and the corresponding sequence $(m_n)$ as an address containing the information necessary to describe how to traverse the tree to get to that point on the boundary.

There are other embeddings of the complete 3-nary tree, including the ‘balloon embedding’ on the right. In this embedding the pro-finite completion is visualized as its (fractal) boundary. This embedding gives another construction of the fractal known as Serpienski’s Triangle.

In this embedding you may think of an ‘address’ of a point on the boundary as given by a sequence of ‘Left’, ‘Right’ and ‘Forward’ directions were you to drive to that point from the root along the edges of the graph. A bijection between $\{0,1,2\}$ and {Left, Right, Forward} will produce the sequence $(m_n)$.

Of course there’s nothing special about the 3-nary tree. We could start with any number of descendants per individual per generation. Indeed, we could let the number of descendants vary either between generations, or within a generation. We will see some examples of this soon.

Balloon embeddings of the complete 2-nary (binary), 4-nary and 5-nary trees. In each case the pro-finite completion is the fractal boundary of these graphs, and not the depicted edges and vertices.

Random Trees

There are lots of ways to make random graphs and trees, but here we will concentrate on a sort of random tree that will arise in the study of prime splitting in towers of number fields. We will suppose we start with a single ancestor (the root), and that each individual in the nth generation has an independent, identically distributed, bounded number of children. Note that the bound on the number of children may grow with generations, but for each generation there is some upper bound on the number of children an individual may have.

A simple example of a random tree where each individual has an equal chance of having 1, 2 or 3 offspring. The images are different embeddings of the same random tree.

Suppose the largest number of children an individual in the $n$the generation may have is $b_n$ (for instance, the random tree above has $b_n=3$ for all $n$). We call the sequence $(b_n)$ the sequence of generation bounds, and we call the tree where each individual in the $n$th generation has exactly $b_n$ children the complete $(b_n)$-nary tree. Every random tree with generation bounds $(b_n)$ can be embedded as a subtree in the complete $(b_n)$-nary tree.

As in the non-random case, the pro-finite completion of a random tree is the address, given as the directions necessary to traverse the tree from the root to a point on the `boundary’. Another way of representing the information given in the address is provided by the list of vertices $(v_n)$ one passes through on the voyage from the root. Here one assumes that the vertices are uniquely labelled. If $(v_n)$ is such a list of vertices we will write $v_m | v_n$ for all $m > n$. Loosely speaking, a vertex $v$ divides the vertex $w$ if $v$ is a vertex further down the tree from $w$. Put even more simply, $v | w$ if $v$ is descended from $w$. We will denote the root of the tree by $v_0$ and note that $v | v_0$ for all vertices $v$.

Let $B$ be the pro-finite completion of our (possibly random) tree as represented by sequences of vertices $(v_n)$, one per generation, with $v_m | v_n$ for all pairs $m > n$. For any vertex $w$ we define $$B(w) = \{ (v_n) \in B : w = v_m \mbox{ for some } m \}.$$ Loosely speaking $B(w)$ is the set of points in the pro-finite completion (boundary of the tree) that are downstream from vertex $w$.

$B(w)$ represents the part of the pro-finite completion that lies in the blue disk (left) or arc (right). Note that these are different representations of the same $B(w)$ on different embeddings of the same random tree.

Note that if $u$ and $w$ are different vertices, then either $B(u)$ and $B(w)$ are disjoint, or one is a subset of the other. It is worth supplying your own proof of this, or at least understanding why it is true from a picture.

$\sigma$-algebras and measures on $B$

We eventually want to talk about measures (and metrics) on the pro-finite completion of a random tree, but first we need a suitable $\sigma$-algebra. As usual, we actually define a nice collection of sets that we want to be in our $\sigma$-algebra and consider the smallest $\sigma$-algebra that does the trick. Dynkin’s $\pi$-$\lambda$ Theorem seems particularly salient here, and we define $\mathcal P$ to be the $\pi$-system given by all $B(w)$ for all vertices of our tree. That is $$\mathcal P = \{ B(v) : v \mbox{ is a vertex} \} \cup \emptyset.$$ We have to throw in the empty set, because a $\pi$-system is a collection of sets closed under intersection, and by our previous remarks, it is possible (common, in fact) for elements of $\mathcal P$ to be disjoint. We set $\mathcal D$ to be the $\sigma$-algebra on $B$ generated by $\mathcal P$. And we take $(B, \mathcal D)$ to be the measurable space in which all calculations occur.

Notice that, since the intersection of any two elements of $\mathcal P$ is again an element of $\mathcal P$ we see that elements of $\mathcal D$ are simply (possibly countable) disjoint unions of elements of $\mathcal P$. That is, for each set $A \in \mathcal D$ there is a (finite or) countable collection of vertices $V$, and a (finite or) countable $X \subset B$ such that $$A = \bigsqcup_{v \in V} B(v) \sqcup \bigsqcup_{x \in X} \{x \}.$$ The disjointness of this union implies we may do this in such a way that for any $u, v \in V$, $u \not | \;\; v$, and for any $x \in X$ and $v \in V$, $x \not \in B(v)$. We call $V$ a reduced set of vertices for $A$.

The $\pi$-$\lambda$ Theorem implies that a measure $\mu$ on $(B, \mathcal D)$ is determined completely by its values on $\mathcal P$. Note that if $w_1, \ldots, w_d$ are the child-vertices of vertex $w$, then $\mu(w) = \mu(B(w_1)) + \cdots + \mu(B(w_d))$ (and conversely, any collection of $\{m_v \in [0, \infty] : v \mbox{ a vertex}\}$ satisfying all consistency conditions of the form $m_w = m_{w_1} + \cdots + m_{w_d}$ will determine a measure on $(B, \mathcal D)$). There may be special measures on $(B, \mathcal D)$ depending on the construction of your tree, but for now we maintain complete generality, and see how various aspects of the measure interact to potentially give a metric on $B$.

Recall that an atom of the measure $\mu$ is a set $A \in \mathcal D$ such that $\mu(A) > 0$ and if $C \in \mathcal D$ is a proper subset of $A$ then $\mu(C) = 0$. By our construction of elements of $\mathcal D$ we see that if $A$ is an atom of $\mu$ then either $A = \{x \}$ for some $x \in B$, or $A = B(v)$ for some vertex $v.$ In fact, we will see that this latter situation is impossible. To see why, suppose the vertices $v_1, \ldots, v_d$ are the immediate descendants of $v$. Then, $$\mu(B(v)) = \mu(B(v_1)) + \cdots + \mu(B(v_d)).$$ If $d > 1$, It is not possible for $\mu(B(v)) > 0$ and $\mu(B(v_n)) = 0$ for all $n=1,\ldots, d$. Hence, if $v$ has more than one immediate descendent, $B(v)$ cannot be an atom of $\mu$. If, on the other hand $d = 1$ then $B(v) =B(v_1)$ and we can repeat our argument to show that either $B(v_1)$ is not an atom, or it only has one immediate descendant. It follows that if $B(v)$ is an atom then each descendent of $v$ has only one immediate descendent. That is, $B(v)$ contains only one $x \in B$, and hence if $B(v)$ is an atom, then in fact $B(v) = \{x \}$.

$B(v)$ can be a singleton only when all descendants of $v$ have only one immediate descendent. The only $B(v)$ that can be atoms of $\mu$ are singletons.

If $\mu$ has no atoms, then it is said to be diffuse. If $\mu(B(v)) > 0$ for all $B(v)$ that are not singletons, then we say $\mu$ is a full measure.

A pseudo-ultrametric formed from $\mu$

A metric on $B$ is a function $\delta: B \times B \rightarrow [0, \infty]$ such that for all $x,y,z \in B$,

  1. $\delta(x,x) = 0$
  2. $\delta(x, y) = 0$ implies $x = y$
  3. $\delta(x,y) = \delta(y,x)$
  4. $\delta(x,z) \leq \delta(x,y) + \delta(y,z)$

Note that we allow the possibility that $\delta$ is infinite. This is a slight generalization of the usual notion of a metric, but it disturbs very little. If we enforce the stronger requirement $$4′. \quad \delta(x,z) \leq \max\{\delta(x,y), \delta(y,z)\}$$ then we say $\delta$ is an ultrametric. If we instead, lose requirement (2), then we say $\delta$ is a pseudometric. Thus a pseudo-ultrametric $\delta$ satisfies for all $x,y,z \in B$,

  • $\delta(x,x) = 0$
  • $\delta(x,y) = \delta(y,x)$
  • $\delta(x,z) \leq \max\{ \delta(x,y), \delta(y,z) \}$

The third condition is called the ultrametric inequality or the strong triangle inequality.

Theorem

Given a measure $\mu$ on $(B, \mathcal D)$, define $\delta : B \times B \rightarrow [0,\infty]$ by $\delta(x,x) = 0$ for all $x \in B$, and for $x \neq y$ , $$\delta(x,y) = \inf\{ \mu(A) : A \in \mathcal P \mbox{ with } x, y \in A\}.$$ Then $\delta$ is a pseudo-ultrametric. Moreover, if $\mu$ is a full measure, then $\delta$ is an ultrametric.

Another way of defining $\delta$ is to first define the least common ancestor vertex of any two $x, y \in B$ by $a(x,y)$ in which case $\delta(x,y) = \mu(B(a(x,y))$. This definition only makes sense when $x \neq y$.

Proof

To show that $\delta$ is a pseudo-ultrametric, the only nontrivial condition to check is the ultrametric inequality. That is, for any $x, y, z \in B$, $\delta(x,z) \leq \max\{ \delta(x,y), \delta(y,z) \}$.

There are two cases. The first (slide up) we have $y \not \in B(a(x,z))$. In this case $B(a(x,z)) \subset B(a(y,z))$ and hence $\delta(x,z) \leq \delta(x,y)$. In the second case (slide down) $y \in B(a(x,z))$ and without loss of generality $B(a(x,z)) = B(a(y,z))$.

In the first case we get $B(a(x,z)) \subset B(a(y,z)) = B(a(x,y))$ and hence $\delta(x,z) \leq \max\{ \delta(x,y), \delta(y,z)\}$. (Note that equality is still possible in this case, because it is possible that $\mu(B(a(x,z)) = \mu(B(a(y,z))$ even when $B(a(x,z))$ is a proper subset of $B(a(y,z))$.

The second of these cases is a bit more delicate, but we see that, changing the labels if necessary, $a(x,z) = a(y,z)$. It follows that $\delta(x,z) = \delta(y,z)$ and $\delta(x,y) \leq \delta(y,z)$ which together yield $\delta(x,z) \leq \max\{\delta(x,y), \delta(y,z)\}$ as desired.

Notice that if $\mu$ is full, then $\delta(x,y) = \mu(B(a(x,y)) > 0$ and hence $\delta$ is in fact an ultrametric. $\square$

We say that $x \in B$ is isolated with respect to $\delta$ if there exists $\epsilon > 0$ such that $\delta(x,y) > \epsilon$ for all $y \neq x$. As the next result shows, isolated points come from atoms of $\mu$.

Lemma

Suppose $x \in B$ is such that $\{x\}$ is an atom of $\mu$. Then $x$ is isolated with respect to $\delta$. In particular, if $y \neq x$ then $\delta(x,y) \geq \mu\{x\}$.

Proof

By definition $x \in B(a(x,y))$. It follows that $\mu\{x\} \leq \mu(B(a(x,y)) = \delta(x,y)$. $\square$

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