An absolute value on a field $K$ is a function $|\cdot| : K \rightarrow [0, \infty)$ such that for any $x, y \in K$,

  • $|x| = 0$ if and only if $x = 0$;
  • $|x y| = |x| |y|$;
  • $|x + y| \leq |x| + |y|$

These properties are called respectively positive definiteness, multiplicativity and the triangle inequality. If the absolute value satisfies the stronger condition (called the strong triangle inequality)

  • $| x + y| \leq \max\{ |x|, |y| \}$

we say it is a non-archimedean absolute value. An absolute value that does not satisfy the strong triangle inequality is called an archimedean absolute value.

The usual absolute values on $\mathbb Q$, $\mathbb R$ and $\mathbb C$ are all archimedean absolute values.

Every field has a trivial absolute value given by $| 0 | = 0$ and $| x | = 1$ for all $x \neq 0$. This absolute value is not very interesting and we will usually concentrate on the non-trivial absolute values of a field.

Equality in the Strong Triangle Inequality

It is often easy to determine when the strong triangle inequality is actually an equality. The proof of the following lemma is easy, but the result is surprisingly powerful.

Lemma

Suppose $| \cdot |$ is a non-archimedean absolute value $x, y \in K$ are such that $| x | < | y |$ then $| x + y | = | y |$. That is, if $| x | \neq | y |$, $|x + y| = \max\{ |x|, |y| \}.$

Proof

Suppose $|x| < |y|$ and $|x + y| < |y|$. Then,$$ |y| = |x + y – x| \leq \max\{|x+y|,|x|\} < |y|;$$ an obvious contradiction.

Absolute Values on $\mathbb Q$

To distinguish the usual absolute value from new ones we may construct we will denote it $| \cdot |_{\infty}$. That is $$ |x|_{\infty} = \left\{ \begin{array}{rl} x & x \geq 0; \\ -x & x < 0. \end{array}\right.$$

If $p$ is a prime integer then define the valuation $v_p : \mathbb Q^{\times} \rightarrow \mathbb Z$ by $$ v_p(x) = v \qquad x = p^v \frac{a}{b}, \quad \mathrm{GCD}(a, b) = 1.$$ That is, we determine the valuation of a rational number by determining the highest power of $p$ that divides it. This power is positive if the numerator has $p$ as a factor, and is negative if the denominator has $p$ as a factor (when written in lowest terms). If the valuation is 0 then the rational number does not have $p$ in its factorization.

The valuation $v_p$ is a homomorphism from $(\mathbb Q^{\times}, \cdot) \rightarrow (\mathbb Z, +)$. That is $v_p$ is additive: $v_p(xy) = v_p(x) + v_p(y)$. It is common to take $v_p(0) = \infty$ with the justification that $0$ is “infinitely divisible” by $p$.

We may write a non-zero rational number $x$ in terms of the various valuations $\{v_p(x) : p \mbox{ prime} \}$ by $$x = \prod_p p^{v_p(x)}.$$ Note that $v_p(x) \neq 0$ for only the primes that appear in the factorization of $x$. It follows that this product is actually a finite product.

We define the $p$-adic absolute value on $\mathbb Q$ by $$| x |_p = p^{-v_p(x)}.$$ Of course, we need to verify that this is an absolute value. Positive definiteness is a matter of definition, multiplicativity comes from the additivity of $v_p$. Only the triangle inequality remains, and we will in fact show that $| \cdot |_p$ satisfies the strong triangle inequality. Suppose $x = p^v a/b$ and $y = p^u c/d$ where $a/b$ and $c/d$ are written in lowest terms. Then $$ x + y = \frac{p^v d a + p^u b c}{bd}.$$ Because $bd$ is relatively prime to $p$, we see that the minimum power of $p$ we can pull out of the numerator is $\min\{u, v\}$. That is $v_p(x + y) \geq \min\{v_p(x), v_p(y) \}$. This is equivalent to the strong triangle inequality.

Note that we actually proved something slightly stronger than the strong triangle inequality here. Written in terms of valuations and absolute values, $$v_p(x + y) > \min\{v_p(x), v_p(y) \} \quad \mbox{only if} \quad v_p(x) = v_p(y)$$ $$|x + y|_p < \max\{|x|_p, |y|_p \} \quad \mbox{only if} \quad |x|_p = |y|_p.$$

Thus, the $p$-adic absolute value is a non-archimedean absolute value.

The Places of $\mathbb Q$

We say two absolute values $| \cdot |_0$ and $| \cdot |_1$ on a field $K$ are equivalent if there is a positive real number $c$ such that $|\cdot |_0 = |\cdot|_1^c$. It is easily verified that this gives an equivalence relation on the set of absolute values of $K$, and we call the equivalence classes the places of $K$. The place corresponding to the trivial absolute value (which is the only representative in its class) is called the trivial place, and is often excluded from attention.

We will eventually talk about how to complete $K$ with respect to an absolute value, using the same methods as when we construct $\mathbb R$ out $\mathbb Q$ with respect to the usual absolute value. We will see that equivalent absolute values produce the same completion, and different places produce different completions. The completion of $\mathbb Q$ with respect to the trivial absolute value is $\mathbb Q$ itself—another hint that nothing interesting happens with trivial absolute values.

The set of non-trivial places of $K$ is denoted $\mathcal M_K$.

Ostrowski’s Theorem

If $| \cdot |$ is a non-trivial absolute value on $\mathbb Q$ then $|\cdot|$ is equivalent to either the usual absolute value $| \cdot |_{\infty}$ or is equivalent to $| \cdot |_p$ for some prime $p$. That is $\mathcal M_{\mathbb Q}$ is in correspondence with $\mathcal P = \{ \mbox{ primes } \} \cup \{ \infty \}$.

For a proof, see https://en.wikipedia.org/wiki/Ostrowski%27s_theorem#Proof

It is hard to understate the importance of this result. It is surprising (though less so after reading the proof) to see a correspondence between the primes and a set of analytic object. It suggests (rightly) that there is progress to be made in understanding the primes by understanding absolute values. Indeed, as we will see this correspondence extends to number fields, and we will see a bijection between non-archimedean absolute values and prime ideals. The archimedean absolute values in that setting correspond to the real and complex embeddings of the number field. It all hangs together very nicely.

The Product Formula

The product formula is the trivial observation that for any rational number $x \neq 0$, $$ \prod_{p \in \mathcal P} | x |_p = 1.$$

This can be verified by calculation: $$ | x |_{\infty} = \prod_{p \mbox{ prime} } p^{v_p(x)} \quad \mbox{and} \quad \prod_{p \mbox{ prime}} | x |_p = \prod_{p \mbox{ prime}} p^{-v_p(x)},$$ and the product formula follows.

In spite of its triviality, it is an important observation and we will return to a version of the product formula for number fields shortly.

Completions

Recall these two definitions from elementary analysis.

Definition

A sequence of rational numbers $(x_n)$ converges to 0 with respect to the absolute value $| \cdot |$ if, for all $\epsilon > 0$ there exists positive integer $N$ such that if $n \geq N$ then $|x_n| < \epsilon$. In this situation we write $\lim x_n = 0$ or $(x_n) \rightarrow 0$.

Definition

A sequence of rational numbers is Cauchy with respect to $| \cdot |$ if, for all $\epsilon > 0$ there exists positive integer $N$ such that if $n, m \geq N$ then $|x_n – x_m| < \epsilon$. We will denote the set of Cauchy sequences by $\mathcal C$.

If you recall from elementary analysis, Cauchy sequences are convergent sequences. However, not every convergent sequence of rational numbers converges to a rational number. That is, taking limits can take you out of $\mathbb Q$. These new limit points live in the completion, and that completion depends on the absolute value (in fact place) used in the definition of convergence and Cauchy.

The equivalence of Cauchy sequences and convergent sequences (once you account for new limit points) is useful because the condition of Cauchyness depends only on the rational numbers in the sequence. That is we can determine if something is Cauchy without having to know what its limit is or where that limit lives.

The other thing that is useful about Cauchy sequences, is that they form a ring under coordinate-wise addition and multiplication. This is essentially equivalent to the limit laws: If $(x_n), (y_n) \in \mathcal C$ then $(x_n + y_n) \in \mathcal C$ and $(x_n y_n) \in \mathcal C$. The identically zero sequence $(0)$ is the additive identity, and $(1)$ is the multiplicative identity. Cauchyness of the coordinate-wise quotient of two sequences also follows from the limit laws, though one has to be careful that the Cauchy sequence in the denominator does not converge to 0, and to “throw out” any quotients in the sequence where the denominator may be 0. We may embed $\mathbb Q \hookrightarrow \mathcal C$ by sending $x$ to the constant sequence $(x)$.

In the real numbers there are many different sequences of rational numbers which converge to the same real number. For instance, much mathematics has been made of discovering interesting rational sequences that converge to some of our favorite irrational numbers, like $\pi$. Notice if we do have two sequence $(x_n)$ and $(y_n)$ and $(x_n) \rightarrow \pi$ and $(y_n) \rightarrow \pi$, and here $\pi$ can be replaced by any real number, then $(x_n – y_n) \rightarrow 0$. Thus we can determine when two convergent sequences converge to the same number, simply by determining whether their difference converges to 0.

Returning to $\mathcal C$, we define an equivalence relation $(x_n) \equiv (y_n)$ if $(x_n – y_n) \rightarrow 0$. In our head we should think of an equivalence class as the set of all Cauchy sequences that converge to the same number, and there is one equivalence class for every possible limit point. And indeed, that is the definition of the completion of $\mathbb Q$ with respect to $| \cdot |$. It is the field formed from the equivalence classes of Cauchy sequences. We add, subtract, multiply and divide by choosing representatives of the equivalence classes and performing the appropriate operation coordinate-wise, and returning the equivalence class of that new Cauchy sequence. The rational number $x$ is represented by the equivalence class of the constant sequence $(x)$.

Let us denote the completion by $\overline{\mathbb Q}$ (this obviously depends on the absolute value, but for the moment we are working the distinguished absolute value $| \cdot |$). Suppose we have another equivalent absolute value $| \cdot |_0 = | \cdot |^c; c > 0$. We wish to argue that both absolute values produce the same completion, that is that the completion is something more appropriately associated to a place rather than a single absolute value. This is done by showing that the absolute values $| \cdot |$ and $| \cdot |_0$ determine the same set of Cauchy sequences, and that they determine the same equivalence relations on those Cauchy sequences.

Suppose $(x_n)$ is Cauchy, and given $\epsilon$ let $N(\epsilon)$ be the guaranteed integer such that $n, m \geq N(\epsilon)$ implies $| x_n – x_m | < \epsilon$. It follows that if we set $N = N(\epsilon^{1/c})$ then $| x_n – x_m |_0 = |x_n – x_m|^c < |\epsilon^{1/c}|^c = \epsilon$. Thus if $(x_n)$ is Cauchy with respect to $| \cdot |$ it is Cauchy with respect to $| \cdot |_0$ the reverse containment is established similarly (but setting $N = N(\epsilon^c)$), and we see both absolute values produce the same Cauchy sequences.

We also need to establish that $(x_n) \rightarrow 0$ with respect to $| \cdot |$ if and only if it does the same with respect to $| \cdot |_0$. The argument is almost identical to that used to show that both absolute values produce the same Cauchy sequences.

The places are in correspondence with $\mathcal P$ and we denote the completion with respect to $| \cdot |_p$ by $\mathbb Q_p$. In particular $\mathbb Q_{\infty}$ is the real numbers.

Extending Absolute Values to $\mathbb Q_p$

Given an element in $\mathbb Q_p$ as represented by Cauchy sequence $x = (x_n)$, then we define $|x|_p = \lim |x_n|_p$ where we take the limit in the real numbers as usual. Upon showing this is well-define, we arrive at an absolute value (which we continue to denote $| \cdot |_p$) on $\mathbb Q_p$ which restricts to $| \cdot |_p$ on $\mathbb Q$.

Suppose $x = (x_n)$ and $y = (y_n)$ are in the same equivalence class—that is $\lim |x_n – y_n|_p = 0$. Then $$| x |_p = \lim | x_n |_p = \lim |y_n + (x_n -y_n)| _p \leq \lim |y_n|_p + \lim |x_n – y_n|_p = |y|_p.$$ But symmetry implies that $|y|_p \leq |x|_p$ so in fact $|x|_p = |y|_p$ and $| \cdot |$ is well defined on $\mathbb Q_p.$

To verify $| \cdot |_p$ is an absolute value on $\mathbb Q_p$, notice that if $| x |_p = 0$ then $(x_n)$ is equivalent to the series $(0)$, that is $x$ is in the zero equivalence class. Multiplicativity follows from the multiplication limit law for series. The triangle inequality likewise follows since $$ | x + y |_p = \lim |x_n + y_n|_p \leq \lim |x_n|_p + \lim|y_n|_p = |x|_p + |y|_p$$ as does the strong inequality (via the continuity of the max function) in the case $| \cdot |_p$ is non-archimedean.

Equipping a field with an absolute value also allows us to define distances, and hence a metric topology. This topology in turn generates a $\sigma$-algebra (which is equal to the Borel $\sigma$-algebra when $p = \infty$) and translation invariant measures (a la Lebesgue measure) on $(\mathbb Q_p, +)$ and $(\mathbb Q^{\times}_p, \cdot)$. I am getting ahead of myself, but the point is that $\mathbb Q_p$ isn’t just a new field with few limit points filled in from $\mathbb Q$, but rather it is a metric space and a measure space and exhibits many features in common with $\mathbb R$.

Different Places Produce Different Completions

Before digging into the topology of $\mathbb Q_p$ we want to justify out claim that the completions of $\mathbb Q$ are in correspondence with $\mathcal P$. So far we have seen that every completion is equal to $\mathbb Q_p$ for some $p \in \mathcal P$, but we have not yet established that different elements in $\mathcal P$ produce different completions.

This is a consequence of the Weak Approximation Theorem for $\mathbb Q_p$, which in our case says that if $| \cdot |$ and $| \cdot |_0$ are non-equivalent absolute values, then you can find an element $x \in \mathbb Q$ such that $| x |$ is large and $| x |_0$ is small (and vice-versa). This will in term imply that different sequences are Cauchy or converge to 0 with respect to these different absolute values.

The Weak Approximation Theorem

Suppose $p_1, \ldots, p_N \in \mathcal P$ index any finite number of places of $\mathbb Q$ and $1 \leq n < N$. Given $a \in \mathbb Q$ and any $\epsilon > 0$ there exists $x \in \mathbb Q$ such that $|x – a|_{p_1}, \ldots, |x – a|_{p_n} \in (0, \epsilon)$ and $|x-a|_{p_{n+1}}, \ldots, |x-a|_{p_N} \in [\epsilon^{-1}, \infty).$

We do this for $a=0$ as follows, first we suppose the $p_1, \ldots, p_N$ are all non-archimedean absolute values and then discuss how to modify the argument if we also want $| x |_{\infty}$ either large or small. Consider $$ x = \frac{p_1^{m_1} \cdots p_n^{m_n}}{p_{n+1}^{m_{n+1}} \cdots p_N^{m_N}}. $$ Then $|x|_{p_1} = p_1^{-m_1}, \ldots, |x|_{p_n} = p_n^{-m_n}$ and these can be made as small as desired by choosing $m_1, \ldots, m_n$ as large as necessary. Similarly, $|x|_{p_{n+1}} = p_{n+1}^{m_{n+1}}, \ldots |x|_{p_N} = p_N^{m_N}$ which can be made as large as desired by choosing $m_{n+1}, \ldots, m_N$ as large as necessary.

How would we simultaneously make $|x|_{\infty}$ as large or small as specified? We may multiply $x$ by a rational number $r$ whose factorization avoids $p_1, p_2, \ldots, p_N$ without changing any of the absolute values $|\cdot|_{p_1}, \ldots, |\cdot|_{p_N}$. That is $|r x|_{p_1} = |x|_{p_1}, \ldots, |r x|_{p_N} = |x|_{p_N}$. Note however, that $|r x|_{\infty} = |x|_{\infty} |r|_{\infty}$, and hence by choosing $r$ sufficiently large or small, the rational number $r x$ will satisfy the conclusions of the theorem when $a=0$. The general case is a consequence of the Chinese Remainder Theorem.

Completions as Topological and Measurable Spaces

Here we are mostly going to assume that $p < \infty$, though we will compare and contrast the situation with the $\mathbb Q_{\infty} = \mathbb R$ case.

The Borel Topology on $\mathbb Q_p$

Once we have an absolute value on a field we can make $\epsilon$-neighborhoods, and these form the basis for a topology. In the case of $x \in \mathbb Q_p$, given $\epsilon > 0$, we define $$B_{\epsilon}(x) = \{ y \in \mathbb Q_p : |y – x|_p < \epsilon \}.$$ When $p = \infty$ these are the usual neighborhoods of $x \in \mathbb R$. The topology generated by all such sets is called the Borel topology. It is easy to see that the collection of open neighborhoods does not depend on which absolute value you use from a particular place. An $\epsilon$-neighborhood with respect to $| \cdot |^c$ is an $\epsilon^{1/c}$-neighborhood of $| \cdot |$ and thus the set of neighborhoods is the same. So from the topological point of view $\mathbb Q_p$ is more naturally associated to a place than to a specific absolute value.

When $p < \infty$ something interesting happens that does not happen in $\mathbb R$. First note that, unlike $| \cdot |_{\infty}$, the non-archimedean absolute values are discrete. Namely $| \cdot |_p$ takes values in $\{p^n : n \in \mathbb Z\}$. This means that any open ball $B_{\epsilon}(x)$ can also be described as a closed ball $\overline{B}_{\epsilon’}(x)$ for some slightly larger $\epsilon’ > \epsilon$. The language sometimes used is that the balls in $\mathbb Q_p$ are clopen.

The balls in $\mathbb Q_p$ are nested in a way that they are not for $\mathbb R$. Namely two balls in $\mathbb Q_p$ are either disjoint, or one is a subset of the other. That is, for $p < \infty$, $\mathbb Q_p$ is totally disconnected. This, like all other differences is driven by the strong triangle inequality. To see this, suppose $B_{\epsilon}(x)$ and $B_{\delta}(y)$ are balls in $\mathbb Q_p$ with $z \in B_{\epsilon}(x) \cap B_{\delta}(y)$. Without loss of generality we may assume $\epsilon \leq \delta$.

First note $x \in B_{\delta}(y)$: $|x – z|_p < \epsilon$ and $|z – y|_p < \delta$. It follows that $|x-y|_p \leq \max\{|x-z|_p, |z-y|_p\} = \delta$.

If $x \not \in B_{\delta}(y)$ then the strong triangle inequality is violated (dotted distance in purple).

Next, if $w \in B_{\epsilon}(x)$ then $|w-y|_p \leq \max\{|w – x|_p, |x – y|_p\} = \delta$ and hence $w \in B_{\delta}(y)$ as claimed. It follows that $B_{\epsilon}(x) \subset B_{\delta}(y)$.

If $w \not \in B_{\delta}(y)$ then the strong triangle inequality is violated (dotted distance in red).

Another property of $\mathbb Q_p$ (this time shared between $p$ finite and infinite) is local compactness. Recall the definition: a space is locally compact if every point $x$ has a neighborhood which is contained in a compact set. It turns out that $\mathbb Q_p$ has the Heine-Borel property, and for any $x$, there is an epsilon such that $x \in B_{\epsilon}(x) \subset \overline B_{\epsilon}$ which does the job.

Haar measure on $\mathbb Q_p$

In the standard manner we set $\mathcal B$ to be the $\sigma$-algebra on $\mathbb Q_p$ generated by all balls. When $p < \infty$ the total disconnectivity of $\mathbb Q_p$ means that the generic sets in $\mathcal B$ are much easier to describe that in the $\mathbb R = \mathbb Q_{\infty}$ situation. Namely, because the countable intersection of balls is either another ball or a singleton (a set containing a single point), we see that a generic set in $\mathcal B$ looks like a countable union of balls and singletons. This is very tidy in comparison the nightmarishness of a general Borel subset of $\mathbb R$.

$(\mathbb Q_p, +)$ and $(\mathbb Q_p^{\times}, \cdot)$ are locally compact abelian groups. Locally compact abelian groups are important kinds of topological and measurable spaces because they can be equipped with a translation invariant measure. Specifically, if $B \in \mathcal B$ is a Borel set and $x \in \mathbb Q_p$ then we call $$ x + B = \{ x + b : b \in B \} \qquad \mbox{and} \qquad x B = \{ x b : b \in B\}$$ the additive and multiplicative translations of $B$ by $x$. A measure $\mu$ on $(\mathbb Q_p, \mathcal B)$ is said to be translation invariant if $\mu(x + B) = \mu(B)$ for all $B \in \mathcal B$ and all $x \in \mathbb Q_p$. In the case of $(\mathbb Q_p^{\times}, \mathcal B^{\times})$ the corresponding condition for the multiplicactive translation invariant measure $\mu^{\times}$ is $\mu^{\times}(x B) = \mu^{\times}(B)$. (I have not formally introduced the $\sigma$-algebra, $\mathcal B^{\times}$ on $\mathbb Q_p^{\times}$ but as as a set of points $\mathbb Q_p^{\times}$ is simply $\mathbb Q_p$ with 0 removed. We may take $\mathcal B’$ to be the $\sigma$-algebra generated by all balls except those containing 0.)

We may make $\mu$ and $\mu^{\times}$ unique by specifying the measure of a single clopen set (or in the case of $\mathbb Q_{\infty} = \mathbb R$ a single (non-singleton) closed interval). Thus we define the measures $\mu_p$ and $\mu^{\times}_p$ to be the unique translation invariant measures on $(\mathbb Q_p, \mathcal B)$ and $(\mathcal Q_p^{\times}, \mathcal B^{\times})$ normalized so that $$ \mu_p \{ x : |x|_p \leq 1 \} = 1 \qquad \mbox{and} \qquad \mu_p^{\times} \{x : |x|_p = 1 \} = \frac{p-1}{p}.$$ These measures are referred to as the (normalized) Haar measures for $\mathbb Q_p$ and $\mathbb Q_p^{\times}$.

The normalization on $\mu_p^{\times}$ may look a little strange. This choice was motivated by the fact that for any $x \in \mathbb Q_p$ and $B \in \mathcal B$, $$\mu_p(x B) = |x|_p \mu_p(B).$$ This implies that $$\mu_p^{\times}(dx) = \frac{\mu_p(dx)}{|x|_p}.$$ Moreover, if we set $C$ to be the closed unit ball, we have $p C = \{ x \in \mathbb Q_p : |x|_p < 1 \}$ and $\mu(pC) = \mu(C)/p$. It follows that $$\mu_p\{ x : |x|_p = 1 \} = \mu_p(C) – \mu_p(pC) = \frac{p-1}{p}.$$ The normalization for $\mu_p^{\times}$ makes it equal to $\mu_p$ on $\{ x : |x|_p = 1 \}$, a situation which can be advantageous when both measures come into play.

Example

Let $B = \{ x : 0< |x|_p < 1\}$ and $\overline B = \{ x : 0< |x|_p \leq 1\}$, and let $U = \overline B \setminus B = \{x : |x|_p = 1\}$. Then $$ \overline B = U \sqcup B \quad \mbox{and} \quad B = p\overline{B}.$$ Induction then implies that $$\overline B = \bigsqcup_{n=0}^{\infty} p^n U; \qquad \mbox{Indeed} \qquad \mathbb Q^{\times}_p = \bigsqcup_{n \in \mathbb Z} p^n U.$$ This is the decomposition of $\mathbb Q^{\times}_p$ into sets of equal absolute value. That is $p^n U$ is exactly the set where on which $|x|_p = p^{-n}$.

Suppose $s > 0$ and consider $$ \int_{\overline B} |x|_p^s \, \mu_p^{\times}(dx) .$$ Using the decomposition, $$\int_{\overline B} |x|_p^s \, \mu_p^{\times}(dx) = \sum_{n=0}^{\infty} \int_{p^n U} |x|_p^s \, \mu_p^{\times}(dx).$$ The integrand is constant (and equal to $p^{-n s}$) on $p^n U$, and $\mu_p^{\times}(p^N U) = \mu_p^{\times}(U) = (p-1)/p$. Hence, $$\int_{\overline B} |x|_p^s \, \mu_p^{\times}(dx) = \sum_{n=0}^{\infty}p^{-ns} \left(\frac{p-1}{p}\right) = \left(\frac{p-1}{p}\right) \frac{1}{1 – p^{-s}}.$$ This is an important calculation in the theory of the Riemann $\zeta$-function.

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