Contents

## Archimedean Absolute Values on $K$

If $K$ is a number field of degree $d$ over $\mathbb Q$, then there are $r_1$ real embeddings $K \hookrightarrow \mathbb R$ and $r_2$ complex embeddings $K \hookrightarrow \mathbb C$ where $r_1 +2 r_2 = d$. Each of these embeddings produces an absolute value on $K$ formed by restricting the usual absolute value on $\mathbb R$ or $\mathbb C$ to the embedded copy of $K$.

Each of these absolute values is incomparable in the sense that given any two of them $| \cdot |_{\sigma}$ and $| \cdot |_{\tau}$ and any $\epsilon > 0$ we can find some $x \in K$ so that $| x |_{\sigma} < \epsilon$ and $| x |_{\tau} > \epsilon^{-1}$. This is a non-trivial fact that follows from the density of $K$ in $\mathbb R^{r_1} \times \mathbb C^{r_2}$ via the diagonal embedding. The $r_1$ real embeddings thus produce $r_1$ distinct real places; similarly there are $r_2$ complex places, and together these form the set of archimedean places of $K$. We denote the set of archimedean places of $K$ by $\mathcal M_{\infty}(K)$ and if $v \in \mathcal M_{\infty}$ we will write $v | \infty$.

###### Theorem

If $| \cdot |$ is an archimedean absolute value on $K$, then $|\cdot|$ is in one of the archimedean places of $K$.

###### Proof

Let $\overline K$ be the closure of $K$ with respect to $| \cdot |$, then $\overline K$ is a complete archimedean field and hence is isomorphic to $\mathbb R$ or $\mathbb C$. $K \hookrightarrow \overline K$ is dense, and so this must be isomorphic to one of the known real or complex embeddings. Since the completion is determined by the place and not the absolute value, $| \cdot |$ must be in the place associated to the embedding $K \hookrightarrow \overline K$.

An archimedean absolute value on $K$ restricts to an archimedean absolute value on $\mathbb Q.$ Given an archimedean place $v$, we choose the representative $\| \cdot \|_v$ to be the restriction of the usual absolute value on $K$ as embedded in $K_v$ ($=\mathbb R$ or $\mathbb C$). We then define another equivalent absolute value by setting $| \cdot |_v = \| \cdot \|_v^{1/d}$ if $v$ is real, and $\| \cdot \|_v^{2/d}$ if $v$ is complex. This latter choice of normalization implies that if $r \in \mathbb Q$, $$ |r|_{\infty} = \prod_{v | \infty} | r |_v.$$

## Non-archimedean Absolute Values on $K$

By Ostrowski’s Theorem the non-archimedean places of $\mathbb Q$ are indexed by the rational primes (see Absolute Values and Completions of $\mathbb Q$). Thus, if $| \cdot |$ is a non-archimedean absolute value on $K$, $| \cdot |$ restricted to $\mathbb Q$ is equivalent to $| \cdot |_p$ for some prime $p$. If $v$ is the place of $| \cdot |$ we will say $v$ *lies above* $p$, and write $v|p$. We denote the set of places of $K$ above $p$ by $\mathcal M_p(K)$.

If $f(x)$ is the minimal polynomial for $K|\mathbb Q$, then $K = \mathbb Q[x] / f(x) \mathbb Q[x]$. Since $\mathbb Q \subseteq \mathbb Q_p$, we may view $f(x)$ as a polynomial with coefficients in $\mathbb Q_p$, and it may be the case that $f(x)$ is no longer irreducible over the larger field $\mathbb Q_p$. Suppose $f(x)$ factors as $f_1(x) \cdots f_{\ell}(x)$, where the $f_i(x) \in \mathbb Q_p[x]$ are irreducible.

If $a(x)$ and $b(x)$ are two polynomials in $\mathbb Q[x]$ with $a(x) – b(x)$ *not* divisible by $f(x)$, then in $\mathbb Q_p[x]$, $a(x) – b(x)$ is not divisible by any of the $f_i(x)$. That is, the map $a(x) + f(x) \mathbb Q[x] \mapsto a(x) + f_i(x) \mathbb Q_p[x]$ is injective for each $i=1,2,\ldots, \ell$. It follows that $K$ embeds in each of the $K_{v_i} := \mathbb Q_p[x] / f_i(x) \mathbb Q_p[x]$.

We can create an absolute value on $K_{v_i}$ by using the norm $N_{K_{v_i}|\mathbb Q_p}$ (the multiplicative homomorphism from $K_{v_i}$ onto $\mathbb Q_p$; see Field Extensions and Number Fields) and then hitting the result with the $p$-adic absolute value. That is, we define $$\|\alpha\|_{v_i} = \left| N_{K_{v_i}|\mathbb Q_p}(\alpha)\right|_p.$$ It remains to show this is a non-archimedean absolute value, and as usual, it is the strong triangle inequality that is the tricky part, and before we prove it we need a couple of results.

### Hensel’s Lemma

Hensel’s Lemma is the name of a set of related results on how factorization in $\mathbb Z_p[x]$ is related to factorization in $\mathbb F_p[x]$ under the map on coefficients given by $\mathbb Z_p / p \mathbb Z_p \cong \mathbb F_p$. Hensel’s Lemma is usually attributed to the result which allows us to (under certain conditions) “lift” roots of the reduced polynomials in $\mathbb F_p[x]$ to roots of the original polynomial in $\mathbb Z_p[x]$. The classic proof is constructive and relies on a variant of Newton’s method for approximating zeros of function based on derivative information. For the moment, we need a less precise (but still useful) version which allows us to test for factorization of polynomials in $\mathbb Z_p[x]$ by checking the factorization of the reduced polynomial in $\mathbb F_p[x]$.

The form of Hensel’s Lemma we prove here replaces the Newton’s method step with the fact that if $\varphi_1$ and $\varphi_2$ are coprime polynomials in $\mathbb F_p[x]$, then the ideal generated by $\varphi_1$ and $\varphi_2$ is all of $F_p[x]$, and hence given any $\eta \in \mathbb F_p[x]$ we can find $\psi_1, \psi_2 \in \mathbb F_p[x]$ such that $$\psi_1 \varphi_2 + \psi_2 \varphi_1 = \eta.$$ The existence of the solution $(\psi_1, \psi_2)$, which can be found constructively using the Division Algorithm, will be necessary for the inductive construction of a factorization for $f \in \mathbb Z_p[x]$ from the factorization of its reduction in $\mathbb F_p[x]$.

Hensel’s Lemma is important in its own right, especially as it gives us a method to looks for factorizations of polynomials in $\mathbb Z_p[x]$ (and $\mathbb Q_p[x]$) by considering a much easier factorization problem in $\mathbb F_p[x]$.

###### Hensel’s Lemma

Suppose $f(x) \in \mathbb Z_p[x]$ and write $\widehat{f}$ for the polynomial in $\mathbb F_p[x]$ formed by reducing the coefficients of $f$ modulo $p \mathbb Z_p$. If there exists a factorization $\widehat f(x) = \varphi_1(x) \varphi_2(x)$ in $\mathbb F_p[x],$ with $\varphi_1$ and $\varphi_2$ coprime, then there exists a factorization $f(x) = f_1(x) f_2(x)$ in $\mathbb Z[x]$ with $\varphi_1 = \widehat f_1$, $\varphi_2 = \widehat f_2$ and $\deg f_1 = \deg \varphi_1$.

###### Proof

We will construct sequences of polynomials $f_1^{(n)}$ and $f_2^{(n)}$ in $\mathbb Z_p[x]$ satisfying

- $\deg f_1^{(n)} = \deg \varphi_1$;
- ${f}_1^{(n)} = \varphi_1 \bmod p \mathbb Z_p$; ${f}_2^{(n)} = \varphi_2 \bmod p \mathbb Z_p$;
- $f_1^{(n+1)} \equiv f_1^{(n)} \bmod p^n \mathbb Z_p$; $f_2^{(n+1)} \equiv f_2^{(n)} \bmod p^n \mathbb Z_p$;
- $f \equiv f_1^{(n)} f_2^{(n)} \bmod p^n \mathbb Z_p$.

Given such sequences $(f_1^{(n)})$ and $(f_2^{(n)})$, $f_1 = \lim f_1^{(n)}$ and $f_2 = \lim f_2^{(n)}$ are defined and satisfy the conditions of the lemma.

Set $f_1^{(1)} = \varphi_1$ and $f_2^{(1)} = \varphi_2$, where we view $a \in \mathbb F_p$ as the constant base-$p$ expansion, $a + 0 \cdot p + 0 \cdot p^2 + \cdots$ in $\mathbb Q_p$.

The inductive hypothesis guarantees the existence of a polynomial $h^{(n)} \in \mathbb Z_p[x]$ such that $$f = f_1^{(n)} f_2^{(n)} + p^n h^{(n)}.$$

We define $f_i^{(n+1)}$ in terms of $f_i^{(n)}$ as $$f_i^{(n+1)} = f_i^{(n)} + p^n g_i^{(n)}, \quad i=1,2;$$ where $g_i^{(n)} \in \mathbb Z_p[x]$ with $\deg g_i^{(n)} < \deg \varphi_i$, and look for $g_i^{(n)}$ so that $(f_i^{(n)})$ satisfies the conclusions of the lemma.

Computing \begin{align*}f_1^{(n+1)} f_2^{(n+1)} &= f_1^{(n)} f_2^{(n)} + p^n(g_1^{(n)} f_2^{(n)} + g_2^{(n)} f_1^{(n)} ) + p^{2n} g_1^{(n)} g_2^{(n)} \\ &= f – p^n(-h^{(n)} + g_1^{(n)} f_2^{(n)} + g_2^{(n)} f_1^{(n)}) + p^{2n} g_1^{(n)} g_2^{(n)}.\end{align*} If we can find $g_1^{(n)}$ and $g_2^{(n)}$ such that $$p \mid \left(-h^{(n)} + g_1^{(n)} f_2^{(n)} + g_2^{(n)} f_1^{(n)}\right),$$ then there is some polynomial in $\mathbb Z_p[x]$, call it $h^{(n+1)}$ such that $$\left(-h^{(n)} + g_1^{(n)} f_2^{(n)} + g_2^{(n)} f_1^{(n)}\right) + p^{2n} g_1^{(n)} g_2^{(n)}= p h^{(n+1)}, $$ and $f = f_1^{(n+1)} f_2^{(n+1)} + p^{n+1} h^{(n+1)}, $ as desired.

To find $g_1^{(n)}$ and $g_2^{(n)}$, Let $\eta = \widehat{h}^{(n)} \in \mathbb F_p[x]$ and consider solutions to the equation $$-\eta + \psi_1 \varphi_2 + \psi_2 \varphi_1 \equiv 0 \bmod p, \qquad \psi_1, \psi_2 \in \mathbb F_p[x].$$ By hypothesis $\varphi_1$ and $\varphi_2$ are relatively prime in $\mathbb F_p[x]$ and therefore there is a non-trivial solution $(\psi_1, \psi_2)$. Let $g_1^{(n)}$ and $g_2^{(n)}$ be any choices of polynomial in $\mathbb Z_p[x]$ with $\widehat{g}_1^{(n)} = \psi_1$ and $\widehat{g}_2^{(n)} = \psi_2$. It follows that $$p \mid \left(-h^{(n)} + g_1^{(n)} f_2^{(n)} + g_2^{(n)} f_1^{(n)}\right),$$ and the lemma is established.

### The Ring of Integers in $K_v$

Fix $v = v_i$ and define $\mathfrak O_v = \{ \alpha \in K_v : \| \alpha \|_v \leq 1 \}$.

###### Theorem

$\mathfrak O_v$ is a ring. Moreover, it is the integral closure of $\mathbb Z_p$ in $K_v$.

###### Proof of Theorem

Suppose $\alpha \in \mathfrak O_v$ and $f(x) \in \mathbb Q_p[x]$ is a monic irreducible polynomial such that $f(\alpha) = 0$. We need to show that, in fact $f(x) \in \mathbb Z_p[x]$. Suppose $$f(x) = \sum_{i=0}^n c_i x^i \qquad c_n = 1.$$ We can find $b \in \mathbb Z_p$ such that $b c_i \in \mathbb Z_p$ for all $i=0,\ldots, n$. Indeed we can do this so that not all the $b c_i$ are in $p \mathbb Z_p$. However, notice that if $f$ has a coefficient not in $\mathbb Z_p$ then the leading coefficient, $b c_n$ is in $p \mathbb Z_p$. It follows that when we reduce the coefficients of $b f(x)$ modulo $p \mathbb Z_p$ we get a polynomial of lower degree. That is, $\widehat{f}(x) = \varphi_1(x) \cdot 1$ where $\varphi_1(x) \in \mathbb F_p[x]$ with $\deg \varphi_1 < \deg f$.

The Weak Hensel’s Lemma then implies that $b f(x)$ factors as $f_1(x) f_2(x)$ in $\mathbb Z_p[x]$ where $\deg f_1(x) = \deg \varphi_1 < \deg $. But then this implies that $f(x)$ is not irreducible in $\mathbb Q_p[x]$. This is a contradiction, and hence $f(x) \in \mathbb Z_p[x]$.

Finally, to see that $\mathfrak O_v$ is a ring, we need to show that it is closed under addition (the other ring axioms are immediate). Let $\alpha, \beta$ be nonzero $\mathfrak O_v$. Then if we denote the linear operators given by multiplication by $\alpha$ and $\beta$ on $K_v \cong \mathbb Q_p^{d_v}$ as $T_{\alpha}$ and $T_{\beta}$, then \begin{align*} \| \alpha – \beta \|_v &= | N_{K_v|\mathbb Q_p}(\alpha – \beta) |_p = | \det(T_{\alpha + \beta}) |_p \\ &= | \det T_{\alpha} + \det T_{\beta} |_p = | \det T_{\beta} |_p |\det (T_{\alpha/\beta} – I)|_p. \end{align*} Note that $\gamma = \alpha/\beta$ is in $\mathfrak O_v$ and hence, the characteristic polynomial of $\gamma$, $f_{\gamma}$ has all coefficients in $\mathbb Z_p$. Note that $f_{\gamma}(1) = \det (T_{\gamma} – I)$, and hence $$\| \alpha – \beta \|_v = \| \beta \|_v |f_{\gamma}(1)|_p. $$ But by assumption $\| \beta \|_v \leq 1$ and $|f_{\gamma}(1)|_p \leq 1$, because $f_{\gamma}(1)$ is the sum of the coefficients of $f_{\gamma}$ all of which have $| \cdot |_p \leq 1$. It follows that $\| \alpha – \beta \|_v \leq 1$, and hence $\mathfrak O_v$ is closed under addition.

### $\| \cdot \|_v$ is an Absolute Value

We finally can prove that $\| \cdot \|_v$ is a non-archimedean absolute value.

###### Theorem

The function $\| \cdot \|_v : K_v \rightarrow [0, \infty)$ given by $\| \alpha \|_v = \left| N_{K_v | \mathbb Q_p}(\alpha)\right|_p$ is a non-archimedean absolute value on $K_v$.

###### Proof

The only non-immediate property is the strong triangle inequality. Suppose $\alpha, \beta \in \mathbb K_v$ and $\| \alpha \|_v \leq \| \beta \|_v$.

Defining $\gamma = \alpha/\beta$ note that $\| \gamma \|_v = \| \alpha \|_v/\|\beta\|_v \leq 1$ and hence $\gamma \in \mathfrak O_v$. Then, $$\det(T_{\beta – \alpha}) = \det(T_{\beta} – T_{\alpha}) = \det T_{\beta} \det(I – T_{\alpha/\beta}) = \det(T_{\beta}) f_{\gamma}(1),$$ where $f_{\gamma}(x) \in \mathbb Z_p[x]$ is the monic characteristic polynomial of $\gamma$. It follows that $\|\beta – \alpha\|_v = \| \beta \|_v |f_{\gamma}(1)|_p \leq \| \beta \|_v$ as desired.

### The Places of $K$ above $p$

The $i$th irreducible factor of $f(x) = f_1(x) \cdots f_{\ell}(x) \in \mathbb Z_p[x]$ gives rise to the absolute value $\| \cdot \|_{v_i}$ on $K$ as restricted from $K_{v_i}$. Define the local degree of $K_{v_i}$ to be $d_i = [K_{v_i} : \mathbb Q_p] = \deg f_i.$ If $r \in \mathbb Q$, then multiplication by $r$ on $K_{v_i} \cong \mathbb Q_p^{d_i}$ is given by the constant matrix $r I$ where $I$ is the $d_i \times d_i$ identity matrix. It follows that $N_{K_{v_i} | \mathbb Q_p} = r^{d_i}$ and hence restricted to $\mathbb Q$, $\| \cdot \|_{v_i} = | \cdot |_{p}^{d_i}$. These absolute values represent all the different different places of $K$ that lie above $p$, a set we denote $\mathcal M_p(K)$. If $v \in \mathcal M_p(K)$ we will write $v|p$.

We choose a canonical absolute value for $v \in \mathcal M_p(K)$ by setting $| \cdot |_{v_i} = \| \cdot \|_{v}^{1/d}$. This choice is motivated by the fact that, if $r \in \mathbb Q$, $$|r|_p = \prod_{v | p} | r |_v. $$

In fact, we have the following prelude to the Product Formula.

###### Lemma

Suppose $\alpha \in K$, then $$\prod_{v \in \mathcal M_p(K)} |\alpha|_v = \left| N_{K|\mathbb Q}(\alpha)\right|^{1/d}_p.$$

###### Proof

Suppose $\alpha \in K$, then multiplication by $\alpha$ gives a linear transformation on $K \cong \mathbb Q^d$. Write $T_{\alpha}$ for this linear transformation. By definition $N_{K|\mathbb Q}(\alpha)$ is the determinant $T_{\alpha}$. Now $T_{\alpha}$ also makes sense as a linear transformation on $\mathbb Q_p[x]/f(x)\mathbb Q_p[x]$, which, as a vector space is isomorphic to $K \otimes_{\mathbb Q} \mathbb Q_p$, and we write $\overline T_{\alpha}$ for the linear transformation on $K \otimes_{\mathbb Q} \mathbb Q_p$. Note that, $\det \overline T_{\alpha} = \det T_{\alpha}$ because any basis for $K$ will lift to a basis for $K \otimes_{\mathbb Q} \mathbb Q_p$, and the matrices for $T_{\alpha}$ and $\overline T_{\alpha}$ are identical for these bases.

The Fundamental Theorem of Finitely Generated Modules over Principal Ideal Domains, means that as vector spaces $$\mathbb Q_p[x]/f(x)\mathbb Q_p[x] \cong \mathbb Q_p[x]/f_1(x) \mathbb Q_p[x] \oplus \cdots \oplus \mathbb Q_p[x]/f_{\ell}(x) \mathbb Q_p[x],$$ or what amounts to the same thing $$K \otimes_{\mathbb Q} \mathbb Q_p \cong K_{v_1} \otimes \cdots \otimes K_{v_{\ell}}.$$ Moreover, as vector subspaces the $K_{v_1}$ are invariant under $\overline T_{\alpha}$. It follows again from the Fundamental Theorem that $\overline T_{\alpha}$ decomposes as a direct sum $\overline T^{(1)}_{\alpha} \oplus \cdots \oplus \overline T^{(\ell)}_{\alpha}$, and $$\det T_{\alpha} = \det \overline T^{(1)}_{\alpha} \cdots \det \overline T^{(\ell)}_{\alpha}.$$ That is, $$N_{K|\mathbb Q}(\alpha) = N_{K_{v_1}|\mathbb Q_p}(\alpha) \cdots N_{K_{v_\ell}|\mathbb Q_p}(\alpha).$$ And hence $$| N_{K|\mathbb Q}(\alpha)|_p = \prod_{v \in \mathcal M_p(K)} |N_{K_{v}|\mathbb Q_p}(\alpha)|_p = \prod_{v \in \mathcal M_p(K)} \|\alpha\|_p = \prod_{v \in \mathcal M_p(K)} |\alpha|^d_p .$$

To calculate $|\alpha|_v$ we note that there is some basis for $K \otimes_{\mathbb Q} \mathbb Q_p$ for which the matrix of $T_{\alpha}$ is given by the Frobenius companion matrix of the associated irreducible factor of $f(x)$ when factored over $\mathbb Q_p$. The determinant of this matrix is the constant coefficient, say $c_0$, of that irreducible factor, and $|\alpha|_v = |c_0|_p^{1/d}$.

## The Product Formula

##### The Product Formula For Number Fields

Let $\mathcal M(K)$ denote the set of places of $K$. Suppose $v \in \mathcal M(K)$, let $d_v = [K_v : \mathbb Q_p]$ be the local degree, and and let $| \cdot |_v$ be the absolute value in $v$ which is equal to $| \cdot |_p^{d_v/d}$ on $\mathbb Q$. Then, for all nonzero $\alpha \in K$, $$\prod_{v \in \mathcal M(K)} | \alpha |_v = 1.$$

###### Proof

The proof reduces to the Product Formula for the rational numbers (see Absolute Values and Completions of $\mathbb Q$). Specifically, $$ \prod_v | \alpha |_v = \prod_{p \in \mathcal M(\mathbb Q)} \prod_{v \in \mathcal M_p(K)} |\alpha|_v = \prod_{p \in \mathcal M(\mathbb Q)} |N_{K|\mathbb Q}(\alpha)|^{1/d}_p = 1,$$ where the penultimate equality is the lemma (and the analogous fact for archimedean places of $K$) and the final equality is the invocation of the Product Formula for $N_{K|\mathbb Q}(\alpha) \in \mathbb Q$.

## The General Setup

So far, in the non-archimedean situation, we have considered $K | \mathbb Q$ with places $v | p$. However we can replace $\mathbb Q$ with some other base field $k$ and an appropriate place $u$ of $k$ with $v | u$ and have many of our previous results still hold. Usually the necessary changes to the proofs to produce these more general results are minor, and so we will not prove all details other than to wave our hands at any necessary changes from the $k = \mathbb Q$, $u = p$ case.

Let $k$ be a number field and suppose $u$ is a place of $k$ lying above $p \in \mathcal M(\mathbb Q)$. If $\| \cdot \|$ is any absolute value on $k$, then we may form the completion $k_u$ by taking the ring of Cauchy sequences in $k$ (Cauchy with respect to $\| \cdot \|$) and modding out by the maximal ideal formed from sequences converging to 0 (again with respect to $\| \cdot \|$). This is a field, and a generic element looks like an equivalence class of Cauchy sequences in $k$ whose elements differ by a sequence which converges to 0. $k$, represented by constant sequences, is dense in $k_u$ and we may extend $\| \cdot \|_u$ to $k_u$. If $u | \infty$ (that is, $\| \cdot \|_u$ is archimedean), then $k_u$ is equal to $\mathbb R$ or $\mathbb C$ and $\| \cdot \|_u$ is some power of the usual absolute value. We will circle back to that case, but for now we will assume that $p$ is a rational prime, and $u$ is a non-archimedean absolute value.

First we associate to $\| \cdot \|$ to a prime ideal $\mf p$ in the ring of integers $\mf o$ of $k$.

###### Theorem

- If $\alpha \in \mf o$, then $\| \alpha \| \leq 1$.
- $\mf p := \{ \alpha \in \mf o : \| \cdot \| < 1 \}$ is a prime ideal in $\mf o$.

###### Proof

The strong triangle inequality implies that $\| c \| \leq 1$ for all $c \in \mathbb Z$. If $\alpha$ is in $\mf o$ then there exists a polynomial $x^n + c_{n-1} x^{n-1} + \cdots + c_1 x + c_0$ with integer coefficients that vanishes at $\alpha$. If $\|\alpha\| > 1$ then $$0 = \| \alpha^n + c_{n-1} \alpha^{n-1} + \cdots + c_1 \alpha + c_0 \| = \| \alpha^n \| > 1,$$ where the penultimate relation follows from the case of equality in the strong triangle inequality because $\| \alpha^n \| > \| c_j \alpha^j\| \geq \|\alpha\|^j$. This is an obvious contradiction, and thus $\| \alpha \| \leq 1$.

For the second statement, it is clear that $\mf p$ is an ideal of $\mf o$; the only questionable axiom is additivity, but as always the strong triangle inequality comes to the rescue. Now, if $\alpha \in \mf p$ and $\alpha = \beta \delta$ for some $\beta, \delta \in \mf o$, then clearly either $\| \beta \| < 1$ or $\| \delta \| < 1$ and hence $\mf p$ is a prime ideal.

It will turn out the this association between non-archimedean places and prime ideals is in fact a bijection, and we will often use $\mf p$ to represent the place associated to the prime ideal. This is, more-or-less, the content of Ostrowski’s Theorem for number fields. In this situation $\mf p | p$ has a meaning in terms and a different meaning in terms of places. However, $\mf p | p$ is simultaneously true or simultaneously false for these different interpretations.

One canonical representative of the place indexed by $\mf p$ is the $\mf p$-adic absolute value given by $$\| \cdot \|_{\mf p} = (\mathbb N \mf p)^{-v_{\mf p}(\cdot)}, $$ where $v_{\mf p}(\alpha)$ is the valuation of $\alpha$; the largest integer $n$ with $\alpha \in \mf p^n$. Here $\mathbb N \mf p$ is the norm of $\mf p$, that is $\mathbb N \mf p = [ \mf o : \mf p ]$. We also define the canonical absolute value $| \cdot |_{\mf p} = \| \cdot \|_{\mf p}^{1/d}$. We will see that this notation is consistent with our previous definitions for $\| \cdot \|_v$ and $| \cdot |_v$ when $v = \mf p$.

Completing $k$ and $\mf o$ with respect to $\mf p$ produces the completions $k_{\mf p}$ and $\mf o_{\mf p}$. The units $U_{\mf p}$ in $\mf o_{\mf p}$ consist of all elements of absolute value 1. In particular, prime ideals not equal to $\mf p$ embed as subsets of $U_{\mf p}$ and hence do not maintain their identity as ideals under the embedding. Many authors continue to use $\mf p \subset \mf o_{\mf p}$ for the maximal ideal formed by completing $\mf p$. However, we will distinguish the completion by denoting it $\mathbf m_{\mf p}$. This is the unique maximal ideal in $\mf o_{\mf p}$. This ideal is principal, and we choose $\pi = \pi_{\mf p}$ to be a generator or uniformizer of $\mf m_{\mf p}$.

The quotient $\mf o_{\mf p} / \mf m_{\mf p}$ is a finite field, and we will denote by $q$ and $f_{\mf p}$ the integers $$q := [\mf o_{\mf p} : \mf m_{\mf p}] = p^{f_{\mf p}}.$$ We identify $\mf o_{\mf p} / \mf m_{\mf p}$ with $\mathbb F_q$. Like in $\mathbb Q_p$ there is a series representation for the elements of $K_{\mf p}$. The proof follows *mutatis mutandis* from the $\mathbb Q_p$ case (see The Algebra and Geometry of $\mathbb Q_p$).

###### Theorem

Suppose $\alpha \in \mf k_{\mf p}$ then $\alpha$ there exists integer $n_0$ and $c_{n_0}, c_{n_0+1}, \ldots, \in \mathbb F_q$ such that $\alpha$ can be represented by the sequence of partial sums of $$\sum_{n=n_0}^{\infty} c_n \pi^n. $$ If $\alpha \in \mf o_{\mf p}$ then $n_0$ can be taken to be 0.

### Notation

##### Number fields

$f(x), g(x), h(x),$ etc. | Polynomials, often in $\mathbb Q[x]$ or $k[x]$ |

$k, K, L$ | Number fields |

$\alpha, \beta, \gamma,$ etc | Generic 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 $\mf O$. |

$\mathbb N \mf a, etc$ etc | The ideal norm $\mathbb N \mf a = [\mf o : \mf a]$. |

##### Absolute Values

$| \cdot |$, $| \cdot |_p$, $| \cdot |_{\infty}$ | A generic absolute value, the $p$-adic absolute value on $\mathbb Q_p$ and the usual absolute value on $\mathbb R$ (respectively). |

$\mathcal M(K)$, $\mathcal M_p(K)$, $\mathcal M_{\infty}(K)$ | The places of $K$, the places of $K$ over rational place $p$, the archimedean places of $K$. |

$v | p, v | \infty$ | Shorthand for $v \in \mathcal M_p(K)$ and $v \in \mathcal M_{\infty}(K)$. |

$\mf p | p$, $\mf p \in \mathcal M_p(K)$ | Non-archimedean places indexed by prime ideals |

$\mathbb Q_p$, $K_v$ | The completion of $\mathbb Q$ with respect to the place $p$, and the completion of $K$ with respect to the place $v$. |

$\mf o_{\mf p}$, $\mf o_v$, $\mf O_{\mf p}$, $\mf O_v$ | Local integers. The completion of the integers $\mf o$ or $\mf O$ with respect to $v$ or $\mf p$. |

$\mf m_{\mf p}$, $\mf m_v$ | The maximal ideal in the local integers. |

$d_v = [K_v : \mathbb Q_p], d = [K : \mathbb Q]$ | The local and global degrees of the place $v \in \mathcal M_p(K)$. |

$\| \cdot \|_v, \| \cdot \|_{\mf p}$ | The absolute value in the place $v = \mf p$ given by $| N_{K_v|\mathbb Q_p} |_p$. |

$| \cdot |_v, | \cdot |_{\mf p}$ | $\mathbb N \mf a,$ etc |

Another theorem of Alexander Ostrowski has it that any field complete with respect to an Archimedean absolute value is isomorphic to either the real or the complex numbers, and the valuation is equivalent to the usual one.

You are absolutely correct! Thanks for the correction.