Hensel’s lemma, valuations, and $p$-adic numbers

1 Hensel’s lemma

Let $p$ be prime and $f(x)\in \mathbb{Z}[x]$. Suppose that , satisfies

$$f(a_0)\equiv 0(modp)$$

and

$$f^{\prime }(a_0)\not\equiv 0(modp).$$

Using the power series expansion

$$f(a_0+h)=f(a_0)+f^{\prime }(a_0)h+\frac{f^{\prime \prime }(a_0)}{2}{h}^2+\mathrm{\cdots },$$

for any $y\in \mathbb{Z}$ we have

$$f(a_0+py)=f(a_0)+f^{\prime }(a_0)py+\frac{f^{\prime \prime }(a_0)}{2}{p}^2{y}^2+\mathrm{\cdots }$$

so

$$\frac{f(a_0+py)}{p}=\frac{f(a_0)}{p}+f^{\prime }(a_0)y+\frac{f^{\prime \prime }(a_0)}{2}p{y}^2+\mathrm{\cdots }.$$

Because $f(a_0)\equiv 0(modp)$, each term on the right-hand side is an integer. Then, $f(a_0+py)\equiv 0(mod{p}^2)$ is equivalent to

$$\frac{f(a_0)}{p}+f^{\prime }(a_0)y+\frac{f^{\prime \prime }(a_0)}{2}p{y}^2+\mathrm{\cdots }\equiv 0(modp),$$

i.e.,

$$f^{\prime }(a_0)y\equiv -\frac{f(a_0)}{p}(modp).$$

Because $f^{\prime }(a_0)\not\equiv 0(modp)$, there is a unique $y(modp)$ that solves the above congruence, so there is a unique $y(modp)$ that solves $f(a_0+py)\equiv 0(mod{p}^2)$. This $y$ is

$$y\equiv -\frac{f(a_0)}{p}{(f^{\prime }(a_0))}^{-1}(modp).$$

Let be $a_1\equiv y(modp)$.

Suppose that

satisfies

$$f(x)\equiv 0(mod{p}^{l-1})$$

and

$$f^{\prime }(x)\not\equiv 0(modp).$$

Using the power series expansion

$$f(x+h)=f(x)+f^{\prime }(x)h+\frac{f^{\prime \prime }(x)}{2}{h}^2+\mathrm{\cdots },$$

for any $y\in \mathbb{Z}$ we have

$$f(x+p^{l-1}y)=f(x)+f^{\prime }(x)p^{l-1}y+\frac{f^{\prime \prime }(x)}{2}{p}^{2l-2}{y}^2+\mathrm{\cdots },$$

i.e.

$$\frac{f(x+p^{l-1}y)}{p^{l-1}}=\frac{f(x)}{p^{l-1}}+f^{\prime }(x)y+\frac{f^{\prime \prime }(x)}{2}{p}^{l-1}{y}^2+\mathrm{\cdots }.$$

Because $f(x)\equiv 0(mod{p}^{l-1})$, each term on the right-hand side is an integer. Then, $f(x+p^{l-1}y)\equiv 0(mod{p}^l)$ is equivalent to

$$\frac{f(x)}{p^{l-1}}+f^{\prime }(x)y+\frac{f^{\prime \prime }(x)}{2}{p}^{l-1}{y}^2+\mathrm{\cdots }\equiv 0(modp),$$

i.e.,

$$f^{\prime }(x)y\equiv -\frac{f(x)}{p^{l-1}}(modp).$$

Because $f^{\prime }(x)\not\equiv 0(modp)$, there is a unique $y(modp)$ that solves the above congruence, so there is a unique $y(modp)$ that solves $f(x+p^{l-1}y)\equiv 0(mod{p}^l)$. This $y$ is

$$y\equiv -\frac{f(x)}{p^{l-1}}{(f^{\prime }(x))}^{-1}(modp).$$

Let be $a_{l-1}\equiv y(modp)$.

We have thus inductively defined a sequence $a_0,a_1,a_2,\mathrm{\dots }$, with , such that for any $l$,

$$f(a_0+a_{1}p+\mathrm{\cdots }+a_{l-1}{p}^{l-1})\equiv 0(mod{p}^l).$$

We wish to make sense of the infinite expression

$$a_0+a_{1}p+a_2{p}^2+a_3{p}^3+\mathrm{\cdots }$$

Calling this $x$, it ought to be the case that $f(x)\equiv 0(modp)$, $f(x)\equiv 0(mod{p}^2)$, $f(x)\equiv 0(mod{p}^3)$, etc.

Example 1.

Take $p=3$ and $f(x)=x^2-7$, $f^{\prime }(x)=2x$. The two conditions $f(x)\equiv 0(modp)$ and $f^{\prime }(x)\not\equiv 0(modp)$ are satisfied both by $a_0=1$ and $a_0=2$. Take $a_0=1$. Then

$$a_1\equiv -\frac{f(1)}{3}{(f^{\prime }(1))}^{-1}\equiv -\frac{-6}{3}{(2)}^{-1}\equiv 1(mod3).$$

So $a_1=1$. Then,

$$a_2\equiv -\frac{f(1+1\cdot 3)}{3^2}{(f^{\prime }(1+1\cdot 3))}^{-1}\equiv -\frac{9}{9}{(8)}^{-1}\equiv -2\equiv 1(mod3).$$

So $a_2=1$. Then,

$$a_3\equiv -\frac{f(1+1\cdot 3+1\cdot 3^2)}{3^3}{(f^{\prime }(1+1\cdot 3+1\cdot 3^2))}^{-1}\equiv -6\cdot 2\equiv 0(mod3).$$

So, $a_3=0$. Then,

$$a_4\equiv -\frac{f(1+1\cdot 3+1\cdot 3^2+0\cdot 3^3)}{3^4}{(f^{\prime }(1+1\cdot 3+1\cdot 3^2+0\cdot 3^3))}^{-1}\equiv -2\cdot 2\equiv 2(mod3).$$

So, $a_4=2$, etc.

2 Absolute values on fields

If $K$ is a field, an absolute value on $K$ is a map $|\cdot |:K\to {ℝ}_{\ge 0}$ such that $|x|=0$ if and only if $x=0$, $|xy|=|x||y|$, and $|x+y|\le |x|+|y|$. The trivial absolute value on $K$ is $|0|=0$ and $|x|=1$ for all nonzero $x\in K$.

If $|\cdot |$ is an absolute value on $K$, then $d(x,y)=|x-y|$ is a metric on $K$. The trivial absolute value yields the discrete metric. Two absolute values $|\cdot {|}_1,|\cdot {|}_2$ on $K$ are said to be equivalent if they induce the same topology on $K$.

The following theorem characterizes equivalent absolute values.

Theorem 2.

Two nontrivial absolute values $\mathrm{|}\mathrm{\cdot }{\mathrm{|}}_{\mathrm{1}}\mathrm{,}\mathrm{|}\mathrm{\cdot }{\mathrm{|}}_{\mathrm{2}}$ are equivalent if and only if there is some real $s\mathrm{>}\mathrm{0}$ such that

$${|x|}_1={|x|}_2^s,x\in K.$$

Proof.

Suppose that $s>0$ and that ${|x|}_1={|x|}_2^s$ for all $x\in K$. Then

$B_{d_1}(x,r)$

$=B_{d_2}(x,r^{1/s}).$

Since the collection of open balls for $d_1$ is equal to the collection of open balls for $d_2$, the absolute values $|\cdot {|}_1,|\cdot {|}_2$ induce the same topology on $K$.

Suppose that $|\cdot {|}_1,|\cdot {|}_2$ are equivalent. If then $d_1(x^n,0)={|x^n|}_1={|x|}_1^n\to 0$ as $n\to \mathrm{\infty }$. Thus $x^n\to 0$ in $d_1$ and hence, because the topologies induced by $|\cdot {|}_1$ and $|\cdot {|}_2$ are equal, $x^n\to 0$ in $d_2$, i.e. ${|x|}_2^n={|x^n|}_2=d_2(x^n,0)\to 0$. Therefore . Thus, if and only if .

Let $y\in K$ such that ${|y|}_1>1$ (there is such an element because $|\cdot {|}_1$ is nontrivial and ${|y^{-1}|}_1={|y|}_1^{-1}$) and let $x\in K$ with ${|x|}_1\ne 0,1$. There is some nonzero $\alpha \in ℝ$ such that ${|x|}_1={|y|}_1^{\alpha }$. Let $\frac{m_i}{n_i}\in \mathbb{Q}$ all be greater than $\alpha$ and converge to $\alpha$. Then, because ${|y|}_1>1$, we have , hence , hence , hence

Because $|\cdot {|}_1$ and $|\cdot {|}_2$ are equivalent,

so . Taking $i\to \mathrm{\infty }$ gives

$${|x|}_2\le {|y|}_2^{\alpha }.$$

Similarly, we check that

$${|x|}_2\ge {|y|}_2^{\alpha }.$$

Therefore,

$${|x|}_2={|y|}_2^{\alpha }.$$

Using this and ${|x|}_1={|y|}_1^{\alpha }$, we have

$$\log {|x|}_1=\alpha \log {|y|}_1,\log {|x|}_2=\alpha \log {|y|}_2,$$

and so, as $\alpha \ne 0$,

$$\frac{\log {|x|}_1}{\log {|x|}_2}=\frac{\log {|y|}_1}{\log {|y|}_2}.$$

This is true for any $x\in K$ with ${|x|}_1\ne 0,1$. We define $s\in ℝ$ to be this common value. The fact that ${|y|}_1>1$ implies, because $|\cdot {|}_1$ and $|\cdot {|}_2$ are equivalent, that ${|y|}_2>1$, and so $s>0$.

Now take $x\in K$. If $x=0$ then ${|x|}_1=0=0^s={|x|}_2^s$. Because $|\cdot {|}_1$ and $|\cdot {|}_2$ are equivalent, ${|x|}_2>1$ implies that ${|x|}_1>1$ and implies that , so if ${|x|}_1=1$ then ${|x|}_2=1$ and hence ${|x|}_1=1=1^s={|x|}_2^s$. If ${|x|}_1\ne 0,1$, then the above shows that

$$\frac{\log {|x|}_1}{\log {|x|}_2}=s,$$

i.e., ${|x|}_1={|x|}_2^s$, proving the claim. ∎

An absolute value $|\cdot |:K\to {ℝ}_{\ge 0}$ is said to be non-Archimedean if

$$|x+y|\le \max \{|x|,|y|\},x,y\in K.$$

An absolute value is called Archimedean if it is not non-Archimedean. For example, the absolute value on the field $ℝ$ is Archimedean, since, for example, $|1+1|=2>\max \{|1|,|1|\}=1$.

Lemma 3.

If $\mathrm{|}\mathrm{\cdot }\mathrm{|}$ is a non-Archimedean absolute value on a field $K$ and $\mathrm{|}x\mathrm{|}\mathrm{\ne }\mathrm{|}y\mathrm{|}$, then

$$|x+y|=\max \{|x|,|y|\}.$$

3 Valuations

A valuation on a field $K$ is a function $v:K\to ℝ\cup \{\mathrm{\infty }\}$ satisfying $v(x)=\mathrm{\infty }$ if and only if $x=0$, $v(xy)=v(x)+v(y)$, and

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

The trivial valuation is $v(x)=0$ for $x\ne 0$ and $v(0)=\mathrm{\infty }$.

Lemma 4.

Let $v$ be a valuation on a field $K$. If $v\mathit{}\mathrm{(}x\mathrm{)}\mathrm{\ne }v\mathit{}\mathrm{(}y\mathrm{)}$, then $v\mathit{}\mathrm{(}x\mathrm{+}y\mathrm{)}\mathrm{=}\min \mathit{}\mathrm{\{}v\mathit{}\mathrm{(}x\mathrm{)}\mathrm{,}v\mathit{}\mathrm{(}y\mathrm{)}\mathrm{\}}$.

Proof.

Take . For $x=0$,

$$v(x+y)=v(y)=\min \{\mathrm{\infty },v(y)\}=\min \{v(x),v(y)\}.$$

For $x\ne 0$, assume by contradiction that $\min \{v(x+y),v(x)\}=v(x)$. Then, since $v(-x)=v(-1\cdot x)=v(-1)+v(x)=v(x)$,

$$v(x)>v(y)=v(x+y-x)\ge \min \{v(x+y),v(x)\}=v(x),$$

a contradiction. Hence $\min \{v(x+y),v(x)\}=v(x+y)$. Then

$v(y)$ $=v(x+y-x)$

$\ge \min \{v(x+y),v(x)\}$

$=v(x+y)$

$\ge \min \{v(x),v(y)\}$

$=v(y).$

Hence $v(x+y)=v(y)=\min \{v(x),v(y)\}$, completing the proof. ∎

Theorem 5.

Let $K$ be a field. If $\mathrm{|}\mathrm{\cdot }\mathrm{|}$ is a non-Archimedean absolute value on $K$ and $s\mathrm{>}\mathrm{0}$, then $v_s\mathrm{:}K\mathrm{\to }\mathrm{R}\mathrm{\cup }\mathrm{\{}\mathrm{\infty }\mathrm{\}}$ defined by $v_s\mathit{}\mathrm{(}x\mathrm{)}\mathrm{=}\mathrm{-}s\mathit{}\log \mathit{}\mathrm{|}x\mathrm{|}$ for $x\mathrm{\ne }\mathrm{0}$ and $v_s\mathit{}\mathrm{(}\mathrm{0}\mathrm{)}\mathrm{=}\mathrm{\infty }$ is a valuation on $K$.

If $v$ is a valuation on $K$ and $q\mathrm{>}\mathrm{1}$, then the function $\mathrm{|}\mathrm{\cdot }{\mathrm{|}}_q\mathrm{:}K\mathrm{\to }{\mathrm{R}}_{\mathrm{\ge }\mathrm{0}}$ defined by ${\mathrm{|}x\mathrm{|}}_q\mathrm{=}{q}^{\mathrm{-}v\mathit{}\mathrm{(}x\mathrm{)}}$ for $x\mathrm{\ne }\mathrm{0}$ and ${\mathrm{|}\mathrm{0}\mathrm{|}}_q\mathrm{=}\mathrm{0}$ is a non-Archimedean absolute value on $K$.

Proof.

Suppose that $|\cdot |$ is a non-Archimedean absolute value on $K$ and that $s>0$. Let $x,y\in K$. If either is $0$, then it is immediate that $v_s(xy)=\mathrm{\infty }=v_s(x)+v_s(y)$. If neither is $0$, then

$$v_s(xy)=-s\log |xy|=-s\log (|x||y|)=-s\log |x|-s\log |y|=v_s(x)+v_s(y).$$

Now, if both $x,y$ are $0$ then

$$v_s(x+y)=v_s(0)=\mathrm{\infty }=\min \{\mathrm{\infty },\mathrm{\infty }\}=\min \{v_s(x),v_s(y)\}.$$

If $x=0$ and $y\ne 0$ then

$$v_s(x+y)=v_s(y)=-s\log |y|=\min \{-s\log |y|,\mathrm{\infty }\}=\min \{v_s(y),v_s(x)\}.$$

If neither $x,y$ is $0$ but $x=-y$, then

$$v_s(x+y)=v_s(0)=\mathrm{\infty }\ge \min \{v_s(x),v_s(y)\}.$$

Finally, if neither $x,y$ is $0$ and $x\ne -y$, then, because $|\cdot |$ is non-Archimedean,

$v_s(x+y)$ $=-s\log |x+y|$

$\ge -s\log (\max \{|x|,|y|\})$

$=\min \{-s\log |x|,-s\log |y|\}$

$=\min \{v_s(x),v_s(y)\}.$

Thus $v_s$ is a valuation on $K$.

Suppose that $v$ is a valuation on $K$ and that $q>1$. If $x,y$ are nonzero, then

$${|xy|}_q=q^{-v(xy)}=q^{-v(x)-v(y)}=q^{-v(x)}{q}^{-v(y)}={|x|}_q{|y|}_q.$$

Let $x,y\in K$. To show that ${|x+y|}_q\le {|x|}_q+{|y|}_q$, it suffices to show that ${|x+y|}_q\le \max \{{|x|}_q,{|y|}_q\}$; proving this will establish that $|\cdot {|}_q$ is an absolute value and furthermore that $|\cdot {|}_q$ is non-Archimedean. If $x,y$ are both $0$, then ${|x+y|}_q={|0|}_q=0=\max \{0,0\}=\max \{{|x|}_q,{|y|}_q\}$. If $x=0$ and $y\ne 0$, then ${|x+y|}_q={|y|}_q=q^{-v(y)}=\max \{q^{-v(y)},0\}=\max \{{|y|}_q,{|x|}_q\}$. If neither $x,y$ is $0$ but $x=-y$, then

$${|x+y|}_q={|0|}_q=0\le \max \{{|x|}_q,{|y|}_q\}.$$

Finally, if neither $x,y$ is $0$ and $x\ne -y$, then

${|x+y|}_q$ $=q^{-v(x+y)}$

$\le q^{-\min \{v(x),v(y)\}}$

$=\max \{q^{-v(x)},q^{-v(y)}\}$

$=\max \{{|x|}_q,{|y|}_q\}.$

∎

Two valuations $v_1,v_2$ on a field $K$ are said to be equivalent if there is some real $s>0$ such that

$$v_1=s{v}_2.$$

A valuation $v$ on a field $K$ is said to be discrete if there is some real $s>0$ such that

$$v(K^{*})=s\mathbb{Z}.$$

A valuation is said to be normalized if

$$v(K^{*})=\mathbb{Z}.$$

4 Valuation rings

Theorem 6.

If $K$ is a field and $v$ is a nontrivial valuation on $K$, then

$${𝒪}_v=\{x\in K:v(x)\ge 0\}$$

is a maximal proper subring of $K$, and for all $x\mathrm{\ne }\mathrm{0}$, $x\mathrm{\in }{\mathrm{O}}_v$ or $x^{\mathrm{-}\mathrm{1}}\mathrm{\in }{\mathrm{O}}_v$. The set

$$\{x\in K:v(x)=0\}$$

is the group of invertible elements of ${\mathrm{O}}_v$, and the set

$${𝔭}_v=\{x\in K:v(x)>0\}$$

is the unique maximal ideal of ${\mathrm{O}}_v$.

Proof.

It is immediate that $0,1\in {𝒪}_v$. For $x\in {𝒪}_v$, $v(-x)=v(x)\ge 0$, so $-x\in {𝒪}_v$. For $x,y\in {𝒪}_v$, $v(xy)=v(x)+v(y)\ge 0$, so $xy\in {𝒪}_v$. And $v(x+y)\ge \min \{v(x),v(y)\}\ge 0$, so $x+y\in {𝒪}_v$. Thus ${𝒪}_v$ is a subring of $K$. For nonzero $x\in K$, if $v(x)\ge 0$ then $x\in {𝒪}_v$, and if then $v(x^{-1})=-v(x)>0$, so $x^{-1}\in {𝒪}_v$.

Since $v$ is nontrivial, there is some $x\in K$ with $v(x)\ne 0,\mathrm{\infty }$. If $x\in {𝒪}_v$ then $v(x)>0$ and so , giving $x^{-1}\notin {𝒪}_v$. Hence ${𝒪}_v\ne K$, showing that ${𝒪}_v$ is a proper subring of $K$.

To show that ${𝒪}_v$ is a maximal proper subring, it suffices to show that if $z\in K\setminus {𝒪}_v$ then ${𝒪}_v[z]=K$, i.e., that the smallest ring containing ${𝒪}_v$ and $z$ is $K$. As $z\notin {𝒪}_v$, . Let $y\in K$. For any positive integer $j$ we have $v(y{z}^{-j})=v(y)-jv(z)$, and because , there is some $j=j(y)$ such that $v(y{z}^{-j})>0$. For this $j$, $y{z}^{-j}\in {𝒪}_v$. Hence $y\in {𝒪}_v[z]$, and so ${𝒪}_v[z]=K$, showing that ${𝒪}_v$ is a maximal proper subring.

Suppose that $x\in {𝒪}_v$ and $x^{-1}\in {𝒪}_v$. If $v(x)>0$, then , contradicting that $x^{-1}\in {𝒪}_v$. Hence $v(x)=0$. If $v(x)=0$, then, as $x^{-1}\in K$, $v(x^{-1})=-v(x)=0$, so $x^{-1}\in {𝒪}_v$, hence $x$ is an element of ${𝒪}_v$ whose inverse is in ${𝒪}_v$.

Let $x,y\in {𝔭}_v$. Then, since $v(x)>0$ and $v(y)>0$,

$$v(x-y)\ge \min \{v(x),v(-y)\}=\min \{v(x),v(y)\}>0,$$

showing that $x-y\in {𝔭}_v$, and thus that ${𝔭}_v$ is an additive subgroup of ${𝒪}_v$. Let $x\in {𝔭}_v$ and $z\in {𝒪}_v$. Then, since $v(z)\ge 0$ and $v(x)>0$,

$$v(zx)=v(z)+v(x)\ge v(x)>0,$$

showing that $zx\in {𝔭}_v$. Therefore ${𝔭}_v$ is an ideal in the ring ${𝒪}_v$. Since $v(1)=0$, $1\notin {𝔭}_v$, so ${𝔭}_v$ is a proper ideal.

The fact that ${𝔭}_v$ is maximal follows from it being the set of noninvertible elements of ${𝒪}_v$. Suppose that $B$ is a maximal ideal $B$ of ${𝒪}_v$. Because $B$ is a proper ideal it contains no invertible elements, and hence is contained in ${𝔭}_v$, the set of noninvertible elements of ${𝒪}_v$. Since $B$ is maximal, it must be that $B={𝔭}_v$. Therefore, any maximal ideal of ${𝒪}_v$ is ${𝔭}_v$, showing that ${𝔭}_v$ is the unique maximal ideal of ${𝒪}_v$. ∎

The above ring ${𝒪}_v$ is called the valuation ring. Generally, a ring that has a unique maximal ideal is called a local ring, and thus the above theorem shows that the valuation ring is a local ring. We call the quotient ${𝒪}_v/{𝔭}_v$ the residue field of ${\mathrm{O}}_v$.

Lemma 7.

If $v$ is a normalized valuation on a field $K$ then for all nonzero $x\mathrm{\in }K$ and $t\mathrm{\in }{\mathrm{p}}_v$, $v\mathit{}\mathrm{(}t\mathrm{)}\mathrm{=}\mathrm{1}$, there is some $u\mathrm{\in }{\mathrm{O}}_v^{\mathrm{*}}$ such that

$$x=u{t}^n,n=v(x).$$

Proof.

Since $x\ne 0$, $v(x)=n\in \mathbb{Z}$. Hence $v(x{t}^{-n})=v(x)-nv(t)=v(x)-n=0$, and therefore $u=x{t}^{-n}\in {𝒪}^{*}$. Then $x=u{t}^n$, completing the proof. ∎

Theorem 8.

If $v$ is a normalized valuation on a field $K$, then ${\mathrm{O}}_v$ is a principal ideal domain. If $A$ is a nonzero ideal of ${\mathrm{O}}_v$, then there is some $t\mathrm{\in }\mathrm{p}$, $v\mathit{}\mathrm{(}t\mathrm{)}\mathrm{=}\mathrm{1}$ and $n\mathrm{\ge }\mathrm{0}$ such that

$$A=t^n{𝒪}_v=\{x\in K:v(x)\ge n\}={𝔭}_v^n,$$

and

$${𝔭}_v^n/{𝔭}_v^{n+1}\cong {𝒪}_v/{𝔭}_v,$$

as ${\mathrm{O}}_v\mathrm{/}{\mathrm{p}}_v$-linear vector spaces.

Proof.

Let $A\ne \{0\}$ be an ideal of ${𝒪}_v$. For any $y\in A$, $v(y)\ge 0$, and we take $x\in A$ such that

$$v(x)=\min \{v(y):y\in A\}.$$ (1)

Since $v(K^{*})=\mathbb{Z}$, there is some $t\in K$ with $v(t)=1$, and because $v(t)>0$, $t\in {𝔭}_v$. By Lemma 7, there is some $u\in {𝒪}^{*}$ such that $x=u{t}^n$, $n=v(x)$. For any $z\in 𝒪$, $xz\in A$ and so $t^{n}z\in A$. Thus $t^n{𝒪}_v\subset A$. On the other hand, let $y\in A$. Then also by Lemma 7 there is some $w\in {𝒪}_v^{*}$ such that $y=w{t}^m$, $m=v(y)$. By (1), $m=v(y)\ge v(x)=n$, so $v(t^{m-n})=(m-n)v(t)=m-n\ge 0$ so $t^{m-n}\in {𝒪}_v$, giving

$$y=w{t}^m=t^n(w{t}^{m-n})\in t^n{𝒪}_v.$$

Therefore $A\subset t^n{𝒪}_v$, and so $A=t^n{𝒪}_v$. That is, $A$ is the principal ideal generated by $t^n$, which shows that ${𝒪}_v$ is a principal ideal domain.

Let $t\in {𝔭}_v$ with $v(t)=1$, and define $\varphi :{𝔭}_v^n\to {𝒪}_v/{𝔭}_v$ by $v(a{t}^n)=a+𝔭$, for $a\in {𝒪}_v$. ∎

Lemma 9.

If $v_{\mathrm{1}}\mathrm{,}{v}_{\mathrm{2}}$ are discrete valuations on a field $K$ such that ${\mathrm{O}}_{v_{\mathrm{1}}}\mathrm{=}{\mathrm{O}}_{v_{\mathrm{2}}}$, then $v_{\mathrm{1}}$ and $v_{\mathrm{2}}$ are equivalent.

5 p-adic valuations

Fix a prime number $p$. For nonzero $a\in \mathbb{Q}$, there are unique integers $n,r,s$ satisfying

$$a=\frac{r}{s}{p}^n,$$

where $r,s$ are coprime, $s>0$, and $p\nmid rs$. We define $v_p(a)=n$. Furthermore, we define $v_p(0)=\mathrm{\infty }$.

Theorem 10.

$v_p:\mathbb{Q}\to ℝ\cup \{\mathrm{\infty }\}$ is a normalized valuation.

Proof.

For nonzero $a,b\in \mathbb{Q}$, write

$$a=\frac{r_1}{s_1}{p}^m,b=\frac{r_2}{s_2}{p}^n,$$

where $\gcd (r_1,s_1)=\gcd (r_2,s_2)=1$, $s_1,s_2>0$, and $p\nmid r_1{s}_1,p\nmid r_2{s}_2$. Then,

$$ab=\frac{r_1{r}_2}{s_1{s}_2}{p}^{m+n},$$

where $p\nmid r_1{s}_1{r}_2{s}_2$; the fraction $\frac{r_1{r}_2}{s_1{s}_2}$ need not be in lowest terms. So $v_p(ab)=m+n=v_p(a)+v_p(n)$.

Suppose that $v_p(a)\le v_p(b)$. Then

$$a+b=\frac{r_1}{s_1}{p}^m+\frac{r_2}{s_2}{p}^n=\left(\frac{r_1}{s_1}+\frac{r_2}{s_2}{p}^{n-m}\right)p^m=\frac{r_1{s}_2+r_2{s}_1{p}^{n-m}}{s_1{s}_2}{p}^m.$$

Since $p\nmid s_1$ and $p\nmid s_2$, then

$$v_p(a+b)\ge m=v_p(a)=\min \{v_p(a),v_p(b)\}.$$

∎

We call $v_p$ the $p$-adic valuation. The valuation ring of $\mathbb{Q}$ corresponding to $v_p$ is

$${𝒪}_p=\{x\in \mathbb{Q}:v_p(x)\ge 0\},$$

in other words, those rational numbers such that in lowest terms, $p$ does not divide their denominator. For example, $\frac{11}{169},-\frac{9}{35}\in {𝒪}_3$, and $\frac{5}{3}\notin {𝒪}_3$. By Theorem 6, the group of units of the valuation ring ${𝒪}_p$ is

$${𝒪}_p^{*}=\{x\in \mathbb{Q}:v_p(x)=0\},$$

in other words, those rational numbers such that in lowest terms, $p$ divides neither their numerator nor their denominator. As well by Theorem 6, ${𝒪}_p$ is a local ring whose unique maximal ideal is

$${𝔭}_p=\{x\in \mathbb{Q}:v_p(x)>0\},$$

in other words, those rational numbers such that in lowest terms, $p$ divides their numerator and does not divide their denominator. We see that $p\in {𝔭}_p$ and $v_p(p)=1$, so the nonzero ideals of ${𝒪}_p$ are of the form

$$p^n{𝒪}_p.$$

Lemma 11.

${𝒪}_p/{𝔭}_p\cong \mathbb{Z}/p\mathbb{Z}$.

6 p-adic absolute values and metrics

We define $|\cdot {|}_p:\mathbb{Q}\to {ℝ}_{\ge 0}$ by ${|a|}_p=p^{-v_p(n)}$ for $a\ne 0$ and ${|0|}_p=0$. This is a non-Archimedean absolute value on $\mathbb{Q}$, which we call the $p$-adic absolute value.

Example 12.

For $p=3$ and $a=-\frac{57}{10}$, we have $n=1,r=-19,s=10$. Thus ${\left|-\frac{57}{10}\right|}_3=3^{-1}$.

For $p=5$ and $a=\frac{28}{75}$, we have $n=-2,r=28,s=3$. Thus ${\left|\frac{28}{75}\right|}_5=5^2$.

We define $d_p(x,y)={|x-y|}_p$. The sequences $x_l=a_0+a_{1}p+a_2{p}^2+\mathrm{\cdots }+a_{l-1}{p}^{l-1}$ constructed when applying Hensel’s lemma satisfy, for ,

$$x_n-x_m=a_m{p}^m+a_{m+1}{p}^{m+1}+\mathrm{\cdots }+a_{n-1}{p}^{n-1}\equiv 0(mod{p}^m),$$

so

$${|x_n-x_m|}_p\le p^{-m},$$

and

$$f(x_n)\equiv 0(mod{p}^n),$$

so

$${|f(x_n)|}_p\le p^{-n}.$$

Thus, $x_n$ is a Cauchy sequence in the $p$-adic metric $d_p(x,y)={|x-y|}_p$, and $f(x_n)\to 0$ as $n\to \mathrm{\infty }$.

Lemma 13.

If $x_n$ and $y_n$ are Cauchy sequences in $\mathrm{(}\mathrm{Q}\mathrm{,}{d}_p\mathrm{)}$, then $x_n\mathrm{+}{y}_n$ and $x_n\mathrm{\cdot }{y}_n$ are Cauchy sequences in $\mathrm{(}\mathrm{Q}\mathrm{,}{d}_p\mathrm{)}$.

Proof.

The claim follows from

$${|x_n+y_n-(x_m+y_m)|}_p\le {|x_n-x_m|}_p+{|y_n-y_m|}_p$$

and

${|x_n\cdot y_n-x_m\cdot y_m|}_p$ $={|x_n\cdot y_n-x_m\cdot y_n+x_m\cdot y_n-x_m\cdot y_m|}_p$

$\le {|x_n-x_m|}_p{|y_n|}_p+{|x_m|}_p{|y_n-y_m|}_p,$

and the fact that $x_n,y_n$ being Cauchy implies that ${|x_n|}_p,{|y_n|}_p$ are bounded. ∎

7 Completions of metric spaces

If $(X,d)$ is a metric space, a completion of $X$ is a complete metric space $(Y,\rho )$ and an isometry $i:X\to Y$ such that for every metric space $(Z,r)$ and isometry $j:X\to Z$, there is a unique isometry $J:Y\to Z$ such that $J\circ i=j$. It is a fact that any metric space has a completion, and that if $(Y_1,{\rho }_1)$ and $(Y_2,{\rho }_2)$ are completions then there is a unique isometric isomorphism $f:Y_1\to Y_2$.

For $p$ prime, let $({\mathbb{Q}}_p,d_p)$ be the completion of $(\mathbb{Q},d_p)$. Elements of ${\mathbb{Q}}_p$ are called $p$-adic numbers. For $x,y\in {\mathbb{Q}}_p$, there are Cauchy sequences $x_n,y_n$ in $(\mathbb{Q},d_p)$ such that $x_n\to x$ and $y_n\to y$ in $({\mathbb{Q}}_p,d_p)$. We define addition and multiplication on the set ${\mathbb{Q}}_p$ by

$$x+y=lim(x_n+y_n),x\cdot y=lim(x_n\cdot y_n);$$

that these limits exists follows from Lemma 13. If $x\in {\mathbb{Q}}_p$, $x\ne 0$, then there is a sequence $x_n\in \mathbb{Q}$, each term of which is $\ne 0$, such that $x_n\to x$ in $({\mathbb{Q}}_p,d_p)$. Then $x_n^{-1}$ is a Cauchy sequence in $(\mathbb{Q},d_p)$ hence converges to some $y\in {\mathbb{Q}}_p$ which satisfies $x\cdot y=1$. Therefore ${\mathbb{Q}}_p$ is a field.

We define $v_p:{\mathbb{Q}}_p\to ℝ\cup \{\mathrm{\infty }\}$

$$v_p(x)=lim{v}_p(x_n),x_n\to x.$$

One proves that $v_p$ is a normalized valuation on the field ${\mathbb{Q}}_p$. We then define $|\cdot {|}_p:{\mathbb{Q}}_p\to {ℝ}_{\ge 0}$ by ${|x|}_p=p^{-v_p(x)}$ for $x\ne 0$ and ${|0|}_p=\mathrm{\infty }$.

8 The exponential function

Lemma 14.

For $a_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{a}_r\mathrm{\in }{\mathrm{Q}}_p$,

$${|a_1+\mathrm{\cdots }+a_r|}_p\le \max \{|a_1|,\mathrm{\dots },|a_r|\}.$$

Lemma 15.

A sequence $a_i\mathrm{\in }{\mathrm{Q}}_p$ is Cauchy if and only if $a_{i\mathrm{+}\mathrm{1}}\mathrm{-}{a}_i\mathrm{\to }\mathrm{0}$ as $i\mathrm{\to }\mathrm{\infty }$.

Proof.

Assume that $a_{i+1}-a_i\to 0$ and let $ϵ>0$. Then there is some $i_0$ such that $i\ge i_0$ implies . For ,

${|a_j-a_i|}_p$ $={|a_j-a_{j-1}+a_{j-1}+\mathrm{\cdots }-a_{i+1}+a_{i+1}-a_i|}_p$

$={|(a_j-a_{j-1})+\mathrm{\cdots }+(a_{i+1}-a_i)|}_p$

$\le \max \{|a_j-a_{j-1}|,\mathrm{\dots },|a_{i+1}-a_i|\}$

∎

The above shows that if $a_i\to 0$ in $({\mathbb{Q}}_p,d_p)$ then the series $\sum a_i$ converges in $({\mathbb{Q}}_p,d_p)$.

Lemma 16 (Exponential power series).

If $v_p\mathit{}\mathrm{(}x\mathrm{)}\mathrm{>}\frac{\mathrm{1}}{p\mathrm{-}\mathrm{1}}$, then

$$\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{n!}$$

converges in $\mathrm{(}{\mathrm{Q}}_p\mathrm{,}{d}_p\mathrm{)}$.

Proof.

$$v_p(n!)=\sum _{j=1}^{\mathrm{\infty }}\left[\frac{n}{p^j}\right]\le \sum _{j=1}^{\mathrm{\infty }}\frac{n}{p^j}=\frac{1}{np}\frac{1}{1-\frac{1}{p}}=\frac{n}{p-1}.$$

Then

$$v_p\left(\frac{x^n}{n!}\right)=n{v}_p(x)-v_p(n!)\ge n{v}_p(x)-\frac{n}{p-1}=n\left(v_p(x)-\frac{1}{p-1}\right).$$

As $n\to \mathrm{\infty }$ this tends to $+\mathrm{\infty }$, hence

$${\left|\frac{x^n}{n!}\right|}_p=p^{-v_p\left(\frac{x^n}{n!}\right)}\to 0,$$

and thus the series ${\sum }_{n=0}^{\mathrm{\infty }}\frac{x^n}{n!}$ converges. ∎

Lemma 17 (Logarithm power series).

If $v_p\mathit{}\mathrm{(}x\mathrm{)}\mathrm{>}\mathrm{0}$, then

$$\sum _{n=1}^{\mathrm{\infty }}{(-1)}^{n+1}\frac{x^n}{n}$$

converges in $\mathrm{(}{\mathrm{Q}}_p\mathrm{,}{d}_p\mathrm{)}$.

Proof.

For $n$ a positive integer we have $v_p(n)\le {\log }_{p}n$. Then,

$$v_p\left(\frac{x^n}{n}\right)=n{v}_p(x)-v_p(n)\ge n{v}_p(x)-{\log }_{p}n.$$

If $v_p(x)>0$ then this tends to $+\mathrm{\infty }$ as $n\to \mathrm{\infty }$. ∎

9 Topology

We define ${\mathbb{Z}}_p$ to be the valuation ring of ${\mathbb{Q}}_p$. Elements of ${\mathbb{Z}}_p$ are called $p$-adic integers. For $x\in {\mathbb{Q}}_p$ and real $r>0$, write

$${\overline{B}}_p(r,x)=\{y\in {\mathbb{Q}}_p:{|x-y|}_p\le r\}=\{y\in {\mathbb{Q}}_p:v_p(x-y)\ge -{\log }_{p}r\}.$$

In particular,

$${\overline{B}}_p(0,1)={\mathbb{Z}}_p.$$

Because $v_p$ is discrete, there is some $ϵ>0$ such that

This shows that ${\overline{B}}_p(x,r)$ is open in the topology induced by $v_p$, and thus is both closed and open. It follows that ${\mathbb{Q}}_p$ is totally disconnected.

Theorem 18.

${\mathbb{Z}}_p$ is totally bounded.

The fact that ${\mathbb{Z}}_p$ is a totally bounded subset of a complete metric space implies that ${\mathbb{Z}}_p$ is compact. Then because

$${\overline{B}}_d(0,p^k)=\{y\in {\mathbb{Q}}_p:{|y|}_p\le p^k\}=\{y\in {\mathbb{Q}}_p:{|p^{k}y|}_p\le 1\}=p^{-k}{\mathbb{Z}}_p$$

and translation is a homeomorphism, any closed ball in ${\mathbb{Q}}_p$ is compact. Therefore ${\mathbb{Q}}_p$ is locally compact.

${\mathbb{Q}}_p$ is a locally compact abelian group under addition, and we take Haar measure on it satisfying $\mu ({\mathbb{Z}}_p)=1$. One can explicitly calculate the characters on ${\mathbb{Q}}_p$.