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

Jordan Bell
November 2, 2014

## 1 Hensel’s lemma

Let $p$ be prime and $f(x)\in\mathbb{Z}[x]$.11 1 Hua Loo Keng, Introduction to Number Theory, Chapter 15, “$p$-adic numbers”. Suppose that $0\leq a_{0}, satisfies

 $f(a_{0})\equiv 0\pmod{p}$

and

 $f^{\prime}(a_{0})\not\equiv 0\pmod{p}.$

Using the power series expansion

 $f(a_{0}+h)=f(a_{0})+f^{\prime}(a_{0})h+\frac{f^{\prime\prime}(a_{0})}{2}h^{2}+\cdots,$

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

 $f(a_{0}+py)=f(a_{0})+f^{\prime}(a_{0})py+\frac{f^{\prime\prime}(a_{0})}{2}p^{2% }y^{2}+\cdots$

so

 $\frac{f(a_{0}+py)}{p}=\frac{f(a_{0})}{p}+f^{\prime}(a_{0})y+\frac{f^{\prime% \prime}(a_{0})}{2}py^{2}+\cdots.$

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

 $\frac{f(a_{0})}{p}+f^{\prime}(a_{0})y+\frac{f^{\prime\prime}(a_{0})}{2}py^{2}+% \cdots\equiv 0\pmod{p},$

i.e.,

 $f^{\prime}(a_{0})y\equiv-\frac{f(a_{0})}{p}\pmod{p}.$

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

 $y\equiv-\frac{f(a_{0})}{p}(f^{\prime}(a_{0}))^{-1}\pmod{p}.$

Let $0\leq a_{1} be $a_{1}\equiv y\pmod{p}$.

Suppose that

 $x=a_{0}+a_{1}p+a_{2}p^{2}+\cdots+a_{l-2}p^{l-2},\qquad 0\leq a_{j}

satisfies

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

and

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

Using the power series expansion

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

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

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

i.e.

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

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

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

i.e.,

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

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

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

Let $0\leq a_{l-1} be $a_{l-1}\equiv y\pmod{p}$.

We have thus inductively defined a sequence $a_{0},a_{1},a_{2},\ldots$, with $0\leq a_{j}, such that for any $l$,

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

We wish to make sense of the infinite expression

 $a_{0}+a_{1}p+a_{2}p^{2}+a_{3}p^{3}+\cdots$

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

###### Example 1.

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

 $a_{1}\equiv-\frac{f(1)}{3}(f^{\prime}(1))^{-1}\equiv-\frac{-6}{3}(2)^{-1}% \equiv 1\pmod{3}.$

So $a_{1}=1$. Then,

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

So $a_{2}=1$. Then,

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

So, $a_{3}=0$. Then,

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

So, $a_{4}=2$, etc.

## 2 Absolute values on fields

If $K$ is a field, an absolute value on $K$ is a map $|\cdot|:K\to\mathbb{R}_{\geq 0}$ such that $|x|=0$ if and only if $x=0$, $|xy|=|x||y|$, and $|x+y|\leq|x|+|y|$. 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.22 2 Absolute values, valuations and completion, https://www.math.ethz.ch/education/bachelor/seminars/fs2008/algebra/Crivelli.pdf

###### Theorem 2.

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

 $|x|_{1}=|x|_{2}^{s},\qquad x\in K.$
###### Proof.

Suppose that $s>0$ and that $|x|_{1}=|x|_{2}^{s}$ for all $x\in K$. Then

 $\displaystyle B_{d_{1}}(x,r)$ $\displaystyle=\{y\in K:|y-x|_{1} $\displaystyle=\{y\in K:|y-x|_{2}^{s} $\displaystyle=\{y\in K:|y-x|_{2} $\displaystyle=B_{d_{2}}(x,r^{1/s}).$

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 $|x|_{1}<1$ then $d_{1}(x^{n},0)=|x^{n}|_{1}=|x|_{1}^{n}\to 0$ as $n\to\infty$. 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 $|x|_{2}<1$. Thus, $|x|_{1}<1$ if and only if $|x|_{2}<1$.

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 $|x|_{1}\neq 0,1$. There is some nonzero $\alpha\in\mathbb{R}$ 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 $|x|_{1}=|y|_{1}^{\alpha}<|y|_{1}^{\frac{m_{i}}{n_{i}}}$, hence $|x|_{1}^{n_{i}}<|y|_{1}^{m_{i}}$, hence $\frac{|x^{n_{i}}|_{1}}{|y^{m_{i}}|_{1}}<1$, hence

 $\left|\frac{x^{n_{i}}}{y^{m_{i}}}\right|_{1}<1.$

Because $|\cdot|_{1}$ and $|\cdot|_{2}$ are equivalent,

 $\frac{|x|_{2}^{n_{i}}}{|y|_{2}^{m_{i}}}=\left|\frac{x^{n_{i}}}{y^{m_{i}}}% \right|_{2}<1,$

so $|x|_{2}<|y|_{2}^{\frac{m_{i}}{n_{i}}}$. Taking $i\to\infty$ gives

 $|x|_{2}\leq|y|_{2}^{\alpha}.$

Similarly, we check that

 $|x|_{2}\geq|y|_{2}^{\alpha}.$

Therefore,

 $|x|_{2}=|y|_{2}^{\alpha}.$

Using this and $|x|_{1}=|y|_{1}^{\alpha}$, we have

 $\log|x|_{1}=\alpha\log|y|_{1},\qquad\log|x|_{2}=\alpha\log|y|_{2},$

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

 $\frac{\log|x|_{1}}{\log|x|_{2}}=\frac{\log|y|_{1}}{\log|y|_{2}}.$

This is true for any $x\in K$ with $|x|_{1}\neq 0,1$. We define $s\in\mathbb{R}$ 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 $|x|_{2}<1$ implies that $|x|_{1}<1$, so if $|x|_{1}=1$ then $|x|_{2}=1$ and hence $|x|_{1}=1=1^{s}=|x|_{2}^{s}$. If $|x|_{1}\neq 0,1$, 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 $|\cdot|:K\to\mathbb{R}_{\geq 0}$ is said to be non-Archimedean if

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

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

###### Lemma 3.

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

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

## 3 Valuations

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

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

The trivial valuation is $v(x)=0$ for $x\neq 0$ and $v(0)=\infty$.

###### Lemma 4.

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

###### Proof.

Take $v(y). For $x=0$,

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

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

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

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

 $\displaystyle v(y)$ $\displaystyle=v(x+y-x)$ $\displaystyle\geq\min\{v(x+y),v(x)\}$ $\displaystyle=v(x+y)$ $\displaystyle\geq\min\{v(x),v(y)\}$ $\displaystyle=v(y).$

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

###### Theorem 5.

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

If $v$ is a valuation on $K$ and $q>1$, then the function $|\cdot|_{q}:K\to\mathbb{R}_{\geq 0}$ defined by $|x|_{q}=q^{-v(x)}$ for $x\neq 0$ and $|0|_{q}=0$ 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 $v_{s}(xy)=\infty=v_{s}(x)+v_{s}(y)$. 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

 $v_{s}(x+y)=v_{s}(0)=\infty=\min\{\infty,\infty\}=\min\{v_{s}(x),v_{s}(y)\}.$

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

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

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

 $v_{s}(x+y)=v_{s}(0)=\infty\geq\min\{v_{s}(x),v_{s}(y)\}.$

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

 $\displaystyle v_{s}(x+y)$ $\displaystyle=-s\log|x+y|$ $\displaystyle\geq-s\log(\max\{|x|,|y|\})$ $\displaystyle=\min\{-s\log|x|,-s\log|y|\}$ $\displaystyle=\min\{v_{s}(x),v_{s}(y)\}.$

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 $|x+y|_{q}\leq|x|_{q}+|y|_{q}$, it suffices to show that $|x+y|_{q}\leq\max\{|x|_{q},|y|_{q}\}$; 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 $y\neq 0$, 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

 $|x+y|_{q}=|0|_{q}=0\leq\max\{|x|_{q},|y|_{q}\}.$

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

 $\displaystyle|x+y|_{q}$ $\displaystyle=q^{-v(x+y)}$ $\displaystyle\leq q^{-\min\{v(x),v(y)\}}$ $\displaystyle=\max\{q^{-v(x)},q^{-v(y)}\}$ $\displaystyle=\max\{|x|_{q},|y|_{q}\}.$

Two valuations $v_{1},v_{2}$ on a field $K$ are said to be equivalent if there is some real $s>0$ such that

 $v_{1}=sv_{2}.$

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

 $\mathcal{O}_{v}=\{x\in K:v(x)\geq 0\}$

is a maximal proper subring of $K$, and for all $x\neq 0$, $x\in\mathcal{O}_{v}$ or $x^{-1}\in\mathcal{O}_{v}$. The set

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

is the group of invertible elements of $\mathcal{O}_{v}$, and the set

 $\mathfrak{p}_{v}=\{x\in K:v(x)>0\}$

is the unique maximal ideal of $\mathcal{O}_{v}$.

###### Proof.

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

Since $v$ is nontrivial, there is some $x\in K$ with $v(x)\neq 0,\infty$. If $x\in\mathcal{O}_{v}$ then $v(x)>0$ and so $v(x^{-1})=-v(x)<0$, giving $x^{-1}\not\in\mathcal{O}_{v}$. Hence $\mathcal{O}_{v}\neq K$, showing that $\mathcal{O}_{v}$ is a proper subring of $K$.

To show that $\mathcal{O}_{v}$ is a maximal proper subring, it suffices to show that if $z\in K\setminus\mathcal{O}_{v}$ then $\mathcal{O}_{v}[z]=K$, i.e., that the smallest ring containing $\mathcal{O}_{v}$ and $z$ is $K$. As $z\not\in\mathcal{O}_{v}$, $v(z)<0$. Let $y\in K$. For any positive integer $j$ we have $v(yz^{-j})=v(y)-jv(z)$, and because $v(z)<0$, there is some $j=j(y)$ such that $v(yz^{-j})>0$. For this $j$, $yz^{-j}\in\mathcal{O}_{v}$. Hence $y\in\mathcal{O}_{v}[z]$, and so $\mathcal{O}_{v}[z]=K$, showing that $\mathcal{O}_{v}$ is a maximal proper subring.

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

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

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

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

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

showing that $zx\in\mathfrak{p}_{v}$. Therefore $\mathfrak{p}_{v}$ is an ideal in the ring $\mathcal{O}_{v}$. Since $v(1)=0$, $1\not\in\mathfrak{p}_{v}$, so $\mathfrak{p}_{v}$ is a proper ideal.

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

The above ring $\mathcal{O}_{v}$ 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 $\mathcal{O}_{v}/\mathfrak{p}_{v}$ the residue field of $\mathcal{O}_{v}$.

###### Lemma 7.

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

 $x=ut^{n},\qquad n=v(x).$
###### Proof.

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

###### Theorem 8.

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

 $A=t^{n}\mathcal{O}_{v}=\{x\in K:v(x)\geq n\}=\mathfrak{p}_{v}^{n},$

and

 $\mathfrak{p}_{v}^{n}/\mathfrak{p}_{v}^{n+1}\cong\mathcal{O}_{v}/\mathfrak{p}_{% v},$

as $\mathcal{O}_{v}/\mathfrak{p}_{v}$-linear vector spaces.

###### Proof.

Let $A\neq\{0\}$ be an ideal of $\mathcal{O}_{v}$. For any $y\in A$, $v(y)\geq 0$, 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 $v(t)>0$, $t\in\mathfrak{p}_{v}$. By Lemma 7, there is some $u\in\mathcal{O}^{*}$ such that $x=ut^{n}$, $n=v(x)$. For any $z\in\mathcal{O}$, $xz\in A$ and so $t^{n}z\in A$. Thus $t^{n}\mathcal{O}_{v}\subset A$. On the other hand, let $y\in A$. Then also by Lemma 7 there is some $w\in\mathcal{O}_{v}^{*}$ such that $y=wt^{m}$, $m=v(y)$. By (1), $m=v(y)\geq v(x)=n$, so $v(t^{m-n})=(m-n)v(t)=m-n\geq 0$ so $t^{m-n}\in\mathcal{O}_{v}$, giving

 $y=wt^{m}=t^{n}(wt^{m-n})\in t^{n}\mathcal{O}_{v}.$

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

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

###### Lemma 9.

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

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 $v_{p}(0)=\infty$.

###### Theorem 10.

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

###### Proof.

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

 $a=\frac{r_{1}}{s_{1}}p^{m},\qquad b=\frac{r_{2}}{s_{2}}p^{n},$

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 $v_{p}(a)\leq v_{p}(b)$. 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

 $v_{p}(a+b)\geq m=v_{p}(a)=\min\{v_{p}(a),v_{p}(b)\}.$

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

 $\mathcal{O}_{p}=\{x\in\mathbb{Q}:v_{p}(x)\geq 0\},$

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

 $\mathcal{O}_{p}^{*}=\{x\in\mathbb{Q}:v_{p}(x)=0\},$

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

 $\mathfrak{p}_{p}=\{x\in\mathbb{Q}:v_{p}(x)>0\},$

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

 $p^{n}\mathcal{O}_{p}.$
###### Lemma 11.

$\mathcal{O}_{p}/\mathfrak{p}_{p}\cong\mathbb{Z}/p\mathbb{Z}$.

## 6 p-adic absolute values and metrics

We define $|\cdot|_{p}:\mathbb{Q}\to\mathbb{R}_{\geq 0}$ by $|a|_{p}=p^{-v_{p}(n)}$ for $a\neq 0$ 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 $x_{l}=a_{0}+a_{1}p+a_{2}p^{2}+\cdots+a_{l-1}p^{l-1}$ constructed when applying Hensel’s lemma satisfy, for $m,

 $x_{n}-x_{m}=a_{m}p^{m}+a_{m+1}p^{m+1}+\cdots+a_{n-1}p^{n-1}\equiv 0\pmod{p^{m}},$

so

 $|x_{n}-x_{m}|_{p}\leq p^{-m},$

and

 $f(x_{n})\equiv 0\pmod{p^{n}},$

so

 $|f(x_{n})|_{p}\leq p^{-n}.$

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 $n\to\infty$.

###### Lemma 13.

If $x_{n}$ and $y_{n}$ are Cauchy sequences in $(\mathbb{Q},d_{p})$, then $x_{n}+y_{n}$ and $x_{n}\cdot y_{n}$ are Cauchy sequences in $(\mathbb{Q},d_{p})$.

###### Proof.

The claim follows from

 $|x_{n}+y_{n}-(x_{m}+y_{m})|_{p}\leq|x_{n}-x_{m}|_{p}+|y_{n}-y_{m}|_{p}$

and

 $\displaystyle|x_{n}\cdot y_{n}-x_{m}\cdot y_{m}|_{p}$ $\displaystyle=|x_{n}\cdot y_{n}-x_{m}\cdot y_{n}+x_{m}\cdot y_{n}-x_{m}\cdot y% _{m}|_{p}$ $\displaystyle\leq|x_{n}-x_{m}|_{p}|y_{n}|_{p}+|x_{m}|_{p}|y_{n}-y_{m}|_{p},$

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

 $x+y=\lim(x_{n}+y_{n}),\qquad x\cdot y=\lim(x_{n}\cdot y_{n});$

that these limits exists follows from Lemma 13. If $x\in\mathbb{Q}_{p}$, $x\neq 0$, then there is a sequence $x_{n}\in\mathbb{Q}$, each term of which is $\neq 0$, 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 $v_{p}:\mathbb{Q}_{p}\to\mathbb{R}\cup\{\infty\}$

 $v_{p}(x)=\lim v_{p}(x_{n}),\qquad x_{n}\to x.$

One proves that $v_{p}$ is a normalized valuation on the field $\mathbb{Q}_{p}$.33 3 cf. Paul Garrett, Classical definitions of $\mathbb{Z}_{p}$ and $\mathbb{A}$, http://www.math.umn.edu/~garrett/m/mfms/notes/05_compare_classical.pdf We then define $|\cdot|_{p}:\mathbb{Q}_{p}\to\mathbb{R}_{\geq 0}$ by $|x|_{p}=p^{-v_{p}(x)}$ for $x\neq 0$ and $|0|_{p}=\infty$.

## 8 The exponential function

###### Lemma 14.

For $a_{1},\ldots,a_{r}\in\mathbb{Q}_{p}$,

 $|a_{1}+\cdots+a_{r}|_{p}\leq\max\{|a_{1}|,\ldots,|a_{r}|\}.$
###### Lemma 15.

A sequence $a_{i}\in\mathbb{Q}_{p}$ is Cauchy if and only if $a_{i+1}-a_{i}\to 0$ as $i\to\infty$.

###### Proof.

Assume that $a_{i+1}-a_{i}\to 0$ and let $\epsilon>0$. Then there is some $i_{0}$ such that $i\geq i_{0}$ implies $|a_{i+1}-a_{i}|_{p}<\epsilon$. For $i_{0}\leq i,

 $\displaystyle|a_{j}-a_{i}|_{p}$ $\displaystyle=|a_{j}-a_{j-1}+a_{j-1}+\cdots-a_{i+1}+a_{i+1}-a_{i}|_{p}$ $\displaystyle=|(a_{j}-a_{j-1})+\cdots+(a_{i+1}-a_{i})|_{p}$ $\displaystyle\leq\max\{|a_{j}-a_{j-1}|,\ldots,|a_{i+1}-a_{i}|\}$ $\displaystyle<\epsilon.$

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 $v_{p}(x)>\frac{1}{p-1}$, then

 $\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$

converges in $(\mathbb{Q}_{p},d_{p})$.

###### Proof.
 $v_{p}(n!)=\sum_{j=1}^{\infty}\left[\frac{n}{p^{j}}\right]\leq\sum_{j=1}^{% \infty}\frac{n}{p^{j}}=\frac{1}{np}\frac{1}{1-\frac{1}{p}}=\frac{n}{p-1}.$

Then

 $v_{p}\left(\frac{x^{n}}{n!}\right)=nv_{p}(x)-v_{p}(n!)\geq nv_{p}(x)-\frac{n}{% p-1}=n\left(v_{p}(x)-\frac{1}{p-1}\right).$

As $n\to\infty$ this tends to $+\infty$, hence

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

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

###### Lemma 17 (Logarithm power series).

If $v_{p}(x)>0$, then

 $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^{n}}{n}$

converges in $(\mathbb{Q}_{p},d_{p})$.

###### Proof.

For $n$ a positive integer we have $v_{p}(n)\leq\log_{p}n$. Then,

 $v_{p}\left(\frac{x^{n}}{n}\right)=nv_{p}(x)-v_{p}(n)\geq nv_{p}(x)-\log_{p}n.$

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

## 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

 $\overline{B}_{p}(r,x)=\{y\in\mathbb{Q}_{p}:|x-y|_{p}\leq r\}=\{y\in\mathbb{Q}_% {p}:v_{p}(x-y)\geq-\log_{p}r\}.$

In particular,

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

Because $v_{p}$ is discrete, there is some $\epsilon>0$ such that

 $\{y\in\mathbb{Q}_{p}:|x-y|_{p}\leq r\}=\{y\in\mathbb{Q}_{p}:|x-y|_{p}

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.44 4 Gerald B. Folland, A Course in Abstract Harmonic Analysis, pp. 34–36.

###### 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

 $\overline{B}_{d}(0,p^{k})=\{y\in\mathbb{Q}_{p}:|y|_{p}\leq p^{k}\}=\{y\in% \mathbb{Q}_{p}:|p^{k}y|_{p}\leq 1\}=p^{-k}\mathbb{Z}_{p}$

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 $\mathbb{Q}_{p}$.55 5 Gerald B. Folland, A Course in Abstract Harmonic Analysis, pp. 91–93, 104. Cf. Keith Conrad, The character group of $\mathbf{Q}$, http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/characterQ.pdf