Explicit construction of the $p$-adic numbers

1 Z_p

Let $p$ be prime, let $N_p=\{0,\mathrm{\dots },p-1\}$, and let ${\mathbb{Z}}_p$ be the set of maps $x:\mathbb{Z}\to N_p$ such that $x(k)=0$ for all .

1.1 Addition

For $x,y\in {\mathbb{Z}}_p$, we define $x+y\in {\mathbb{Z}}_p$ by induction. Define

$$(x+y)(0)\equiv x(0)+y(0)(modp),(x+y)(0)\in N_p.$$

Assume for $k\ge 0$ that there is some $A_k\in \mathbb{Z}$ such that

$$\sum _{j=0}^k(x+y)(j)p^j=A_k{p}^{k+1}+\sum _{j=0}^k(x(j)+y(j))p^j.$$

Define

$$(x+y)(k+1)\equiv -A_k+x(k+1)+y(k+1)(modp),(x+y)(k+1)\in N_p,$$

and then define $A_{k+1}\in \mathbb{Z}$ by

$$(x+y)(k+1)=A_{k+1}p-A_k+x(k+1)+y(k+1).$$

Then

${\displaystyle \sum _{j=0}^{k+1}}(x+y)(j)p^j$ $=(x+y)(k+1)p^{k+1}+{\displaystyle \sum _{j=0}^k}(x+y)(j)p^j$

$=A_{k+1}{p}^{k+2}-A_k{p}^{k+1}+(x(k+1)+y(k+1))p^{k+1}$

$+A_k{p}^{k+1}+{\displaystyle \sum _{j=0}^k}(x(j)+y(j))p^j$

$=A_{k+1}{p}^{k+2}+{\displaystyle \sum _{j=0}^{k+1}}(x(j)+y(j))p^j.$

Thus, for each $k\ge 0$, $(x+y)(k)\in N_p$ and

$$\sum _{j=0}^k(x+y)(j)p^j\equiv \sum _{j=0}^k(x(j)+y(j))p^j(mod{p}^{k+1}).$$ (1)

It is immediate that $x+y=y+x$.

Lemma 1.

If $x,y\in {\mathbb{Z}}_p$ and for each $k\ge 0$,

$$\sum _{j=0}^{k}x(j)p^j\equiv \sum _{j=0}^{k}y(j)p^j(mod{p}^{k+1}),$$

then $x=y$.

Proof.

Suppose by contradiction that $x\ne y$. Now, $x(0)\equiv y(0)(modp)$ and $x(0),y(0)\in N_p$ so $x(0)=y(0)$. As $x\ne y$, there is a minimal $k\ge 0$ such that $x(k+1)\ne y(k+1)$. On the one hand,

$$\sum _{j=0}^{k+1}x(j)p^j=x(k+1)p^{k+1}+\sum _{j=0}^{k}y(j)p^j,$$

and on the other hand,

$$\sum _{j=0}^{k+1}x(j)p^j\equiv \sum _{j=0}^{k+1}y(j)p^j(mod{p}^{k+2}).$$

Then there is some $B$ such that

$$x(k+1)p^{k+1}=C{p}^{k+2}+y(k+1)p^{k+1}.$$

so $x(k+1)-y(k+1)=Bp$. But $-p+1\le x(k+1)-y(k+1)\le p-1$, so $B=0$ and hence $x(k+1)=y(k+1)$, a contradiction and thus $x=y$. ∎

Therefore, if $t\in {\mathbb{Z}}_p$ satisfies, for all $k\ge 0$,

$$\sum _{j=0}^{k}t(j)p^j\equiv \sum _{j=0}^k(x(j)+y(j))p^j(mod{p}^{k+1}).$$

then $t=x+y$. Now let $x,y,z\in {\mathbb{Z}}_p$. For $k\ge 0$,

${\displaystyle \sum _{j=0}^k}(x+(y+z))(j)p^j$ $\equiv {\displaystyle \sum _{j=0}^k}(x(j)+(y+z)(j))p^j(mod{p}^{k+1})$

$={\displaystyle \sum _{j=0}^k}(x(j)+y(j)+z(j))p^j(mod{p}^{k+1})$

$\equiv {\displaystyle \sum _{j=0}^k}((x+y)(j)+z(j))p^j(mod{p}^{k+1}),$

which shows that $x+(y+z)=(x+y)+z$.

Define $t\in {\mathbb{Z}}_p$ by $t(k)=0$ for all $k\ge 0$. It is immediate that for $x\in {\mathbb{Z}}_p$, $x+t=x$, $t+x=x$. If $x\ne 0$, let $m\ge 0$ be minimal such that $x(m)\ne 0$, and define $y\in {\mathbb{Z}}_p$ by

This makes sense because $1\le x(m)\le p-1$. Then $x(k)+y(k)=0$ for , $x(m)+y(m)=p$, and $x(k)+y(k)=p-1$ for $k>m$. For $k>m$,

${\displaystyle \sum _{j=0}^k}(x(j)+y(j))p^j$ $=p\cdot p^m+{\displaystyle \sum _{j=m+1}^k}(p-1)p^j$

$=p^{m+1}+(p-1)\cdot {\displaystyle \frac{p^{k+1}-p^{m+1}}{p-1}}$

$=p^{k+1},$

so

$$\sum _{j=0}^k(x(j)+y(j))p^j\equiv \sum _{j=0}^{k}0\cdot p^j(mod{p}^{k+1}),$$

and it follows that $x+y=0$, $y+x=0$, namely $y=-x$.

We have established that $({\mathbb{Z}}_p,+)$ is an abelian group whose identity is $k\mapsto 0$, $k\ge 0$.

Lemma 2.

For $x\in {\mathbb{Z}}_p$ and $m\ge 1$,

Proof.

For $x\in {\mathbb{Z}}_p$ and $m\ge 1$ define $y(j)=0$ for and $y(j)=x(j-m)$ for $j\ge m$. By (1), for $k\ge m$,

${\displaystyle \sum _{j=0}^k}(p^{m}x)(j)p^j$ $\equiv {\displaystyle \sum _{j=0}^k}{p}^{m}x(j)p^j(mod{p}^{k+1})$

$\equiv {\displaystyle \sum _{j=0}^k}x(j)p^{j+m}(mod{p}^{k+1})$

$\equiv {\displaystyle \sum _{j=m}^{m+k}}x(j-m)p^j(mod{p}^{k+1})$

$\equiv {\displaystyle \sum _{j=m}^k}x(j-m)p^j(mod{p}^{k+1})$

$\equiv {\displaystyle \sum _{j=0}^k}y(j)p^j(mod{p}^{k+1}).$

∎

The following lemma shows that if $x(k)=0$ for then it makes sense to talk about $p^{-m}x\in {\mathbb{Z}}_p$. That is, if $x(k)=0$ for then there is a unique $y\in {\mathbb{Z}}_p$ such that $p^{m}y=x$. (For comparison, it is false that for any $z\in ℂ$ there is a unique $z^{1/2}\in ℂ$, or that for any $n\in \mathbb{Z}$ there is a unique $p^{-1}n\in \mathbb{Z}$.)

Lemma 3.

Let $x\in {\mathbb{Z}}_p$ with $x(0)=0$. If $y\in {\mathbb{Z}}_p$ and $py=x$ then $y(k)=x(k+1)$ for $k\ge 0$.

Proof.

By Lemma 2, $(py)(0)=0$ and $(py)(k)=y(k-1)$ for $k\ge 1$, and as $py=x$ this means $x(0)=0$ and $x(k)=y(k-1)$ for $k\ge 1$, i.e. $x(k+1)=y(k)$ for $k\ge 0$. ∎

1.2 Multiplication

For $x,y\in {\mathbb{Z}}_p$, we define $xy\in {\mathbb{Z}}_p$ by induction. Define

$$(xy)(0)\equiv x(0)y(0)(modp),(xy)(0)\in N_p.$$

Assume for $k\ge 0$ that there is some $A_k\in \mathbb{Z}$ such that

$$\sum _{j=0}^k(xy)(j)p^j=A_k{p}^{k+1}+\left(\sum _{j=0}^{k}x(j)p^j\right)\left(\sum _{j=0}^{k}y(j)p^j\right).$$

There is some $B\in \mathbb{Z}$ such that

Hence

$\left({\displaystyle \sum _{j=0}^{k+1}}x(j)p^j\right)\left({\displaystyle \sum _{j=0}^{k+1}}y(j)p^j\right)$ $=B{p}^{k+2}+x(k+1)y(0)p^{k+1}+x(0)y(k+1)p^{k+1}$

$+{\displaystyle \sum _{j=0}^k}(xy)(j)p^j-A_k{p}^{k+1}.$

Now define

$$(xy)(k+1)\equiv x(k+1)y(0)+x(0)y(k+1)-A_k(modp),(xy)(k+1)\in N_p,$$

and let $C\in \mathbb{Z}$ such that

$$(xy)(k+1)=Cp+x(k+1)y(0)+x(0)y(k+1)-A_k,$$

whence, taking $A_{k+1}=B-C$,

$\left({\displaystyle \sum _{j=0}^{k+1}}x(j)p^j\right)\left({\displaystyle \sum _{j=0}^{k+1}}y(j)p^j\right)$ $=B{p}^{k+2}+(xy)(k+1)p^{k+1}-C{p}^{k+2}+A_k{p}^{k+1}$

$+{\displaystyle \sum _{j=0}^k}(xy)(j)p^j-A_k{p}^{k+1}$

$=A_{k+1}{p}^{k+2}+{\displaystyle \sum _{j=0}^{k+1}}(xy)(j)p^j.$

Thus, for each $k\ge 0$, $(xy)(k)\in N_p$ and

$$\sum _{j=0}^k(xy)(j)p^j\equiv \left(\sum _{j=0}^{k}x(j)p^j\right)\left(\sum _{j=0}^{k}y(j)p^j\right)(mod{p}^{k+1}).$$ (2)

It is immediate that $xy=yz$.

For $t\in {\mathbb{Z}}_p$, if for each $k\ge 0$,

$$\sum _{j=0}^{k}t(j)p^j\equiv \left(\sum _{j=0}^{k}x(j)p^j\right)\left(\sum _{j=0}^{k}y(j)p^j\right)(mod{p}^{k+1}).$$

then $t=xy$. Now let $x,y,z\in {\mathbb{Z}}_p$. For $k\ge 0$,

${\displaystyle \sum _{j=0}^k}(x(yz))(j)p^j$ $\equiv \left({\displaystyle \sum _{j=0}^k}x(j)p^j\right)\left({\displaystyle \sum _{j=0}^k}(yz)(j)p^j\right)(mod{p}^{k+1})$

$\equiv \left({\displaystyle \sum _{j=0}^k}x(j)p^j\right)\left({\displaystyle \sum _{j=0}^k}y(j)p^j\right)\left({\displaystyle \sum _{j=0}^k}z(j)p^j\right)(mod{p}^{k+1})$

$\equiv \left({\displaystyle \sum _{j=0}^k}(xy)(j)p^j\right)\left({\displaystyle \sum _{j=0}^k}z(j)p^j\right)(mod{p}^{k+1})$

$\equiv {\displaystyle \sum _{j=0}^k}((xy)z)(j)p^j(mod{p}^{k+1}),$

which shows that $x(yz)=(xy)z$.

Define $u\in {\mathbb{Z}}_p$ by $u(0)=1$, $u(k)=0$ for $k\ge 1$. It is apparent that for $x\in {\mathbb{Z}}_p$, $xu=x$ and $ux=x$.

1.3 Ring

For $x,y,z\in {\mathbb{Z}}_p$ and for $k\ge 0$, using (1) and (2),

${\displaystyle \sum _{j=0}^k}(x(y+z))(j)p^j$ $\equiv \left({\displaystyle \sum _{j=0}^k}x(j)p^j\right)\left({\displaystyle \sum _{j=0}^k}(y+z)(j)p^j\right)(mod{p}^{k+1})$

$\equiv \left({\displaystyle \sum _{j=0}^k}x(j)p^j\right)\left({\displaystyle \sum _{j=0}^k}(y(j)+z(j))p^j\right)(mod{p}^{k+1})$

$\equiv \left({\displaystyle \sum _{j=0}^k}x(j)p^j\right)\left({\displaystyle \sum _{j=0}^k}y(j)p^j\right)$

$+\left({\displaystyle \sum _{j=0}^k}x(j)p^j\right)\left({\displaystyle \sum _{j=0}^k}z(j)p^j\right)(mod{p}^{k+1})$

$\equiv {\displaystyle \sum _{j=0}^k}(xy)(j)p^j+{\displaystyle \sum _{j=0}^k}(xz)(j)p^j(mod{p}^{k+1})$

$\equiv {\displaystyle \sum _{j=0}^k}(xy+xz)(j)p^j(mod{p}^{k+1}),$

which shows that $x(y+z)=xy+xz$. Therefore ${\mathbb{Z}}_p$ is a commutative ring with unity $0\mapsto 1$, $k\mapsto 0$ for $k\ge 1$.

1.4 Integral domain

Let ${\mathbb{Z}}_p^{*}$ be the set of those $x\in {\mathbb{Z}}_p$ for which there is some $y\in {\mathbb{Z}}_p$ such that $xy=1$, namely the set of invertible elements of ${\mathbb{Z}}_p$.

Lemma 4.

Let $x\in {\mathbb{Z}}_p$. $x\in {\mathbb{Z}}_p^{*}$ if and only if $x(0)\ne 0$.

Proof.

If $x(0)=0$ and $y\in {\mathbb{Z}}_p$ then $(xy)(0)\equiv x(0)y(0)\equiv 0(modp)$ while $1(0)\equiv 1(modp)$, so $xy\ne 1$ and therefore $x\notin {\mathbb{Z}}_p^{*}$.

If $x(0)\ne 0$, we define $y\in {\mathbb{Z}}_p$ by induction. As $x(0)\ne 0$, it makes sense to define

$$y(0)x(0)\equiv 1(modp),y(0)\in N_p.$$

We use (2) and the fact that $1(0)=1$, $1(k)=0$ for $k\ge 1$. Suppose for $k\ge 0$ that there is some $A_k\in \mathbb{Z}$ such that

$$\left(\sum _{j=0}^{k}x(j)p^j\right)\left(\sum _{j=0}^{k}y(j)p^j\right)=A_k{p}^{k+1}+1.$$

Because $x(0)\ne 0$, it makes sense to define

$$y(k+1)x(0)+x(k+1)y(0)\equiv -A_k(modp).$$

Then

$\left({\displaystyle \sum _{j=0}^{k+1}}x(j)p^j\right)\left({\displaystyle \sum _{j=0}^{k+1}}y(j)p^j\right)$ $\equiv x(k+1)y(0)p^{k+1}+y(k+1)x(0)p^{k+1}$

$\left({\displaystyle \sum _{j=0}^k}x(j)p^j\right)\left({\displaystyle \sum _{j=0}^k}y(j)p^j\right)(mod{p}^{k+2})$

$\equiv -A_k{p}^{k+1}+A_k{p}^{k+1}+1(mod{p}^{k+2})$

$\equiv 1(mod{p}^{k+2}).$

This shows that $xy=1$, thus $x\in {\mathbb{Z}}_p^{*}$ and $y=x^{-1}$. ∎

Theorem 5.

${\mathbb{Z}}_p$ is an integral domain.

Proof.

Let $x,y\in {\mathbb{Z}}_p$ be nonzero. Let $m\ge 0$ be minimal such that $x(m)\ne 0$ and let $n\ge 0$ be minimal such that $y(n)\ne 0$. Then $(p^{-m}x)(0)\ne 0$ and $(p^{-n}y)(0)\ne 0$, and using $p^{-m-n}(xy)=p^{-m}x\cdot p^{-n}y$,

$(xy)(m+n)$ $\equiv (p^{-m-n}(xy))(0)(modp)$

$\equiv (p^{-m}x)(0)\cdot (p^{-n}y)(0)(modp)$

$\not\equiv 0(modp),$

thus $xy\ne 0$. ∎

1.5 p-adic valuation

For $x\in {\mathbb{Z}}_p$, let

$$v_p(x)=inf\{k\ge 0:x(k)\ne 0\}.$$

$x(k)=0$ for . $v_p(x)=\mathrm{\infty }$ if and only if $x=0$.

Lemma 6.

For $x,y\in {\mathbb{Z}}_p$,

$$v_p(xy)=v_p(x)+v_p(y)$$

and

$$v_p(x+y)\ge \min (v_p(x),v_p(y)).$$

Lemma 4 says that for $x\in {\mathbb{Z}}_p$, $x\in {\mathbb{Z}}_p^{*}$ if and only if $x(0)\ne 0$. In other words,

$${\mathbb{Z}}_p^{*}=\{x\in {\mathbb{Z}}_p:v_p(x)=0\}=\{x\in {\mathbb{Z}}_p:{|x|}_p=1\}.$$

For $n\ge 1$, define ${\pi }_n:{\mathbb{Z}}_p\to \mathbb{Z}/p^n\mathbb{Z}$ by

$${\pi }_n(x)=\sum _{k=0}^{n-1}x(k)p^k+p^n\mathbb{Z}.$$

It is apparent that ${\pi }_n$ is onto.

Lemma 7.

${\pi }_n:{\mathbb{Z}}_p\to \mathbb{Z}/p^n\mathbb{Z}$ is a ring homomorphism, and

$$\ker {\pi }_n=\{x\in {\mathbb{Z}}_p:v_p(x)\ge n\}=p^n{\mathbb{Z}}_p.$$

Proof.

Let $x,y\in {\mathbb{Z}}_p$. By (1),

$$\sum _{k=0}^{n-1}(x+y)(k)p^k+p^n\mathbb{Z}=\sum _{k=0}^{n-1}x(k)p^k+\sum _{k=0}^{n-1}y(k)p^k+p^n\mathbb{Z},$$

i.e.

$${\pi }_n(x+y)={\pi }_n(x)+{\pi }_n(y).$$

By (2),

$$\sum _{k=0}^{n-1}(xy)(k)p^k+p^n\mathbb{Z}=\left(\sum _{k=0}^{n-1}x(k)p^k+p^n\mathbb{Z}\right)\left(\sum _{k=0}^{n-1}y(k)p^k+p^n\mathbb{Z}\right),$$

i.e.

$${\pi }_n(xy)={\pi }_n(x){\pi }_n(y).$$

For $1\in {\mathbb{Z}}_p$, $1(0)=1$, $1(k)=0$ for $k\ge 1$, so

$${\pi }_n(1)=1+p^n\mathbb{Z},$$

which is the unity of $\mathbb{Z}/p^n\mathbb{Z}$. Therefore ${\pi }_n$ is a ring homomorphism.

${\pi }_n(x)=0$ means

$$\sum _{k=0}^{n-1}x(k)p^k\in p^n\mathbb{Z}.$$

But , so ${\pi }_n(x)=0$ if and only if $x(k)=0$ for $0\le k\le n-1$. ∎

Then for $n\ge 1$,

${\mathbb{Z}}_p$ $={\displaystyle \bigcup _{j=0}^{p^n-1}}(j+p^n{\mathbb{Z}}_p)$

$={\displaystyle \bigcup _{j=0}^{p^n-1}}\{x\in {\mathbb{Z}}_p:v_p(x-j)\ge n\}$

$={\displaystyle \bigcup _{j=0}^{p^n-1}}\{x\in {\mathbb{Z}}_p:{|x-j|}_p\le p^{-n}\}$

Because $\mathbb{Z}/p\mathbb{Z}$ is a field and ${\pi }_1:{\mathbb{Z}}_p\to \mathbb{Z}/p\mathbb{Z}$ is an onto ring homomorphism,

$$\ker {\pi }_1=p{\mathbb{Z}}_p$$

is a maximal ideal in ${\mathbb{Z}}_p$.

Theorem 8.

If $I$ is an ideal in ${\mathbb{Z}}_p$ and $I\ne \{0\}$, then there is some $n\ge 0$ such that $I=p^n{\mathbb{Z}}_p$.

Proof.

There is some $a\in I$ with minimal $v_p(a)\ge 0$, and as $I\ne \{0\}$, $v_p(a)\ne \mathrm{\infty }$. Then $(p^{-v_p(a)}a)(0)=a(v_p(a))\ne 0$, so by Lemma 4, $p^{-v_p(a)}a\in {\mathbb{Z}}_p^{*}$. Hence there is some $u\in {\mathbb{Z}}_p^{*}$ such that $p^{-v_p(a)}a=u$, i.e. $p^{v_p(a)}=u^{-1}a$. But $I$ is an ideal and $a\in I$, so $p^{v_p(a)}\in I$, which shows that $p^{v_p(a)}{\mathbb{Z}}_p\subset I$. Let $x\in I$, $x\ne 0$. Then there is some $v\in {\mathbb{Z}}_p^{*}$ such that $p^{-v_p(x)}x=v$, i.e. $x=p^{v_p(x)}v$. Because $v_p(a)$ is minimal, $v_p(x)\ge v_p(a)$ and so

$$x=p^{v_p(x)}v=p^{v_p(a)}\cdot p^{v_p(x)-v_p(a)}\in p^{v_p(a)}{\mathbb{Z}}_p.$$

Therefore $I=p^{v_p(a)}{\mathbb{Z}}_p$. ∎

2 Q_p

Let ${\mathbb{Q}}_p$ be the set of maps $x:\mathbb{Z}\to N_p$ such that for some $m\in \mathbb{Z}$, $x(k)=0$ for all . For $x\in {\mathbb{Q}}_p$ define

$$v_p(x)=inf\{k\in \mathbb{Z}:x(k)\ne 0\}.$$

$x(k)=0$ for , $k\in \mathbb{Z}$. $v_p(x)=\mathrm{\infty }$ if and only if $x=0$.

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

For $m\in \mathbb{Z}$ and $x\in {\mathbb{Q}}_p$, define

$$(T_{m}x)(k)=x(k+m),k\in \mathbb{Z}.$$

For $x\in {\mathbb{Q}}_p$ with $x(k)=0$ for , if then and so

$$(T_{m}x)(k)=x(k+m)=0,$$

which means that $T_{m}x\in {\mathbb{Z}}_p$. For $x,y\in {\mathbb{Q}}_p$ with $x(k)=0$ and $y(k)=0$ for , $T_{m}x,T_{m}y\in {\mathbb{Z}}_p$ and $T_{m}x+T_{m}y\in {\mathbb{Z}}_p$. Define

$$x+y=T_{-m}(T_{m}x+T_{m}y)\in {\mathbb{Q}}_p.$$

Check that this makes sense. Likewise, $T_{m}x\cdot T_{m}y\in {\mathbb{Z}}_p$, and define

$$xy=T_{-m}(T_{m}x\cdot T_{m}y)\in {\mathbb{Q}}_p.$$

Check that this makes sense. Check that ${\mathbb{Q}}_p$ is a commutative ring with additive identity $k\mapsto 0$ for $k\in \mathbb{Z}$. and unity $0\mapsto 1$, $k\mapsto 0$ for $k\ne 0$. Finally,

$$T_{m}x=p^{-m}x.$$

Theorem 9.

${\mathbb{Q}}_p$ is a field, of characteristic $0$.

3 Metric

For $x\in {\mathbb{Q}}_p$ define

$${|x|}_p=p^{-v_p(x)}.$$

${|x|}_p=0$ if and only if $x=0$. For $x,y\in {\mathbb{Q}}_p$ define

$$d_p(x,y)={|x-y|}_p.$$

$d_p$ is an ultrametric:

$$d_p(x,z)\le \max (d_p(x,y),d_p(y,z)).$$

Theorem 10.

${\mathbb{Q}}_p$ is a topological field.

Proof.

For $(x,y),(u,v)\in {\mathbb{Q}}_p\times {\mathbb{Q}}_p$ let

$$\rho ((x,y),(u,v))=\max (d_p(x,u),d_p(y,v)).$$

$$d_p(x+y,u+v)={|(x-u)+(y-v)|}_p=\max ({|x-u|}_p,{|y-v|}_p)=\rho ((x,y),(u,v)),$$

which shows that $(x,y)\mapsto x+y$ is continuous ${\mathbb{Q}}_p\times {\mathbb{Q}}_p\to {\mathbb{Q}}_p$. And

$$d_p(-x,-y)={|-x-y|}_p={|-1|}_p{|x+y|}_p={|x+y|}_p=d_p(x,y),$$

which shows that $x\mapsto -x$ is continuous ${\mathbb{Q}}_p\to {\mathbb{Q}}_p$. For $\rho ((x,y),(u,v))\le \delta$, ${|x-u|}_p\le \delta$ so ${|u|}_p\le {|x|}_p+\delta$ and

$d_p(xy,uv)$ $={|xy-uv|}_p$

$={|xy-uy+uy-uv|}_p$

$=\max ({|xy-uy|}_p,{|uy-uv|}_p)$

$=\max ({|y|}_p{|x-u|}_p,{|u|}_p{|y-v|}_p)$

$\le \max ({|y|}_p\delta ,({|x|}_p+\delta )\delta ),$

which shows that $(x,y)\mapsto xy$ is continuous ${\mathbb{Q}}_p\times {\mathbb{Q}}_p\to {\mathbb{Q}}_p$. Finally, for $x,y\ne 0$,

$$d_p(x^{-1},y^{-1})={|x^{-1}-y^{-1}|}_p={|xy|}_p^{-1}{|y-x|}_p,$$

which shows that $x\mapsto x^{-1}$ is continuous ${\mathbb{Q}}_p\setminus \{0\}\to {\mathbb{Q}}_p\setminus \{0\}$. ∎

For $x\in {\mathbb{Q}}_p$ and $r>0$, write

Thus, for $x\in {\mathbb{Q}}_p$ and $n\ge 0$,

$$x+p^n\mathbb{Z}=B_{\le p^{-n}}(x).$$

Lemma 11.

For $x\in {\mathbb{Q}}_p$,

$$\{x+p^n{\mathbb{Z}}_p:n\ge 0\}$$

is a local base at $x$.

Proof.

For $ϵ>0$, let , $n\ge 0$, namely $n>\frac{1}{\log p}\log \frac{1}{ϵ}$. For this $n$,

∎

Theorem 12.

${\mathbb{Z}}_p$ is a compact subspace of ${\mathbb{Q}}_p$.

Proof.

Let $x_n\in {\mathbb{Z}}_p$ be a sequence. Because $x_n(0)\in N_p$, $n\ge 0$, there is some $a(0)\in N_p$ and an infinite subset $I_0$ of $\{n\ge 0\}$ such that $x_n(0)=a(0)$ for $n\in I_0$. Suppose by induction that for some $N\ge 0$ there are $a(0),\mathrm{\dots },a(N)\in N_p$ and an infinite set $I_N\subset \{n\ge 0\}$ such that

$$x_n(k)=a(k),0\le k\le N,n\in I_N.$$

But for each $x\in I_N$, $x_n(N+1)$ belongs to the finite set $N_p$, and because $I_N$ is infinite there is some $a(N+1)\in N_p$ and an infinite set $I_{N+1}\subset I_N$ such that $x_n(N+1)=a(N+1)$ for $n\in I_{N+1}$. We have thus defined $a\in {\mathbb{Z}}_p$.

Let ${\alpha }_0\in I_0$, and by induction let ${\alpha }_n>{\alpha }_{n-1}$, ${\alpha }_n\in I_n$; in particular as ${\alpha }_0\ge 0$ we have ${\alpha }_n\ge n$. Then for any $n\ge 0$, $x_{{\alpha }_n}(k)=a(k)$ for $0\le k\le n$. Take $ϵ>0$ and let . For $n\ge m$,

which shows that the sequence $x_{{\alpha }_n}$ tends to $a$. This means that ${\mathbb{Z}}_p$ is sequentially compact and therefore compact. ∎

For $x,y\in {\mathbb{Q}}_p$,

$$d_p(px,py)={|px-py|}_p={|p|}_p{|x-y|}_p=p^{-1}{|x-y|}_p,$$

which shows that $x\mapsto px$ is continuous ${\mathbb{Q}}_p\to {\mathbb{Q}}_p$. Therefore, the fact that ${\mathbb{Z}}_p$ is compact implies that for $n\ge 0$, $p^n{\mathbb{Z}}_p$ is compact. Then by Lemma 11 we get the following.

Theorem 13.

${\mathbb{Q}}_p$ is locally compact.

Theorem 14.

${\mathbb{Q}}_p$ is a complete metric space.

A topological space $X$ is zero-dimensional if there is a base for its topology each element of which is clopen. In a Hausdorff space, a compact set is closed, and because the sets $p^n{\mathbb{Z}}_p$ are compact, $n\ge 0$, from Lemma 11 we get the following.

Lemma 15.

${\mathbb{Q}}_p$ is zero-dimensional.

It is a fact that if a Hausdorff space is zero-dimensional then it is totally disconnected, so by the above, ${\mathbb{Q}}_p$ is totally disconnected.

4 p-adic fractional part

For $x\in {\mathbb{Q}}_p$, let

$${[x]}_p=\sum _{k\ge 0}x(k)p^k\in {\mathbb{Z}}_p$$

and

We call ${\{x\}}_p$ the $p$-adic fractional part of $x$. Then

$$x={[x]}_p+{\{x\}}_p\in {\mathbb{Q}}_p.$$

Furthermore, as $x(k)\to 0$ as $k\to -\mathrm{\infty }$,

therefore for $x\in {\mathbb{Q}}_p$,

$${\{x\}}_p\in [0,1)\cap \mathbb{Z}[1/p].$$

Define the Prüfer $p$-group

$$\mathbb{Z}(p^{\mathrm{\infty }})=\{e^{2\pi im{p}^{-n}}:m,n\ge 0\}.$$

We assign the Prüfer $p$-group the discrete topology.

Define ${\psi }_p:{\mathbb{Q}}_p\to S^1$ by

$${\psi }_p(x)=e^{2\pi i{\{x\}}_p}.$$

We prove that this is a homomorphism from the locally compact group ${\mathbb{Q}}_p$ whose image is the Prüfer $p$-group and whose kernel is ${\mathbb{Z}}_p$.

Theorem 16.

${\psi }_p:{\mathbb{Q}}_p\to S^1$ is a homomorphism of locally compact groups. ${\psi }_p({\mathbb{Q}}_p)=\mathbb{Z}(p^{\mathrm{\infty }})$, and $\ker {\psi }_p={\mathbb{Z}}_p$.

Proof.

For $x,y\in {\mathbb{Q}}_p$,

$\mathrm{\{}x+y\}{}_p-\{x\}{}_p-\{y\}{}_p$ $=x+y-{[x+y]}_p-x+{[x]}_p-y+{[y]}_p$

$={[x]}_p+{[y]}_p-{[x+y]}_p\in {\mathbb{Z}}_p.$

Check that $\mathbb{Z}[1/p]\cap {\mathbb{Z}}_p=\mathbb{Z}$. It then follows that

$${\{x+y\}}_p-{\{x\}}_p-{\{y\}}_p\in \mathbb{Z},$$

therefore $e^{2\pi i({\{x+y\}}_p-{\{x\}}_p-{\{y\}}_p)}=1$, i.e.

$${\psi }_p(x+y)=e^{2\pi i{\{x+y\}}_p}=e^{2\pi i{\{x\}}_p}{e}^{2\pi i{\{y\}}_p}={\psi }_p(x){\psi }_p(y),x,y\in {\mathbb{Q}}_p,$$

namely ${\psi }_p$ is a homomorphism.

${\psi }_p(x)=1$ if and only if $e^{2\pi i{\{x\}}_p}=1$ if and only if ${\{x\}}_p\in \mathbb{Z}$. But ${\{x\}}_p\in [0,1)$, so ${\psi }_p(x)=1$ if and only if ${\{x\}}_p=0$, hence ${\psi }_p(x)=1$ if and only if $x\in {\mathbb{Z}}_p$, namely

$$\ker {\psi }_p={\mathbb{Z}}_p.$$

Let $x\in {\mathbb{Q}}_p$. As ${\{x\}}_p\in \mathbb{Z}[1/p]$, there is some $n\ge 0$ such that $p^n{\{x\}}_p\in \mathbb{Z}$, so ${\psi }_p{(x)}^{p^n}=1$, which means that ${\psi }_p(x)\in \mathbb{Z}[p^{\mathrm{\infty }}]$. Let $e^{2\pi im{p}^{-n}}\in \mathbb{Z}[p^{\mathrm{\infty }}]$, $n,m\ge 0$. But $p^{-n}\in {\mathbb{Q}}_p$ and, whether or not $n>0$,

$${\psi }_p(p^{-n})=e^{2\pi i{\{p^{-n}\}}_p}=e^{2\pi i{p}^{-n}},$$

and $m{p}^{-n}\in {\mathbb{Q}}_p$, and using that ${\psi }_p$ is a homomorphism,

$${\psi }_p(m{p}^{-n})={\psi }_p{(p^{-n})}^m=e^{2\pi im{p}^{-n}}.$$

This shows that ${\psi }_p({\mathbb{Q}}_p)=\mathbb{Z}[p^{\mathrm{\infty }}]$.

Finally, let $x\in {\mathbb{Q}}_p$. For $y\in B_{\le 1}(x)=x+{\mathbb{Z}}_p$, so there is some $w\in {\mathbb{Z}}_p$ such that $y=x+w$. But ${\psi }_p(x+w)={\psi }_p(x){\psi }_p(w)={\psi }_p(x)$, so

$$|{\psi }_p(y)-{\psi }_p(x)|=|{\psi }_p(x)-{\psi }_p(x)|=0,$$

showing that ${\psi }_p$ is continuous at $x$. ∎

Because $\mathbb{Z}[p^{\mathrm{\infty }}]$ is discrete, it is immediate that ${\psi }_p$ is an open map. The first isomorphism theorem for topological groups states that if $G$ and $H$ are locally compact groups, $f:G\to H$ is a homomorphism of topological groups that is onto and open, then $G/\ker f$ and $H$ are isomorphic as topological groups. Therefore the quotient group ${\mathbb{Q}}_p/{\mathbb{Z}}_p$ and the Prüfer group $\mathbb{Z}[p^{\mathrm{\infty }}]$ are isomorphic as topological groups.