Explicit construction of the p-adic numbers

Jordan Bell
March 17, 2016

1 Zp

Let p be prime, let Np={0,,p-1}, and let p be the set of maps x:Np such that x(k)=0 for all k<0.

1.1 Addition

For x,yp, we define x+yp by induction. Define


Assume for k0 that there is some Ak such that




and then define Ak+1 by



j=0k+1(x+y)(j)pj =(x+y)(k+1)pk+1+j=0k(x+y)(j)pj

Thus, for each k0, (x+y)(k)Np and

j=0k(x+y)(j)pjj=0k(x(j)+y(j))pj(modpk+1). (1)

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

Lemma 1.

If x,yp and for each k0,


then x=y.


Suppose by contradiction that xy. Now, x(0)y(0)(modp) and x(0),y(0)Np so x(0)=y(0). As xy, there is a minimal k0 such that x(k+1)y(k+1). On the one hand,


and on the other hand,


Then there is some B such that


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

Therefore, if tp satisfies, for all k0,


then t=x+y. Now let x,y,zp. For k0,

j=0k(x+(y+z))(j)pj j=0k(x(j)+(y+z)(j))pj(modpk+1)

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

Define tp by t(k)=0 for all k0. It is immediate that for xp, x+t=x, t+x=x. If x0, let m0 be minimal such that x(m)0, and define yp by


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

j=0k(x(j)+y(j))pj =ppm+j=m+1k(p-1)pj



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

We have established that (p,+) is an abelian group whose identity is k0, k0.

Lemma 2.

For xp and m1,


For xp and m1 define y(j)=0 for 0j<m and y(j)=x(j-m) for jm. By (1), for km,

j=0k(pmx)(j)pj j=0kpmx(j)pj(modpk+1)

The following lemma shows that if x(k)=0 for k<m then it makes sense to talk about p-mxp. That is, if x(k)=0 for k<m then there is a unique yp such that pmy=x. (For comparison, it is false that for any z there is a unique z1/2, or that for any n there is a unique p-1n.)

Lemma 3.

Let xp with x(0)=0. If yp and py=x then y(k)=x(k+1) for k0.


By Lemma 2, (py)(0)=0 and (py)(k)=y(k-1) for k1, and as py=x this means x(0)=0 and x(k)=y(k-1) for k1, i.e. x(k+1)=y(k) for k0. ∎

1.2 Multiplication

For x,yp, we define xyp by induction. Define


Assume for k0 that there is some Ak such that


There is some B such that



(j=0k+1x(j)pj)(j=0k+1y(j)pj) =Bpk+2+x(k+1)y(0)pk+1+x(0)y(k+1)pk+1

Now define


and let C such that


whence, taking Ak+1=B-C,

(j=0k+1x(j)pj)(j=0k+1y(j)pj) =Bpk+2+(xy)(k+1)pk+1-Cpk+2+Akpk+1

Thus, for each k0, (xy)(k)Np and

j=0k(xy)(j)pj(j=0kx(j)pj)(j=0ky(j)pj)(modpk+1). (2)

It is immediate that xy=yz.

For tp, if for each k0,


then t=xy. Now let x,y,zp. For k0,

j=0k(x(yz))(j)pj (j=0kx(j)pj)(j=0k(yz)(j)pj)(modpk+1)

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

Define up by u(0)=1, u(k)=0 for k1. It is apparent that for xp, xu=x and ux=x.

1.3 Ring

For x,y,zp and for k0, using (1) and (2),

j=0k(x(y+z))(j)pj (j=0kx(j)pj)(j=0k(y+z)(j)pj)(modpk+1)

which shows that x(y+z)=xy+xz. Therefore p is a commutative ring with unity 01, k0 for k1.

1.4 Integral domain

Let p* be the set of those xp for which there is some yp such that xy=1, namely the set of invertible elements of p.

Lemma 4.

Let xp. xp* if and only if x(0)0.


If x(0)=0 and yp then (xy)(0)x(0)y(0)0(modp) while 1(0)1(modp), so xy1 and therefore xp*.

If x(0)0, we define yp by induction. As x(0)0, it makes sense to define


We use (2) and the fact that 1(0)=1, 1(k)=0 for k1. Suppose for k0 that there is some Ak such that


Because x(0)0, it makes sense to define



(j=0k+1x(j)pj)(j=0k+1y(j)pj) x(k+1)y(0)pk+1+y(k+1)x(0)pk+1

This shows that xy=1, thus xp* and y=x-1. ∎

Theorem 5.

p is an integral domain.


Let x,yp be nonzero. Let m0 be minimal such that x(m)0 and let n0 be minimal such that y(n)0. Then (p-mx)(0)0 and (p-ny)(0)0, and using p-m-n(xy)=p-mxp-ny,

(xy)(m+n) (p-m-n(xy))(0)(modp)

thus xy0. ∎

1.5 p-adic valuation

For xp, let


x(k)=0 for 0k<vp(x). vp(x)= if and only if x=0.

Lemma 6.

For x,yp,




Lemma 4 says that for xp, xp* if and only if x(0)0. In other words,


For n1, define πn:p/pn by


It is apparent that πn is onto.

Lemma 7.

πn:p/pn is a ring homomorphism, and


Let x,yp. By (1),




By (2),




For 1p, 1(0)=1, 1(k)=0 for k1, so


which is the unity of /pn. Therefore πn is a ring homomorphism.

πn(x)=0 means


But 0k=0n-1x(k)pk<k=0n-1(p-1)pk=pn-1, so πn(x)=0 if and only if x(k)=0 for 0kn-1. ∎

Then for n1,

p =j=0pn-1(j+pnp)

Because /p is a field and π1:p/p is an onto ring homomorphism,


is a maximal ideal in p.

Theorem 8.

If I is an ideal in p and I{0}, then there is some n0 such that I=pnp.


There is some aI with minimal vp(a)0, and as I{0}, vp(a). Then (p-vp(a)a)(0)=a(vp(a))0, so by Lemma 4, p-vp(a)ap*. Hence there is some up* such that p-vp(a)a=u, i.e. pvp(a)=u-1a. But I is an ideal and aI, so pvp(a)I, which shows that pvp(a)pI. Let xI, x0. Then there is some vp* such that p-vp(x)x=v, i.e. x=pvp(x)v. Because vp(a) is minimal, vp(x)vp(a) and so


Therefore I=pvp(a)p. ∎

2 Qp

Let p be the set of maps x:Np such that for some m, x(k)=0 for all k<m. For xp define


x(k)=0 for k<vp(x), k. vp(x)= if and only if x=0.


For m and xp, define


For xp with x(k)=0 for k<m, if k<0 then k+m<m and so


which means that Tmxp. For x,yp with x(k)=0 and y(k)=0 for k<m, Tmx,Tmyp and Tmx+Tmyp. Define


Check that this makes sense. Likewise, TmxTmyp, and define


Check that this makes sense. Check that p is a commutative ring with additive identity k0 for k. and unity 01, k0 for k0. Finally,11 1 For a ring R with xR, px=k=1px. It does not make sense to talk about px before we have x+y, and it is nonsense to talk about p-mx for xp before have defined addition on p. This is why I defined Tm rather than initially using xp-mx; it is incorrect and a sloppy habit to use properties of an object before showing that it exists.

Theorem 9.

p is a field, of characteristic 0.

3 Metric

For xp define


|x|p=0 if and only if x=0. For x,yp define


dp is an ultrametric:

Theorem 10.

p is a topological field.


For (x,y),(u,v)p×p let


which shows that (x,y)x+y is continuous p×pp. And


which shows that x-x is continuous pp. For ρ((x,y),(u,v))δ, |x-u|pδ so |u|p|x|p+δ and

dp(xy,uv) =|xy-uv|p

which shows that (x,y)xy is continuous p×pp. Finally, for x,y0,


which shows that xx-1 is continuous p{0}p{0}. ∎

For xp and r>0, write


Thus, for xp and n0,

Lemma 11.

For xp,


is a local base at x.


For ϵ>0, let p-n<ϵ, n0, namely n>1logplog1ϵ. For this n,


Theorem 12.

p is a compact subspace of p.


Let xnp be a sequence. Because xn(0)Np, n0, there is some a(0)Np and an infinite subset I0 of {n0} such that xn(0)=a(0) for nI0. Suppose by induction that for some N0 there are a(0),,a(N)Np and an infinite set IN{n0} such that


But for each xIN, xn(N+1) belongs to the finite set Np, and because IN is infinite there is some a(N+1)Np and an infinite set IN+1IN such that xn(N+1)=a(N+1) for nIN+1. We have thus defined ap.

Let α0I0, and by induction let αn>αn-1, αnIn; in particular as α00 we have αnn. Then for any n0, xαn(k)=a(k) for 0kn. Take ϵ>0 and let p-m-1<ϵ. For nm,


which shows that the sequence xαn tends to a. This means that p is sequentially compact and therefore compact. ∎

For x,yp,


which shows that xpx is continuous pp. Therefore, the fact that p is compact implies that for n0, pnp is compact. Then by Lemma 11 we get the following.

Theorem 13.

p is locally compact.

Theorem 14.

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 pnp are compact, n0, from Lemma 11 we get the following.

Lemma 15.

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, p is totally disconnected.

4 p-adic fractional part

For xp, let




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


Furthermore, as x(k)0 as k-,


therefore for xp,


Define the Prüfer p-group


We assign the Prüfer p-group the discrete topology.

Define ψp:pS1 by


We prove that this is a homomorphism from the locally compact group p whose image is the Prüfer p-group and whose kernel is p.22 2 Alain M. Robert, A Course in p-adic Analysis, p. 42, Proposition 5.4.

Theorem 16.

ψp:pS1 is a homomorphism of locally compact groups. ψp(p)=(p), and kerψp=p.


For x,yp,

{x+y}p-{x}p-{y}p =x+y-[x+y]p-x+[x]p-y+[y]p

Check that [1/p]p=. It then follows that


therefore e2πi({x+y}p-{x}p-{y}p)=1, i.e.


namely ψp is a homomorphism.

ψp(x)=1 if and only if e2πi{x}p=1 if and only if {x}p. But {x}p[0,1), so ψp(x)=1 if and only if {x}p=0, hence ψp(x)=1 if and only if xp, namely


Let xp. As {x}p[1/p], there is some n0 such that pn{x}p, so ψp(x)pn=1, which means that ψp(x)[p]. Let e2πimp-n[p], n,m0. But p-np and, whether or not n>0,


and mp-np, and using that ψp is a homomorphism,


This shows that ψp(p)=[p].

Finally, let xp. For yB1(x)=x+p, so there is some wp such that y=x+w. But ψp(x+w)=ψp(x)ψp(w)=ψp(x), so


showing that ψp is continuous at x. ∎

Because [p] is discrete, it is immediate that ψp is an open map. The first isomorphism theorem for topological groups states that if G and H are locally compact groups, f:GH is a homomorphism of topological groups that is onto and open, then G/kerf and H are isomorphic as topological groups. Therefore the quotient group p/p and the Prüfer group [p] are isomorphic as topological groups.