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

Jordan Bell
November 2, 2014

1 Hensel’s lemma

Let p be prime and f(x)[x].11 1 Hua Loo Keng, Introduction to Number Theory, Chapter 15, “p-adic numbers”. Suppose that 0a0<p, satisfies




Using the power series expansion


for any y we have




Because f(a0)0(modp), each term on the right-hand side is an integer. Then, f(a0+py)0(modp2) is equivalent to




Because f(a0)0(modp), there is a unique y(modp) that solves the above congruence, so there is a unique y(modp) that solves f(a0+py)0(modp2). This y is


Let 0a1<p be a1y(modp).

Suppose that






Using the power series expansion


for any y we have




Because f(x)0(modpl-1), each term on the right-hand side is an integer. Then, f(x+pl-1y)0(modpl) is equivalent to




Because f(x)0(modp), there is a unique y(modp) that solves the above congruence, so there is a unique y(modp) that solves f(x+pl-1y)0(modpl). This y is


Let 0al-1<p be al-1y(modp).

We have thus inductively defined a sequence a0,a1,a2,, with 0aj<p, such that for any l,


We wish to make sense of the infinite expression


Calling this x, it ought to be the case that f(x)0(modp), f(x)0(modp2), f(x)0(modp3), etc.

Example 1.

Take p=3 and f(x)=x2-7, f(x)=2x. The two conditions f(x)0(modp) and f(x)0(modp) are satisfied both by a0=1 and a0=2. Take a0=1. Then


So a1=1. Then,


So a2=1. Then,


So, a3=0. Then,


So, a4=2, etc.

2 Absolute values on fields

If K is a field, an absolute value on K is a map ||:K0 such that |x|=0 if and only if x=0, |xy|=|x||y|, and |x+y||x|+|y|. The trivial absolute value on K is |0|=0 and |x|=1 for all nonzero xK.

If || 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 ||1,||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 ||1,||2 are equivalent if and only if there is some real s>0 such that


Suppose that s>0 and that |x|1=|x|2s for all xK. Then

Bd1(x,r) ={yK:|y-x|1<r}

Since the collection of open balls for d1 is equal to the collection of open balls for d2, the absolute values ||1,||2 induce the same topology on K.

Suppose that ||1,||2 are equivalent. If |x|1<1 then d1(xn,0)=|xn|1=|x|1n0 as n. Thus xn0 in d1 and hence, because the topologies induced by ||1 and ||2 are equal, xn0 in d2, i.e. |x|2n=|xn|2=d2(xn,0)0. Therefore |x|2<1. Thus, |x|1<1 if and only if |x|2<1.

Let yK such that |y|1>1 (there is such an element because ||1 is nontrivial and |y-1|1=|y|1-1) and let xK with |x|10,1. There is some nonzero α such that |x|1=|y|1α. Let mini all be greater than α and converge to α. Then, because |y|1>1, we have |x|1=|y|1α<|y|1mini, hence |x|1ni<|y|1mi, hence |xni|1|ymi|1<1, hence


Because ||1 and ||2 are equivalent,


so |x|2<|y|2mini. Taking i gives


Similarly, we check that




Using this and |x|1=|y|1α, we have


and so, as α0,


This is true for any xK with |x|10,1. We define s to be this common value. The fact that |y|1>1 implies, because ||1 and ||2 are equivalent, that |y|2>1, and so s>0.

Now take xK. If x=0 then |x|1=0=0s=|x|2s. Because ||1 and ||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=1s=|x|2s. If |x|10,1, then the above shows that


i.e., |x|1=|x|2s, proving the claim. ∎

An absolute value ||:K0 is said to be non-Archimedean if


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

Lemma 3.

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


3 Valuations

A valuation on a field K is a function v:K{} satisfying v(x)= if and only if x=0, v(xy)=v(x)+v(y), and


The trivial valuation is v(x)=0 for x0 and v(0)=.

Lemma 4.

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


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


For x0, assume by contradiction that min{v(x+y),v(x)}=v(x). Then, since v(-x)=v(-1x)=v(-1)+v(x)=v(x),


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

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

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

Theorem 5.

Let K be a field. If || is a non-Archimedean absolute value on K and s>0, then vs:KR{} defined by vs(x)=-slog|x| for x0 and vs(0)= is a valuation on K.

If v is a valuation on K and q>1, then the function ||q:KR0 defined by |x|q=q-v(x) for x0 and |0|q=0 is a non-Archimedean absolute value on K.


Suppose that || is a non-Archimedean absolute value on K and that s>0. Let x,yK. If either is 0, then it is immediate that vs(xy)==vs(x)+vs(y). If neither is 0, then


Now, if both x,y are 0 then


If x=0 and y0 then


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


Finally, if neither x,y is 0 and x-y, then, because || is non-Archimedean,

vs(x+y) =-slog|x+y|

Thus vs is a valuation on K.

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


Let x,yK. To show that |x+y|q|x|q+|y|q, it suffices to show that |x+y|qmax{|x|q,|y|q}; proving this will establish that ||q is an absolute value and furthermore that ||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 y0, 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


Finally, if neither x,y is 0 and x-y, then

|x+y|q =q-v(x+y)

Two valuations v1,v2 on a field K are said to be equivalent if there is some real s>0 such that


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


A valuation is said to be normalized if


4 Valuation rings

Theorem 6.

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


is a maximal proper subring of K, and for all x0, xOv or x-1Ov. The set


is the group of invertible elements of Ov, and the set


is the unique maximal ideal of Ov.


It is immediate that 0,1𝒪v. For x𝒪v, v(-x)=v(x)0, so -x𝒪v. For x,y𝒪v, v(xy)=v(x)+v(y)0, so xy𝒪v. And v(x+y)min{v(x),v(y)}0, so x+y𝒪v. Thus 𝒪v is a subring of K. For nonzero xK, if v(x)0 then x𝒪v, and if v(x)<0 then v(x-1)=-v(x)>0, so x-1𝒪v.

Since v is nontrivial, there is some xK with v(x)0,. If x𝒪v then v(x)>0 and so v(x-1)=-v(x)<0, giving x-1𝒪v. Hence 𝒪vK, showing that 𝒪v is a proper subring of K.

To show that 𝒪v is a maximal proper subring, it suffices to show that if zK𝒪v then 𝒪v[z]=K, i.e., that the smallest ring containing 𝒪v and z is K. As z𝒪v, v(z)<0. Let yK. 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-j𝒪v. Hence y𝒪v[z], and so 𝒪v[z]=K, showing that 𝒪v is a maximal proper subring.

Suppose that x𝒪v and x-1𝒪v. If v(x)>0, then v(x-1=-v(x)<0, contradicting that x-1𝒪v. Hence v(x)=0. If v(x)=0, then, as x-1K, v(x-1)=-v(x)=0, so x-1𝒪v, hence x is an element of 𝒪v whose inverse is in 𝒪v.

Let x,y𝔭v. Then, since v(x)>0 and v(y)>0,


showing that x-y𝔭v, and thus that 𝔭v is an additive subgroup of 𝒪v. Let x𝔭v and z𝒪v. Then, since v(z)0 and v(x)>0,


showing that zx𝔭v. Therefore 𝔭v is an ideal in the ring 𝒪v. Since v(1)=0, 1𝔭v, so 𝔭v is a proper ideal.

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

The above ring 𝒪v 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 𝒪v/𝔭v the residue field of Ov.

Lemma 7.

If v is a normalized valuation on a field K then for all nonzero xK and tpv, v(t)=1, there is some uOv* such that


Since x0, v(x)=n. Hence v(xt-n)=v(x)-nv(t)=v(x)-n=0, and therefore u=xt-n𝒪*. Then x=utn, completing the proof. ∎

Theorem 8.

If v is a normalized valuation on a field K, then Ov is a principal ideal domain. If A is a nonzero ideal of Ov, then there is some tp, v(t)=1 and n0 such that




as Ov/pv-linear vector spaces.


Let A{0} be an ideal of 𝒪v. For any yA, v(y)0, and we take xA such that

v(x)=min{v(y):yA}. (1)

Since v(K*)=, there is some tK with v(t)=1, and because v(t)>0, t𝔭v. By Lemma 7, there is some u𝒪* such that x=utn, n=v(x). For any z𝒪, xzA and so tnzA. Thus tn𝒪vA. On the other hand, let yA. Then also by Lemma 7 there is some w𝒪v* such that y=wtm, m=v(y). By (1), m=v(y)v(x)=n, so v(tm-n)=(m-n)v(t)=m-n0 so tm-n𝒪v, giving


Therefore Atn𝒪v, and so A=tn𝒪v. That is, A is the principal ideal generated by tn, which shows that 𝒪v is a principal ideal domain.

Let t𝔭v with v(t)=1, and define ϕ:𝔭vn𝒪v/𝔭v by v(atn)=a+𝔭, for a𝒪v. ∎

Lemma 9.

If v1,v2 are discrete valuations on a field K such that Ov1=Ov2, then v1 and v2 are equivalent.

5 p-adic valuations

Fix a prime number p. For nonzero a, there are unique integers n,r,s satisfying


where r,s are coprime, s>0, and prs. We define vp(a)=n. Furthermore, we define vp(0)=.

Theorem 10.

vp:{} is a normalized valuation.


For nonzero a,b, write


where gcd(r1,s1)=gcd(r2,s2)=1, s1,s2>0, and pr1s1,pr2s2. Then,


where pr1s1r2s2; the fraction r1r2s1s2 need not be in lowest terms. So vp(ab)=m+n=vp(a)+vp(n).

Suppose that vp(a)vp(b). Then


Since ps1 and ps2, then


We call vp the p-adic valuation. The valuation ring of corresponding to vp is


in other words, those rational numbers such that in lowest terms, p does not divide their denominator. For example, 11169,-935𝒪3, and 53𝒪3. By Theorem 6, the group of units of the valuation ring 𝒪p is


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


in other words, those rational numbers such that in lowest terms, p divides their numerator and does not divide their denominator. We see that p𝔭p and vp(p)=1, so the nonzero ideals of 𝒪p are of the form

Lemma 11.


6 p-adic absolute values and metrics

We define ||p:0 by |a|p=p-vp(n) for a0 and |0|p=0. This is a non-Archimedean absolute value on , which we call the p-adic absolute value.

Example 12.

For p=3 and a=-5710, we have n=1,r=-19,s=10. Thus |-5710|3=3-1.

For p=5 and a=2875, we have n=-2,r=28,s=3. Thus |2875|5=52.

We define dp(x,y)=|x-y|p. The sequences xl=a0+a1p+a2p2++al-1pl-1 constructed when applying Hensel’s lemma satisfy, for m<n,








Thus, xn is a Cauchy sequence in the p-adic metric dp(x,y)=|x-y|p, and f(xn)0 as n.

Lemma 13.

If xn and yn are Cauchy sequences in (Q,dp), then xn+yn and xnyn are Cauchy sequences in (Q,dp).


The claim follows from



|xnyn-xmym|p =|xnyn-xmyn+xmyn-xmym|p

and the fact that xn,yn being Cauchy implies that |xn|p,|yn|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,ρ) and an isometry i:XY such that for every metric space (Z,r) and isometry j:XZ, there is a unique isometry J:YZ such that Ji=j. It is a fact that any metric space has a completion, and that if (Y1,ρ1) and (Y2,ρ2) are completions then there is a unique isometric isomorphism f:Y1Y2.

For p prime, let (p,dp) be the completion of (,dp). Elements of p are called p-adic numbers. For x,yp, there are Cauchy sequences xn,yn in (,dp) such that xnx and yny in (p,dp). We define addition and multiplication on the set p by


that these limits exists follows from Lemma 13. If xp, x0, then there is a sequence xn, each term of which is 0, such that xnx in (p,dp). Then xn-1 is a Cauchy sequence in (,dp) hence converges to some yp which satisfies xy=1. Therefore p is a field.

We define vp:p{}


One proves that vp is a normalized valuation on the field p.33 3 cf. Paul Garrett, Classical definitions of Zp and A, http://www.math.umn.edu/~garrett/m/mfms/notes/05_compare_classical.pdf We then define ||p:p0 by |x|p=p-vp(x) for x0 and |0|p=.

8 The exponential function

Lemma 14.

For a1,,arQp,

Lemma 15.

A sequence aiQp is Cauchy if and only if ai+1-ai0 as i.


Assume that ai+1-ai0 and let ϵ>0. Then there is some i0 such that ii0 implies |ai+1-ai|p<ϵ. For i0i<j,

|aj-ai|p =|aj-aj-1+aj-1+-ai+1+ai+1-ai|p

The above shows that if ai0 in (p,dp) then the series ai converges in (p,dp).

Lemma 16 (Exponential power series).

If vp(x)>1p-1, then


converges in (Qp,dp).




As n this tends to +, hence


and thus the series n=0xnn! converges. ∎

Lemma 17 (Logarithm power series).

If vp(x)>0, then


converges in (Qp,dp).


For n a positive integer we have vp(n)logpn. Then,


If vp(x)>0 then this tends to + as n. ∎

9 Topology

We define p to be the valuation ring of p. Elements of p are called p-adic integers. For xp and real r>0, write


In particular,


Because vp is discrete, there is some ϵ>0 such that


This shows that B¯p(x,r) is open in the topology induced by vp, and thus is both closed and open. It follows that p is totally disconnected.44 4 Gerald B. Folland, A Course in Abstract Harmonic Analysis, pp. 34–36.

Theorem 18.

p is totally bounded.

The fact that p is a totally bounded subset of a complete metric space implies that p is compact. Then because


and translation is a homeomorphism, any closed ball in p is compact. Therefore p is locally compact.

p is a locally compact abelian group under addition, and we take Haar measure on it satisfying μ(p)=1. One can explicitly calculate the characters on p.55 5 Gerald B. Folland, A Course in Abstract Harmonic Analysis, pp. 91–93, 104. Cf. Keith Conrad, The character group of Q, http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/characterQ.pdf