p-adic analysis 1

Shu Sasaki

25th January 2011

Abstract

Lecture notes for p-adic analysis, 17/01/2011. In the lecture (17/01/2011), I actually proved much less than what I had prepared and wrote here partly because I was running out of time.

1 Introduction

REFERENCES that I find useful *for the course* are:

J.W.S. Cassels, Local fields, CUP.

A.Fröhlich & M.Taylor, Algebraic number theory, CUP

F.Gouvea, p-adic numbers, Springer-Verlag, Universitext.

J.P.Serre, Cours d’arithmétique, PUF.

J.P.Serre, Lie algebras and Lie groups, Springer-Verlag, LNM.

P.Schneider, Non-archimedean functional analysis, Springer-Verlag, Springer Monograph in Mathematics.

BGR, Non-archimedean analysis, Springer-Verlag.

My main reference is Cassels’ book. It’s because I grew up with it. “Puzzles” it offers were irresistible when I first read it.

2 p-adic numbers and non-archimedean fields

2.1 Absolute values

Let F be a field. An absolute value on F is a function || || : F → R_{≥0} which satisfies the conditions:

(1) ||x|| = 0 if and only if x = 0, (2) ||x|| ||y|| = ||xy|| for x, y ∈ F, (3) ||x + y|| ≤ ||x|| + ||y|| for every x, y ∈ F. An absolute value is non-archimedean if it satisfies the condition: (4) ||x+y|| ≤ max{||x||, ||y||} for every x, y ∈ F. Otherwise it is archimedean.

EXAMPLES: the trivial absolute value ||x|| = 1 for every non-zero x ∈ F and ||0|| = 0, NON-ARCHIMEDEAN:

(1) Let F = Q. The p-adic absolute value ||x|| = (1/p)^{v_{p}(x)} where v_{p}(x) is defined by x = p^{v_{p}(x)}(y/z) where p divides neither y nor z. This is in fact non-archimedean. In order to check it is, show that v_{p}(xy) = v_{p}(x) + v_{p}(y) and v_{p}(x + y) ≥ min{v_{p}(x), v_{p}(y)}.

(2) Let F be a field. A discrete valuation is a function v : F^{×} → Z such that (1) it is surjective (2) it is a homomorphism: v(x + y) = v(x) + v(y) and (3) v(x + y) ≥ inf{v(x), v(y)}. Normally, v is extended to F by v(0) = ∞. Discrete valuations are “non-archimedean” in the sense that: let v denote a discrete valuation on a field F. Choose a constant c ∈ R such that 0 < c < 1. Then ||x|| = c^{v(x)} defines a non-archimedean absolute value. In fact, || || is discrete in the sense that ||F^{×}|| is a discrete subgroup of R_{>0}. Indeed ||F^{×}|| = c^{Z}.

(3) Let F = k((T)) = {f/g | f, g ∈ k[[T]], g ≠ 0}, where k is a field. Fix c > 1. For f = f_{0} + f_{1}T + ⋯ + f_{n}T^{n}, f_{n} ≠ 0, define ||f|| = c^{n}. Then for f = f/g ∈ k((T)), define ||h|| = ||f||/||g||. Check that it satisfies the “ultrametric inequality” (||f +g|| ≤ max{||f||, ||g||}) which is evidently stronger than the “triangle inequality” ||f + g|| ≤ ||f|| + ||g||.

(4) Let F = k((T)). Fix 0 < d < 1 and an irreducible polynomial e ∈ k[T]. For h ∈ k((T)), define ||h|| = d^{v} where v ∈ Z, which depends only on h (and e), is defined by h = e^{v}(f/g) where f, g ∈ k[T], both of which not divisible by e, and ||0|| = 0.

ARCHIMEDEAN: Let σ be an (real) embedding of a field F into R. Then ||x|| = ||σ(x)||_{R}, where || ||_{R} denote the standard absolute value on R, is an archimedean absolute value on F. If σ is an “imaginary” (or “complex” as opposed to “real”) embedding of a field F into C in the sense that σ ≠ c ∘ σ where c ∈ Gal(C/R) is the complex conjugation, then ||x|| = ||σ(x)||_{C}, where || ||_{C} denote the standard absolute value on C: ||a + bi||_{C} = \sqrt{a^{2} + b^{2}}.

2.2 Absolute values in the sense of E.Artin

Let F be a field. An absolute value (valuation, according to Cassels) on F is a function || || : F → R_{≥0} which satisfies the conditions: there exists C ∈ R such that (1) ||x|| = 0 if and only if x = 0, (2) ||x|| ||y|| = ||xy|| for x, y ∈ F, (3)’ ||x|| ≤ 1 implies ||1 + x|| ≤ C. This is substantially simpler than the “traiangle inequality” in checking.

PROPOSITION. ||1|| = 1. Proof. ||1||^{2} = ||1||||1|| by (2). On the other hand 1^{2} = 1 so ||1^{2}|| = ||1||. Combining, ||1||^{2} = ||1||. Since 1 ≠ 0, by (1), ||1|| = 1. □

PROPOSITION. If ||x^{n}|| = 1 then ||x|| = 1. Proof. ||x||^{n} = 1. □

PROPOSITION. || − 1|| = 1. Proof. ||(−1)(−1)|| = || − 1||^{2}. OTOH, ||(−1)(−1)|| = ||1|| = 1 □ COROLLARY. || − x|| = ||x||. Proof. || − x|| = ||(−1)x|| = ||(−1)||||x|| = 1||x||. □ COROLLARY. If F is a finite field, an absolute value || || on F is necessarily trivial. PROPOSITION. If || ||_{1} is a valuation on F and e ∈ R_{>0}, then ||x||_{2} = ||x||_{1}^{e} is also an absolute

value.

In which case, we say || ||_{1} and || ||_{2} are equivalent (in the sense of Artin) (For || ||_{2}, let C_{2} = C_{1}^{e}).

LEMMA. An absolute value, in the sense of Artin, satisfies the triangle inequality (||x + y|| ≤ ||x|| + ||y||) if and only if one can take C to be 2. Proof. Suppose || || satisfies the triangle inequality. If ||x|| ≤ 1, then ||1 + x|| ≤ ||1|| + ||x|| ≤ 2. Conversely, suppose || || is “Artin” with C = 2. If x_{1}, x_{2} ∈ F, WLOG ||x_{1}|| ≤ ||x_{2}||, and may suppose x_{1} = xx_{2} with x ∈ F and ||x|| ≤ 1. Then ||x_{1} + x_{2}|| = ||x_{1}(1 + x)|| = ||x_{1}||||1 + x|| ≤ ||x_{1}||2. So ||x_{1} + x_{2}|| ≤ 2max{||x_{1}||, ||x_{2}||}. By induction, ||x_{1} + ⋯ + x_{2^{n}}|| ≤ 2^{n}max_{1≤i≤2^{n}}{||x_{i}||} for any n ∈ Z_{≥0}. Suppose we are given x_{1}, …, x_{N} ∈ F. Choose n such that 2^{n−1} < N ≤ 2^{n}, and let x_{N+1} = x_{N+2} = ⋯ = x_{2^{n}} = 0. Then ||x_{1} + ⋯ + x_{N}|| ≤ 2^{n}max_{1≤i≤N}{||x_{i}||} < 2Nmax_{1≤i≤N}{||x_{i}||}. Setting x_{1} = x_{2} = ⋯ = x_{N} = 1,

we deduce ||N|| ≤ 2N. For x, y ∈ F, ||x + y||^{n} = ||(x + y)^{n}|| = || ∑_{i=0}^{n} ( \begin{matrix}n \\ i \end{matrix}) x^{i}y^{n−i}|| ≤ 2(n +

1)max_{1≤i≤n}|| ( \begin{matrix}n \\ i \end{matrix}) x^{i}y^{n−i}|| = 2(n+1)max_{1≤i≤n}|| ( \begin{matrix}n \\ i \end{matrix}) ||||x||^{i}||y||^{n−i} ≤ 4(n+1)max_{1≤i≤n} ( \begin{matrix}n \\ i \end{matrix}) ||x||^{i}||y||^{n−i} ≤

4(n + 1) ∑_{i=0}^{n} ( \begin{matrix}n \\ i \end{matrix}) ||x||^{i}||y||^{n−i} = 4(n + 1)(||x|| + ||y||)^{n}. Taking the n-th root and letting n tend to ∞, we obtain ||x + y|| ≤ ||x|| = ||y|| as required.□

COROLLARY. Every “Artin” absolute value is equivalent to an absolute value in the old sense.

An absolute value || || : F → R is non-archimedean if one can take C to be 1. An absolute value || || is non-archimedean in the sense of Artin if and only if it satisfies the ultrametric inequality (||x + y|| ≤ max{||x||, ||y||}). Indeed, suppose ||x|| ≤ 1. Then by the ultrametric inequality, ||1 + x|| ≤ max{||1||, ||x||} ≤ 1. Conversely, suppose ||x|| ≤ ||y||. If y = 0, then ||y|| = 0 and therefore ||x|| = 0. The ultrametric inequality clearly holds for x = 0, y = 0. Otherwise y ≠ 0 and ||x/y|| = ||x||/||y|| ≤ 1. By (3)’, ||1 + x/y|| ≤ C = 1 and therefore ||x + y|| ≤ ||y|| = {max{||x||, ||y||}.

Also an absolute value, which is equivalent to a non-archimedean absolute value, is itself nonarchimedean.

KEY LEMMA. Let || || be an non-archimedean absolute value. Suppose ||y|| < ||x||. Then ||x + y|| = ||x||(= max{||x||, ||y||}). Proof. Since || || is non-archimedean, ||x+y|| ≤ ||x||. OTOH, x = (x+y)+(−y) and ||x|| ≤ max{||x + y||, ||y||}. Then ||x + y|| = ||x||. Since otherwise the first implies ||x + y|| < ||x|| and, combined with the second, ||x|| ≤ max{||x + y||, ||y||} < ||x|| and contradiction. □

USEFUL LEMMA. Let || || be an absolute value. Then || || is non-archimedean if and only if ||x|| ≤ 1 for all x in the ring “Z” generated by 1 in F. Proof. Suppose || || is non-archimedean. ||1|| = 1. Suppose ||n|| ≤ 1 for n ∈ Z. Then, by (3)’, ||1 + n|| ≤ C = 1 since || || is non-archimedean in the sense Artin. Conversely, WLOG suppose that || || satisfies the triangle inequality. For x, y ∈ F,

||x + y||^{n} ≤ ∑_{0≤m≤n} || ( \begin{matrix}n \\ m \end{matrix}) ||||x||^{m}||y||^{n−m} ≤ ∑_{0≤m≤n} ||x||^{m}||y||^{n−m} ≤ (n + 1)(max\{||x||, ||y||\})^{n}

and thus ||x + y|| ≤ (1 + n)^{1/n}max{||x||, ||y||}. Letting n → ∞, we are home. □

COROLLARY. Let F be a finite extension of f. Let || || be an absolute value on F. Then || || is non-archimedean if and only if || || when restricted to f is non-archimedean.

COROLLARY. If the characteristic of F is p > 0, then every absolute value is non-archimedean. THEOREM (Ostrowski) every non-trivial absolute value on Q is equivalent to the p-adic absolute

value || ||_{p} for p either a (finite) prime p or (the infinite prime) ∞. Cassels Theorem 2.1, p.16. THEOREM (the product formula): ∏_{p} ||x||_{p} = 1 for any x ∈ Q^{×}.

2.3 Counting absolute values on number fields

ARCHIMEDEAN: Let F be an algebraic number field. There are r_{1} + r_{2} (equivalent) archimedean absolute values on F where r_{1}=the number #Hom(F, R) of the real embeddings of F, and r_{2}=the number of the (conjugate) pairs of complex embeddings of F, so that r_{1} + 2r_{2} = [F : Q]. Equivalent archimedean absolute values on F are called the infinite places of F.

NON-ARCHIMEDEAN: Let F be an algebraic number field. There are bijections between:

(1) the prime ideals of the ring O_{F} of integers;

(2) the discrete valuations on F;

(3) the equivalent classes of (non-archimedean) discrete absolute values on F.

REMARK If F is any field of positive characteristic, any absolute value on F is discrete and nonarchimedean and, as a result, there is a bijection between the discrete valuations on F and the equivalence classes of absolute values on F. OTOH, if F is any field of characteristic zero, it follows from Fröhlich-Taylor (2.19), p.70 that “discrete” implies “non-archimedean” and there is a bijection between the discrete valuations on F and the equivalence classes of discrete absolute values on F. We assume F to be a number field for (1).

Given a prime ideal 𝔭 of F, define a discrete valuation v on F by 𝔭^{v(x)}||xO_{F} for every x ∈ F^{×}. The converse is much harder! Given a discrete valuation v on F, let O = {x ∈ F |v(x) ≥ 0} and 𝔪 = {x ∈ F |v(x) > 0}.

LEMMA 1. O is a PID. Proof. let 𝔞 be a non-zero ideal of O. Then we can and will choose a ∈ 𝔞 such that v(a) is minimum in {v(x)|x ∈ 𝔞}. Then 𝔞 = aO. Indeed, clearly aO ⊆ 𝔞. Conversely, if b ∈ 𝔞, then v(b) ≥ v(a) thus v(a^{−1}b) ≥ 0. Which implies a^{−1}b ∈ O and b = a(a^{−1}b) ∈ aO. □

LEMMA 2. 𝔪 is a maximal ideal of O. Proof. STP O^{×} = O − 𝔪. x ∈ O^{×} if and only if x ∈ O and x^{−1} ∈ O if and only if v(x) ≥ 0 and v(x^{−1}) ≥ 0, i.e., v(x) ≤ 0 if and only if v(x) = 0, i.e., x ∈ O − 𝔪. □

LEMMA 3. O ⊃ Z. Proof. v(−1) + v(−1) = v((−1)(−1)) = v(1) = 0. □

LEMMA 4. O ⊃ O_{F}. Proof. for every x ∈ O_{F}, there exists c_{0}, …, c_{n−1} ∈ Z such that c_{0} + c_{1}x + ⋯ + c_{n−1}x^{n−1} + x^{n} = 0. It then follows that nv(x) ≥ min_{0≤i≤n−1}v(c_{i}x^{i}) = min_{0≤i≤n−1}v(c_{i}) + iv(x) ≥ min_{0≤i≤n−1}iv(x). This clearly implies that v(x) ≥ 0 and therefore x ∈ O. □

Finally let 𝔭 = 𝔪 ∩ O_{F}. It is a prime ideal of O_{F} (Check!).

LEMMA 5. v is exactly the discrete valuation V associated to 𝔭. Proof. Note that v(x) = 0 if and only if V (x) = 0. Furthermore 𝔭O = 𝔪^{e} for some e ≥ 1. Say x is a non-zero element of O_{F}. Then xO_{F} = 𝔭^{V (x)}𝔮 with 𝔭 not dividing 𝔮 (in particular 𝔮O = O). Then xO = (𝔭O)^{V (x)}(𝔮O) = 𝔪^{eV (x)}. Therefore v(x) = V (x). □

OTOH, given a discrete valuation v, define ||x|| = c^{v(x)} for (any) 0 < c < 1. Clearly changing c does not change its equivalence class. This is clearly a non-archimedean absolute value (since the “exponent” v is somehow “non-archimedean”). Since v is discrete, || || is discrete. Conversely, given an equivalent class || || of non-archimedean discrete absolute value, since it is discrete, there exists 0 < c < 1 such that ||F^{×}|| = c^{Z} (Fröhlich-Taylor (2.9e), p.63). Then define v by ||x|| = c^{v(x)}. This is a valuation by Fröhlich-Taylor (2.20), p.70.

REMARK. We may identify a prime ideal 𝔭 with the equivalent class of absolute values || || on F. We call it a finite place.

2.4 Topology

Let F be a field and || || be an absolute value on F. We define d(x, y) = ||x − y||, which makes F into a metric topological space. The topology is discrete, i.e., all sets are open!, if and only if || || is (equivalent to) the trivial absolute value. Suppose that || || is equivalent to the trivial absolute value. It suffices to check that the trivial absolute value defines the discrete topology. Which follows since {x} ⊆ {y ∈ F |||x − y|| < 1/2} ⊆ {x}, therefore {x} = {y ∈ F |||x − y|| < 1/2} is open. This clearly implies that any subset of F is open. Conversely, if || || defines the discrete topology, || || is necessarily equivalent to the trivial absolute value.

For x ∈ F and r ∈ R_{≥0}, let B(r)(x) = {y ∈ F |d(x, y) < r} and C(r)(x) = {y ∈ F |d(x, y) ≤ r}.

EXERCISES:

(1) for any y ∈ B(r)(x), B(r)(x) = B(r)(y);

(2) for any y ∈ C(r)(x), C(r)(x) = C(r)(y);

(3) B(r)(x) is both open and closed;

(4) C(r)(x), r ≠ 0, is both open and closed;

(5) for r, s ∈ R_{>0}, B(r)(x)∩B(s)(y) ≠ ∅ if and only if either B(r)(x) ⊆ B(s)(y) or B(r)(x) ⊃ B(s)(y);

(6) for r, s ∈ R_{>0}, C(r)(x)∩C(s)(y) ≠ ∅ if and only if either C(r)(x) ⊆ C(s)(y) or C(r)(x) ⊃ C(s)(y).

VERY IMPORTANT: if F is a field with non-archimedean absolute value || ||. With respect to the metric topology, F is a totally disconnected (i.e., the “connected component” of every x ∈ F is {x}) topological space. Indeed, suppose that S is a set containing x. If there existed an element y ≠ x in S, then S would not be closed; S ∩ B(r/2)(x) and S ∩ (F − B(r/2)(x)) where r = ||x − y|| are non-empty (since x in the former while y in the latter) open (since B(r/2)(x) is closed, the complement F − B(r/2)(x) is open, and S ∩ (F − B(r/2)(x)) is therefore open) cover of S.

LEMMA. Let || ||_{1} and || ||_{2} are absolute values on F. Suppose that || ||_{1} is not trivial, and that ||x||_{1} < 1 implies that ||x||_{2} < 1 for every x ∈ F. Then || ||_{1} and || ||_{2} are equivalent. Proof. Firstly, observe that ||x||_{1} > 1 implies ||x||_{2} > 1 holds. Secondly we prove that ||x||_{1} = 1 implies ||x||_{2} = 1. Suppose that there existed x ∈ F such that ||x||_{1} = 1 but ||x||_{2} ≠ 1. WLOG suppose ||x||_{2} > 1. Since || ||_{1} is non-trivial, there exist y ∈ F such that ||y||_{1} < 1. Then ||yx^{n}||_{1} = ||y||_{1}||x||_{1}^{n} = ||y||_{1} < 1 for every n ≥ 0 while ||yx^{n}||_{2} = ||y||_{2}||x||_{2}^{n} > 1 for sufficiently large n, contradicting the assumption. Therefore ||x||_{1} = 1 implies ||x||_{2} = 1 for every x ∈ F.

Let y, z ∈ F. Substitute x = y^{n}z^{m}, n, m ∈ Z, get mlog||y||_{1} + nlog||z||_{1} > 0 (resp. = 0, resp. < 0) implies mlog||y||_{2} + nlog||z||_{2} > 0 (resp. = 0, resp. < 0) for every n, m ∈ Z. Let e > 0 such that mlog||y||_{1}+nlog||z||_{1} = e(mlog||y||_{2}+nlog||z||_{2}). Then m(log||y||_{1}−elog||y||_{2}) = −n(log||z||_{1}−elog||z||_{2}) for every n, m ∈ Z. Therefore, choosing z so that ||z||_{2} ≠ 1, log||y||_{1} = elog||y||_{2}, i.e., ||y||_{1} = ||y||_{2}^{e} with e = log||z||_{1}/log||z||_{2}. □

LEMMA. || ||_{1} and || ||_{2} are equivalent if and only if || ||_{1} and || ||_{2} define the same topology. Proof. If they are equivalent, they clearly define the same topology. Conversely, if || ||_{1} and || ||_{2} define the discrete topology, they *both* are necessarily trivial. Assume || ||_{1} and || ||_{2} are not trivial. Then ||x||_{1} < 1 if and only if ||x^{n}||_{1} = ||x||_{1}^{n} → 0 as n → ∞ if only if x^{n} tends to 0 with respect to the metric topology if and only if ||x^{n}||_{2} = ||x||_{2}^{n} → 0 as n → ∞ if and only if ||x||_{2} < 1. By the preceding lemma, || ||_{1} and || ||_{2} are then equivalent. □

CHINESE REMAINDER THEOREM. Let || ||_{1}, …, ||_{n} be pairwise inequivalent absolute values on F. For x_{1}, …, x_{n} ∈ F and ϵ > 0, there exists x ∈ F such that ||x − x_{i}|| ≤ ϵ for every 1 ≤ i ≤ n.

2.5 Completions

Let F be a field and || || an absolute value on F. A sequence of elements x_{n} ∈ F, n ∈ Z is called a Cauchy sequence if for every ϵ, there exists N such that for any n, m ≥ N, ||x_{n} − x_{m}|| < ϵ. The field F is complete (with respect to || ||) if any Cauchy sequence converges.

Say a two Cauchy sequence (x_{n}) and (y_{n}) of F are equivalent if (x_{n} − y_{n}) converges to 0 with respect to || ||. Then the equivalent classes of Cauchy sequences of F are defined to be the completion of F with respect to || ||. The sequence (.., x, x, x, ...) clearly define a Cauchy sequence and therefore F embeds into the completion of F.

2.5.1 F = Q and the p-adic numbers

If F = Q, we write Q_{p} for its completion. The p-adic absolute value || || on Q extends to a nonarchimedean absolute value on Q_{p}: Suppose (x_{n}) represents an equivalence class x of Cauchy sequences in Q. If the class is non-trivial, (x_{n}) does not converge to zero, and therefore there exist c > 0 such that there exists N such that for any n ≥ N, ||x_{n}|| ≥ c. OTOH, since (x_{n}) is Cauchy, there exists M such that for n, m ≥ M, ||x_{n} − x_{m}|| < c (by letting “ϵ”= c). Then for n, m ≥ max{N, M}, ||x_{n}|| = ||(x_{n} − x_{m}) + x_{m}|| = ||x_{m}|| since ||x_{n} − x_{m}|| < c ≤ ||x_{m}||. Therefore the sequence (||x_{n}||) is stationary, i.e., there exists N such that ||x_{N}|| = ||x_{N+1}|| = · · ·. Then define ||x|| to be ||x_{N}||. If (x_{n}) represent the trivial class, then define ||x|| = 0. In other words, ||x|| = lim_{n→∞} ||x_{n}|| for a Cauchy sequence representing x.

EXERCISE. This is well-defined: if (x_{n}) and (y_{n}) are Cauchy sequence in Q which are equivalent, then lim_{n→∞} ||x_{n}|| = lim_{n→∞} ||y_{n}||.

EXERCISE. || || on Q_{p} is non-archimedean.

EXERCISE. || || on Q_{p} extends || || on Q: for x ∈ Q, ||(x)|| = lim_{n→∞} ||x|| = ||x||.

PROPOSITION. The image of Q is dense in Q_{p}. Proof. Let x ∈ Q_{p}. For any r > 0, B(r)(x) = {y ∈ Q_{p}|||y − x|| < r} contains a class represented by a “constant” Cauchy sequence in Q. The proof is constructive. Choose 0 < s < r. Let (x_{n}) be a Cauchy sequence representing x. Then there exists N such that for n, m ≥ N, ||x_{n} − x_{m}|| < s. For brevity, let y denote the class defined by the constant sequence (x_{N}); we are home if ||x − y|| < r, in other words, lim_{n→∞} ||x_{n} − x_{N}||. However for n ≥ N, ||x_{n} − X_{N}|| < s < r, and therefore lim_{n→∞} ||x_{n} − x_{N}|| ≤ s < r as required. □

PROPOSITION. Q_{p} is complete with respect to || ||_{Q_{p}}. Proof. Let x_{1}, x_{2}, … be a Cauchy sequence in Q_{p} with respect to || ||_{Q_{p}}. Since Q is dense in Q_{p}, there exists a sequence y_{1}, y_{2}, … in Q such that ||x_{n} − (…, y_{n}, y_{n}, y_{n} …)||_{Q_{p}} → 0 as n → ∞. Firstly observe that the y_{n} form a Caucy sequence in Q: for ϵ > 0, there exists N such that for any n ≥ N, ||x_{n} −(y_{n})||_{Q_{p}} < ϵ/3. OTOH, there exists M such that for n, m ≥ M, ||x_{n} − x_{m}||_{Q_{p}} < ϵ/3. Then for any n, m ≥ max{N, M}, ||y_{n} − y_{m}|| = ||(y_{n}) − (y_{m})||_{Q_{p}} = ||(y_{n})−x_{n}+x_{n}−x_{m}+x_{m}−(y_{m})||_{Q_{p}} ≤ ||(y_{n})−x_{n}||_{Q_{p}}+||x_{n}−x_{m}||_{Q_{p}}+||x_{m}−(y_{m})||_{Q_{p}} < ϵ/3+ϵ/3+ϵ/3 = ϵ. Since y_{1}, y_{2}, … is Cauchy in Q, it define an element, denoted by y, in Q_{p}. In particular, with respect to || ||_{Q_{p}}, lim_{n→∞}(y_{n}) = y. Then the x_{n} converges to y with respect to || ||_{Q_{p}}: as above, there exists N such that for any n ≥ N, ||x_{n} − (y_{n})|| ≤ ϵ/2. OTOH, there exists M such that for any n ≥ M, ||y − (y_{n})||_{Q_{p}} < ϵ/2. Then for any n ≥ max{N, M}, ||x_{n} − y||_{Q_{p}} = ||x_{n} − (y_{n}) + (y_{n}) − y||_{Q_{p}} ≤ ||x_{n} − (y_{n})||_{Q_{p}} + ||(y_{n}) − y||_{Q_{p}} < ϵ/2 + ϵ/2 = ϵ. □

By the ring of p-adic integers, we mean Z_{p} = {x ∈ Q_{p}|||x|| ≤ 1}. It is a local ring with maximal ideal pZ_{p} = {x ∈ Q_{p}|||x|| < 1}.

Q ↪ Q_{p} and Z_{p} ↪ Q_{p}. The “fibre” Q ∩ Z_{p} is precisely the localisation Z_{(p)} of Z at the maximal ideal (p), i.e., the set of elements r/s ∈ Q such that p does not divide s. Clearly Z ↪ Z_{(p)}. The image of Z ↪ Z_{p} is dense. In fact Z_{p} is the completion of Z with respect to || || on Z.