A p-adic integer is defined as an infinite sequence of elements with the property that a_n+1 ≡ a_n (mod pⁿ) for every . Each a_n, being an element of , can be represented as a (rational) integer unique modulo pⁿ. Thus, if b_n, , define another sequence of integers with b_n ≡ a_n (mod pⁿ) for every n, the p-adic integers (a_n) and (b_n) are treated the same. In particular, if 0 ≤ b_n < pⁿ for every n, then (b_n) is called the canonical representation of (a_n). The set of all p-adic integers is denoted by .^[20] A sequence (a_n) of integers with a_n+1 ≡ a_n (mod pⁿ) for every n is called a p-coherent sequence.

^[20] Well! We are now in a mess of notations. We have for every . In particular, for we have which is a field that we planned to denote also by . It is superfluous to have two notations for the same thing. Many authors, therefore, prefer to avoid the hat and call as . For them, our is and/or written explicitly. Let us stick to our old conventions and use hats to remove ambiguities.

Let (a_n) and (b_n) be two p-adic integers. Define:

One can easily check that these operations are well-defined, that is, independent of the choice of the representatives of a_n and b_n. It also follows easily that these operations make a ring with additive identity and with multiplicative identity . The additive inverse of (a_n) is –(a_n) = (–a_n). Moreover is an injective ring homomorphism . In view of this, one often identifies the rational integer a with the p-adic integer . We will also do so, provided that we do not expect to face a danger of confusion. Also note that for the l-fold sum l(a_n) is the same as (l)(a_n) = (la_n). Thus in this context the two interpretations of l remain perfectly consistent.

For , the following conditions are equivalent:

Proof

[(a)⇒(b)] Let (a_n)(b_n) = (a_nb_n) = 1 = (1) for some . Then for every we have a_nb_n ≡ 1 (mod pⁿ), that is, a_n is invertible modulo pⁿ and hence modulo p as well, that is, p  a_n.

[(b)⇒(c)] Obvious.

[(c)⇒(a)] Let us construct a p-coherent sequence b_n, , of (rational) integers with a_nb_n ≡ 1 (mod pⁿ). This (b_n) would be the desired inverse of (a_n) in . Since p  a₁ and a_n ≡ a₁ (mod p), it follows that p  a_n as well and, therefore, the congruence a_nx ≡ 1 (mod pⁿ) has a unique solution modulo pⁿ, namely (mod pⁿ).

We also have a_n+1b_n+1 ≡ 1 (mod pⁿ), that is, a_nb_n+1 ≡ 1 (mod pⁿ), that is, .

Every can be written uniquely as x = p^ry for some and for some .

Proof

If p  a₁, take r := 0 and y := x. So assume that p|a₁. Choose such that [a_n]_pⁿ = [0]_pⁿ for 1 ≤ n ≤ r, whereas [a_r+1]_p^r+1 ≠ [0]_p^r+1. Such an r exists, since x ≠ 0 by hypothesis. For , we have a_r+n ≡ a_r ≡ 0 (mod p^r), that is, p^r|a_r+n, whereas a_r+n ≡ a_r+1 ≢ 0 (mod p^r+1), that is, p^r+1  a_r+n, that is, v_p(a_r+n) = r. Define b_n := a_r+n/p^r. Since a_r+n+1 ≡ a_r+n (mod p^r+n), division by p^r gives b_n+1 ≡ b_n (mod pⁿ), that is, . Moreover, p^rb_n = a_r+n ≡ a_n (mod pⁿ), that is, x = p^ry. Finally, since p  b₁, we have . This establishes the existence of a factorization x = p^ry. The uniqueness of this factorization is left to the reader as an easy exercise.

is an integral domain.

Proof

Let x₁ and x₂ be non-zero elements of . By Proposition 2.53, we can write x₁ = p^r₁ y₁ and x₂ = p^r₂ y₂ with r₁, and y₁, . Then (a_n) := x₁x₂ = p^r₁+r₂ y₁y₂. Now and hence no b_n is divisible by p. Therefore, a_r1+r2+1 = p^r₁+r₂ b_r1+r2+1 ≢ 0 (mod p^r₁+r₂+1), that is, (a_n) = x₁x₂ ≠ 0.

The quotient field of is called the field of p-adic numbers.

Every non-zero can be expressed uniquely as x = p^ry with and .

Proof

One can write x = a/b for some a, . Then a = p^sc and b = p^td for some s, , c, and so x = p^s–t(c/d) with . The proof for the uniqueness is left to the reader.

A metric on a set S is a map such that for every x, y, we have:

A set S together with a metric d is called a metric space (with metric d).

A norm on a field K is a map such that for all x, we have:

It is an easy check that for a norm ‖ ‖ on K the function , d(x, y) := ‖x – y‖, defines a metric on K.

A norm ‖ ‖ on a field K is called non-Archimedean (or a finite valuation), if ‖x + y‖ ≤ max(‖x‖, ‖y‖) for all x, (a condition stronger than the triangle inequality). A norm which is not non-Archimedean is called Archimedean (or an infinite valuation).

The p-adic norm on is defined as:

The p-adic norm | |_p is a non-Archimedean norm on .

Proof

Non-negative-ness, non-degeneracy and multiplicativity of | |_p are immediate. For proving the triangle inequality, it is sufficient to prove the non-Archimedean condition. Take x, . If x = 0 or y = 0 or x + y = 0, we clearly have |x + y|_p ≤ max(|x|_p, |y|_p). So assume that each of x, y and x + y is non-zero. Write x = p^ru and y = p^sv with r, and u, . Without loss of generality, we may assume that r ≥ s. Then, x + y = p^sz, where . Since x + y ≠ 0, we have z ≠ 0; so we can write z = p^tw for some and . But then |x + y|_p = p^–(s+t) ≤ p^–s = max(p^–r, p^–s) = max(|x|_p, |y|_p).

Two metrics d₁ and d₂ on a metric space S are called equivalent if a sequence (x_n) from S is Cauchy with respect to d₁ if and only if it is Cauchy with respect to d₂. Two norms on a field are called equivalent if they induce equivalent metrics.

Every non-trivial norm on is equivalent to | |_p for some .

Let x₁, x₂, . . . be a sequence of elements of . We say that this sequence converges to a limit , if given there exists such that |x_n – x|_p ≤ p^–M for all n ≥ N. We write this as x = lim x_n or as x_n → x.

Consider the partial sums for each . If there exists with s_n → s, we say that the sum converges to s and write .

A sequence x₁, x₂, . . . of elements of is said to be a Cauchy sequence if for every , there exists an such that |x_m – x_n|_p ≤ p^–M for all m, n ≥ N.

A field K is called complete under a norm ‖ ‖ if every sequence of elements of K, which is Cauchy under ‖ ‖, converges to an element in K.

A sequence (a_n) of p-adic numbers is a Cauchy sequence if and only if the sequence (a_n+1 – a_n) converges to 0.

Proof

[if] Take any . Since a_n+1–a_n → 0 by hypothesis, there exists such that |a_n+1 –a_n|_p ≤ p^–M for all n ≥ N. But then for all m, n ≥ N with m = n+k, , we have .

Thus (a_n) is a Cauchy sequence.

[only if] Take any . Since (a_n) is a Cauchy sequence by hypothesis, there exists such that |a_m – a_n|_p ≤ p^–M for all m, n ≥ N. In particular, |a_n+1 – a_n|_p ≤ p^–M for all n ≥ N, that is, a_n+1 – a_n → 0.

The field is complete with respect to | |_p.

Proof

Let (a_n) be a Cauchy sequence in . By Lemma 2.18, a_n+1 – a_n → 0. Therefore, there exists such that |a_n+1 – a_n|_p ≤ 1 for all n ≥ N. For n = N + k, , we have

It then follows that |a_n|_p ≤ p^–m for all , where satisfies p^–m = max(1, |a₁|_p, . . . , |a_N |_p). If m ≥ 0, then each (Exercise 2.148). Otherwise consider the sequence (p^–ma_n) which is clearly Cauchy and in which each , since |p^–ma_n|_p ≤ p^mp^–m = 1. Thus, without loss of generality, we may assume that the given sequence (a_n) itself is one of p-adic integers.

Let a_n = a_n,0+a_n,1p+a_n,2p²+· · · be the p-adic expansion of a_n (Exercise 2.145). Since (a_n) is Cauchy, for every there exists such that |a_m – a_n|_p ≤ p^–(M+1) for all m, n ≥ N_M: that is, a_{n, i} = a_{m, i} for 0 ≤ i ≤ M, m, n ≥ N_M. Define x_M := a_{n, M} for any n ≥ N_M and . It then follows that a_n → x.

is the completion of with respect to the norm | |_p.

Proof

Let C denote the ring of Cauchy sequences from (under the p-adic norm), the maximal ideal of C consisting of sequences that converge to 0, and . We now show that .

If has the p-adic expansion a = a_–rp^–r +· · ·+a_–1p^–1 +a₀+a₁p+a₂p² + · · · (Exercise 2.145), then α_n := a_–rp^–r + · · · + a_–1p^–1 + a₀ + a₁p + · · · + a_npⁿ, , define a sequence of elements of . We have |α_n – a|_p ≤ p^–(n+1), that is, α_n → a. Moreover, the sequence (α_n) of rational numbers is Cauchy with respect to | |_p, since for every we have |α_m – α_n|_p ≤ p^–(M+1) for all m, n≥ M. Thus , , is a well-defined field homomorphism. Being a field homomorphism is injective.

What remains is to show that the map is surjective. Take any . Since (β_n) is a Cauchy sequence, by Theorem 2.61 it converges to a point . We construct the sequence (α_n) corresponding to a as described in the last paragraph. Then α_n → a as well and hence using the triangle inequality (or the non-Archimedean condition) we have α_n – β_n = (α_n – a) – (β_n – a) → 0, that is, , that is, .

The p-adic series (with ) converges if and only if |a_n|_p → 0.

Proof

The only if part is obvious. For the if part, take a sequence (a_n) of p-adic numbers with |a_n|_p → 0. Define . Since a_n+1 = s_n+1 – s_n → 0 by hypothesis, Lemma 2.18 guarantees that (s_n) is a Cauchy sequence, that is, (s_n) converges in .

Let . Suppose that there exist and satisfying:

Then there exists a unique such that f(α) = 0 and |α – α₀|_p ≤ p^–(M+1) (that is, α ≡ α₀ (mod p^M+1)).

Proof

Let us inductively construct a sequence α₀, α₁, α₂, · · · of p-adic integers with the properties that |f(α_n)|_p ≤ p^–(2M+n+1) and |f′(α_n)|_p = p^–M for every . The given α₀ provides the starting point (induction basis). For the inductive step, assume that n ≥ 1 and that α₀, α₁, . . . , α_n–1 have been constructed with the desired properties. we now explain how to construct α_n from α_n–1. Put

αn := αn–1 + knpM + n

for some .

We want to find a suitable k_n for which |f(α_n)|_p ≤ p^–(2M+n+1). Taylor expansion gives f(α_n) = f(α_n–1) + k_np^M+nf′(α_n–1) + c_np^2(M+n) for some . Since by induction hypothesis p^2M+n |f(α_n–1) and p^M |f′(α_{n – 1}), we can write

Since p^M+1  f′(α_n–1), the element and, therefore, there is a unique solution for k_n of the congruence

This value of k_n yields

f (α_n) = p^{2M + n}(b_np + c_npⁿ) ≡ 0 (mod p^2M+n+1)

for some . The Taylor expansion of f′ gives f′(α_n) = f′(α_n–1) + d_np^M+n (for some ) which implies that f′(α_n) ≡ f′(α_n–1) (mod p^M), that is, |f′(α_n)|_p = p^–M.

Since |α_n – α_n–1|_p ≤ p^–(M+n), it follows that α_n – α_n–1 → 0, that is, (α_n) is a Cauchy sequence (under | |_p). By the completeness of , we then have an such that α_n → α. Similarly f(α_n) – f(α_n–1) → 0, that is, the sequence (f(α_n)) is Cauchy and hence converges to f(α). Also |f(α_n)|_p ≤ p^–(2M+n+1), that is, f(α_n) → 0, that is, f(α) = 0. Finally, each α_n ≡ α₀ (mod p^M+1), so that α ≡ α₀ (mod p^M+1). This establishes the existence of a desired .

For proving the uniqueness of α, let satisfy f(β) = 0 and |β – α₀|_p ≤ p^–(M+1). By Taylor expansion, f(β) = f(α) + (β – α)f′(α) + (β – α)²c for some , that is, (β – α)(f′(α) + (β – α)c) = 0. Now β – α = (β – α₀) – (α – α₀) and so |β – α|_p ≤ max(|β – α₀|_p, |α – α₀|_p) ≤ p^–(M+1), whereas f′(α_n) → f′(α), so that |f′(α)|_p = p^–M. Therefore, f′(α)+(β –α)c ≢ 0 (mod p^M+1) and, in particular, f′(α) + (β – α)c ≠ 0. Thus we must have β – α = 0.

Let . Suppose that there exists an satisfying:

Then there exists a unique such that f(α) = 0 and |α – α₀|_p < 1 (that is, α ≡ α₀ (mod p)).

is canonically embedded in and so is in . Thus it makes sense to carry out the lifting process for a polynomial and for some solution α₀ of f(X) ≡ 0 (mod p) in . One solves Formula (2.21) in and obtains each . The limit α belongs to and is a solution of f(X) = 0 in .

For example, let p be an odd prime and . Let be a solution of X² ≡ a (mod p). Here f(X) = X² – a, so that f′(X) = 2X, that is, f′(α₀) = 2α₀ ≢ 0 (mod p). Thus the conditions of Corollary 2.30 are satisfied and we get a unique square root of α in with α ≡ α₀ (mod p). This α has a p-adic expansion of the form α = α₀ + k₁p + k₂p² + k₃p³ + · · ·.

As a specific numerical example, take p = 7, a = 2 and α₀ = 3. Using Formula (2.21), we compute k₁ = 1, α₁ = 10, k₂ = 2, α₂ = 108, k₃ = 6, α₃ = 2166, and so on. Thus a square root of 2 in is 3 + 1 × 7 + 2 × 7² + 6 × 7³ + · · ·. The other square root of α in can be obtained by starting with α₀ = 4.