LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

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Literature-sourced — the exact statement appears in a cited source; only wording and notation differ.
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For a prime $p$ and a nonzero integer $a$ : $p^{v_p(a)} \mid a$ and $p^{v_p(a)+1} \nmid a$ ; $p^{k} \mid a$ holds exactly for $k \le v_p(a)$ ; $v_p(a) \ge 1$ exactly when $p \mid a$ ; $v_p(1) = v_p(-1) = 0$ ; and $v_p(p) = 1$

Statement

Let $p$ be a prime (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ), let $a \in \mathbb{Z}$ with $a \ne 0$ , and write $v := v_p(a)$ (The $p$ -adic valuation $v_p(a)$ of a nonzero integer: the greatest $k \in \mathbb{N}$ with $p^{k} \mid a$ ). Powers are the natural powers of Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ in the commutative monoid $(\mathbb{Z},\cdot,1)$ ( $(\mathbb{Z}, \cdot, 1)$ is a commutative monoid whose group of units is $\{1, -1\}$ ; equivalently $u \mid 1$ holds exactly for $u = 1$ and $u = -1$ ), and $k + 1$ means $\sigma(k)$ . Then:

$p^{k} \ge 1$ for every $k \in \mathbb{N}$ ; in particular $p^{k} > 0$ and $p^{k} \ne 0$ ;
$p^{v} \mid a$ and $p^{v+1} \nmid a$ ;
for $k \in \mathbb{N}$ : $p^{k} \mid a$ if and only if $k \le v$ ;
$a = p^{v} a'$ for some $a' \in \mathbb{Z}$ with $a' \ne 0$ and $p \nmid a'$ ;
$v \ge 1$ if and only if $p \mid a$ ;
$v_p(1) = v_p(-1) = 0$ and $v_p(p) = 1$ .

Facts & Assumptions

Given: A prime $p$ , a nonzero integer $a$ , and $v := v_p(a)$ .

[L1]

$v_p(a)$ is the greatest $k \in \mathbb{N}$ with $p^{k} \mid a$ ; in particular $p^{v} \mid a$ , and $p^{k} \mid a$ fails for every $k > v$ (The $p$ -adic valuation $v_p(a)$ of a nonzero integer: the greatest $k \in \mathbb{N}$ with $p^{k} \mid a$ ).

[L2]

$p^{0} = 1$ and $p^{\sigma(k)} = p^{k} \cdot p$ (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ , Semigroup and monoid, $(\mathbb{Z}, \cdot, 1)$ is a commutative monoid whose group of units is $\{1, -1\}$ ; equivalently $u \mid 1$ holds exactly for $u = 1$ and $u = -1$ ).

[L3]

Exponent law in a monoid, for natural exponents: $g^{m+n} = g^{m} g^{n}$ (Exponent laws in a group: $g^{m+n} = g^{m}g^{n}$ and $(g^{m})^{n} = g^{mn}$ for all $m, n \in \mathbb{Z}$ , and $(gh)^{n} = g^{n}h^{n}$ when $g$ and $h$ commute).

[L4]

Divisibility is reflexive and transitive; $d \mid u$ implies $d \mid uc$ ; and $d \mid u$ , $d \mid -u$ , $-d \mid u$ are equivalent (Divisibility is reflexive and transitive on $\mathbb{Z}$ , and is linear: if $d \mid a$ and $d \mid b$ then $d \mid ax + by$ for all integers $x, y$ ; also $d \mid a$ implies $d \mid ac$ , $-d \mid a$ and $d \mid -a$ ).

[L5]

$d \mid u$ means $u = dc$ for some $c \in \mathbb{Z}$ (Divisibility in $\mathbb{Z}$ : $d \mid a$ when $a = dq$ for some integer $q$ ).

[L6]

$u \mid 1$ exactly when $u = 1$ or $u = -1$ ( $(\mathbb{Z}, \cdot, 1)$ is a commutative monoid whose group of units is $\{1, -1\}$ ; equivalently $u \mid 1$ holds exactly for $u = 1$ and $u = -1$ ).

[L7]

Every prime satisfies $p > 1$ (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

[L8]

If $xz = yz$ and $z \ne 0$ then $x = y$ ; and a product of two nonzero integers is nonzero (The integers have no zero divisors; multiplicative cancellation).

[L9]

On $\mathbb{N}$ : $m \le n$ means $m + c = n$ for some $c$ (Order on the natural numbers); $\sigma(k) = k + 1$ (Addition of natural numbers, The natural numbers $\mathbb{N}$ (von Neumann)); $k < \sigma(k)$ and $m < \sigma(n)$ exactly when $m \le n$ (On $\mathbb{N}$ the order is membership: $m < n \iff m \in n$ ); $m < n$ exactly when $\sigma(m) \le n$ (Discreteness: $\sigma(n)$ is the immediate successor); and $\le$ is a linear order, so it is antisymmetric and total ( $\le$ is a linear order on $\mathbb{N}$ ).

[L10]

$\mathbb{Z}$ is a commutative ring, its order is total, antisymmetric and transitive and compatible with addition, and $\iota : \mathbb{N} \to \mathbb{Z}$ is injective, order preserving, with image the nonnegative integers and $\iota(0) = 0$ , $\iota(1) = 1$ (The integers form a commutative ring, Arithmetic on the integers, The integers as equivalence classes of pairs of naturals, The integers form a totally ordered ring, Order on the integers, The naturals embed in the integers).

[L11]

Induction on $\mathbb{N}$ : a subset containing $0$ and closed under $\sigma$ is all of $\mathbb{N}$ (The principle of mathematical induction).

Proof

technique · direct

1.1

$p^{1} = p^{\sigma(0)} = p^{0} \cdot p = 1 \cdot p = p$ .

L2L10

1.2

$0 < 1 < p$ in $\mathbb{Z}$ , since $1 = \iota(1)$ is nonnegative and differs from $0$ ; hence $p \ne 0$ , $p \ne 1$ and $p \ne -1$ .

L7L10

1.3

Clause 1. $p^{v} \mid a$ by [L1], and $v < \sigma(v) = v + 1$ , so $p^{v+1} \mid a$ would contradict the maximality of $v$ ; hence $p^{v+1} \nmid a$ .

L1L9

1.4

Clause 2, the other direction. If $p^{k} \mid a$ then $k$ belongs to the set of which $v$ is the greatest element, so $k \le v$ .

L1L9

2.1

Clause 0: $p^{k} \ge 1$ for every $k \in \mathbb{N}$ . The set of such $k$ contains $0$ , since $p^{0} = 1$ ; and if $p^{k} \ge 1$ then $p^{\sigma(k)} = p^{k} p$ has both factors $\ge 1 > 0$ , so the product is positive, hence $\ge 1$ by discreteness of the order on $\mathbb{Z}$ (a positive integer is $\iota(j)$ with $j \ne 0$ , so $1 = \sigma(0) \le j$ and $1 \le \iota(j)$ ). By induction the set is all of $\mathbb{N}$ , and in particular $p^{k} > 0$ and $p^{k} \ne 0$ .

step 1.1step 1.2L2L9L10L11

2.2

Clause 2, one direction. Let $k \le v$ and write $v = k + c$ with $c \in \mathbb{N}$ . Then $p^{v} = p^{k+c} = p^{k} p^{c}$ , so $p^{k} \mid p^{v}$ ; with $p^{v} \mid a$ and transitivity, $p^{k} \mid a$ .

step 1.3L1L3L4L5L9

2.3

Clause 3. By [L1] fix $a'$ with $a = p^{v} a'$ . Then $a' \ne 0$ , since $a \ne 0$ . If $p \mid a'$ , say $a' = pb$ , then $a = p^{v}(pb) = (p^{v} p) b = p^{v+1} b$ , so $p^{v+1} \mid a$ , contradicting step 1.3. Hence $p \nmid a'$ .

step 1.3L1L2L5L9L10choose

2.4

If $k \ge 1$ then $p \mid p^{k}$ : write $k = 1 + c$ , so $p^{k} = p^{1} p^{c} = p \cdot p^{c}$ .

step 1.1L3L5L9

3.1

Clause 4. If $v \ge 1$ then $p = p^{1} \mid a$ by step 2.2. Conversely if $p \mid a$ then $p^{1} \mid a$ , so $1 \le v$ by step 1.4.

step 1.1step 2.2step 1.4

3.2

$v_p(1) = 0$ . If $p^{k} \mid 1$ with $k \ge 1$ , then $p \mid p^{k} \mid 1$ by step 2.4 and transitivity, so $p = 1$ or $p = -1$ , contradicting step 1.2. So no $k \ge 1$ satisfies $p^{k} \mid 1$ , while $p^{0} = 1 \mid 1$ ; the greatest such $k$ is $0$ .

step 1.2step 2.4L1L2L4L6L9

3.3

$v_p(p) = 1$ . First $p^{1} = p \mid p$ , so $1 \le v_p(p)$ by step 1.4 applied to $a := p$ . Next $p^{2} \nmid p$ , where $2 = \sigma(1)$ : otherwise $p = p^{2}c = p(pc)$ for some $c$ , and cancelling $p \ne 0$ gives $1 = pc$ , so $p \mid 1$ , contradicting step 1.2. Hence $v_p(p) < 2$ by step 2.2, that is $v_p(p) \le 1$ , and antisymmetry gives $v_p(p) = 1$ .

step 1.1step 1.2step 2.2step 1.4L2L5L8L9

4.1

$v_p(-1) = 0$ , because $p^{k} \mid -1$ holds exactly when $p^{k} \mid 1$ , so the two sets of exponents coincide.

step 3.2L1L4

5.1

Clauses 0 to 5 are established.

step 1.3step 2.1step 2.2step 1.4step 2.3step 3.1step 3.2step 4.1step 3.3∎

Remarks

Clause 3 is the working form. Every later proof on this page writes a nonzero integer as $a = p^{v_p(a)} a'$ with $p \nmid a'$ and then argues about $a'$ ; that is how $v_p(ab) = v_p(a) + v_p(b)$ for nonzero integers $a, b$ , and $v_p(a+b) \ge \min\{v_p(a), v_p(b)\}$ whenever $a$ , $b$ and $a+b$ are all nonzero gets additivity, and it is where Euclid's lemma enters.
The hypothesis $a \ne 0$ is carried everywhere because $v_p(0)$ is not defined: every power of $p$ divides $0$ , so there is no greatest exponent (The $p$ -adic valuation $v_p(a)$ of a nonzero integer: the greatest $k \in \mathbb{N}$ with $p^{k} \mid a$ ).
Clause 2 says the exponents form an initial segment, which is what makes "the greatest $k$ with $p^{k} \mid a$ " the same as "the number of times $p$ divides $a$ ". Without it the maximum could in principle skip values.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 76 results over 25 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

P-adic valuation (Wikipedia) (standard reference, not scraped)
University of Chicago REU notes: p-adic numbers (standard reference, not scraped)