LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)

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$v_p(ab) = v_p(a) + v_p(b)$ 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

Statement

Let $p$ be a prime (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ) and let $a, b \in \mathbb{Z}$ be nonzero, with $v_p$ as in The $p$ -adic valuation $v_p(a)$ of a nonzero integer: the greatest $k \in \mathbb{N}$ with $p^{k} \mid a$ . Then $ab \ne 0$ and

$v_p(ab) \;=\; v_p(a) + v_p(b),$

the sum taken in $\mathbb{N}$ (Addition of natural numbers). If moreover $a + b \ne 0$ , then

$v_p(a+b) \;\ge\; \min\{\, v_p(a),\, v_p(b) \,\},$

the minimum of two natural numbers, which exists because the order on $\mathbb{N}$ is total ( $\le$ is a linear order on $\mathbb{N}$ ).

Facts & Assumptions

Given: A prime $p$ and nonzero integers $a, b$ ; $\alpha := v_p(a)$ and $\beta := v_p(b)$ .

[L1]

For a nonzero $u$ : $p^{v_p(u)} \mid u$ , and $u = p^{v_p(u)} u'$ for some nonzero $u'$ with $p \nmid u'$ ; moreover $p^{k} \mid u$ holds exactly for $k \le v_p(u)$ (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 $v_p(p) = 1$ , The $p$ -adic valuation $v_p(a)$ of a nonzero integer: the greatest $k \in \mathbb{N}$ with $p^{k} \mid a$ ).

[L2]

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, 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]

If $p$ is prime and $p \mid uv$ then $p \mid u$ or $p \mid v$ (Euclid's lemma: if $p$ is prime and $p \mid ab$ then $p \mid a$ or $p \mid b$ ).

[L4]

Divisibility is linear: $d \mid u$ and $d \mid v$ give $d \mid ux + vy$ , in particular $d \mid u + v$ (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 $d \mid -a$ ); $d \mid u$ means $u = dc$ for some $c$ (Divisibility in $\mathbb{Z}$ : $d \mid a$ when $a = dq$ for some integer $q$ ).

[L5]

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

[L6]

$\mathbb{Z}$ is a commutative ring: multiplication is associative and commutative and $x \cdot 1 = x$ (The integers form a commutative ring, Arithmetic on the integers, The integers as equivalence classes of pairs of naturals); its order is total, antisymmetric and transitive (The integers form a totally ordered ring, Order on the integers, The naturals embed in the integers).

[L7]

On $\mathbb{N}$ : $\le$ is a linear order, so any two naturals are comparable and have a minimum ( $\le$ is a linear order 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)); $m < n$ exactly when $\sigma(m) \le n$ (Discreteness: $\sigma(n)$ is the immediate successor), and $m < \sigma(n)$ exactly when $m \le n$ (On $\mathbb{N}$ the order is membership: $m < n \iff m \in n$ ).

Proof

technique · direct

1.1

$ab \ne 0$ , so $v_p(ab)$ is defined.

1.2

Fix $a'$ and $b'$ with $a = p^{\alpha} a'$ , $b = p^{\beta} b'$ , both nonzero, and $p \nmid a'$ , $p \nmid b'$ .

L1choose

1.3

Now assume also $a + b \ne 0$ , and put $m := \min\{\alpha,\beta\}$ , which exists because $\le$ is total on $\mathbb{N}$ ; then $m \le \alpha$ and $m \le \beta$ .

2.1

$ab = (p^{\alpha} a')(p^{\beta} b') = (p^{\alpha} p^{\beta})(a' b') = p^{\alpha+\beta}(a'b')$ , using commutativity, associativity and the exponent law.

step 1.2L2L6

2.2

$p \nmid a'b'$ : otherwise [L3] would give $p \mid a'$ or $p \mid b'$ , both excluded by step 1.2.

step 1.2L3

2.3

By [L1], $p^{m} \mid a$ and $p^{m} \mid b$ , so $p^{m} \mid a + b$ by linearity; since $a + b \ne 0$ , [L1] applied to $a+b$ gives $m \le v_p(a+b)$ , which is the second assertion.

step 1.3L1L4

3.1

$p^{\alpha+\beta} \ne 0$ , since $ab \ne 0$ and $ab = p^{\alpha+\beta}(a'b')$ would otherwise be $0$ .

step 1.1step 2.1L6

3.2

$p^{\alpha+\beta} \mid ab$ by step 2.1, so $\alpha + \beta \le v_p(ab)$ .

step 1.1step 2.1L1L4

4.1

Suppose $\alpha + \beta < v_p(ab)$ . Then $\alpha + \beta + 1 \le v_p(ab)$ , so $p^{\alpha+\beta+1} \mid ab$ ; fix $c$ with $ab = p^{\alpha+\beta+1}c = p^{\alpha+\beta}(pc)$ , using the exponent law. Cancelling $p^{\alpha+\beta} \ne 0$ against step 2.1 gives $a'b' = pc$ , that is $p \mid a'b'$ , contradicting step 2.2.

step 2.1step 2.2step 3.1L1L2L4L5L7

5.1

Hence $v_p(ab) = \alpha + \beta$ by totality and antisymmetry of the order on $\mathbb{N}$ , which is the first assertion.

step 3.2step 4.1L7

6.1

Both assertions are established.

step 5.1step 2.3∎

Remarks

Primality is used exactly once, at step 2.2, and it is used through Euclid's lemma: if $p$ is prime and $p \mid ab$ then $p \mid a$ or $p \mid b$ . That is the whole reason the additivity of $v_p$ is a theorem about primes and not about arbitrary integers greater than $1$ : for a composite base the analogue fails, and the companion page of the divisibility page already carries the failure of the product principle ( $6 \mid 4 \cdot 9$ while $6 \nmid 4$ and $6 \nmid 9$ : dividing a product does not force dividing a factor, and the coprimality hypothesis is what fails).
The second clause is an inequality and cannot be an equality. Take $p = 2$ , $a = 2$ , $b = 2$ : then $v_2(a) = v_2(b) = 1$ while $v_2(a+b) = v_2(4) = 2$ . The companion page's worked example computes such valuations explicitly ( $360 = 2^{3} \cdot 3^{2} \cdot 5$ and $84 = 2^{2} \cdot 3 \cdot 7$ , with $\gcd(360,84) = 12$ and $\operatorname{lcm}(360,84) = 2520$ read off the exponents ↗).
The minimum is taken in $\mathbb{N}$ , through $\le$ is a linear order on $\mathbb{N}$ , not through the library's Maximum and minimum of a set, which is stated for real numbers. The values of $v_p$ are natural numbers, so no passage to $\mathbb{R}$ is needed or wanted here.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 85 results over 27 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)
Valuation (algebra) (Wikipedia) (standard reference, not scraped)
University of Chicago REU notes: p-adic numbers (standard reference, not scraped)
Jürgen Neukirch, Algebraic Number Theory (standard reference, not scraped)

vp(ab)=vp(a)+vp(b)v_p(ab) = v_p(a) + v_p(b)vp​(ab)=vp​(a)+vp​(b) for nonzero integers a,ba, ba,b, and vp(a+b)≥min⁡{vp(a),vp(b)}v_p(a+b) \ge \min\{v_p(a), v_p(b)\}vp​(a+b)≥min{vp​(a),vp​(b)} whenever aaa, bbb and a+ba+ba+b are all nonzero