LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-11

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Content is the positive common divisor of the coefficients divisible by every common divisor

Statement

Let $0 \neq f = a_{0} + \dots + a_{n} x^{n} \in Z [x]$ . Its content is positive, divides every coefficient, and is divisible by every integer that divides every coefficient. Consequently, $f$ is primitive exactly when no prime divides all of its coefficients.

Facts & Assumptions

Given: A nonzero integer polynomial with trimmed coefficient list $(a_{0}, \dots, a_{n})$ and iterated gcds $c_{0} = 0$ , $c_{k + 1} = \gcd (c_{k}, a_{k})$ .

[L1]

The final iterated gcd is the content, and primitiveness means content $1$ (Content and primitive integer polynomials).

[L2]

Every common divisor of two integers divides their gcd, and that gcd is the nonnegative common divisor with this universal property (Every common divisor of $a$ and $b$ divides $\gcd (a, b)$ ; consequently $d = \gcd (a, b)$ exactly when $d \geq 0$ , $d ∣ a$ , $d ∣ b$ , and every common divisor of $a$ and $b$ divides $d$ — a characterisation that holds at $(a, b) = (0, 0)$ as well).

[L3]

The induction principle proves a natural-number property from its base and successor cases (The principle of mathematical induction).

[L4]

Every positive integer is a finite product of primes, with the empty product occurring only at $1$ (The fundamental theorem of arithmetic: every integer $n \geq 1$ is a product of primes, and the factorisation is unique up to order — if $\prod_{i < r} p_{i} = \prod_{j < s} q_{j}$ with every $p_{i}$ and $q_{j}$ prime, then $r = s$ and $q_{i} = p_{π (i)}$ for some $π \in Sym (r)$ ).

Proof

technique · induction

1.1

By induction on $k$ , [L2] shows that $c_{k + 1}$ divides $a_{0}, \dots, a_{k}$ and that every common divisor of those coefficients divides $c_{k + 1}$ ; the base $k = 0$ uses $\gcd (0, a_{0}) = ∣ a_{0} ∣$ .

basegivenL2L3

2.1

The induction step replaces the universal common divisor of $a_{0}, \dots, a_{k - 1}$ by $c_{k}$ and applies [L2] to $(c_{k}, a_{k})$ , so [L3] and [L1] give the asserted characterization of $cont (f)$ ; positivity follows because some coefficient is nonzero.

step 1.1ihL1L2L3discharge-induction

3.1

If the content exceeds $1$ , [L4] supplies a prime divisor of it, which step 2.1 makes a divisor of every coefficient; conversely, any prime dividing every coefficient divides the content by step 2.1 and prevents it from being $1$ .

step 2.1L1L4discharge-induction∎

Depends on

Used by

Cited to discharge well-definedness by Content and primitive integer polynomials.

Dependency tree · two levels

36 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Keith Conrad, Irreducibility Tests in Q[T], Appendix A.1 (standard reference, not scraped)