LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11

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

Statement

Let $0\ne f=a_0+\cdots+a_nx^n\in\mathbb Z[x]$ . 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 $(a_0,\ldots,a_n)$ 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 $d \ge 0$ , $d \mid a$ , $d \mid b$ , 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 $n \ge 1$ 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 $q_i = p_{\pi(i)}$ for some $\pi \in \operatorname{Sym}(r)$ ).

Proof

technique · induction

1.1

By induction on $k$ , [L2] shows that $c_{k+1}$ divides $a_0,\ldots,a_k$ 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 $a_0,\ldots,a_{k-1}$ by $c_k$ and applies [L2] to $(c_k,a_k)$ , so [L3] and [L1] give the asserted characterization of $\operatorname{cont}(f)$ ; 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

The product of primitive integer polynomials is primitive, and contents multiply Lemma

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

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 87 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

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