Gauss's Lemma (Polynomial)

Statement

Let $R$ be a UFD and $K = \operatorname{Frac}(R)$ its fraction field. A polynomial $f \in R[x]$ is primitive if the ideal generated by its coefficients equals $R$ (equivalently, no prime of $R$ divides all coefficients).

Gauss's Lemma (content form): The product of two primitive polynomials is primitive.

Gauss's Lemma (irreducibility form): A primitive polynomial $f \in R[x]$ is irreducible over $R$ if and only if it is irreducible over $K$.

In particular, if $f \in \mathbb{Z}[x]$ is primitive and has no factorisation over $\mathbb{Z}$, it has no factorisation over $\mathbb{Q}$ either.

Visualization

Primitive part extraction. Given $f(x) = 2x^2 + 4x + 6$, extract the content:

f(x) = 2x^2 + 4x + 6
     = 2 · (x^2 + 2x + 3)
       ^^    ^^^^^^^^^^^^
    content   primitive part

Content: $\gcd(2, 4, 6) = 2$. Primitive part: $x^2 + 2x + 3$ (coefficients have $\gcd = 1$).

Product of primitives stays primitive. Let $f = x + 1$ and $g = x - 1$, both primitive.

Prime $p$	$p \mid$ all coeffs of $f$?	$p \mid$ all coeffs of $g$?	$p \mid$ all coeffs of $fg = x^2-1$?
$2$	No ($1 \nmid 2$)	No	No ($1$ is odd)
$3$	No	No	No
any $p$	No	No	No

No prime divides all coefficients of $fg$, so $fg$ is primitive.

Proof Sketch

1. Content is multiplicative. For $f \in R[x]$, define $\operatorname{cont}(f) = \gcd$ of all coefficients. The key identity is $\operatorname{cont}(fg) = \operatorname{cont}(f)\cdot\operatorname{cont}(g)$ up to units.

2. Primitive times primitive. If $f$ and $g$ are primitive and $p$ is any prime, reduce $f$ and $g$ modulo $p$. Since $(R/pR)[x]$ is an integral domain (as $pR$ is prime), $\bar{f}\,\bar{g} \ne 0$. So $p$ does not divide all coefficients of $fg$.

3. Irreducibility transfer. If $f$ is primitive and $f = g \cdot h$ over $K$, clear denominators: write $g = (1/d_1)g_1$, $h = (1/d_2)h_2$ with $g_1, h_2 \in R[x]$ primitive. Then $d_1 d_2 f = g_1 h_2$, and taking contents: $d_1 d_2 \cdot \operatorname{cont}(f) = \operatorname{cont}(g_1)\operatorname{cont}(h_2) = 1$. So $d_1 d_2$ is a unit, meaning $f = g_1 h_2$ is a factorisation in $R[x]$.

4. Contrapositive. If $f$ is irreducible over $R$, then any $K$-factorisation descends to an $R$-factorisation — but one factor must be a unit in $R[x]$, hence a unit in $K[x]$ too. So $f$ is irreducible over $K$.

Connections

Gauss's Lemma is the bridge that makes Eisenstein's CriterionEisenstein's Criterion$$p \mid a_0,\dots,a_{n-1},\; p^2 \nmid a_0,\; p \nmid a_n \implies f \text{ irreducible}$$A prime-divisibility condition on coefficients that certifies irreducibility of a polynomial over ℤ or any UFD.Read more → work over $\mathbb{Q}$: one proves irreducibility over $\mathbb{Z}$ and Gauss transfers it. The result generalises to any UFD, and UFDs arise from the chain PID Implies UFDPID Implies UFD$$\text{PID} \implies \text{UFD}$$Every principal ideal domain is a unique factorisation domain: elements factor uniquely into irreducibles up to units and order.Read more →.

-- Mathlib: IsPrimitive.irreducible_iff_irreducible_map_fraction_map
-- in Mathlib.RingTheory.Polynomial.GaussLemma
open Polynomial in
/-- Gauss's Lemma: a primitive polynomial over ℤ is irreducible iff
    its image in ℚ[X] is irreducible. -/
theorem gauss_lemma_int (p : ℤ[X]) (hp : p.IsPrimitive) :
    Irreducible p ↔ Irreducible (p.map (algebraMap ℤ ℚ)) :=
  hp.irreducible_iff_irreducible_map_fraction_map