Gauss's lemma

Statement

Let $R$ be a unique factorisation domain with field of fractions $K$ and let $f \in R[x]$ be a primitive polynomial. Then $f$ is irreducible in $R[x]$ if and only if $f$ is irreducible in $K[x]$.

Remarks

The key step: if $f = g h$ in $K[x]$, then one can scale $g$ and $h$ by elements of $K$ so that they are in $R[x]$ and $f = g h$ still holds. This uses that the product of primitive polynomials is primitive. For $R = \mathbb{Z}$: a primitive integer polynomial that factors over $\mathbb{Q}$ already factors over $\mathbb{Z}$, into factors of the same degrees. The hypothesis that $f$ is primitive is necessary: $2x$ is irreducible in $\mathbb{Q}[x]$, but $2x = 2 \cdot x$ is a product of two nonunits in $\mathbb{Z}[x]$.

Explanations

L1 human

To show that an integer polynomial has no factorisation over the rationals, it is enough to check factorisations over the integers. Fractions do not help to factor a primitive polynomial. This is why criteria such as Eisenstein's work with integer coefficients.