Gauss' Lemma

Lemma: A polynomial in \(\mathbb{Z}[x]\) is irreducible if and only if it is irreducible over \(\mathbb{Q}[x]\).

Proof: Let \(m,n\) be the gcd’s of the coefficients of \(f,g \in \mathbb{Z}[x]\). Then \(m n\) divides the gcd of the coefficients of \(f g\). We wish to show that this is in fact an equality.

Divide \(f\) by \(m\) and \(g\) by \(n\), so that we need only consider the case \(m = n = 1\). It suffices to show that the gcd \(d\) of the coefficients of \(f g\) must be 1. If \(d\gt 1\), then let \(p\) be some prime dividing \(d\). Consider the equation \(f g = 0 \pmod{p}\). Since \(\mathbb{Z}_p[x]\) is an integral domain, this means \(f = 0\) or \(g=0\), implying that \(p\) divides all the coefficients of \(f\) or \(g\), which is a contradiction. Thus \(d = 1\).

Now suppose \(f = g h\) over \(\mathbb{Q}[x]\). Find \(m,n \in \mathbb{Q}\) such that \(m g, n h \in\mathbb{Z}[x]\), and the gcd’s of the coefficients of \(m g, n h\) are 1. Then we have \((m g)(n h) = d f\) for some \(d\).


Ben Lynn blynn@cs.stanford.edu 💡