Gauss's lemma: factorisation in $\QQ[x]$ versus $\ZZ[x]$

What does factorization over $\QQ[x]$ say about factorization over $\ZZ[x]$?

Solution

For $f\in\ZZ[x]$, let $\operatorname{cont}(f)$ be the gcd of its coefficients. A polynomial is primitive if its content is $1$.

Gauss’s lemma says that the product of primitive polynomials in $\ZZ[x]$ is primitive. Consequently, if a primitive polynomial $f\in\ZZ[x]$ factors in $\QQ[x]$, \[ f=gh,\qquad g,h\in\QQ[x], \] with both factors nonconstant, then after clearing denominators and dividing out contents one obtains a factorization \[ f=g_0h_0 \] with nonconstant primitive $g_0,h_0\in\ZZ[x]$.

Therefore a primitive polynomial in $\ZZ[x]$ is irreducible in $\ZZ[x]$ if and only if it is irreducible in $\QQ[x]$.

For arbitrary $f\in\ZZ[x]$, write \[ f=c(f)f_0 \] with $f_0$ primitive. The nonconstant factorization theory of $f$ over $\QQ[x]$ is exactly that of its primitive part $f_0$ over $\ZZ[x]$; only the integer content changes.

Source Collections

Unsorted algebra problems, problem 651