Monic polynomial reducible over rationals 
Let $P(x)\in \mathbb{Z}[x]$, $Q(x),R(x)\in \mathbb{Q}[x]$, and all three polynomials are monic. Suppose $P(x)=Q(x)R(x)$. Is it true that $Q(x),R(x)\in\mathbb{Z}[x]$?

Gauss's Lemma says that since $P(x)$ is reducible over $\mathbb{Q}[x]$, it must be reducible over $\mathbb{Z}[x]$. But still it doesn't say that $Q,R\in\mathbb{Z}[x]$.
 A: That is the original form of Gauss's Lemma. A proof follows easily from said modern form GL.
Write $\,Q = \dfrac{f}c,\ R = \dfrac{g}d\, $ for primitive $\,f,g\in\Bbb Z[x],\,$ and $\,0 < c,d\in\Bbb Z,\,$ and suppose that $\, h = QR = fg/(cd) \in\Bbb Z[x],\,$ so $fg = cd\,h.\,$ By GL: $\,f,g\,$ primitive $\Rightarrow$ $fg$ primitive, so $\,c,d = 1,\,$ hence $\,Q =  f,\ R =  g,\,$ so both are $\in \Bbb Z[x].$
Remark $\ $ More generally the following is true. If $f,g\in\Bbb Q[x]$ and $\,fg\in\Bbb Z[x]\,$ then $\,f_i g_j\in \Bbb Z$ for all coefficients $f_i,g_j$ of $f,g.\,$ In particular, if $f$ is monic, taking $f_i = 1$ = leading coefficient, shows that every $\,g_j\in \Bbb Z,\,$ and, similarly every $f_i\in \Bbb Z,\,$ yielding the above special case.
This general form is true over any integrally-closed domain (it is equivalent to integral closure). This is sometimes called the Gauss-Kronecker Lemma, or Dedekind's Prague Theorem. Dedekind discovered a form applying to algebraic integers after studying a divisor-theoretic form that appeared in Kronecker's seminal work on divisor theory. For much further discussion, both historical and mathematical, see Harold Edwards, Divisor Theory.
Below is Gauss's original form of Gauss Lemma, from Art. $42$ of Disq. Arith. Notice that is of the form mentioned above, not the reformulated modern form using primitivity or content.


