Prove that if $\rm p(x)$ and $\rm q(x)$ are polynomials with rational coefficients whose product $\rm f(x)g(x)$ has integer coefficients , then the product of any coefficient of $\rm g(x)$ with any coefficient of $\rm f(x)$ is an integer.
I tried to prove this as follows:
Let $\rm G(x):=p(x)q(x)$. It is given that $\rm p(x),q(x)\in \mathbb Q[x]$.
If $\rm deg \, p(x)=0 $ or $\rm \, deg\, q(x)=0$, then the statement is true. So suppose that $\rm p(x), q(x)$ are of $\rm degree\gt 0$.
By Gauss' Lemma (stated below) $\rm G(x)$ is reducible in $\rm \mathbb Z[x]$ and there exist $\rm r,s\in \mathbb Q\setminus \{0\}$ such that $\rm rp(x), sq(x)\in \mathbb Z[x]$ and $\rm G(x)=(rp(x))(sq(x)).$ It follows that $\rm rs=1$.
Let coefficients of $\rm p(x), q(x)$ be denoted by $\rm p_i,q_i$'s respectively. For any $\rm i\in \{0,1,\cdots,\deg p(x)\}, j\in \{0,1,\cdots,\deg q(x)\} $, it follows that $\rm p_iq_j=1.p_iq_j=rs.p_iq_j=\overbrace {rp_i}^{\rm\in \mathbb Z}.\overbrace{sq_j}^{\rm\in \mathbb Z}\in \mathbb Z\quad\quad \square$
Is my proof correct? Thanks.
Gauss' Lemma denotes the following: If $\rm R$ is a Unique Factorization Domain with field of fractions $\rm F$, and $\rm p(x)\in R[x]$ is reducible in $\rm F[x]$ then $\rm p(x)$ is reducible in $\rm R[x]$. More precisely, if $\rm p(x)= A(x) B(x)$ for some non-constant polynomials in $\rm F[x]$ then there are non zero elements $\rm r,s\in R$ such that $\rm r A(x), sB(x)\in R[x]$ and $\rm p(x)=(r A(x))(sB(x)).$
