Let $F$ be a field, and $P(X_1,\dots,X_m)$, $Q(X_1,\dots,X_m) \in F[X_1,\dots,X_m]$ two relatively prime polynomials. Now consider $n$ new polynomials $R_1(Y_{1 1},\dots,Y_{1 n}) \in F[Y_{11},\dots,Y_{1n}]$, $\dots$, $R_m(Y_{m 1},\dots,Y_{m n}) \in F[Y_{m1},\dots,Y_{mn}]$ in the new $mn$ indeterminates $Y_{11},\dots,Y_{1n},\dots,Y_{m1},\dots,Y_{mn}$, and assume that $R_1,\dots,R_m$ all have positive degree. My question is: are the polynomials $P(R_1(Y_{1 1},\dots,Y_{1 n}),\dots,R_m(Y_{m 1},\dots,Y_{m n}) )$ and $Q(R_1(Y_{1 1},\dots,Y_{1 n}),\dots,R_m(Y_{m 1},\dots,Y_{m n}))$, obtained from $P$ and $Q$ by substituting $R_i$ to each $X_i$ for $i=1,\dots,m$, relatively prime?
I don't know the answer in general. Thank you very much in advance for your kind attention.
NOTE The answer is positive for $m=n=1$. This is an easy consequence of following simple
Theorem. Let $F$ be a field, $\bar{F}$ its algebraic closure, and $P(X), Q(X) \in F[X]$. $P$ and $Q$ are relatively prime in F[X] if and only if they have no common root in $\bar{F}$.
Proof. The theorem follows immediately from the simple result proved in the post Given fields $K\subseteq L$, why does $f,g$ relatively prime in $K[x]$ imply relatively prime in $L[x]$? QED
Now suppose that $P(R(Y)$ and $Q(R(Y))$ are not relatively prime in $F[Y]$. Then they have a common root $y \in \bar{F}$, so that $x=R(y)$ turns out be a common root of $P(X)$ and $Q(X)$ in $\bar{F}$, which implies that $P(X)$ and $Q(X)$ are not relatively prime, contradicting the hypothesis.
