When some polynomials in $\mathbb Z[X]$ determine a regular sequence in $\mathbb Z[X_1,\dots,X_n]$? 
Let $f_1,\dots,f_n\in\mathbb Z[X]$ be non-constant polynomials (not necessarily distinct). Is it true that $f_1(X_1),\dots,f_n(X_n)$ is a regular sequence in $\mathbb Z[X_1,\dots,X_n]$? 

The trivial case (that suggested me this question) is $f_1(X)=\cdots=f_n(X)=X$ when we get the regular sequence $X_1,\dots,X_n$. 
 A: The answer is yes if $f_i(X_i)\in\mathbb Z[X_i]$ is non-zero and primitive (gcd of the coefficients is $1$) for all $i$.  

Lemma. Let $R$ be a commutative ring. Let $f(X)\in R[X]$ be a regular element
  such that $R[X]/(f(X))$ is flat over $R$ (e.g. if $f(X)$ is monic, but this is not a necessary condition). Then for any flat $R$-algebra $A$, $f(X)$ (viewed in $A[X]$) is a regular element of $A[X]$ and $A[X]/(f(X))$ is flat over $R$.  

Proof: The first part comes from the flatness of $A[X]$ over $R[X]$. The second part comes from the fact $A[X]/(f(X))=A\otimes_R R[X]/(f(X))$ is flat over $R[X]/(f(X))$. 

Corollary. Let $f_i(X)\in R[X]$, $i=1, \dots, n$, satisfy the conditions of the above lemma. Then $f_1(X_1), \dots, f_n(X_n)$ is a regular sequence in $R[X_1, \dots, X_n]$. 

Proof: Let $A_n:=R[X_1,\dots, X_n]/(f_1(X_1), \dots, f_n(X_n))$ is flat over $R$. If $n\ge 2$, we have $A_n=A_{n-1}[X_n]/(f_n(X_n))$. The above lemma implies (by induction on $n$) that $A_{n-1}$ is flat over $R$ and $f_n(X_n)$ is a regular element in $A_n$. 
Now in the case $R=\mathbb Z$, $f_i(X)\ne 0$ is equivalent to $f_i(X)$ regular, and $f_i(X)$ primitive is equivalent to $\mathbb Z[X]/(f(X))$ torsion-free (equivalently, flat) over $\mathbb Z$.
Remark. Clearly one can not remove the condition being primitive (consider $2X$ and $2$). But if we suppose all $f_i(X)$ are irreducible, and no prime number occurs twice in the sequence, then I think we still have a regular sequence (I didn't check the details). 
Remark'.  As for the flatness of $R[X]/(f(X))$ over an arbitrary $R$, if $f(X)$ is primitive (in the sense that its coefficients generate the unit ideal of $R$), then $R[X]/(f(X))$ is flat over $R$. Conversely, if $\mathrm{Spec}(R)$ is connected, then the flatness implies $f=0$ or $f$ is primitive.
