Let $R$ be a ring. Take the polynomial ring over $R$


$$R[x\_1,\dots, x\_n]$$


nonzerodivisors $f,g\in R[x\_1,\dots, x\_n]$ such that $f$ is a polynomial in the first $k$ indeterminates, $g$ a polynomial in the last $n-k$, $0\le k\le n$.


Suppose both $R[x\_1,\dots, x\_n]/(f)$ and $R[x\_1,\dots, x\_n]/(g)$ are flat over $R$. 


Is $R[x\_1,\dots, x\_n]/(f,g)$ still flat over $R$?


Is $Tor\_1^{R[x\_1,\dots, x\_n]}(R[x\_1,\dots, x\_n]/(f), R[x\_1,\dots, x\_n]/(g)) = 0$?


If $R$ is a field this is easily seen. Thank you



Tor independence
With the flatness hypothesis, it looks both statements are true.


The first being true implies the morphism


$$\text{Spec}(R[x\_1,\dots, x\_n]/(f,g)) \to \text{Spec}(R)$$


is a flat lci. But then the ideal $(f,g)\subset R[x\_1,\dots, x\_n]$ has to be regular, hence, calling $A := R[x\_1,\dots, x\_n]$, $B := A/(f)$, the Koszul complex of $(f,g)$ is $K\_{\bullet}(f)\otimes\_B B/(g)$, whose first homology (which vanishes) is $\text{Tor}\_1^{A}(A/(f), A/(g))$.


