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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

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  • 8
    $\begingroup$ Denote $R[x_{k+1},\dots,x_n]/\langle g \rangle$ by $S$. Since the short exact sequence $0\to f\cdot R[x_1,\dots,x_k] \to R[x_1,\dots,x_k] \to R[x_1\dots,x_k]/\langle f \rangle \to 0$ is a short exact sequence of flat $R$-modules, so is the sequence obtained by applying $-\otimes_R S$. Thus $R[x_1,\dots,x_n]/\langle f,g \rangle$ is flat over $S$, which is flat over $R$. Your second question is equivalent to asking whether $\langle f \rangle \cap \langle g \rangle$ equals $\langle fg \rangle$. This is false when $R=\mathbb{Z}$, $f=4x_1$ and $g=6x_2$ (without the flatness hypotheses). $\endgroup$ – Jason Starr Aug 7 '16 at 10:21
  • $\begingroup$ Great, thanks Don't know how I didn't see it. Too early in morning $\endgroup$ – user95222 Aug 7 '16 at 10:34
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$R[x_1,\dots, x_n]/(f,g)=R[x_1,\dots, x_k]/(f)\otimes_RR[x_{k+1},\dots, x_n]/(g)$, so the answer to the first question is yes.

For the second, Tor$_1^A(A/I,A/J)=(I\cap J)/IJ$.

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  • 4
    $\begingroup$ Another way to see the answer to the second question (after reading the one by Merlin) is to use Tor$_1^A(A/I,A/J)=(I\cap J)/IJ$ and apply Bourbaki, Commutative Algebra, chap.I.2.6, Proposition 7 with $A=R,E=R[x_1,\dots, x_k], E'=(f), F=R[x_{k+1},\dots, x_n], F'=(g)$ to conlcude that $I\cap J=IJ$ under your hypothesis (even less). $\endgroup$ – A.G Aug 7 '16 at 15:41
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    $\begingroup$ Nice! I didn't remember that $\endgroup$ – user87684 Aug 7 '16 at 20:47
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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 $\text{Tor}_1^{A}(A/(f), A/(g))$.

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  • 6
    $\begingroup$ Ah. You're saying: since it's true that $Tor_1^{R[x_1,\dots, x_n]}(R[x_1,\dots, x_n]/(f), R[x_1,\dots, x_n]/(g)) = 0$ when $R$ is a field, then it's true that $Tor_1^{\kappa(\mathfrak{p})[x_1,\dots, x_n]}(\kappa(\mathfrak{p})[x_1,\dots, x_n]/(f), \kappa(\mathfrak{p})[x_1,\dots, x_n]/(g)) = 0$ at all points $\mathfrak{p}\in Spec(R)$. Then $Spec(A/(f,g)) \to Spec(R)$ is flat and its fibers are all lci maps, because $\kappa(\mathfrak{p})\to \kappa(\mathfrak{p})[x_1,\dots, x_n]\to \kappa(\mathfrak{p})[x_1,\dots, x_n]/(f,g)$ is lci. Then it's a flat lci, and $(f,g)$ has to be a regular ideal of $A$ $\endgroup$ – user95222 Aug 7 '16 at 11:18

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