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Let $R$ be a commutative (noetherian, if needed) ring, let $f_1，\ldots，f_r\in R[x_1，\ldots，x_n]$ and $A=R[x_1，…，x_n]/(f_1，\ldots，f_r)$, when is $A$ flat over $R$?

I found a nice answer for the case $n=r=1$ here, but I don't know how to formulate a characterisation in general, or if it is solved. Any idea or reference is welcome.

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There is a nice criterion that is usually applied in this situation.

A is flat over $R$ if the Krull dimension of the fiber ring $A_{\mathfrak{P}}/\mathfrak{P}$ for all prime ideals $\mathfrak{P} \subset R$ is $n-r$ (or the fiber is empty). In scheme theoretic language, the fibers either have the expected dimension (are complete intersections) or are empty. This is what is called a relative global complete intersection here https://stacks.math.columbia.edu/tag/00SK. Flatness follows from https://stacks.math.columbia.edu/tag/00SW.

This is a generalization of the criterion you cited when $r=1$. In that case the fibers of $Spec(R[x_1, \ldots, x_n]) \to Spec(R)$ are affine spaces over a field (so integral) and the condition ensures that the restriction of the polynomial $f$ to every fiber over some closed and open subset $U \subset Spec(R)$ is nonzero, so it must cut a closed subset of the fiber of dimension $n-1$ (or empty if it restricts to a constant). The idempotent comes from the fact that if $Spec(R)$ is not connected, then we can allow the preimage over the complement of this open and closed to be just isomorphic to the whole affine space $\mathbb{A}^n_{Spec(R) \setminus U}$, ($f$ is allowed to be identically $0$ on $Spec(R) \setminus U$).

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  • $\begingroup$ This is a great answer, thanks a lot! $\endgroup$
    – Lao-tzu
    Nov 24, 2021 at 16:47
  • $\begingroup$ Thinking more carefully I think fiber to be the whole affine $n$-space is allowed, so: every nonempty fibre of Spec(A)→Spec(R) has dimension n−r or n. $\endgroup$
    – Lao-tzu
    Nov 24, 2021 at 17:06
  • $\begingroup$ If you allow the dimension to be $n$ in some fibers, it should be over an open an closed subset as in the second paragraph above (i.e. if you allow two dimensions $n$ and $n-r$, you want fiber dimension to be locally constant on the base). You don't want things like $R = k[x]$ and $A = k[x][y]/(xy)$, where the fiber dimension is $n-1 =0$ away from $x=0$ and jumps to $1$ at $x=0$. $\endgroup$
    – afh
    Nov 24, 2021 at 17:17
  • $\begingroup$ I agree, we need fiber dimension to be locally constant on the base. And I want to have a criterion like the one I cited for any $n, r\geq 1$. $\endgroup$
    – Lao-tzu
    Nov 24, 2021 at 17:24
  • $\begingroup$ Ok, I am just providing a possible generalization. Probably the best way to think about this is that you can check flatness on each connected component: on some components you can apply the global complete intersection result above, and in other components you could have all $f_i$ be identically $0$, which is clearly yields a flat ring. I hope that clarifies what I was trying to convey in the answer. The answer applies to any $n$ and $r$. $\endgroup$
    – afh
    Nov 24, 2021 at 17:28

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