Let $R \subseteq S$ be two commutative $\mathbb{C}$-algebras such that:

(1) $R$ and $S$ are integral domains.

(2) $Q(R)=Q(S)$, namely, their fields of fractions are equal.

(3) $S=R[w]$, for some $w \in S$.

(4) $S$ is separable over $R$, namely, $S$ is a projective $S \otimes_R S$-module via $f: S \otimes_R S \to S$ given by: $f(s_1 \otimes_R s_2)=s_1s_2$.

Should such $S$ be flat over $R$? I guess no, so please it would be nice to see a counterexample.

Is there a fifth condition that would guarantee flatness of $R \subseteq S$?

Perhaps adding a fifth condition (5) $R$ is a UFD (or at least integrally closed) would guarantee flatness of $R \subseteq S$? (I am not sure).

The above is (almost) question 3 of this question. Also asked here without comments.

Thank you very much!


1 Answer 1


Take $R$ to be the coordinate ring of the nodal curve $\mathbb{C}[t^2-1, t(t^2-1)]$ and $S$ to be its normalization $\mathbb{C}[t]$. It satisfies (1), ..., (4): The first three are immediate. For (4), note that since $R$ is noetherian, the projectivity of $S$ as an $S \otimes_R S$-module is equivalent to $S$ being unramified over $R$ (Theorem 2.5 in Auslander-Buchsbaum, On Ramification Theory in Noetherian Rings, American Journal of Mathematics, 1959). It is enough to check that maximal ideals are unramified. Since we are in equi-characteristic zero, it suffices to show that for every maximal ideal $\mathfrak{q}$ of $S$, $(\mathfrak{q} \cap R) S_{\mathfrak{q}} = \mathfrak{q} S_{\mathfrak{q}}$. Let $\mathfrak{q} = (t-\alpha)$, $\alpha \in \mathbb{C}$. Then $(\mathfrak{q} \cap R) S = (t^2 - \alpha^2, t(t^2-1)- \alpha(\alpha^2-1))$. If $\alpha \neq 0$, then $t+\alpha \not \in \mathfrak{q}$ so $t-\alpha \in (\mathfrak{q} \cap R)S_{\mathfrak{q}}$. If $\alpha = 0$, then $t^2-1 \not \in \mathfrak{q}$ so $t \in (\mathfrak{q} \cap R)S_{\mathfrak{q}}$. Either way, $(\mathfrak{q} \cap R)S_{\mathfrak{q}} = \mathfrak{q}S_{\mathfrak{q}}$. However, $S$ is not a flat $R$-module.

Further comments:

  1. If $R$ is noetherian and $R \rightarrow S$ is a finite (as it is in the example), then $S$ is $R$-flat if and only if it is $R$-projective; since $\mathrm{Spec} R$ is connected, for every prime $R$-ideal $\mathfrak{p}$, the fibre $\dim_{\kappa(\mathfrak{p})}\kappa(\mathfrak{p}) \otimes_R S$ does not depend on $\mathfrak{p}$, so $\kappa(\mathfrak{p}) \otimes_R S = \kappa(\mathfrak{p})$, since $Q(R) = Q(S)$. Hence for maximal $R$-ideals $\mathfrak{p}$, the map $R/\mathfrak{p} \to S/\mathfrak{p}S$ is an isomorphism; by the Nakayama lemma, $S/R = 0$, i.e., $S=R$.

  2. If $w = \frac{1}{r}$ for some $r \in R$, then $S$ is flat.

  3. If $R$ is integrally closed, but $S$ is not obtained by inverting one element of $r$, then I don't know an example where $(4)$ holds. For example, let $R = \mathbb{C}[x,y]$ and $S = \mathbb{C}[x, \frac{y}{x}]$. This ring map comes from blowing up $\mathbb{C}^2$ at the origin. Note that $xS$ defines the exceptional divisor in the affine open set of the blow-up given by $\mathrm{Spec} S$. If $\mathfrak{q}$ is a maximal $S$-ideal containing $xS$, then $\mathfrak{q} \cap R = (x,y)R$ and $(\mathfrak{q} \cap R)S = xS$. Hence $S$ is not unramified over $R$.

  • $\begingroup$ Thank you very much! Interesting. Please, do you think it is possible to find a mild condition that will guarantee flatness of $R \subseteq S$ satisfying (1)-(4)? $\endgroup$
    – user237522
    Sep 22, 2020 at 11:19
  • $\begingroup$ I have expanded my earlier answer with some comments. $\endgroup$ Sep 22, 2020 at 13:08
  • $\begingroup$ Thank you very much for expanding your answer! $\endgroup$
    – user237522
    Sep 22, 2020 at 13:26
  • $\begingroup$ Please, could you take a look at my recent question, which also deals with separability: mathoverflow.net/questions/371948/… Any ideas are welcome! Thank you. $\endgroup$
    – user237522
    Jan 24, 2021 at 13:41
  • $\begingroup$ In particular, I do not mind to restrict the above quoted question to the following: Assume that $R=\mathbb{C}+(h) \subseteq \mathbb{C}[x]$ is separable.(hence the four conditions of my current question are satisfied). Notice that $R$ is integrally closed iff $R=\mathbb{C}[x]$, so let's assume that $R$ is not integrally closed (this holds for any $h$ of degree $\geq 2$). Then is it possible to find an exact form of such $h$? $\endgroup$
    – user237522
    Jan 24, 2021 at 13:49

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