# Separable non-flat simple ring extension

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!

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.

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 $$R$$.
• 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)? Sep 22 '20 at 11:19
• 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 $h$? Jan 24 at 13:49