Flatness of certain subrings

Question

The following question appears, more or less, here:

Let $k$ be an algebraically closed field of characteristic zero and let $S$ be a commutative $k$-algebra (I do not mind to further assume that $S$ is an integral domain).

Let $R_1,R_2 \subseteq S$ be two (probably different) $k$-subalgebras of $S$ such that $R_i \subseteq S$, $1 \leq i \leq 2$ is: finitely generated, flat and separable (in other words, $R_i \subseteq S$ is etale).

Denote the $k$-algebra generated by $R_1$ and $R_2$ by $R$.

Is it true that $R \subseteq S$ is flat?

Clearly, $R \subseteq S$ is finitely generated and separable (hence unramified). But what about flatness? I guess (?) that there exists a counterexample in dimension two.

A non-counterexample: $R_1=k[x^2], R_2=k[x^3], S=k[x]$; here $R=k[x^2,x^3] \subseteq k[x]$ is not flat, but also $R_i \subseteq S$ are not separable (althouh they are finitely generated and free). See this question, which explains that $k[x^2,x^3] \subseteq k[x]$ is not separable, hence $k[x^2] \subseteq k[x]$ and $k[x^3] \subseteq k[x]$ are not separable, since if at least one of them were, then $k[x^2,x^3] \subseteq k[x]$ was separable, by a result concerning separability).

Perhaps this question is relevant.

Any hints and comments are welcome; thank you very much!

Answer 1

This is again false. The geometric interpretation is as follows: write $Y = \operatorname{Spec} S$ and $X_i = \operatorname{Spec} R_i$. Given étale morphisms $f_1 \colon Y \to X_1$ and $f_2 \colon Y \to X_2$ of affine schemes, the image factorisation $$R_1 \underset k\otimes R_2 \twoheadrightarrow R \hookrightarrow S$$ corresponds geometrically to taking the (scheme-theoretic) image $Y \to X \hookrightarrow X_1 \times X_2$ of the product map $f \colon Y \to X_1 \times X_2$. If $Y$ is smooth, then so are the $X_i$, but $X$ need not be smooth. If $f$ is surjective, it cannot be flat when $X$ is singular [Tags 07NG and 00HQ].

We can turn this around to make a counterexample: start with a map $f \colon Y \to X_1 \times X_2$ of smooth schemes such that the image is singular but both projections $f_i \colon Y \to X_i$ are étale.

Example. This is the typical picture of a plane nodal curve: let $Y_0 = X_1 = X_2 = \mathbf A^1$, and consider the morphism $f \colon Y_0 \to X_1 \times X_2$ given by $t \mapsto (t^2-1,t^3-t)$. The scheme-theoretic image is the nodal curve $X_0 = V(y^2 - x^2(x+1))$, and the map $Y_0 \to X_0$ is surjective.

The projections $f_i \colon Y_0 \to X_i$ are not étale, but they become so after removing the points $0$ and $\pm\tfrac{\sqrt 3}{3}$ from $Y_0$. Let $Y \subseteq Y_0$ (resp. $X \subseteq X_0$) be the complement of these points (resp. their images in $X$). This gives the desired counterexample, since we did not remove the singular point $(0,0) = f(-1) = f(1)$ from $X$.

(This parametrisation is explained on this page of Donu Arapura, which is literally the first result that my favourite search engine produces when I type 'nodal curve'.)

Translated back into algebra, this means $S = k\bigl[t,\tfrac{1}{t},\tfrac{1}{3t^2-1}\bigr]$, with $R_1 = k[t^2-1] \subseteq S$ and $R_2 = k[t^3-t] \subseteq S$.