Return to 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).

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

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!

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).

Any hints and comments are welcome; thank you!

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).

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