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Let $(R,m)$ and $(S,n)$ be two local rings, $R \subseteq S$, $R$ regular, $S$ a finitely generated and flat $R$-algebra, and $mS=n$.

In comments to this question it was claimed that in such situation necessarily $R \subseteq S$ is unramified; however, I am not able to prove this. If I am not wrong (but I may be wrong), we need to show that $A/m \subseteq B/n$ is a (finite?) separable field extension.

Could one please help or just refer to the relevant known results that help prove that $R \subseteq S$ is unramified?

Also see this question, where the above question was discussed but not solved.

Thank you very much!

Edit: Assume that $A$ and $B$ are $k$-algebras, where $k$ is a perfect field. Then:

(a) If $S$ is a finitely generated $R$-module, then $R/m \subseteq S/n$ is a field extension of finite degree (hence algebraic). Then the field extension is separable.

(b) If $S$ is an algebraic $R$-algebra, then $R/m \subseteq S/n$ is an algebraic field extension. Then again the field extension is separable. (A special case when $S$ is algebraic over $R$ is when $S=R[w]$, where $w \in S$ is algebraic over $R$).

Let $(R,m)$ and $(S,n)$ be two local rings, $R \subseteq S$, $R$ regular, $S$ a finitely generated and flat $R$-algebra, and $mS=n$.

In comments to this question it was claimed that in such situation necessarily $R \subseteq S$ is unramified; however, I am not able to prove this. If I am not wrong (but I may be wrong), we need to show that $A/m \subseteq B/n$ is a (finite?) separable field extension.

Could one please help or just refer to the relevant known results that help prove that $R \subseteq S$ is unramified?

Also see this question, where the above question was discussed but not solved.

Thank you very much!

Edit: Assume (in addition to the above conditions) that $A$ and $B$ are $k$-algebras, where $k$ is a perfect field. Then:

(a) If $S$ is a finitely generated $R$-module, then $R/m \subseteq S/n$ is a field extension of finite degree (hence algebraic). Then the field extension is separable.

(b) If $S$ is an algebraic $R$-algebra, then $R/m \subseteq S/n$ is an algebraic field extension. Then again the field extension is separable. (A special case when $S$ is algebraic over $R$ is when $S=R[w]$, where $w \in S$ is algebraic over $R$).

Also, I now see that there is no need to assume regularity of $R$.

Summarizing, we have:

Let $(R,m)$ and $(S,n)$ be two local rings which are $k$-algebras, $k$ is a perfect field, $R \subseteq S$, $S$ is an algebraic $R$-algebra (= every element of $S$ satisfies a polynomial over $R$), and $mS=n$. Then $R \subseteq S$ is unramified.

Please, am I missing something or am I right? Any comments are welcome.

Let $(R,m)$ and $(S,n)$ be two local rings, $R \subseteq S$, $R$ regular, $S$ a finitely generated and flat $R$-algebra, and $mS=n$.

In comments to this question it was claimed that in such situation necessarily $R \subseteq S$ is unramified; however, I am not able to prove this. If I am not wrong (but I may be wrong), we need to show that $A/m \subseteq B/n$ is a (finite?) separable field extension.

Could one please help or just refer to the relevant known results that help prove that $R \subseteq S$ is unramified?

Also see this question, where the above question was discussed but not solved.

Thank you very much!

Edit: Special cases: (a) If $S$ is a finitely generated $R$-module, then $R/m \subseteq S/n$ is a field extension of finite degree. Therefore, if we assume that $A$ and $B$ are $k$-algebras, where $k$ is a perfect field, then the field extension is separable.  (b) If $S$ is an algebraic $R$-algebra, then $R/m \subseteq S/n$ is an algebraic field extension. Therefore, if we assume that $A$ and $B$ are $k$-algebras, where $k$ is algebraically closed, then $R/m = S/n$, which is trivially separable.

Let $(R,m)$ and $(S,n)$ be two local rings, $R \subseteq S$, $R$ regular, $S$ a finitely generated and flat $R$-algebra, and $mS=n$.

In comments to this question it was claimed that in such situation necessarily $R \subseteq S$ is unramified; however, I am not able to prove this. If I am not wrong (but I may be wrong), we need to show that $A/m \subseteq B/n$ is a (finite?) separable field extension.

Could one please help or just refer to the relevant known results that help prove that $R \subseteq S$ is unramified?

Also see this question, where the above question was discussed but not solved.

Thank you very much!

Edit: Assume that $A$ and $B$ are $k$-algebras, where $k$ is a perfect field. Then:

(a) If $S$ is a finitely generated $R$-module, then $R/m \subseteq S/n$ is a field extension of finite degree (hence algebraic). Then the field extension is separable. 

(b) If $S$ is an algebraic $R$-algebra, then $R/m \subseteq S/n$ is an algebraic field extension. Then again the field extension is separable. (A special case when $S$ is algebraic over $R$ is when $S=R[w]$, where $w \in S$ is algebraic over $R$).

Let $(R,m)$ and $(S,n)$ be two local rings, $R \subseteq S$, $R$ regular, $S$ a finitely generated and flat $R$-algebra, and $mS=n$.

In comments to this question it was claimed that in such situation necessarily $R \subseteq S$ is unramified; however, I am not able to prove this. If I am not wrong (but I may be wrong), we need to show that $A/m \subseteq B/n$ is a (finite?) separable field extension.

Could one please help or just refer to the relevant known results that help prove that $R \subseteq S$ is unramified?

Also see this question, where the above question was discussed but not solved.

Thank you very much!

Edit: If $R \subseteq S$ is of finite type, then the relevant field extension $R/m \subseteq S/n$ is of finite degree, so unramifiedness is satisfied, at least if $A$ and $B$ are $k$-algebras, where $k$ is a perfect field (and actually I do not mind to assume that $k=\mathbb{C}$); please, am I right?

Let $(R,m)$ and $(S,n)$ be two local rings, $R \subseteq S$, $R$ regular, $S$ a finitely generated and flat $R$-algebra, and $mS=n$.

In comments to this question it was claimed that in such situation necessarily $R \subseteq S$ is unramified; however, I am not able to prove this. If I am not wrong (but I may be wrong), we need to show that $A/m \subseteq B/n$ is a (finite?) separable field extension.

Could one please help or just refer to the relevant known results that help prove that $R \subseteq S$ is unramified?

Also see this question, where the above question was discussed but not solved.

Thank you very much!

Edit: Special cases: (a) If $S$ is a finitely generated $R$-module, then $R/m \subseteq S/n$ is a field extension of finite degree. Therefore, if we assume that $A$ and $B$ are $k$-algebras, where $k$ is a perfect field, then the field extension is separable. (b) If $S$ is an algebraic $R$-algebra, then $R/m \subseteq S/n$ is an algebraic field extension. Therefore, if we assume that $A$ and $B$ are $k$-algebras, where $k$ is algebraically closed, then $R/m = S/n$, which is trivially separable.