# Smoothness of a subalgebra of a smooth algebra

Given commutative rings $A \subseteq B \subseteq C$ with $C$ a smooth $A$-algebra, I am interested to know if there are some "mild" conditions that make $B$ a smooth $A$-algebra.

For example: Assume $A \subseteq B \subseteq C$ are commutative noetherian domains of zero characteristic, $C$ is a f.g. $B$-module, $B$ is a f.p. $A$-algebra ($B$ is not necessarily a f.g. $A$-module), $C$ is a flat and separable $A$-algebra (hence, $C$ is a smooth $A$-algebra) and $B$ is a flat $A$-algebra.

Is it true that $B$ is a smooth $A$-algebra?; equivalently, is it true that $fd_{B \otimes_A B}(B) < \infty$? (please see Corollary 2).

• I have tried to show formal smoothness of $B$ over $A$ (since $B$ is f.p. over $A$, smoothness is equivalent to formal smoothness), namely: For each $A$-algebra $E$, and each ideal $J$ in $E$ with $J^2=0$, the canonical homomorphism $Hom_{A−alg}(B,E) \to Hom_{A−alg}(B,E/J)$ is surjective. However, I had a problem when, given an $A$-algebras homomorphism $f: B \to E/J$ to extend it to an $A$-algebras homomorphism $F: C \to E/J$ (since, if we have such $F$, then $F$ has a preimage $G: C \to E$, and then we can restrict $G$ to $B$, and get a preimage for $f$, if I am not wrong).

• I suspect there exists a counterexample.

Remarks: (1) This answer is not applicable in the current "example", since there $B$ is not a domain. (Also, I prefer to assume that $A$ is not a field, and here $C$ is not assumed to be a finitely generated $A$-module). (2) This question asks about smoothness of $C$ over $B$. (3) See also this question and comment, which is now almost answered.

• Do the algebras $A = \mathbb C$, $B = A[t^2,t^3] \subset C = A[t]$ not satisfy your hypotheses? – Peter Samuelson Sep 9 '15 at 23:32
• If I am not wrong, $\mathbb{C}[t]$ is not separable over $\mathbb{C}$; math.stackexchange.com/questions/1317810/… – user237522 Sep 10 '15 at 6:04
• @PeterSamuelson, anyway, thanks for trying to help. – user237522 Sep 10 '15 at 15:40

Anyhoo, let's take $A = k[x, x^{-1}]$ and $C = k[y, y^{-1}]$ where $k$ is a field of characteristic not equal to $2$. Let $A \subset C$ be given by mapping $x$ to $y^2$. Now just pick any ring $B$ strictly in between $A$ and $C$. For example $B = \{g(y) \in C \mid g(1) = g(-1)\}$. Since $\text{Spec}(B)$ is the glueing of two points of $\text{Spec}(C)$ it cannot be smooth over a smooth variety like $\text{Spec}(A)$.
The $A$-Algebra $B$ will not be smooth in general. Basic examples are of the form $A = k$ a field, $C = k[X_1,\ldots ,X_n]$ and $B = C^G$ the ring of invariants of a finite group $G$ acting on $C$ by algebra automorphisms. For instance take $k = \mathbb{C}$, $n = 2$ and $G = \{\pm\}$ acting on $\mathbb{C}[X_1, X_2]$ via $\sigma(f) = f(-X_1,-X_2)$, $\sigma \neq 1$. Then $B = \mathbb{C}[X_1^2, X_1 X_2, X_2^2] \simeq \mathbb{C}[X,Y,Z]/(XZ - Y^2)$ corresponds to a surface in $\mathbb{A}^3$ which is singular at the origin.
• Thank you. However, if I am not wrong, your $C$ is not separable over $A$...I guess I was not clear enough, but I meant that people here will help me to prove or disprove my "example". – user237522 Sep 10 '15 at 6:10