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.

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

OK, this is an answer for the question with "separable algebra" replaced by "finite etale algebra". I am too lazy to look up separable algebra, but it seems to me that finite etale ring maps are the most beautiful rings maps you could ever want to look at (well, besides identity maps perhaps), so they should be separable algebras. I am also assuming f.p. and f.g. means what I think they mean and that flat means what I think it means. And so on and so forth. Now what I am trying to say here is that this is an answer to the question as I see it.

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.

  • $\begingroup$ 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". $\endgroup$ – user237522 Sep 10 '15 at 6:10
  • $\begingroup$ Sorry, misread you. $\endgroup$ – Nicolas Schmidt Sep 10 '15 at 15:30
  • $\begingroup$ It's really ok; I should have been more clear. Anyway, I have learned something new from your answer, so it was not in vain. $\endgroup$ – user237522 Sep 10 '15 at 15:36

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