Smoothness and field of fractions

Question

If $k$ is an integral domain and $A$ is a Noetherian finitely presented $k$-algebra for which $A \otimes_k Q(k)$ is a smooth $Q(k)$ algebra, then can it be deduced that $A$ was initially smooth over $k$?

Answer 1

To expand on Zhen Lin's comment:

Let $R$ be an integral domain, $A$ a finitely presented $R$-algebra, $Q$ the fraction field of $R$.

Then:

• to say that $A$ is smooth over $R$ can be reinterpreted geometrically as saying that the morphism of affine schemes $\operatorname{Spec} A \rightarrow \operatorname{Spec }R$ has nonsingular schemes as fibres over each point.
• to say that $A \otimes_R Q$ is smooth over $Q$ means geometrically that the fibre of $\operatorname{Spec} A \rightarrow \operatorname{Spec }R$ over the generic point $\operatorname{Spec} Q$ of $\operatorname{Spec }R$ is nonsingular.

The second does not imply the first.

As a simple example, let $R=k[t]$, $A=k[x,y,t]/\left<xy-t\right>$. Geometrically, $\operatorname{Spec} A \rightarrow \operatorname{Spec} R$ is a family of affine conics in $\mathbf A^2$ (with coordinates $x,y$) parametrised by the affine line $\mathbf A^1$ (with coordinate $t$). The generic member of this family is indeed a nonsingular conic. However, the fibre over the point $0 \in \mathbf A^1$ is the conic $xy=0$, which is singular.

Answer 2

$\mathbb{Z} \to \mathbb{Z}[\sqrt{2}]$ is not smooth (for example since $\Omega^1_{\mathbb{Z}[\sqrt{2}]/\mathbb{Z}} \cong \mathbb{Z}[\sqrt{2}]/(2)$ is torsion), but $\mathbb{Q} \to \mathbb{Z}[\sqrt{2}] \otimes_\mathbb{Z} \mathbb{Q} = \mathbb{Q}(\sqrt{2})$ is smooth (in fact, étale).