Why is the multiplicative group $\operatorname{Spec} (\mathbb Z [T,T^{-1}])$ smooth over $\operatorname{Spec}(\mathbb Z)$?

Does one have a handy criterion for affine morphisms of schemes to be smooth? I haven't found anything like this in the standard books, but perhaps I overlooked it.

This would be nice as I also would like to prove that for any scheme $S$ and a locally free sheaf $\mathcal E$ on it, the associated vector bundle $V(\mathcal E)$ is smooth over $S$.


A morphism of schemes $f : X \to Y$ is smooth iff it it is locally of finite presentation and it is formally smooth; the latter means that for every affine scheme $\mathrm{Spec}(R)$ over $Y$ and every ideal $I \subseteq R$ with $I^2=0$ the canonical map $\hom_Y(\mathrm{Spec}(R),X) \to \hom_Y(\mathrm{Spec}(R/I),X)$ is surjective. Geometrically this means that every point can be thickened to an infinitesimal point. For a proof, see EGA IV, 17.

In particular, $A \to B$ is a smooth ring homomorphism iff it is of finite type and for every $A$-algebra $R$ and every ideal $I \subseteq R$ with $I^2=0$ the canonical map $\hom_A(B,R) \to \hom_A(B,R/I)$ is surjective. For $\mathbb{Z} \to \mathbb{Z}[T,T^{-1}]$ this map corresponds to $R^* \to (R/I)^*$. And yes, it is surjective (elementary calculation using that $I$ is nilpotent).

Of course, there are other ways to see that this is smooth. Every affine n-space $\mathbb{A}^n_S$ is smooth over $S$ (follows directly from the Jacobi definition or the characterization above). Also every open immersion is a smooth (even étale) morphism. Smooth morphisms are closed under composition. Thus every open subscheme of $\mathbb{A}^n_S$ is smooth over $S$. In particular, $\mathrm{Spec}(\mathbb{Z}[T,T^{-1}])$ is smooth.

As for your other question: Smoothness can be checked locally on the base. Therefore you may assume that your vector bundle is trivial, thereby reducing to the smoothness of the affine n-space.

  • 1
    $\begingroup$ Dear Martin, I would say that every honest point can be thickened to an infinitesimal point. Say $Y=Spec(k)$ (and forget about it ! ) . I consider $P=Spec (k)$ to be a simple point and $D=Spec (k[\epsilon]$ to be a thickened, double point. Smoothness implies that honest points $P\to X$ extend to infinitesimal thickenings $D\to X$. The point(!) is that if $R=k[\epsilon ]$ and $I=(\epsilon)$ , then despite appearances $R/I$ corresponds to the honest point and $R$ to the thickened one. $\endgroup$ – Georges Elencwajg Dec 13 '11 at 13:40
  • $\begingroup$ You're right, thanks. I've edited the answer. $\endgroup$ – Martin Brandenburg Dec 13 '11 at 14:28

Morally, this must be a consequence of the fact that a hyperbola is a smooth curve, over any field. Officially, however, we need to do some computations with these arithmetic schemes...

Let $A = \mathbb{Z}[T, T^{-1}]$, $X = \operatorname{Spec} A$, $Y = \operatorname{Spec} \mathbb{Z}$. We need to show:

  1. The morphism $X \to Y$ is flat.
  2. The morphism $X \to Y$ is locally finitely presented.
  3. For each $\mathfrak{p} \in Y$, the fibre $Y_{\mathfrak{p}}$ of $Y$ over $\mathfrak{p}$ is a smooth scheme over the residue field $\kappa(\mathfrak{p})$, i.e. $Y_\mathfrak{p} \mathbin{\times_{\kappa(\mathfrak{p})}} \overline{\kappa(\mathfrak{p})}$ is a regular scheme.

It is clear that $A$ is a flat $\mathbb{Z}$-algebra: after all, it is a free $\mathbb{Z}$-module. It is also clear that $A$ is finitely presented. For the last part, observe that $Y_\mathfrak{p} \cong \operatorname{Spec} \kappa(\mathfrak{p})[T, T^{-1}]$; but for any field $k$, by the Jacobian criterion, we find that $\operatorname{Spec} k[T, T^{-1}] \cong \operatorname{Spec} k[S, T]/(ST - 1)$ is a regular scheme. So we are done.


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