A curiosity: "supersmooth" varieties

I want to share a curious condition on varieties for which I have found no use. Let $k$ be a field and let $X$ be a locally finite type $k$ -scheme. Recall that $X$ is said to be smooth if, for every Artin local $k$ -algebra $A$ with a proper ideal $I \subset A$ and every $k$ -morphism $\Spec A/I \to X$ , there exists a $k$ -morphism $\Spec A \to X$ extending it, i.e., the map $X(A) \to X(A/I)$ is surjective. (It takes real work to show that this is equivalent to other more concrete notions.) It is an amusing exercise to check that this is equivalent to the following statement: for every complete noetherian local $k$ -algebra $R$ and every ideal $I \subset R$ such that $R/I$ is Artin, the map $X(R) \to X(R/I)$ is surjective. Motivated by this, we will say that a locally finite type $k$ -scheme $X$ is supersmooth if for any noetherian local $k$ -algebra $R$ and any ideal $I \subset R$ such that $R/I$ is Artin, the map $X(R) \to X(R/I)$ is surjective. Of course, by the above discussion every supersmooth $k$ -scheme is smooth. One could define a relative notion of supersmoothness, but it is not a “good” condition because it cannot be checked etale-locally, see below.

Supersmoothness will turn out to be a rather rigid condition; to show that our results are not vacuous, let us first give some examples.

Examples. (1) Affine $n$ -space $\bA^n = \Spec k[t_1, \dots, t_n]$ is supersmooth: if $R$ and $I$ are as in the definition, then a morphism $\Spec R/I \to \bA^n$ is equivalent to the data of $n$ elements of the ring $R/I$ . Since $R \to R/I$ is surjective, it is clear that $\bA^n(R) \to \bA^n(R/I)$ is surjective.

(2) Because the rings involved in the definition are local, any open subscheme of a supersmooth $k$ -scheme is supersmooth, and supersmoothness may be checked Zariski-locally on any locally finite type $k$ -scheme. Moreover, supersmoothness is preserved by field extension (although it may not be checked after field extension, see Example (5)). The product of supersmooth varieties is clearly supersmooth.

(3) If $G$ is a split reductive group, then using the existence of the open cell and the fact that its translates by rational points cover $G$ , it follows from (1) and (2) that $G$ is supersmooth. One sees furthermore that $G/P$ is supersmooth whenever $P$ is a parabolic $k$ -subgroup of $G$ . For example, $\mathrm{GL}_n$ , flag varieties, and Grassmannians (in particular, projective spaces) are all supersmooth.

(4) Let $\ell/k$ be a finite separable field extension, and suppose that $X$ is a quasi-projective $\ell$ -scheme satisfying the condition for supersmoothness whenever $R$ is merely a semi-local noetherian local ring (e.g., $X = \bA^n$ ). Then the Weil restriction $\mathrm{R}_{\ell/k}(X)$ is supersmooth, as the reader familiar with Weil restrictions may easily check.

(5) If $X = \Spec \ell$ for some finite field extension $\ell/k$ , then $X$ is supersmooth if and only if $\ell = k$ . Since in particular $\ell$ is smooth over $k$ , the extension must be separable and thus $\ell = k(\alpha)$ for some $\alpha$ by the primitive element theorem. Let $f(t) \in k[t]$ denote the minimal polynomial of $\alpha$ over $k$ , so that $\ell \cong k[t]/(f(t))$ . Thus by supersmoothness there is a $k$ -homomorphism $\ell \to k[t]_{(f(t))}$ , so that $\ell = k$ because the fraction field of $k[t]_{(f(t))}$ is $k(t)$ , which does not contain any nontrivial finite extension of $k$ . This example shows already that supersmoothness is not a “geometric” condition, in the sense that it may not be checked after extending $k$ : note that if $\ell/k$ is separable then $\ell \otimes \overline{k} \cong \overline{k}^n$ for $n = [\ell: k]$ , and this is evidently supersmooth over $\overline{k}$ .

(6) If $X$ is a supersmooth integral $k$ -scheme of dimension $\geq 1$ , then $X$ admits a nonconstant rational map from $\bP^1$ . Indeed, since $X$ is smooth, $X(k_s) \neq \emptyset$ , so there is some closed point $x \in X$ whose residue field $\ell$ is finite separable over $k$ , say $\ell = k(\alpha)$ as in the previous example. Let $f(t) \in k[t]$ be the minimal polynomial of $\alpha$ over $k$ , and let $R = k[t]_{(f(t))}$ , $I = f(t)^2R$ . Since $\dim X \geq 1$ , there is a surjective homomorphism of $\ell$ -algebras $\cO_{X, x}/\fm_x^2 \to R/I$ , and by supersmoothness this extends to an $R$ -point of $X$ . Since the map on cotangent spaces is nontrivial, the generic point of $R$ maps to a nonclosed point of $X$ , and thus there is indeed a nonconstant rational map from $\bP^1$ . In particular, any integral $k$ -scheme of dimension $\geq 1$ admitting no nonconstant rational maps from $\bP^1$ is not supersmooth: for example, the only supersmooth integral curves are open in $\bP^1$ , and abelian varieties are never supersmooth.

Note that all of the examples we have given are rational. The last two examples in particular suggest that supersmoothness is a rather restrictive condition, and that it is somewhat similar to rationality. (We will see below that there exist non-rational examples.) Let $X$ be a locally finite type $k$ -scheme. We call $X$ nearly rational if the following condition holds: for every point $x \in X$ , there is some open subscheme $U$ of an affine space over $k$ and a smooth $k$ -morphism $f: U \to X$ such that $x$ lies in the image of $f$ and there exists a section $\Spec k(x) \to U$ .

Proposition. If $X$ is nearly rational, then it is supersmooth. If $k$ is of characteristic $0$ , then the converse is true.

Proof. First, suppose $X$ is nearly rational. Let $R$ be a noetherian local $k$ -algebra and let $I \subset R$ be an ideal such that $R/I$ is Artin. Suppose given a $k$ -morphism $\overline{f}: \Spec R/I \to X$ , and let $x \in X$ denote the unique point in the image of this morphism. Let $U$ be some smooth neighborhood of $x$ as in the statement of the Proposition, and let $u \in U$ be a $k(x)$ -rational point mapping to $x$ . If $\fm$ denotes the maximal ideal of $R$ , then this means that the morphism $\Spec R/\fm \to X$ lifts to morphism $\Spec R/\fm \to U$ with image $\{u\}$ . Since $U \to X$ is smooth, it follows from the infinitesimal criterion for smoothness that $\overline{f}$ lifts to a morphism $\Spec R/I \to U$ . By Examples (1) and (2) above, $U$ is supersmooth and thus there is an extension of this latter morphism to some map $\Spec R \to U$ . The composition of this map with $U \to X$ is then the desired extension of $\overline{f}$ .

For the other direction, we need two Lemmas.

Lemma 1. Let $K/k$ be a finitely generated extension of fields. There exists a positive integer $N$ and a point $y \in \bA^N$ such that $k(y) \cong K$ .

Proof of Lemma 1. Let $A$ be a finite type domain over $k$ with fraction field $K$ . There is some surjective map $f: k[t_1, \dots, t_N] \to A$ , and if $\fp$ is the kernel of $f$ then $k(\fp) = K$ .

Lemma 2. Suppose that $A$ is a smooth $k$ -algebra. If $\fp \subset A$ is a prime ideal such that $k(\fp)$ is separable over $k$ , then there is a natural split short exact sequence

$0 \to (\fp/\fp^2 \otimes_{A/\fp} k(\fp)) \to \Omega_{A/k} \otimes_A k(\fp) \to \Omega_{k(\fp)/k} \to 0.$

Proof of Lemma 2. By separability of $k(\fp)$ over $k$ , note that $A/\fp$ is generically smooth over $k$ , so by localizing (using the good behavior of $\Omega$ with respect to localization of $A$ ), Neron Models, 2.2, Prop. 7 shows that there is a natural split short exact sequence

$0 \to \fp/\fp^2 \to \Omega_{A/k} \otimes_A A/\fp \to \Omega_{(A/\fp)/k} \to 0.$

Localizing $A/\fp$ at the generic point then gives the desired result.

Now we finish the proof of the Proposition. Suppose that $X$ is supersmooth over the field $k$ of characteristic $0$ , and let $x \in X$ . Let $K = k(x)$ , so that there exist $N$ and $y$ as in Lemma 1. By increasing $N$ as necessary, we may assume that there is an injective homomorphism $\cO_{X, x}/\fm_x^2 \to \cO_{\bA^N, y}/\fm_y^2$ inducing an isomorphism on residue fields. By supersmoothness, the induced map $\Spec \cO_{\bA^N, y}/\fm_y^2 \to X$ extends to a map $\Spec \cO_{\bA^N, y} \to X$ . Since $X$ is locally finite type, this is the localization of some map $f: U \to X$ , where $U$ is an open subscheme of $\bA^N$ containing $y$ . Note that this map induces an isomorphism of residue fields $k(x) \cong k(y)$ and moreover the map $\fm_x/\fm_x^2 \to \fm_y/\fm_y^2$ is injective. Since $k$ is of characteristic $0$ (and thus all extension fields of $k$ are perfect), it follows from Lemma 2 that the map $f^*\Omega_{X/k} \otimes k(y) \to \Omega_{U/k} \otimes k(y)$ is injective. Since $U$ and $X$ are smooth over $k$ , the map $U \to X$ is therefore smooth at $y$ by Neron Models, 2.2, Prop. 8. By openness of the smooth locus we may shrink $U$ around $y$ to assume that in fact $U \to X$ is smooth. This $U$ satisfies all the conditions in the Proposition, and we have completed the proof.

Note that this implies that any supersmooth $k$ -scheme is unirational, and hence over an infinite field it has a dense set of rational points.

This Proposition can be used to give more examples of supersmooth varieties. For example, we will use it to show that any stably rational variety is generically supersmooth (i.e., that it admits a dense open subscheme which is supersmooth). Recall that an integral scheme $X$ locally of finite type over $k$ is called stably rational if $X \times_{\Spec k} \bA^m$ is rational for some $m \geq 0$ . The fact that non-rational stably rational varieties exist was first proven in Variétés stablement rationnelles non rationnelles by Beauville–Colliot-Thélène–Sansuc–Swinnerton-Dyer. A field-theoretic formulation of this condition is the following: if $K$ is the function field of $X$ , then there exists some $m \geq 0$ such that the purely transcendental extension $K(t_1, \dots, t_m)$ is $k$ -isomorphic to $k(u_1, \dots, u_n)$ for some $n \geq 0$ .

In general, suppose given an integral scheme $X$ of finite type over $k$ with function field $K$ and a finitely generated separable extension $L/K$ such that $L/k$ is purely transcendental. If $K/k$ is finite (i.e., $\dim X = 0$ ) then clearly $K = k$ and $X$ is supersmooth, so assume $\dim X \geq 1$ . We can spread out these extensions to obtain an open subscheme $U$ of some affine space over $k$ and a smooth $k$ -morphism $U \to X$ whose generic fiber has function field $L$ . If the generic fiber admits a $K$ -point, then we may spread this out to a section of $U \to X$ over some dense open $V$ of $X$ , and this $V$ is supersmooth by the Proposition. Since $\dim X \geq 1$ , $K$ is infinite and thus the generic fiber admits a rational point whenever $L/K$ is purely transcendental (since the $K$ -points are dense in $\bA_K^m$ ). Such $L$ exists when $X$ is stably rational, so indeed stably rational varieties are supersmooth.

I don’t know whether the second statement in the Proposition is true when the characteristic of $k$ is $p > 0$ . It would be interesting to know whether there are any counterexamples.

Thuses

A curiosity: “supersmooth” varieties