Number of points modulo q is a stable birational invariant

This post is about a (very weak) shadow in characteristic p of the Larsen–Lunts theorem. See my previous post for the statement and sketch of the proof of Larsen–Lunts.

Remark. In characteristic p, we do not even know the weakest form of resolution of singularities (e.g. find a smooth proper model for any function field). Thus, we certainly do not know the Larsen–Lunts theorem. However, we can still try to prove corollaries (and if they fail, we know that resolution must fail).

Today, I want to talk about the following statement:

Theorem. (Ekedahl) Let k = \mathbb F_q. Let X and Y be smooth proper varieties, and assume X and Y are stably birational. Then |X(k)| \equiv |Y(k)| \pmod{q}.

Remark. This would follow immediately from Larsen–Lunts if we knew a sufficiently strong form of resolution of singularities. Indeed, the map

    \[K_0(\operatorname{Var}_k) \to \Z/q\Z\]

given by counting \F_q-points modulo q factors through K_0(\operatorname{Var}_k)/(\mathbb L) since |\mathbb A^1(\F_q)| = q. Hence, by Larsen–Lunts, it factors through \mathbb Z[\operatorname{SB}].

It turns out that the theorem is true without assuming resolution of singularities, and the proof is due to Ekedahl (although in his paper he never explicitly states it in this form). The reader should definitely check out Ekedahl’s article (see references below), because his proof is more beautiful than the one I present here, and actually proves a bit more.

We will need one fairly deep theorem:

Theorem. Let X be a variety of dimension n over k = \F_q. Let \alpha be an eigenvalue of Frobenius on H^i_c(\bar X\et, \Q_\ell). Then \alpha and q^n\alpha^{-1} are both algebraic integers.

The first part (integrality of \alpha) is fairly well-known. For the second part (integrality of q^n\alpha^{-1}), see SGA 7_{\text{II}}, Exp. XXI, Corollary 5.5.3(iii).

The statement that appears in Ekedahl’s article is the following:

Theorem. (Ekedahl’s version) Let k = \mathbb F_q. Let X and Y be smooth connected varieties (not necessarily proper!), and assume X and Y are birational. If \alpha is an eigenvalue of Frobenius on H^i(\bar X, \Q_\ell) which is not an eigenvalue on H^i(\bar Y, \Q_\ell), then \alpha is divisible by q.

This statement should be taken to include multiplicities; e.g. a double eigenvalue for X which is a simple eigenvalue for Y is also divisible by q. By symmetry, we also get the opposite statement (with X and Y swapped). Thus, the eigenvalues (with multiplicities) that are not divisible by q are the same for X and Y.

Proof. We immediately reduce to the case where X \sbq Y is an open immersion, with complement Z. We have a long exact sequence for étale cohomology with compact support:

    \[\cdots \to H^{i-1}_c(Z) \to H^i_c(X) \to H^i_c(Y) \to H^i_c(Z) \to H^{i+1}(X) \to \cdots.\]

If \alpha is an eigenvalue on some H^i_c(Z), then q^{n-1}\alpha^{-1} is an algebraic integer (see above). Hence, for any valuation v on \bar \Q with v(q) = 1, we have v(\alpha) \leq n-1. We conclude that the eigenvalues for which some valuation is > n-1 on H^i_c(X) and H^i_c(Y) agree. Hence, by Poincaré duality, the eigenvalues of H^{2n-i}(X) and H^{2n-i}(Y) for which some valuation is < 1 agree. These are exactly the ones that are not divisible by q. \qedsymbol

The theorem I stated above immediately follows from this one:

Proof. Since |X\times\P^n(\F_q)| = (q^n + \ldots + 1) |X(\F_q)|, we may replace X by X \times \P^n. Thus, we can assume X and Y are birational; both of dimension n.

By the Weil conjectures, we know that

    \[|X(\F_q)| = \sum_{i=0}^{2n} \sum_\alpha \alpha^i,\]

where the inner sum runs over all eigenvalues of Frobenius. If we reduce mod q, then we only need to consider eigenvalues that are not divisible by q. By Ekedahl’s version of the theorem, the set (with multiplicities) of such \alpha are the same for X and Y. \qedsymbol

Historical remark. Although the theorem above was essentially proven in 1983 (but not explicitly stated), a separate proof for threefolds appeared in a paper by Gilles Lachaud and Marc Perret in 2000. It uses Abhyankar’s results on resolution of singularities, and is much closer to the proof of Larsen–Lunts than Ekedahl’s proof was. In 2002, Bruno Kahn provided a different proof for the general case using some (fairly advanced?) motive machinery (‘almost without cohomology’).


Torsten Ekedahl, Sur le groupe fondamental d’une variété unirationelle. Comptes rendus de l’académie des sciences de Paris, Serie I: mathématiques, 297(12), p. 627-629 (1983).

Bruno Kahn, Number of points of function fields over finite fields. arXiv:math/0210202

Gilles Lachaud and Marc Perret, Un invariant birationnel des variétés de dimension 3 sur un corps fini. Journal of Algebraic Geometry 9 (2000), p. 451-458.

The Larsen–Lunts theorem

The Larsen–Lunts theorem is one of the most beautiful theorems I know. But first, let me recall some definitions.

Definition. The Grothendieck ring of varieties K_0(\operatorname{Var}_k) over a field k is the free abelian generated by (formal) symbols [X] for X a variety over k (which I will take to mean a geometrically reduced, separated scheme of finite type over k), subject to the relations

    \[[X] = [U] + [Z]\]

whenever Z \to X is a closed immersion and U = X \setminus Z. It becomes a ring by setting [X] \cdot [Y] = [X \times Y] (exercise: show that this is well-defined). The class \mathbb L = [\A^1] is called the Lefschetz motif.

Remark. Recall that a rational map f \colon X \dashrightarrow Y is a morphism U \to Y defined on some dense open U \subseteq X. Varieties with rational morphisms form a category, and f is called a birational map if it is an isomorphism in this category. We say that X and Y are birational if there exists a birational map f \colon X \stackrel\sim\dashrightarrow Y. If X and Y are integral, this is equivalent to the equality of function fields K(X) = K(Y).

We say that X and Y are stably birational if X \times \P^n is birational to Y \times \P^m for some m, n \in \Z_{\geq 0}. This is equivalent to the existence of an isomorphism

    \[K(X)(x_1,\ldots,x_n) \cong K(Y)(y_1,\ldots,y_m).\]

There are examples of stably birational varieties that are not birational.

Definition. Write \operatorname{SB} for the set of stable birational classes of smooth proper varieties over k. To avoid confusion, I shall denote the class of X by (X). This set becomes a commutative monoid by setting (X) \cdot (Y) = (X \times Y) (again: show that this is well-defined).

Theorem. (Larsen–Lunts) Let k = \C. There exists a unique ring homomorphism

    \[\phi \colon K_0(\operatorname{Var}_\C) \to \Z[\operatorname{SB}]\]

such that for any smooth proper X, the image of [X] is (X). Moreover, the kernel of \phi is the ideal generated by \mathbb L.

Proof (sketch). The map \phi is constructed by induction on the dimension. For smooth proper X, it is clear what \phi(X) should be (namely (X)). If X is smooth, we can find a smooth compactification X \to \bar X (using resolution of singularities). Then we set \phi(X) := (\bar X) - \phi(\bar X \setminus X), where the right-hand side is defined by the induction hypothesis.

To check that it is independent of the compactification chosen, we need a strong form of weak factorisation: any two compactifications X \to \bar X_1, \bar X_2 differ by a series of blow-ups and blow-downs along smooth centres disjoint from X. Now if X \to Y is the blow-up along a smooth centre Z \sbq Y with exceptional divisor E \sbq X, then E is a \P^r-bundle over Z for some r; thus Z and E are stably birational. Now well-definedness of the map on lower-dimensional varieties proves independence on the smooth compactification.

Finally if X is singular, we simply set \phi(X) = \phi(X^{\operatorname{nonsing}}) + \phi(X^{\operatorname{sing}}). After some further checks (like additivity and multiplicativity), this finishes the construction of \phi.

Now clearly \mathbb L \in \ker \phi, since (\P^1) = (\P^0), and \mathbb L = [\P^1] - [\P^0]. Conversely, let \alpha \in \ker \phi. We can write any \alpha as

    \[\alpha = \sum_{i=1}^n [X_i] - \sum_{j=1}^m [Y_j]\]

for certain X_i, Y_j smooth proper (we again use resolution here). Since \Z[\operatorname{SB}] is the free algebra on \operatorname{SB}, we conclude that n = m and (X_i) = (Y_i) after renumbering. Thus it suffices to consider the case \alpha = [X] - [Y] for X and Y smooth proper and stably birational (to each other). We may replace X by X \times \P^n since their difference is [X] \cdot (\mathbb L + \ldots + \mathbb L^n), which is in the kernel. Thus, we may assume X and Y are birational.

Now by weak factorisation, we reduce to the case of a blow-up X \to Y in a smooth centre Z \sbq Y. Let E be the exceptional divisor, which is a \P^r-bundle over Z. Thus Z and E differ by a multiple of \mathbb L, since [\P^r] - [\P^0] = \mathbb L + \ldots + \mathbb L^r. \qedsymbol

Remark. The hard part of the theorem is the definition of the map. In order to define \phi(X) for X not necessarily smooth and proper, we need to assume resolution of singularities (for this, a very mild version of resolution suffices). To check that it is independent of choices, we need the weak factorisation theorem (which in turn uses a very strong version of resolution of singularities). The computation of the kernel again uses resolution of singularities and weak factorisation.

This is why we restrict ourselves to k = \C. I suspect that it is also fine for arbitrary algebraically closed fields of characteristic 0.

Corollary. Let X and Y be smooth proper. Then X and Y are stably birational if and only if [X] \equiv [Y] \pmod{\mathbb L}.

Proof. Since \Z[\operatorname{SB}] is the free algebra on \operatorname{SB}, we have (X) = (Y) if and only if X and Y are stably birational. The result is now immediate from the theorem. \qedsymbol

Remark. If we knew weak factorisation (without knowing resolution), then one implication would follow immediately: if X and Y are stably birational, then X \times \P^n \stackrel\sim\dashrightarrow Y \times \P^m for some m, n. Clearly [X \times \P^n] - [X] = (\mathbb L + \ldots + \mathbb L^n) \cdot [X] is divisible by \mathbb L, so we may assume X \stackrel\sim\dashrightarrow Y. Now by weak factorisation, a birational map factors as a chain of blow-ups and blow-downs along smooth centres, so we reduce to that case. But if X = \operatorname{Bl}_Z Y has exceptional divisor E, then E is a \mathbb P^r-bundle over Z for some r, hence [X] - [Y] = [E] - [Z] is divisible by \mathbb L.

However, for the other implication there is no direct proof even if we knew weak factorisation.

In my next post, I will address a statement in positive characteristic (where neither resolution of singularities nor weak factorisation are currently known) that is related to the corollary (but much weaker).

Properness and completeness of curves

In this post, I want to show an application of fpqc descent (specifically, pro-Zariski descent) to a classical lemma about properness. Recall (EGA II, Thm 7.3.8) the valuative criterion of properness:

Theorem. Let f \colon X \ra Y be a finite type morphism of locally Noetherian schemes. Then f is proper if and only if for every commutative diagram

Rendered by

where A is a discrete valuation ring with fraction field K, there exists a unique morphism \Spec A \ra X making commutative the diagram

Rendered by

Lemma. Suppose f \colon X \ra Y is a finite type morphism of locally Noetherian schemes. Then f is proper if and only if for every Dedekind scheme C \in \ob(\Sch/Y) and every closed point p \in C, every Y-morphism g \colon C\setminus\{p\} \ra X extends uniquely to C \ra X.

Proof. If A is a discrete valuation ring with fraction field K and maximal ideal \fr m, then \Spec A is a Dedekind scheme, and (\Spec A)\setminus\{\fr m\} = \Spec K. Thus, the condition of the lemma clearly implies properness, by the theorem above.

Conversely, suppose f is proper, and let C be a Dedekind scheme over Y, and p \in C a closed point. Write U = C\setminus\{p\}, and let V = \Spec \O_{C,p}. Let \eta be the generic point of C, and K = \O_{C,\eta}.

The valuative criterion shows that the the induced map g|_{\eta} \colon \Spec K \ra X extends uniquely to a Y-morphism \tilde{g} \colon V \ra X. Moreover, since U \sbq C is an open immersion, the fibre product U \times_C V is the open \Spec K \sbq V.

Now \{U, V\} is an fpqc cover of C (in fact, a pro-Zariski cover). The above shows that g and \tilde{g} have the same restriction to U \times_C V. Since representable presheaves are sheaves for the fpqc topology (Tag 03O3), we thus see that they glue to a unique map C \ra X. \qedsymbol

Remark. Of the course, the classical proof of the lemma goes by noting that the morphism V \ra X factors through some Zariski-open V' containing p, since X is of finite type over Y. The only thing that we changed is that we didn’t pass from the pro-Zariski to the Zariski covering, but instead argued directly using fpqc descent.

Relative Frobenius

This is the third in a three-part post about a proof that I contributed to the Stacks project. The result was already there, but I found a slightly easier proof. The proof is given in my previous post. In this post, I will present the application that caused me to look at the result in the first place.

Remark. Recall that a morphism f \colon X \ra Y of schemes is smooth of relative dimension r if all of the following hold:

  • f is locally of finite presentation;
  • f is flat;
  • all nonempty fibres have dimension r;
  • \Omega_{X/Y} is locally free of rank r.

If f is smooth of relative dimension 0, then f is étale. In this case, the third condition follows from the other ones.

Example. To show that the third condition is really necessary, consider any finite inseparable field extension. This is clearly flat of finite presentation. Moreover, \Omega_{L/K} is a vector space of dimension r > 0, with basis given by a p-basis of L/K. Yet the (unique) fibre has dimension 0.

Definition. Let S be a scheme of prime characteristic p > 0. Then the absolute Frobenius on S is given by the morphism \Frob_S \colon S \rA S which is the identity on the underlying topological space, and is given by x \rm x^p on \O_S. This definition makes sense because for a ring A of characteristic p, the Frobenius \Frob_A \colon A \ra A induces the identity on \Spec A.

Definition. Suppose that X \ra S is a morphism of schemes of characteristic p. Then the absolute Frobenius \Frob_X factors through \Frob_S, and therefore induces a morphism \Frob_{X/S} \colon X \ra X^{(p)} in the following diagram

Rendered by

where the square is a pullback (i.e. X^{(p)} = X \times_S S, where S is viewed as an S-scheme along \Frob_S). The morphism \Frob_{X/S} is called the relative Frobenius of X over S.

Lemma. Assume f \colon X \ra S is étale, with S a scheme of characteristic p. Then \Frob_{X/S} is an isomorphism. In other words, the square

Rendered by

is a pullback.

Proof. Note that \Frob_S is universally bijective, hence so is g \colon X^{(p)} \ra X. Similarly, \Frob_X is universally bijective. Therefore so is \Frob_{X/S}, since g \circ \Frob_{X/S} = \Frob_X.

On the other hand, f is étale, hence by base change so is X^{(p)} \ra S. But any map between schemes étale over S is étale (see Tag 02GW, or for a nice geometric proof taken from Milne’s book on étale cohomology, see Corollary 1.1.9 of my Master’s Thesis), so in particular \Frob_{X/S} is étale.

Now \Frob_{X/S} is étale and universally bijective, so the result follows from my previous post. \qedsymbol

Remark. Recall (see Tag 054L) that if f \colon X \ra S is smooth of relative dimension r, then around every x \in X there exist ‘smooth coordinates’ in the following sense: there exist affine opens U \sbq X, V \sbq Y with f(U) \sbq V, such that f|_U \colon U \ra V factors as

Rendered by

where \pi is étale. In particular, this forces \Omega_{U/V} = \bigoplus_{i=1}^r dx_i \O_U, by the first fundamental exact sequence.

Corollary. Assume f \colon X \ra S is smooth of relative dimension r, with S a scheme of characteristic p. Then \Frob_{X/S} is locally free of rank p^r.

Proof. The question is local on both X and S. By the remark above, we may assume X is étale over \A^r_S, with both X and S affine. We have a diagram

Rendered by

where the horizontal compositions are the absolute Frobenii on X and \A^r_S respectively. Here, \pi^{(p)} denotes the unique map making the top right square commutative. (Exercise: use the various universal properties to show that the top left square commutes).

The bottom right square and the right large rectangle are pullback squares, hence so is the top right square. The top large rectangle is a pullback by the lemma above. Hence, since the top right square is a pullback, so is the top left square. Hence, it suffices to prove the case X = \A^r_S, since the result is stable under base change.

But in this case, if S = \Spec A, then X = \Spec A[x_1, \ldots, x_r], and X^{(p)} = \Spec A[y_1,\ldots,y_r], with the relative Frobenius given by the A-linear (!) map

    \begin{align*} A[y_1,\ldots,y_r] &\rA A[x_1,\ldots,x_r]\\ y_i &\rM x_i^p. \end{align*}

But in this case the result is clear: an explicit basis is

    \[ \{x_i^j\ |\ i\in\{1,\ldots,r\}, j\in\{0,\ldots,p-1\}\}. \]


Étale and universally injective

This is the second in a three-part post about a proof that I contributed to the Stacks project. The result was already there, but I found a slightly easier proof. This post contains the actual result, and the next post will contain a fun application.

Remark. Recall that a morphism f \colon X \ra Y of schemes is étale if f is flat and locally of finite presentation, and \Omega_{X/Y} = 0.

Lemma. Suppose f \colon X \ra S is étale and universally injective. Then f is an open immersion.

Proof. Since étale morphisms are open and surjectivity is stable under base change, we may assume that f is universally bijective; then we have to prove that f is an isomorphism. Since étale morphisms are open, f is in fact a universal homeomorphism. By Tag 04DE, f is affine.

The question is local on S, so we may assume S is affine, and hence so is X. Say f is induced by g \colon A \ra B. Now f is proper and affine, hence finite. Moreover, since B is finitely presented and finite as A-algebra, and B is a finitely presented B-module, it is also a finitely presented A-module (Tag 0564).

Now B is flat of finite presentation over A, hence locally free (actually, we need the slightly stronger result that I mention in the first remark; see Tag 00NX for statement and proof). Since the question is local, we may assume B is free of rank r.

Now let \bar{s} \ra S be a geometric point; that is, let A \ra \bar{K} be a map to an algebraically closed field. Then the tensor product B \tens_A \bar{K} is étale of dimension r over \bar{K}. Hence, X_{\bar s} = X \times_{S} \bar{s} is a union of r points. Since f is universally bijective, we have r = 1. Then the result follows from my previous post. \qedsymbol

Locally free algebras

This is the first in a three-part post about a proof that I contributed to the Stacks project. The result was already there, but I found a slightly easier proof. This post contains a preliminary lemma; the second post contains the result; and the third one contains the application that I was interested in.

Remark. If M is a (locally) free A-module of rank 1, then \End_A(M) = A. Multiplication by a \in A is injective on M if and only if a is not a zero-divisor, and it is surjective if and only if a \in A\x. In particular, if it is surjective, it is also injective.

Lemma. Suppose f \colon A \ra B is a ring homomorphism, such that B is locally free of rank 1 over A. Then f is an isomorphism.

Proof. The question is local on A, so (after replacing A with a suitable localisation) we may assume that B is free of rank 1. Let b be a basis element.

Then we can write 1 = f(a)b for some a \in A, hence b \in B\x. Also, we can write b^2 = f(c)b for some c \in A, hence b = f(c). Therefore, f is surjective, so by the remark above, it is an isomorphism. \qedsymbol

Using a different argument, we can also prove:

Lemma. Suppose f \colon A \ra B is a ring homomorphism, such that B is locally free of rank r > 0 over A. Then f is injective.

Proof. Since B is locally free of rank r, it is faithfully flat over A. Thus it suffices to prove that f \tens 1 \colon B \ra B \tens_A B is injective. But this map admits a contraction B \tens_A B \ra B given by b_1 \tens b_2 \rm b_1b_2. \qedsymbol

Proper + Affine = Finite

Another argument I really like. We first present two auxiliary lemmata.

Lemma 1. Let f \colon A \ra B be an injective ring homomorphism such that the associated map \phi \colon \Spec B \ra \Spec A is closed. Then f^{-1}(B\x) = A\x.

Proof. Since f is injective, the map \phi is dominant. Hence, it is a closed surjective map. For such a map, a closed set Z \sbq \Spec A is empty if and only if \phi^{-1}(Z) is empty. Applying this to Z = V(a) for a \in A, we get V(a) = \varnothing if and only if V(f(a)) = \varnothing, i.e. a \in A\x if and only if f(a) \in B\x. \qedsymbol

Lemma 2. Let f \colon A \ra B be an injective ring homomorphism such that the associated map \phi \colon \A^1_B \ra \A^1_A (induced by A[x] \ra B[x]) is closed. Then f is integral.

Proof. Let b \in B. Consider the ideal I = (xb - 1) \sbq B[x], and let J = I \cap A[x]. Note that B[x]/I = B[\tfrac{1}{b}], and A[x]/J is the image C of the composite map A[x] \ra B[x] \ra B[\tfrac{1}{b}] (first isomorphism theorem). Write g for the inclusion C \ra B[\tfrac{1}{b}]. Note that the map

    \[ \psi \colon \Spec B\left[\tfrac{1}{b}\right] \ra \Spec C \]

induced by g is just the restriction of \phi to the closed subsets V(I) \ra V(J), hence it is a closed map. Since g is injective, Lemma 1 asserts that

    \[ g^{-1}\left(B\left[\tfrac{1}{b}\right]\x\right) = C\x. \]

But \tfrac{1}{b} is invertible in B[\tfrac{1}{b}] (its inverse is b), and it lies in C since it is the image of x under A[x] \ra B[\tfrac{1}{b}]. Hence, it is invertible in C, i.e. b \in C. That is, we can write

    \[ b = a_0 + a_1 \left(\frac{1}{b}\right) + \ldots + a_n \left(\frac{1}{b}\right)^n \]

for certain a_0, \ldots, a_n \in A. Hence, b^{n+1} - a_0 b^n - \ldots - a_n = 0 in B\left[\tfrac{1}{b}\right]. Hence, some multiple b^m \left(b^{n+1} - a_0 b^n - \ldots - a_n\right) is 0 in B, proving that b is integral over A. \qedsymbol

Corollary. Let f \colon A \ra B be a ring homomorphism such that the associated map \Spec B \ra \Spec A is proper. Then f is finite.

Proof. Let I be the kernel of f; then \Spec B \ra \Spec A lands in \Spec A/I. Then \Spec B \ra \Spec A/I is proper (Hartshorne, Corollary II.4.8(e)). Moreover, the map

    \[ \bar{f} \colon A/I \ra B \]

is injective, hence integral by Lemma 2. Since \Spec B \ra \Spec A/I is proper, it is of finite type. Hence, \bar{f} is integral and of finite type, hence finite. Hence so is f. \qedsymbol

A more geometric version is the following:

Theorem. Let \phi \colon X \ra Y be a morphism of schemes that is both affine and proper. Then \phi is finite.

Proof. Let U = \Spec A be an affine open in Y. Then V = \phi^{-1}(U) is affine (since \phi is an affine morphism); say V = \Spec B. Then the restriction \phi|_V \colon V \ra U is proper (properness is local on the target), hence B is finite over A by the theorem above. This proves that \phi is finite. \qedsymbol

Remark. The converse is also true, and in fact much easier: a finite morphism is affine by definition, and proper since in the affine case it is in fact projective. The question whether finite morphisms are always projective turns out to be somewhat subtle: it is true in the EGA sense of projective (it is given by the relative Proj of some sheaf of graded algebras, i.e. it embeds into some projective bundle), but not in the sense of Hartshorne (it embeds into \P^n_Y for some n, i.e. it embeds into a trivial projective bundle).

I believe that the following is an example of the last statement (i.e. a finite morphism which is not H-projective). Let K be a field of infinite degree of imperfection, i.e. \char K = p and K is an infinite extension of K^p. Then Theorem 2 in [B-McL] says that for each n \in \N there exists a finite field extension L_n of K which cannot be generated by fewer than n elements¹. Correspondingly, \Spec L_n \ra \Spec K cannot be embedded into \P^m_K for m < n. Then consider Y an infinite disjoint union of \Spec K (labelled by \N), and X the disjoint union of \Spec L_n for all n. Then Y is certainly projective in EGA’s sense (although any projective bundle it embeds into cannot have constant dimension). Yet it cannot be embedded into any \P^n_X.

[B-McL] M.F. Becker, S. MacLane, The minimum number of generators for inseparable algebraic extensions. Bull. Amer. Math. Soc. Volume 46-2 (1940), p. 182-186.

¹It is probably not very hard to actually come up with an example of such a field K with extensions L_n. I think that K = \F_p(x_1, x_2, \ldots) and L_n = K(\sqrt[\uproot{3}p]{x_1}, \ldots, \sqrt[\uproot{3}p]{x_n}) should do the trick.