The Hermite-Joubert problem over p-fields

Matthew Brassil and Zinovy Reichstein

Abstract. Motivated by the classical theorems of Ch. Hermite and P. Joubert, we give a necessary and sufficient condition for an integer n, a field F_{0} and a prime p to have the following property:

Every étale algebra E/F of degree n, where F is a p-field containing F_{0}, has an element 0 ≠ a ∈ E such that F [a] = E and tr(a) = tr(a^{p}) = 0.

1. Introduction

An 1861 theorem of Ch. Hermite [He] asserts that for every étale algebra E/F of degree 5 there exists an element 0 ≠ a ∈ E whose characteristic polynomial is of the form

f(x) = x^{5} + b_{2}x^{3} + b_{4}x + b_{5} .

An easy application of Newton’s formulas shows that this is equivalent to tr_{E/F} (a) = tr_{E/F} (a^{3}) = 0; see, e.g., [Co_{2}, section 1]. A similar result for étale algebras of degree 6 was proved by P. Joubert in 1867; see [Jo]. For modern proofs of these results, see [Co_{2}, Kr]. (Here we are assuming that F is an infinite field of characteristic ≠ 2 or 3. As usual, by an étale algebra E/F of degree n we mean a direct product E := E_{1} × … × E_{r}, where each E_{i} is a separable field extension of F and [E_{1} : F] + … + [E_{r} : F] = n.)

It is natural to ask if the above-mentioned theorems of Hermite and Joubert can be extended to n ⩾ 7; cf., e.g. [Co_{2}, Section 4]. The answer is “no” if n is of the form 3^{k} or 3^{k_{1}} + 3^{k_{2}}, where k_{1} > k_{2} ⩾ 0; see [Re_{1}, Theorem 1.3] or [RY_{2}, Corollary 1.7(a) and Theorem 1.8]. For other values of n (in particular, for n = 7), this question remains open. One can also ask a similar (even more difficult) question for an arbitrary prime p.

Hermite-Joubert Problem 1.1. Let n ⩾ 2 be an integer, p be a prime, and F_{0} be a base field. Which triples (F_{0}, p, n) have the property that every étale algebra E/F of degree n, with F_{0} ⊂ F, has an element 0 ≠ a ∈ E such that tr(a) = tr(a^{p}) = 0?

We will usually want to choose the element a ∈ E above so that F [a] = E, i.e., a generates E as an F -algebra. We will also consider a variant of this problem, where a is only required to satisfy tr(a^{p}) = 0, rather than tr(a) = tr(a^{p}) = 0.

2010 Mathematics Subject Classification. 12F10, 14J70, 14G05.

Key words and phrases. Hermite-Joubert problem, étale algebra, hypersurface, rational point, p-field.

The second author has been partially supported by a Discovery Grant from the National Science and Engineering Board of Canada.

In this paper we will show that these questions become tractable if we restrict our attention to the case, where F is a p-field. Recall that a field F is called a p-field if the degree of every finite field extension of F is a power of p; see Section 3. Equivalently, for an arbitrary field F and an étale algebra E/F of degree n, we are asking if there is a finite a field extension F′/F of degree prime to p and an element 0 ≠ a ∈ E′ := E ⊗_{F} F′ such that tr_{E′/F′}(a) = tr_{E′/F′}(a^{p}) = 0; see Lemma 3.2.

Before stating our main results, we recall the definition of the “general field extension” E_{n}/F_{n} of degree n. Let F_{0} be a base field and x_{1}, …, x_{n} be independent variables. Set L_{n} := F_{0}(x_{1}, …, x_{n}), F_{n} := L_{n}^{S_{n}} and E_{n} := L_{n}^{S_{n−1}} = F_{n}(x_{1}), where S_{n} acts on L_{n} by permuting x_{1}, …, x_{n} and S_{n−1} by permuting x_{2}, …, x_{n}.

Theorem 1.2. Let p be a prime, F_{0} be a field of characteristic ≠ p containing a primitive pth root of unity ζ_{p}, n ⩾ 3 be an integer, and n = p^{k_{1}} + … + p^{k_{m}} be the base p expansion of n. Here, as usual, each power of p appears in the sum at most p − 1 times. Then the following conditions are equivalent.

(1) For every p-field F containing F_{0} and every n-dimensional étale algebra E/F, there exists an element 0 ≠ a ∈ E such that tr_{E/F} (a^{p}) = 0.

(2) There exists a finite field extension F′/F_{n} of degree prime to p and an element 0 ≠ a ∈ E′ := E_{n} ⊗_{F} F′ such that tr_{E′/F′}(a^{p}) = 0. Here E_{n}/F_{n} is the general field extension of degree n defined above.

(3) The equation

(1.1)  p^{k_{1}}y_{1}^{p} + … + p^{k_{m}}y_{m}^{p} = 0

has a solution y = (y_{1} : …: y_{m}) ∈ ℙ^{n−1}(F_{0}). (∗) Moreover, if (3) holds, then the element a in parts (1) and (2) can be chosen so that E = F [a] and E′ = F′[a], respectively.

The implication (1) ⟹ (2) readily follows from the definition of a p-field. The proof of the implication (2) ⟹ (3), based on the fixed point method, is implicit in [RY_{2}, Sections 6] (where m is assumed to be 2). We will present a self-contained argument in Section 4. Our proof of the implication (3) ⟹ (1) in Section 5 is based on ideas from [DR, Section 8].

Theorem 1.3. Let F_{0}, p and n = p^{k_{1}} +…+p^{k_{m}} be as in the statement of Theorem 1.2. Then the following conditions are equivalent.

(1) For every p-field F containing F_{0} and every n-dimensional étale algebra E/F, there exists an element 0 ≠ a ∈ E such that tr_{E/F} (a) = tr_{E/F} (a^{p}) = 0.

(2) There exists a finite field extension F′/F_{n} of degree prime to p and an element 0 ≠ a ∈ E′ := E_{n} ⊗_{F} F′ such that tr_{E′/F′}(a) = tr_{E′/F′}(a^{p}) = 0.

(3) The system of equations

(1.2)  \{ \begin{matrix}p^{k_{1}}y_{1} + … + p^{k_{m}}y_{m} = 0 \\ p^{k_{1}}y_{1}^{p} + … + p^{k_{m}}y_{m}^{p} = 0 \end{matrix}

has a solution y = (y_{1} : …: y_{m}) ∈ ℙ^{m−1}(F_{0}).

(∗∗) Moreover, assume (3) holds, (char(F_{0}), p, n) ≠ (2, 3, 3), (2, 3, 4), or (3, 2, 3), and one of the following conditions is met:

(i) y ≠ (1 : …: 1),

(ii) char(F_{0}) does not divide n, or

(iii) p > 2 and char(F_{0}) ≠ 2. Then the element a in parts (1) and (2) can be chosen so that E = F [a] and E′ = F′[a], respectively.

We will prove Theorems 1.3 in Section 6 by modifying our proof of Theorem 1.2. For p = 2 Springer’s theorem about rational points of quadric hypersurfaces allows

us to remove the requirement that F is a 2-field in part (1) of Theorems 1.2 and 1.3. This leads to a solution of the Hermite-Joubert Problem 1.1 for p = 2; see Corollary 8.1. The same arguments go through for p = 3, if we assume a long-standing conjecture of J. W. S. Cassels and P. Swinnerton-Dyer about rational points on cubic hypersurfaces; see Section 9.

It is natural to ask for which n, p and F_{0} there exist non-trivial solutions to equation (1.1) and the system (1.2). Some partial answers to this question are given in Section 10. In particular, we show that for p ⩾ 3 the system (1.2) has a solution over any field F_{0}, if in the base p presentation [n_{d}n_{d−1} … n_{0}]_{p} of n one of the digits n_{k} is ⩾ 2, or if the number of non-zero digits is ⩾ p + 3; see Lemma 10.3. (Here n_{k} is the number of times p^{k} occurs in the presentation n = p^{k_{1}} + … + p^{k_{m}} in Theorems 1.2 and 1.3. If each n_{k} is 0 or 1, then the number of non-zero digits is m.) This implies, in particular, that for “most” n the system (1.2) has a non-trivial solution over any base field F_{0}. That is, if we fix p ⩾ 3 and let S_{N} be the set of integers n ∈ {1, …, N} such that the system (1.2) has a non-trivial solution over every base field F_{0}, then |S_{N}|/N will rapidly converge to 1, as N → ∞.

On the other hand, it is easy to see that the system (1.2) has no non-trivial solutions if n = p^{k} for any k ⩾ 1 or n = p^{k_{1}} + p^{k_{2}}, where k_{1} > k_{2} ⩾ 0 and char(F_{0}) does not divide p^{(k_{1}−k_{2})(p−1)} + (−1)^{p}. This way we recover most of [Re_{1}, Theorem 1.3]. In Section 11 we will extend this result as follows (for p = 3 only).

Theorem 1.4. Let E_{n}/F_{n} be the general field extension of degree n, over the base field F_{0} = ℚ. Suppose n = 3^{k_{1}} + 3^{k_{2}} + 3^{k_{3}}, where k_{1}, k_{2}, k_{3} ⩾ 0 are distinct integers such that k_{1} + k_{2} + k_{3} ≡ 0 or 1 (mod 3). Then for any finite field extension F′/F_{n} of degree prime to 3 there does not exist an element 0 ≠ a ∈ E′_{n} := E_{n} ⊗_{F} F′ such that tr_{E′/F′}(a) = tr_{E′/F′}(a^{3}) = 0.

This yields new examples, where the Hermite-Joubert Problem 1.1 has a negative answer in the classical setting (i.e., for p = 3 and F_{0} = ℚ). The smallest of these are n = 13 = 3^{2} +3^{1} +3^{0} and n = 31 = 3^{3} +3^{1} +3^{0}. We conjecture that Theorem 1.4 remains valid for all triples k_{1}, k_{2}, k_{3} of distinct non-negative integers; see Conjecture 12.1. Some evidence in support of this conjecture is presented in Section 12. In particular, we show that the Hermite-Joubert Problem 1.1 (again, for p = 3 and F_{0} = ℚ) has a negative answer in the case, where n = 37 = 3^{3} + 3^{2} + 3^{0}, which is not covered by Theorem 1.4.

We conclude this section with two remarks, which aim to place Theorems 1.2 and 1.3 into a broader context.

Remark 1.5. Our approach to the Hermite-Joubert Problem 1.1 is to subdivide it into two parts. First we restrict our attention to p-fields F. In the language of [Re_{2}, Section 5], this is a Type 1 problem. The present paper is devoted to solving this Type 1 problem. In those cases, where this Type 1 problem has a negative solution, so does the Hermite-Joubert Problem 1.1 (e.g., as in Theorem 1.4).

The remaining question is as follows. If condition (1) of Theorem 1.3 is satisfied for some n, p, and F_{0}, does it continue to hold if we allow F to range over all fields containing F_{0}, not just p-fields? This is a Type 2 problem, and it remains open, except in a few special cases (the case considered in Section 8, where p = 2, or the cases studied by Hermite and Joubert, where p = 3 and n = 5 or 6).

Many questions concerning algebraic objects over fields F, can be subdivided into two parts in a similar manner: a Type 1 problem, where F is assumed to be a p-field for some prime p, and a Type 2 problem (the rest, in those cases, where the Type 1 problem has a positive solution). Existing techniques are often effective in addressing Type 1 problems but Type 2 problems tend to be out of reach, except in a few special cases. For a discussion of this phenomenon and numerous examples, see [Re_{2}, Section 5].

Remark 1.6. Consider the hypersurfaces

X_{n,p} := {(x_{1} : …: x_{n}) | x_{1}^{p} + … + x_{n}^{p} = 0} ⊂ ℙ^{n−1}

and

Y_{n,p} := {(x_{1} : …: x_{n}) | x_{1} + … + x_{n} = x_{1}^{p} + … + x_{n}^{p} = 0} ⊂ ℙ^{n−2}

defined over the base field F_{0}. Here ℙ^{n−2} denotes the hyperplane x_{1} + … + x_{n} = 0 in ℙ^{n−1}. The symmetric group S_{n} acts on both X_{n,p} and Y_{n,p} by permuting x_{1}, …, x_{n}. Let us assume that Y_{n,p} is not a cone. This is a very mild assumption on n, p, and F_{0}; see Lemma 2.1(h) below.

By [DR, Theorem 1.1] the Hermite-Joubert Problem 1.1 is equivalent to the following question: Is the S_{n}-action on Y_{n,p} weakly versal? (For the definition of various types of versality for group actions on algebraic varieties, see [DR, Introduction].) Theorems 1.2 and 1.3 tell us when the S_{n}-action on X_{n,p} and Y_{n,p} is p-versal; see [DR, Section 8]. In particular, the S_{n}-action on Y_{n,p} is p-versal if and only if the system (1.2) has a non-trivial solution. If the S_{n}-action on Y_{n,p} is p-versal (which, as we saw above, happens for “most” values of n), we do not know whether or not it is weakly versal, except in a small number of special cases. This is the Type 2 problem we mentioned in Remark 1.5.

2. Geometry of the hypersurfaces X_{n,p} and Y_{n,p}

In this section we will prove some simple geometric properties of the hypersurfaces X_{n,p} ⊂ ℙ^{n−1} and Y_{n,p} ⊂ ℙ^{n−2} defined in Remark 1.6. We will continue to denote the base field by F_{0}.

Recall that a closed subvariety V of projective space is called a cone over a point c ∈ V if V contains the line through c and c′ for every c ≠ c′ ∈ V. We will say that V is a cone if it is a cone over one of its points.

Let Δ_{n} be the union of the “diagonal” hyperplanes x_{i} = x_{j}, over all 1 ⩽ i < j ⩽ n.

Lemma 2.1. Assume char(F_{0}) ≠ p. Then (a) X_{n,p} is smooth.

(b) The singular locus of Y_{n,p} is Y_{n,p} ∩ {(x_{1} : …: x_{n}) |x_{1}^{p−1} = … = x_{n}^{p−1} = 1}.

(c) X_{n,p} is absolutely irreducible if n ⩾ 3.

(d) Y_{n,p} is absolutely irreducible if n ⩾ 5.

(e) X_{n,p} is not contained in Δ_{n} for any n ⩾ 3.

(f) Y_{n,p} is not contained in Δ_{n}, if n ⩾ 3 and (char(F_{0}), p, n) ≠ (2, 3, 3), (2, 3, 4) or (3, 2, 3).

(g) Let (1 : …: 1) ≠ c ∈ Y_{n,p}. Then Y_{n,p} is not a cone over c.

(h) Y_{n,p} is not a cone if one of the following conditions holds: (i) char(F_{0}) does not divide n or (ii) p > 2 and char(F_{0}) ≠ 2.

Proof. In order to prove the lemma we may, without loss of generality, pass to the algebraic closure of F_{0}, i.e., assume that F_{0} is algebraically closed.

(a) and (b) readily follow from the Jacobian criterion.

(c) Assume the contrary. Then X_{n,p} has at least two irreducible components, X_{1} and X_{2}. Since X_{n,p} is a hypersurface in ℙ^{n−1}, dim(X_{1}) = dim(X_{2}) = n−2, and dim(X_{1}∩X_{2}) = n − 3. Since we are assuming that n ⩾ 3, this implies that X_{1} ∩ X_{2} ≠ ∅. On the other hand, every point of X_{1} ∩ X_{2} is singular in X, contradicting (a).

(d) Assume the contrary: Y_{n,p} has at least two irreducible components, Y_{1} and Y_{2}. Arguing as in (c), we see that Y_{1} ∩ Y_{2} is a closed subvariety of the singular locus of Y, and dim(Y_{1} ∩ Y_{2}) = n − 4. On the other hand, by part (b), the singular locus of Y is 0-dimensional. Thus n − 4 ⩽ 0, as desired.

(e) Assume the contrary. Then by part (c), X_{n,p} is contained in one of the hyperplanes H_{ij} ⊂ ℙ^{n−1} given by x_{i} = x_{j}. Since X_{n,p} is invariant under the action of S_{n}, it is contained in every hyperplane of this form. That is,

X_{n,p} ⊂ ⋂_{1⩽i<j⩽n} H_{ij} = {(1 : …: 1)} ,

which is impossible, since dim(X_{n,p}) = n − 2 ⩾ 1.

(f) First assume n ⩾ 5. Here Y_{n,p} is irreducible by part (d). Assume the contrary: Y_{n,p} is contained in Δ_{n}. Then Y_{n,p} is contained in one of the linear subspaces H′_{ij} given by intersecting the hyperplanes x_{i} = x_{j} and x_{1} + ⋯ + x_{n} = 0. Arguing as in part (e), we see that

Y_{n,p} ⊂ ⋂_{1⩽i<j⩽n} H′_{ij} = ∅ ,

a contradiction.

In the remaining cases, where n = 3 or 4, but (char(F_{0}), p, n) ≠ (2, 3, 3), (2, 3, 4) or (3, 2, 3), we will exhibit a point y ∈ Y_{n,p} which does not lie in Δ_{n}.

If n = 4 and char(F_{0}) ≠ 2, we can take y := (1 : ζ_{4} : ζ_{4}^{2} : ζ_{4}^{3}). Here ζ_{4} ∈ F_{0} is a primitive 4th root of unity. (Recall that we are assuming F_{0} to be algebraically closed.) If char(F_{0}) = 2 but p ≠ 3, set y := (1 : ζ_{3} : ζ_{3}^{2} : 0).

Now suppose n = 3. If char(F_{0}) ≠ 2 and p ≠ 2, then we can take y := (1 : −1 : 0). If char(F_{0}) ≠ 3 and p ≠ 3, set y := (1 : ζ_{3} : ζ_{3}^{2}). This covers all pairs (char(F_{0}), p), except for (2, 3) and (3, 2). (Recall that we are assuming that char(F_{0}) ≠ p throughout.)

(g) Assume the contrary. Since S_{n} acts on Y_{n,p} by permuting coordinates, Y_{n,p} is a cone over g · c for every g ∈ S_{n}. Now it is easy to see that Y_{n,p} contains the linear span of {g · c | g ∈ S_{n}}. Denote this linear span by L. Then L is an S_{n}-invariant linear subspace of ℙ^{n−2}. If char(F_{0}) does not divide n, then the S_{n}-representation on F_{0}^{n−1} is irreducible. Hence, the only S_{n}-invariant subspace of ℙ^{n−2} is ℙ^{n−2} itself. Thus ℙ^{n−2} = L ⊂ Y_{n,p}, a contradiction. If char(F_{0}) divides n, then the only other possibility is L = {(1 : 1 …: 1)}. This is ruled out by our assumption that c ≠ (1 : …: 1).

(h) Assume Y_{n,p} is a cone over some point c ∈ Y_{n,p}. By part (d), c = (1 : …: 1). If char(F_{0}) does not divide n, then (1 : …: 1) ∉ Y_{n,p}, a contradiction. This completes the proof in case (i).

To prove (ii), assume that p > 2. We want to show that if char(F_{0}) ≠ 2 then Y_{n,p} is not a cone over c = (1 : …: 1). Assume the contrary: whenever Y_{n,p} contains a point y = (y_{1} : …: y_{n}), it contains the entire line through c and y. That is,

(1 + ty_{1})^{p} + … + (1 + ty_{n})^{p} = 0

as a polynomial in t. In particular, p(y_{1}^{p−1} + … + y_{n}^{p−1}), which is the coefficient of t^{p−1} in this polynomial, should be equal to 0, for every y = (y_{1} : …: y_{n}) ∈ Y_{n,p}. Taking y = (−1 : 1 : 0 : …: 0), we obtain 2p = 0, contradicting our assumptions that char(F_{0}) ≠ 2 or p. □

3. Preliminaries on p-fields

A field F is called a p-field if the degree of every finite field extension of L is a power of p; see [Pf, Definition 4.1.11].

Lemma 3.1. Let F be a p-field of characteristic ≠ p.

(a) (J.-L. Colliot-Thélène) Suppose a smooth irreducible algebraic variety X has an F -point c ∈ X(F). Then F -points are dense in X.

(b) Suppose Y ⊂ ℙ^{l} is a projective variety of degree ≤ p defined over a p-field F and c ∈ Y (F) is an F -point of Y. Assume Y is not a cone over c. Then F -points are dense in Y.

Note that in part (b) we do not assume that Y is irreducible, either over F or over the algebraic closure \overline{F}.

Proof. For a proof of part (a), see [CT, p. 360].

(b) Case 1: Y is a hypersurface of degree d < p. Note that effective zero cycles of degree d are dense in Y (these can be obtained by intersecting Y with lines defined over F in ℙ^{l}). Since F is a p-field, every effective zero cycle of degree d < p splits over F (i.e., is a sum of d F -points). Consequently, F -points are dense in Y.

Case 2: Y is reducible over F, i.e. its irreducible components, Y_{1}, …, Y_{r} are defined over F and r ⩾ 2. Here each Y_{i} is a hypersurface of degree < p. By Case 1, F -points are dense in each Y_{i}; hence, they are dense in Y.

Case 3: Y is irreducible over F but reducible over \overline{F}. Note that since F is a p-field, and char(F) ≠ p, F is perfect. Hence, the irreducible components Y_{1}, …, Y_{r} of Y are transitively permuted by the Galois group Gal(\overline{F} /F), which is a pro-p group. Thus r ⩾ 2 is a power of p. Moreover, since deg(Y) = deg(Y_{1}) + … + deg(Y_{r}) ⩽ p, we conclude that r = p and deg(Y_{1}) = … = deg(Y_{p}) = 1. In other words, Y is a union of hyperplanes Y_{1}, …, Y_{p}. Now observe that c ∈ Y (F) is fixed by Gal(\overline{F} /F). After relabeling the components, we may assume that c ∈ Y_{1}. Translating Y_{1} by Gal(\overline{F} /F), we see that c lies on every translate of Y_{1}, i.e., on every Y_{i} for i = 1, …, r. Since each Y_{i} is a hyperplane, we conclude that Y is a cone over c, contradicting our assumption.

Case 4: Y is absolutely irreducible. Choose a hyperplane H ≃ ℙ^{l−1} in ℙ^{l} such that H is defined over F and c ∉ H. Let π: Y −{c} → H be projection from c. Since Y is not a cone over c, this map is dominant. In particular, there is a dense open subset U ⊂ H such that π is finite over U. The preimage π^{−1}(u) of any F -point u ∈ U(F) is then an effective 0-cycle of degree ≤ p − 1. Once again, every such 0-cycle splits over F, i.e., π^{−1}(u) is a union of F -points. Taking the union of π^{−1}(u), as u varies over U(F), we obtain a dense set of F -points in Y. □

Now recall that for every field F, there exists an algebraic extension F ⊂ F^{(p)}, such that F^{(p)} is p-field and, for every finite subextension F ⊂ F′ ⊂ F^{(p)}, the degree [F′ : F] is prime to p. The field F^{(p)} satisfying these conditions is unique up to F -isomorphism. We will refer to it as a p-closure of F. For details, see [EKM, Proposition 101.16].

Lemma 3.2. Let E/F be an étale algebra of degree n. Then

(a) every element a ∈ E ⊗_{F} F^{(p)} lies in the image of the natural map

ϕ: E ⊗_{F} F′ ↪ E ⊗_{F} F^{(p)}

for some intermediate field F ⊂ F′ ⊂ F^{(p)} (depending on a), such that [F′ : F] is finite (and thus automatically prime to p).

(b) x ∈ E′ := E ⊗_{F} F′ generates E′ over F′ (i.e., E′ := F′[x]) if and only if ϕ(x) generates E ⊗_{F} F^{(p)} over F^{(p)}.

Proof. (a) Let b_{1}, …, b_{n} be a basis of E, viewed as an F -vector space. Then

a = f_{1}(b_{1} ⊗ 1) + … + f_{n}(b_{n} ⊗ 1)

in E ⊗_{F} F′ for some f_{1}, …, f_{n} ∈ F^{(p)}, and we can take F′ = F (f_{1}, …, f_{n}).

(b) Working in the basis 1 ⊗ b_{1}, …, 1 ⊗ b_{n}, one readily checks that 1, x, …, x^{n−1} are linearly dependent over F′ if and only if 1, ϕ(x), …, ϕ(x)^{n−1} are linearly dependent over F^{(p)}. □

4. Proof of Theorem 1.2: (1) ⟹ (2) ⟹ (3)

(1) ⟹ (2): Applying (1) to the étale algebra E_{n} ⊗_{F_{n}} F_{n}^{(p)}/F_{n}^{(p)} we see that there exists an element a ∈ E_{n} ⊗_{F_{n}} F_{n}^{(p)} such that tr(a^{p}) = 0. By Lemma 3.2(a), this element descends to E_{n} ⊗_{F_{n}} F′ for some intermediate field F ⊂ F′ ⊂ F^{(p)} such that [F′ : F] is finite (and hence, prime to p). □

Remark 4.1. Suppose ϕ(a′) = a. By Lemma 3.2(b), if a generates E_{n} ⊗_{F_{n}} F_{n}^{(p)} as an algebra over F^{(p)}, then a′ to generates E_{n} ⊗_{F_{n}} F′ over F′.

The rest of this section will be devoted to proving the implication (2) ⟹ (3). Choose F′ and a as in (2), and let d := [F′ : F]. Then L′ := L_{n} ⊗_{F_{n}} F′ is an S_{n}-Galois algebra over F′ and E′ := (L′)^{S^{n−1}} is an étale algebra of degree n over F′.

Let Z be birational model for the S_{n}-Galois algebra L′, i.e., an F_{0}-variety with a S_{n}- action, whose F_{0}-algebra of rational functions F_{0}(Z) is S_{n}-equivariantly isomorphic to L′. (Note that Z is not necessarily irreducible. If L′ is the direct product of r field extensions of F′, then Z has r irreducible components.) The S_{n}-equivariant inclusion

L_{n} ↪ L′ = L_{n} ⊗_{F_{n}} F′

gives rise to a dominant S_{n}-equivariant rational map Z −−→ 𝔸^{n} of degree d = [F′ : F].

Now the element a gives rise to a S_{n}-equivariant rational map f_{a}: Z −−→ ℙ^{n−1} defined as follows. Choose representatives h_{1}, …, h_{n} of the left cosets of S_{n−1} in S_{n}, so that h_{i}(1) = i, and set

\begin{aligned}f_{a}: & Z & −−→ & ℙ^{n−1} \\ & z & ↦ & (h_{1}(a)(z), …, h_{n}(a)(z)) . \end{aligned}

Note that h_{1}(a) = a, h_{2}(a), …, h_{n}(a) are the conjugates of a in L′. Since a ∈ E′ := (L′)^{S^{n−1}}, h_{i}(a) ∈ L′ depends only on the coset h_{i} S_{n−1} (i.e., only on i) and not on the particular choice of h_{i} in this coset. Moreover, h_{1}(a)^{p} + … + h_{n}(a)^{p} = tr_{L′/F′}(a^{p}) = 0, so the image of f_{a} lies in the hypersurface X_{n,p} ⊂ ℙ^{n−1}, given by x_{1}^{p} + … + x_{n}^{p} = 0, as in Section 2. In summary, we have the following diagram of S_{n}-equivariant rational maps:

(4.1)  \xymatrix{Z \ar@{-->}[dr]^{f_{a}} \ar@{-->}[d]_{generically d : 1} \\ 𝔸^{n} & X_{n,p} \ar@{^{(}->}[r] & ℙ^{n−1}}

Note that since Z is only defined up to an S_{n}-equivariant birational isomorphism, we may assume without loss of generality that Z is projective.

Now consider the abelian subgroup

A := (ℤ/pℤ)^{k_{1}} × … × (ℤ/pℤ)^{k_{m}}

of S_{n}. Recall from the statement of Theorem 1.2 that n := p^{k_{1}} + … + p^{k_{m}} is the base p presentation of n. We view A as a subgroup of S_{n} by embedding each factor (ℤ/pℤ)^{k_{i}} into S_{p^{k_{i}}} via the regular representation. We now observe that that the origin is a smooth A-fixed F_{0}-point in 𝔸^{n}. (In fact, this point is fixed by all of S_{n}.) Hence, by the “going up” theorem of J. Kollár and E. Szabó [RY_{1}, Proposition A.2], X_{n,p} also has an A-fixed F_{0}-point^{1}. In order to complete the proof of the implication (2) ⟹ (3) of Theorem 1.2, it remains to establish the following lemma.

Lemma 4.2. X_{n,p} has an A-fixed point defined over F_{0} if and only if equation (1.1) has a non-trivial solution in ℙ^{m−1}(F_{0}).

Proof. An A-fixed point of ℙ^{n−1} is the same thing as a 1-dimensional A-invariant linear subspace of 𝔸^{n}. To find all possible 1-dimensional A-invariant linear subspaces, we will decompose the natural representation of A on F_{0}^{n} as a direct sum of irreducibles. First decompose F_{0}^{n} as a direct sum of A-invariant subspaces

1Note that [RY_{1}, Proposition A.2] assumes F_{0} is algebraically closed. However, in the case where A is a finite abelian group of exponent p, the proof only requires that ζ_{p} ∈ F_{0}; see [RY_{1}, Remark A.7].

F_{0}^{n} = F_{0}^{p^{k_{1}}} ⊕ … ⊕ F_{0}^{p^{km}} ,

where A acts on F_{0}^{p^{k_{i}}} by composing the natural projection A → (ℤ/pℤ)^{k_{i}} with the regular representation of (ℤ/pℤ)^{k_{i}}. It is natural to label the coordinates of F_{0}^{p^{k_{i}}} by the elements g_{1}, …, g_{p^{k_{i}}} of (ℤ/pℤ)^{k_{i}}, rather than by the numbers 1, 2, …, p^{k_{i}}. In this notation, F_{0}^{p^{k_{i}}} decomposes as a direct product of 1-dimensional invariant subspaces Span_{F_{0}}(R_{χ}), one for each character χ: (ℤ/pℤ)^{k_{i}} → F_{0}^{∗}, where R_{χ} = (χ(g_{1}), …, χ(g_{p^{k_{i}}})). Note that since we are assuming that ζ_{p} ∈ F_{0}, every character χ and every vector R_{χ} are defined over F_{0}. We conclude that the irreducible decomposition of the natural representation of A ⊂ S_{n} on F_{0}^{n} is as follows:

(4.2) F_{0}^{n} = V_{0} ⊕ (⨁_{χ_{1}} Span_{F_{0}}(R_{χ_{1}}, 0, …, 0)) ⊕ … ⊕ (⨁_{χ_{m}} Span_{F_{0}}(0, …, 0, R_{χ_{m}})) ,

where χ_{i} ranges over the non-trivial characters of (ℤ/pℤ)^{k_{i}} → F_{0}^{∗}. Here V_{0} is the mdimensional subspace with trivial associated character,

V_{0} := {(x_{1}, …, x_{1}, x_{2}, …, x_{2}, …, x_{m}, …, x_{m}) | each x_{i} ∈ F_{0} repeats p^{k_{i}} times} ⊂ F_{0}^{n}.

Note that A acts on the 1-dimensional subspace Span_{F_{0}}(0, …, 0, R_{χ_{i}}, 0, …, 0) by the character A → (ℤ/pℤ)^{k_{i}} ⟶^{χ_{i}} F_{0}^{∗}, so the 1-dimensional summands in the sum (4.2) are pairwise non-isomorphic. We conclude that the A-fixed points of ℙ^{n−1} are either of the form

(4.3)  (y_{1} : …: y_{1} : y_{2} : …: y_{2} : …: y_{m} : …: y_{m}) ∈ ℙ(V_{0})

for some (y_{1} : …: y_{m}) ∈ ℙ^{m−1} or of the form

(4.4)  (0 : …: 0 : R_{χ_{i}} : 0 …: 0)

for some non-trivial character χ_{i}: (ℤ/pℤ)^{k_{i}} → F_{0}^{∗}.

A point of ℙ^{n−1} the form (4.4) has exactly p^{k_{i}} non-zero coordinates, and each of these non-zero coordinates is a pth root of unity. Hence the sum of the pth powers of the coordinates of this point is p^{k_{i}}, which is non-zero in F_{0}. Thus no A-fixed point of ℙ^{n−1} of the form (4.4) lies on X_{n,p}. We conclude that every A-fixed point of X_{n,p} is necessarily of the form (4.3). That is, X_{n,p} has an A-fixed point defined over F_{0} if and only if X_{n,p} has an F_{0}-point of the form (4.3) or, equivalently, if and only if equation (1.1) has a solution in ℙ^{m−1}(F_{0}). This completes the proof of Lemma 4.2 and thus of the implication (2) ⟹ (3) of Theorem 1.2. □

5. Conclusion of the proof of Theorem 1.2

(3) ⟹ (1): Assume that (3) holds. That is, there exists y = (y_{1} : …: y_{m}) ∈ ℙ^{m−1}(F_{0}) such that p^{k_{1}}y_{1}^{p} + … + p^{k_{m}}y_{m}^{p} = 0.

Let E/F be an étale algebra of degree n, such that F_{0} ⊂ F and F is a p-field. Let X_{E/F, p} be the degree p hypersurface in ℙ(E) = ℙ_{F}^{n−1} given by tr_{E/F} (x^{p}) = 0.

To prove (1), we need to show that X_{E/F, p} has an F -point. Our solution y to the system (1.2) gives rise to an F_{0}-point

c = (y_{1} : …: y_{1} : y_{2} : …: y_{2} : …: y_{m} : … y_{m}) ∈ ℙ^{n−1} ,

on the hypersurface

X_{n,p} := {(x_{1} : …: x_{n}) | x_{1}^{p} + … + x_{n}^{p} = 0} ⊂ ℙ^{n−1},

which we considered in Section 2. Note that X_{n,p} := X_{F^{n}/F,p}, where F^{n} is the split étale algebra of degree n over F, and X_{E/F, p} is the twist of X_{n,p} by E/F. For the definition of the twisting operation, a discussion of its properties and further references, see [DR, Section 3].

Now observe that the stabilizer of c in S_{n} contains S_{p^{k_{1}}} × … × S_{p^{km}}, which, in turn, contains a Sylow p-subgroup of S_{n}; see, e.g., [Ro, 1.6.19(ii), p. 41]. Hence the orbit S_{n} ·c is a zero cycle in X_{n,p}, whose degree d = [S_{n} : Stab_{S_{n}}(c)] is prime to p. This zero cycle is defined over F_{0} (and thus over F). Twisting the inclusion morphism

S_{n} ·c ↪ X_{n,p}

by the étale algebra E/F, we see that X_{E/F, p} contains the zero cycle ^{E/F} (S_{n} ·c) of degree d defined over F. Since d is prime to p and F is p-closed, we conclude that X_{E/F} has an F - point, as claimed. This completes the proof of the implication (3) ⟹ (1) of Theorem 1.2.

We now proceed with the proof of Assertion (∗) of Theorem 1.2. Assume (3) holds. Our goal is to show that a can be chosen (i) so that E = F [a] in part (1), and (ii) so that E′ = F′[a] in part (2). In fact, only (i) needs to be proved; (ii) follows from (i) by Remark 4.1.

To prove (i), let Δ_{E/F} be the discriminant locus in ℙ(E), i.e., the closed subvariety of ℙ(E) whose F′-points correspond to elements of a ∈ E′ = E ⊗_{F} F′ with fewer than n Galois conjugates, or equivalently, to elements a ∈ E′ such that F′[a] ≠ E′. (Here F′ denotes an arbitrary field extension of F.) We need to show that X_{E/F, p} has an F -point away from Δ_{E/F}. Note that Δ_{E/F} is the twist to Δ_{n} by E/F. By Lemma 2.1(e) X_{n,p} is not contained in Δ_{n}. Hence. X_{E/F, p} is not contained in Δ_{E/F}. Thus it suffices to show that F -points are dense in X_{E/F, p}.

By Lemma 2.1(a) and (c), X_{n,p} is smooth and absolutely irreducible. (Here we use the assumption that n ⩾ 3.) Hence, so is the twist X_{E/F, p} of X_{n,p}. We have just seen that X_{E/F, p} has an F -point. Hence, by Lemma 3.1(a), F -points are dense in X_{E/F, p}, as desired. □

6. Proof of Theorem 1.3

The proof of Theorem 1.3 is largely similar to the proof of Theorem 1.2. We will outline the necessary modifications below. The most significant of these are in the proof of Assertion (∗∗).

Once again, the implication (1) ⟹ (2) readily follows from Lemma 3.2(b).

(2) ⟹ (3). Assumption (2) gives rise to a diagram

\xymatrix{Z \ar@{-->}[dr]^{f_{a}} \ar@{-->}[d]_{generically d : 1} \\ 𝔸^{n} & Y_{n,p} \ar@{^{(}->}[r] & ℙ^{n−1}}

of S_{n}-equivariant dominant rational maps. Here d = [F′ : F_{n}] is prime to p. The only difference, compared to (4.1), is that we have replaced X_{n,p} by Y_{n,p}. The “going up theorem” of theorem of Kollár and Szabó tells us that Y_{n,p} has an A-fixed point defined over F_{0}. Here

A := (ℤ/pℤ)^{k_{1}} × … × (ℤ/pℤ)^{k_{m}} ⊂ S_{n},

as in Section 4. As we saw in the proof of Lemma 4.2, every A-fixed point of X_{n,p} (and hence, of Y_{n,p}) defined over F_{0} is of the form

(y_{1} : …: y_{1} : y_{2} : …: y_{2} : …: y_{m} : …: y_{m})

for some (y_{1} : …: y_{m}) ∈ ℙ^{m−1}(F_{0}). Thus a point of this form has to lie on Y_{n,p}. Equivalently, the system (1.2) has a non-trivial solution over F_{0}, as desired.

(3) ⟹ (1): Let E/F be an étale algebra of degree n, such that F_{0} ⊂ F and F is a pfield. Let Y_{E/F, p} be the degree p hypersurface in ℙ^{n−2} given by tr_{E/F} (x) = tr_{E/F} (x^{p}) = 0. Here by ℙ^{n−2}, we mean the hyperplane tr_{E/F} (x) = 0 in ℙ(E).

To prove (1), we need to show that Y_{E/F, p} has an F -point. A non-zero solution (y_{1} : …: y_{m}) of the system (1.2) gives rise to a point

(6.1)  c = (y_{1} : …: y_{1} : y_{2} : …: y_{2} : …: y_{m} : … y_{m}) ∈ ℙ^{n−1}(F_{0}) ,

whose orbit S_{n} ·c is a zero cycle in Y_{n,p} of degree d = \frac{n!}{Stab_{S_{n}}(c)}, prime to p. Twisting the

inclusion

S_{n} ·c ↪ Y_{n,p}

by an étale algebra E/F, we see that Y_{E/F, p} contains the zero cycle ^{E/F} (S_{n} ·c) of degree d defined over F. Since d is prime to p and F is p-closed, we conclude that Y_{E/F, p} has an F -point, as desired. For future reference, let us denote this F -point of Y_{E/F, p} by c′.

Proof of Assertion (∗∗) of Theorem 1.3: Note that Y_{E/F, p} is an F -form of Y_{n,p}. Since we are assuming that (char(F_{0}), p, n) ≠ (2, 3, 3), (2, 3, 4) or (3, 2, 3), Lemma 2.1(f) implies that Y_{E/F, p} is not contained in Δ_{E/F}. Thus it suffices to show that F -points are dense if Y_{E/F, p}, as in the proof of Assertion (∗) of Theorem 1.2. If we can prove this, then we will be certain to find an F -point away from Δ_{E/F}, and Assertion (∗∗) will follow.

Now recall that we have constructed an F -point c′ of Y_{E/F, p}. We claim that if one of the conditions (i), (ii) , (iii) holds, then Y_{E/F, p} is not a cone over c′. If we prove this claim, then we will be able to conclude that F -points are dense in Y_{E/F, p} by Lemma 3.1(b), and we will be done.

To prove the claim, assume the contrary: Y_{E/F, p} is a cone over c′. Then Y_{E/F, p} will remain a cone over c′ when we pass to the algebraic closure \overline{F} of F. Over \overline{F}, Y_{E/F, p} becomes isomorphic to Y_{n,p}. If conditions (ii) or (iii) hold, i.e., if char(F_{0}) = char(\overline{F}) does not divide n or if p > 2 and char(F_{0}) ≠ 2, then by Lemma 2.1(h), Y_{n,p} is not a cone over any of its points. This contradiction proves the claim in cases (ii) and (iii).

It remains to prove the claim in case (i), where we assume that

(y_{1} : …: y_{m}) ≠ (1 : …: 1)

in ℙ^{m−1} and thus c ≠ (1 : …: 1) in ℙ^{n−1}; see (6.1). Recall that we constructed the F -point c′ ∈ Y_{E/F, p} by twisting the 0-cycle S_{n} ·c by E/F. Once we pass to \overline{F}, we split this cycle once again. Thus c′ is one of the points in the S_{n}-orbit of c (now viewed as an \overline{F} -point). In particular, c′ ≠ (1 : …: 1). Lemma 2.1(g) now tells us that Y_{n,p} is not a cone over c′. This completes the proof of the claim and thus of Theorem 1.3. □

7. Remarks on Theorems 1.2 and 1.3

Remark 7.1. The requirement that char(F_{0}) ≠ p is harmless. In characteristic p, tr(a^{p}) = tr(a)^{p}. In this setting the Hermite-Joubert Problem 1.1 amounts to finding an element 0 ≠ a ∈ E of trace zero, which is always possible (assuming n ⩾ 2).

Remark 7.2. Note that condition (1) in either theorem holds if and only if it holds after we replace F by F^{(p)} (or by any any finite extension F_{1} such that [F_{1} : F] is prime to p). In particular, if F does not contain ζ_{p}, we are free to replace F by F (ζ_{p}). Similarly for condition (2). Consequently, the assumption that ζ_{p} ∈ F_{0} in both theorems can be dropped if we ask that y_{1}, …, y_{m} lie in F_{0}(ζ_{p}), rather than F_{0}, in part (3).

Remark 7.3. The requirement that one of the conditions (i), (ii), (iii) should hold at the end of Theorem 1.3 cannot be dropped.

Indeed, let us consider the following example: n = 6 = 3^{1}+3^{1}, p = 3 and char(F_{0}) = 2. In this case the system (1.2) reduces to

\{ \begin{matrix}3y_{1} + 3y_{2} = 0 \\ 3y_{1}^{3} + 3y_{2}^{3} = 0. \end{matrix}

Conditions (1), (2) and (3) of Theorem 1.3 are satisfied in this example; we can take a = 1_{F} in part (1), F′ := F_{6} and a = 1_{F_{6}} in part (2), and y = (1 : 1) in part (3). On the other hand, it is shown in [Re_{3}] that no element a ∈ E′−F′ satisfies

tr_{E′/F′}(a) = tr_{E′/F′}(a^{3}) = 0

in part (2), for any finite extension F′/F_{6} of degree prime to 3; see [Re_{3}, Theorem 2 and Remark (3) in Section 8]. Thus we cannot choose F′ and a ∈ E′ in part (2), so that E′ = F′[a]. Note that conditions (i), (ii) and (iii) of Theorem 1.3 all fail here. □

8. The Hermite-Joubert problem for p = 2

For p = 2, Theorems 1.2 and 1.3 can be strengthened to give the following answer to the Hermite-Joubert Problem 1.1.

Corollary 8.1. Let F_{0} be a field of characteristic ≠ 2, and n = 2^{k_{1}} + … + 2^{k_{m}} ⩾ 3, where the exponents k_{1}, …, k_{m} ⩾ 0 are distinct integers.

(a) Conditions (1), (2) and (3) of Theorem 1.2 (with p = 2) are equivalent to:

(4) For every field F containing F_{0} and every n-dimensional étale algebra E/F, there exists an element 0 ≠ a ∈ E such that tr_{E/F} (a^{2}) = 0.

(b) Moreover, if (4) holds, and F is an infinite field, then the element a ∈ E in (4) can be chosen so that E = F [a].

(c) Conditions (1), (2) and (3) of Theorem 1.3 (with p = 2) are equivalent to: (4′) For every field F containing F_{0} and every n-dimensional étale algebra E/F, there exists an element 0 ≠ a ∈ E such that tr_{E/F} (a) = tr_{E/F} (a^{2}) = 0.

(d) Moreover, if (4′) holds, char(F_{0}) does not divide n, and F is an infinite field, then the element a ∈ E in (4′) can be chosen so that E = F [a].

Proof. By a theorem of Springer, a quadric hypersurface in ℙ^{l} defined over a field F of characteristic ≠ 2, has an F -point if and only if it has an F^{(2)}-point; see, e.g., [Lam, Theorem VII.2.7] or [Pf, Theorem 6.1.12]. Applying this to the hypersurfaces X_{E/F, 2} ⊂ ℙ_{F}^{n−1} and Y_{E/F, 2} ⊂ ℙ_{F}^{n−2} given by tr(x^{2}) = 0 and tr_{E/F} (x) = tr_{E/F} (x^{2}) = 0, respectively, we see that (1) ⟺ (4) in part (a) and (1) ⟺ (4′) in part (c).

Proof of part (b). We begin by establishing the following claim. Let E/F be an étale algebra of degree n. Assume X_{E/F, 2} has an F -point. Then F -points are dense in X_{E/F, 2}.

Indeed, by Lemma 2.1(a), X_{n,2} is a smooth quadric hypersurface in ℙ^{n−1}, and hence, so is X_{E/F, 2}. Thus the existence of an F -point on X_{E/F, 2} implies that X_{E/F, 2} is rational over F. Since we are assuming that F is infinite, this tells us that F -points are dense in X_{E/F, 2}. This proves the claim.

Now observe that by Lemma 2.1(e), X_{n,p} is not contained in Δ_{n} and thus X_{E/F, 2} is not contained in the discriminant locus Δ_{E/F}. The claim tells us that there exists an F -point of X_{E/F, 2} away form Δ_{E/F}. This F -point is represented by an element a ∈ E such that tr_{E/F} (a^{2}) = 0 and F [a] = E, as desired.

Finally, we turn to the proof of part (d). Since we are assuming that char(F_{0}) ≠ 2 and does not divide n, Lemma 2.1(b) tells us that Y_{n,2} is a smooth quadric hypersurface in ℙ^{n−2}, and hence, so is any of its twisted forms Y_{E/F, 2}. Moreover, by Lemma 2.1(f), Y_{n,2} is not contained in Δ_{n} and hence, Y_{E/F, 2} is not contained in Δ_{E/F}.

Now, arguing as in the proof of part (b) above, we see that if (4) holds, then Y_{E/F, 2} is rational and hence, F -points are dense in Y_{E/F, p} (recall that F is assumed to be an infinite field). In particular, there there exists an F -point of Y_{E/F, 2} away from the discriminant locus Δ_{E/F}, and part (d) follows. This completes the proof of Corollary 8.1. □

Remark 8.2. Suppose p = 2. Let us arrange the exponents k_{1}, …, k_{m} in Corollary 8.1 so that k_{1}, …, k_{s} are even and k_{s+1}, …, k_{m} are odd. (Here k_{1}, …, k_{m} are distinct nonnegative integers; we do not require that k_{1} > … > k_{m}.) The quadratic form 2^{k_{1}}y_{1}^{2} +…+ 2^{k_{m}}y_{m}^{2} is then equivalent to q(z_{1}, …, z_{n}) = z_{1}^{2} + … + z_{s}^{2} + 2(z_{s+1}^{2} + … + z_{m}^{2}). Condition (3) of Theorem 1.2 amounts to requiring q to be isotropic over F_{0}. Condition (3) of Theorem 1.3 is equivalent to saying that q has an isotropic vector in the hyperplane given by

(8.1)  2^{k_{1}/2}z_{1} + … + 2^{k_{s}/2}z_{s} + 2^{(k_{s+1}−1)/2}z_{1} + … + 2^{(k_{m}−1)/2}z_{m} = 0

in ℙ^{m−1}. Note that condition (3) of Theorem 1.2 fails if F_{0} is formally real. On the other hand, condition (3) of Theorem 1.3 holds if the Witt index of q is ≥ 2. Indeed, in this case the quadric hypersurface in ℙ^{m−1} given by q = 0 has a line defined over F_{0}; see [Lam, Theorem II.4.3]. Intersecting this line with the hyperplane (8.1) we obtain a desired isotropic vector defined over F_{0}. □

9. The Hermite-Joubert problem for p = 3

Springer’s theorem has the following conjectural analogue for p = 3.

Conjecture 9.1. (J. W. S. Cassels, P. Swinnerton-Dyer [Co_{1}, p. 267])

Let X be a cubic hypersurface in ℙ^{l} defined over a field F. If X(F_{1}) ≠ 0 for some finite extension F_{1}/F and [F_{1} : F] is prime to 3, then X(F) ≠ ∅. In other words, if X has an F^{(3)}-point, then X has an F -point.

Remark 9.2. This long-standing conjecture remains largely open; to the best of our knowledge, the partial results published by D. Coray [Co_{1}] in 1976 remain state of the art. One special case, where the conjecture is known (and easy to prove) is the following:

Let X be a cubic hypersurface in ℙ^{l} defined over a field F. If [F_{1} : F] = 2 and X(F_{1}) ≠ ∅, then X(F) ≠ ∅; see [Co_{1}, Proposition 2.2].

Remark 9.3. If p = 3, then the assumption that ζ_{p} ∈ F_{0} in the statements of Theorems 1.2 and 1.3 can be dropped.

To prove this, let us assume that ζ_{3} ∉ F_{0} and see what happens if we replace F_{0} by F_{0}(ζ_{3}). As we explained in Remark 7.2, the validity of conditions (1) and (2) will not change. Since [F_{0}(ζ_{3}) : F_{0}] ⩽ 2, Remark 9.2 tells us that the validity of condition (3) will not change either. □

In view of Corollary 8.1 and Conjecture 9.1, it is natural to expect the following answer to the Hermite-Joubert Problem 1.1 for p = 3.

Conjecture 9.4. Let F_{0} be a field of characteristic ≠ 3, n ⩾ 3 be an integer, and n = 3^{k_{1}} + … + 3^{k_{m}} be the base 3 expansion of n.

(a) Conditions (1), (2) and (3) of Theorem 1.2 (with p = 3) are equivalent to:

(4) For every field F containing F_{0} and every n-dimensional étale algebra E/F, there exists an element 0 ≠ a ∈ E such that tr_{E/F} (a^{3}) = 0.

(b) Moreover, if (4) holds, n ⩾ 4, and F is an infinite field, then the element a ∈ E in (4) can be chosen so that E = F [a].

(c) Conditions (1), (2) and (3) of Theorem 1.3 (with p = 3) are equivalent to: (4′) For every field F containing F_{0} and every n-dimensional étale algebra E/F, there exists an element 0 ≠ a ∈ E such that tr_{E/F} (a) = tr_{E/F} (a^{3}) = 0.

(d) Moreover, if (4′) holds, n ⩾ 5, char(F_{0}) ≠ 2, and F is an infinite field, then the element a in (4′) can be chosen so that E = F [a].

Proposition 9.5. Conjecture 9.4 follows from Conjecture 9.1.

Proof. Recall that by Remark 9.3, for p = 3, Theorems 1.2 and 1.3 remain valid even if F_{0} does not contain ζ_{3}.

The proof of the equivalences (1) ⟺ (4) in part (a) and (1) ⟺ (4′) in part (c) is now exactly the same as in Corollary 8.1, with Conjecture 9.1 used in place of Springer’s theorem.

To prove part (b), let E/F be an étale algebra of degree n. By (4), X_{E/F, 3} has an F -point. By Lemma 2.1, X_{E/F, 3} is a smooth absolutely irreducible cubic hypersurface of dimension n − 2 ⩾ 2. Since it has an F -point, [Ko, Theorem 1.2] tells us that X_{E/F, 3} is unirational over F. Since we are assuming that F is infinite, this implies that F -points are dense in X_{E/F, 3}. On the other hand, by Lemma 2.1(e), X_{n,3} is not contained in Δ_{n}. Hence, X_{E/F, 3} is not contained in Δ_{E/F}. Therefore, we can find an F -point of X_{E/F, 3} away from Δ_{E/F}, and part (b) follows.

We now turn to part (d). Since n ⩾ 5, Lemma 2.1(f) tells us that Y_{n,p} is not contained in Δ_{n}. Hence, Y_{E/F, 3} is not contained in Δ_{E/F}. By (4’) Y_{E/F, 3} has an F -point. Thus it suffices to show that F -points are dense in Y_{E/F, 3}.

Since we are assuming that n ⩾ 5, Lemma 2.1(e) tells us that Y_{n,3} is an absolutely irreducible cubic hypersurface of dimension ⩾ 2, and hence, so is Y_{E/F, 3}. Moreover, since char(F_{0}) ≠ 2, Y_{n,3} is not a cone by Lemma 2.1(h). Hence, neither is Y_{E/F}. Thus by [Ko, Theorem 1.3] the existence of an F -point on Y_{E/F, 3} implies that Y_{E/F, 3} is unirational^{2}. In particular, F -points are dense in Y_{E/F, 3}. This completes the proof of Proposition 9.5. □

Remark 9.6. Let Z_{m,p} and W_{m,p} be the degree p hypersurfaces cut out by the equation (1.1) and the system (1.2) in ℙ^{m−1} and ℙ^{m−2}, respectively.

If ζ_{p} ∈ F_{0}, it follows from Theorem 1.2 (respectively, Theorem 1.3) that Z_{m,p} (respectively, W_{m,p}) has an F_{0}-point if and only if it has an F_{0}^{(p)}-point. Indeed, as we noted in Remark 7.2, the validity of conditions (1) and (2) does not change when we replace F_{0} by F_{0}^{(p)}. Hence, neither does the validity of (3).

In particular, this shows that Conjecture 9.1 is true for the cubic hypersurfaces Z_{m,3} and W_{m,3} defined over F_{0}. Note also that for p = 3 the requirement that ζ_{3} ∈ F_{0} can be dropped; see Remark 9.3.

10. When are there solutions to (1.1) and (1.2)?

Lemma 10.1. Let F_{0} be a field of characteristic ≠ p. Equation (1.1) has a solution y = (y_{1}, …, y_{m}) ∈ ℙ^{m−1}(F_{0}) if one of the following conditions holds:

(a) \sqrt[p]{−p^{k_{i}−k_{j}}} lies in F_{0}, for some 1 ⩽ i < j ⩽ m.

(b) k_{i} ≡ k_{j} (mod p) for some 1 ⩽ i < j ⩽ m and either p is odd or p = 2 and \sqrt{−1} ∈ F_{0}.

(c) m ⩾ p + 1.

Proof. (a) Set y_{i} := 1, y_{j} := \sqrt[p]{−p^{k_{i}−k_{j}}}, and y_{h} = 0 for every h ≠ i, j. Then y = (y_{1} : …: y_{m}) is a solution to (1.1).

(b) If k_{i} ≡ k_{j} (mod p), then \sqrt[p]{−p^{k_{i}−k_{j}}} ∈ F_{0}.

(c) If m ⩾ p + 1, then k_{1}, …, k_{m} cannot all be distinct modulo p, and part (b) applies. □

2[Ko, Theorem 1.3] assumes that the field F is perfect. For the case, where F is an imperfect field of characteristic ≠ 2, 3, see the remark after the statement of Theorem 1.3 in [Ko].

We will now prove the converse to Lemma 10.1(b) for F_{0} = ℚ(ζ_{p}).

Proposition 10.2. Let p be an odd prime and F_{0} = ℚ(ζ_{p}). Then the following conditions are equivalent.

(a) Equation (1.1) has no solutions in ℙ^{m−1}(F_{0}).

(b) The integers k_{1}, …, k_{m} are distinct modulo p.

Proof. The implication (a) ⟹ (b) follows from Lemma 10.1(b).

(b) ⟹ (a): Assume (y_{1} : …: y_{m}) ∈ ℙ^{m−1}(F_{0}) is a solution to (1.1), i.e.,

(10.1)  p^{k_{1}}y_{1}^{p} + … + p^{k_{m}}y_{m}^{p} = 0

The p-adic valuation ν_{p}: ℚ^{∗} → ℤ can be extended to ν_{p}: ℚ(ζ_{p})^{∗} → Γ, where Γ is a subgroup of ℚ such that [Γ : ℤ] ⩽ p − 1; see [Lang, Theorem XII4.1 and Proposition

XII.4.2]. In fact, we can take Γ = \frac{1}{p − 1} ℤ, but we will not need this in the sequel.

From (10.1) we see that

ν_{p}(p^{k_{i}}y_{i}^{p}) = ν_{p}(p^{k_{j}}y_{j}^{p})

for some 1 ⩽ i < j ⩽ m such that y_{i}, y_{j} ≠ 0. It remains to show that k_{i} − k_{j} is divisible by p. Indeed, assume the contrary. Then k_{i} + pν_{p}(y_{i}) = k_{j} + pν_{p}(y_{j}), and

\frac{k_{i} − k_{j}}{p} = ν_{p}(y_{j}) − ν_{p}(y_{i}) ∈ Γ .

Thus [Γ : Z] ⩾ [ \frac{1}{p} ℤ : ℤ] = p, a contradiction. □

Lemma 10.3. Let F_{0} be a field of characteristic ≠ p. The system (1.2) has a solution in ℙ^{m−1}(F_{0}) if one of the following conditions holds:

(a) p is odd and k_{i} = k_{j} for some i ≠ j,

(b) \sqrt[p]{−p^{k_{i}−k_{j}}} and \sqrt[p]{−p^{k_{i′}−k_{j′}}} both lie in F_{0}, for some 4-tuple of distinct integers i, j, i′, j′ between 1 and m,

(c) m ⩾ p + 3, and either p is odd or p = 2 and \sqrt{−1} ∈ F_{0},

(d) m ⩾ p + 2 and char(F_{0}) > 0.

Proof. (a) Set y_{i} := 1, y_{j} := −1, and y_{h} = 0 for any h ≠ i, j. Then y = (y_{1} : …: y_{m}) is a solution to (1.2).

(b) The hypersurface Z_{m,p} ⊂ ℙ^{m−1} given by p^{k_{1}}y_{1}^{p} + … + p^{k_{m}}y_{m}^{p} = 0, contains the line through y := (y_{1} : …: y_{m}) and y′ := (y′_{1} : …: y′_{m}), where y_{i} := 1, y_{j} := \sqrt[p]{−p^{k_{i}−k_{j}}} and y_{h} = 0 for every h ≠ i, j, and similarly y′_{i′} := 1, y′_{j′} := \sqrt[p]{−p^{k_{i′}−k_{j′}}} and y_{h′} = 0 for every h′ ≠ i′, j′. Intersecting this line with the hyperplane p^{k_{1}}y_{1} + … + p^{k_{m}}y_{m} = 0, we obtain a solution to (1.2).

(c) Assume m ⩾ p + 3. Then there exist 1 ⩽ i < j ⩽ m such that k_{i} ≡ k_{j} (mod p). Since m − 2 ⩾ p + 1, after removing k_{i} and k_{j} from the sequence k_{1}, …, k_{m}, we will find two other distinct subscripts i′ and j′ such that k_{i′} ≡ k_{j′} (mod p). The desired conclusion now follows from part (b).

(d) Let 𝔽 be the prime subfield of F_{0}. By Chevalley’s theorem 𝔽 is a C_{1}-field; see [Pf, Theorem 5.2.1]. Note that the coefficients p^{k_{i}} of the system (1.2) all lie in 𝔽. Since we are assuming that m − 1 > p, the C_{1}-property of 𝔽 guarantees that the system (1.2) has a solution in ℙ^{m−2}(𝔽) and hence, in ℙ^{m−2}(F_{0}). □

11. Proof of Theorem 1.4

By Theorem 1.3 it suffices to show that the system (1.2) has no non-trivial solutions in ℚ. (Recall that by Remark 9.3, for p = 3, Theorem 1.3 is valid for F_{0} = ℚ, even though ζ_{3} ∉ ℚ.)

We will say that two triples, (k_{1}, k_{2}, k_{3}) and (k′_{1}, k′_{2}, k′_{3}) ∈ ℤ^{3}, are equivalent if

(k′_{1}, k′_{2}, k′_{3}) = (k_{σ(1)} + c, k_{σ(2)} + c, k_{σ(3)} + c) ,

for some σ ∈ S_{3} and c ∈ ℤ. For each triple of integers, (k_{1}, k_{2}, k_{3}) we would like to know whether or not the system

(11.1)  \{ \begin{matrix}3^{k_{1}}y_{1} + 3^{k_{2}}y_{2} + 3^{k_{3}}y_{3} = 0 \\ 3^{k_{1}}y_{1}^{3} + 3^{k_{2}}y_{2}^{3} + 3^{k_{3}}y_{3}^{3} = 0 \end{matrix}

has a non-trivial solution in ℚ. For the purpose of proving Theorem 1.4, we may replace (k_{1}, k_{2}, k_{3}) by an equivalent triple (k′_{1}, k′_{2}, k′_{3}). This will cause the system (11.1) to be replaced by an equivalent system. Moreover, k_{1} + k_{2} + k_{3} ≡ k′_{1} + k′_{2} + k′_{3} (mod 3) and if k_{1}, k_{2}, k_{3} are distinct, then so are k′_{1}, k′_{2}, k′_{3}.

One easily checks that any triple (k_{1}, k_{2}, k_{3}) with k_{1} + k_{2} + k_{3} ≡ 0 or 1(mod 3), is equivalent to some (k′_{1}, k′_{2}, k′_{3}), where (k′_{1}, k′_{2}, k′_{3}) ≡ (0, 0, 0), (0, 1, 2) or (0, 0, 1) (mod 3). Thus it suffices to show that our system has no non-zero solutions over ℚ in each of these three cases.

Case 1: k_{1} = 3e_{1}, k_{2} = 3e_{2}, k_{3} = 3e_{3}, where e_{1}, e_{2}, and e_{3} are distinct integers. Substituting z_{i} := 3^{e_{i}}y_{i}, we obtain

\{ \begin{matrix}3^{2e_{1}}z_{1} + 3^{2e_{2}}z_{2} + 3^{2e_{3}}z_{3} = 0 \\ z_{1}^{3} + z_{2}^{3} + z_{3}^{3} = 0. \end{matrix}

By Fermat’s last theorem, the only solutions to the second equation in ℙ^{2}(ℚ) are

(1 : −1 : 0), (1 : −1 : 0) and (1 : 0 : −1).

None of them satisfy the first equation.

Case 2: k_{1} = 3e_{1}, k_{2} = 3e_{2} + 1, k_{3} = 3e_{3} + 2. In this case equation (1.1) has no non-trivial solutions over ℚ by Proposition 10.2. Hence, neither does the system (1.2).

Case 3: k_{1} = 3e_{1}, k_{2} = 3e_{2}, and k_{3} = 3e_{3} + 1, where e_{1} ≠ e_{2}. Once again, setting z_{i} := 3^{e_{i}}y_{i}, we reduce our system to

\{ \begin{matrix}3^{2e_{1}}z_{1} + 3^{2e_{2}}z_{2} + 3^{2e_{3}+1}z_{3} = 0 \\ z_{1}^{3} + z_{2}^{3} + 3z_{3}^{3} = 0. \end{matrix}

By [Sel, Theorem VIII, p. 301], the only solution (z_{1} : z_{2} : z_{3}) ∈ ℙ^{2}(ℚ) to the second equation is (1 : −1 : 0). Since e_{1} ≠ e_{2}, this point does not satisfy the first equation. This completes the proof of Theorem 1.4. □

12. Beyond Theorem 1.4

Conjecture 12.1. Theorem 1.4 remains true for all triples k_{1}, k_{2}, k_{3} of distinct nonnegative integers.

We offer the following partial result in support of Conjecture 12.1.

Proposition 12.2. Theorem 1.4 remains valid for any n = 3^{k_{1}} + 3^{k_{2}} + 3^{k_{3}} such that k_{1} > k_{2} > k_{3} ⩾ 0 and k_{1} ≢ k_{2} (mod 3).

In particular, the Hermite-Joubert Problem 1.1 (with p = 3 and F_{0} = ℚ) has a negative answer for n = 3^{k_{1}}+3^{k_{2}}+3^{k_{3}}, where k_{1} > k_{2} > k_{3} ⩾ 0 and k_{1} ≡ k_{3} ≡ 0 (mod 3), and k_{2} ≡ 2 (mod 3) or alternatively, if k_{2} ≡ k_{3} ≡ 0 (mod 3) and k_{1} ≡ 2 (mod 3). These cases are not covered by Theorem 1.4. The smallest of these new examples is n = 3^{3} + 3^{2} + 3^{0} = 37.

Proof of Proposition 12.2. By Theorem 1.3 it suffices to show that the system

\{ \begin{matrix}3^{k_{1}}y_{1} + 3^{k_{2}}y_{2} + 3^{k_{3}}y_{3} = 0 \\ 3^{k_{1}}y_{1}^{3} + 3^{k_{2}}y_{2}^{3} + 3^{k_{3}}y_{3}^{3} = 0 \end{matrix}

does not have a solution (y_{1}, y_{2}, y_{3}) ≠ (0, 0, 0) with y_{1}, y_{2}, y_{3} ∈ ℚ. Assume the contrary. After dividing both equations by 3^{k_{3}}, and replacing k_{1}, k_{2} by k_{1}−k_{3} and k_{2}−k_{3} respectively, we may assume without loss of generality that k_{3} = 0. Substituting y_{3} = −3^{k_{1}}y_{1} − 3^{k_{2}}y_{2} into the second equation, we obtain

(12.1)  3^{k_{1}}y_{1}^{3} + 3^{k_{2}}y_{2}^{3} − 3^{3k_{1}}y_{1}^{3} − 3^{2k_{1}+k_{2}+1}y_{1}^{2}y_{2} − 3^{k_{1}+2k_{2}+1}y_{1}y_{2}^{2} − 3^{3k_{2}}y_{2}^{3} = 0 .

Clearly y_{1}, y_{2} ≠ 0. Set

M_{1} := ν_{3}(3^{k_{1}}y_{1}^{3}) = k_{1} + 3ν_{3}(y_{1}),
M_{2} := ν_{3}(3^{k_{2}}y_{2}^{3}) = k_{2} + 3ν_{3}(y_{2}), and
M := min(M_{1}, M_{2}) .

Here ν_{3} denotes the 3-adic valuation. Since k_{1} ≢ k_{2} (mod 3), we have M_{1} ≠ M_{2}.

We claim that the 3-adic valuation of each of the last four terms on the left hand side of (12.1) is > M. If we manage to prove this claim, then we will be able to conclude that

ν_{3}(3^{k_{1}}y_{1}^{3} + 3^{k_{2}}y_{2}^{3} − 3^{3k_{1}}y_{2}^{3} − 3^{2k_{1}+k_{2}+1}y_{1}^{3}y_{2} − 3^{k_{1}+2k_{2}+1}y_{1}y_{2}^{2} − 3^{3k_{2}}y_{2}^{3}) = M,

contradicting (12.1), and Proposition 12.2 will follow.

To prove the claim, we will consider each term separately:

(i) ν_{3}(3^{3k_{1}}y_{1}^{3}) = 3k_{1} + 3ν_{3}(y_{1}) > M_{1} ⩾ M.

(ii) ν_{3}(3^{2k_{1}+k_{2}+1}y_{1}^{2}y_{2}) = 2k_{1}+k_{2}+2ν_{3}(y_{1})+ν_{3}(y_{2})+1 > \frac{2}{3}M_{1}+ \frac{1}{3}M_{2} > \frac{2}{3}M + \frac{1}{3}M = M.

(iii) ν_{3}(3^{k_{1}+2k_{2}+1}y_{1}y_{2}^{2}) = k_{1}+2k_{2}+ν_{3}(y_{1})+2ν_{3}(y_{2})+1 > \frac{1}{3}M_{1}+ \frac{2}{3}M_{2} > \frac{2}{3}M + \frac{1}{3}M = M.

(iv) ν_{3}(3^{3k_{2}}y_{2}^{3}) = 3k_{2} + 3ν_{3}(y_{2}) > M_{2} ⩾ M. This completes the proof of the claim and thus of Proposition 12.2. □

Using Proposition 12.2, one readily checks that Conjecture 12.1 follows from Conjecture 12.3 below.

Conjecture 12.3. Let (q_{1} : q_{2} : q_{3}) be a ℚ-point of the curve C ⊂ ℙ^{2} given by x_{1}^{3} + x_{2}^{3} + 9x_{3}^{3} = 0. Then 3^{a}q_{1} + 3^{b}q_{2} + q_{3} ≠ 0 for any integers a > b > 0.

Note that if we view C as an elliptic curve with the origin at (1 : −1 : 0), then the group C(ℚ) of rational points is cyclic, generated by (1 : 2 : −1); see [Sel, p. 357].

Acknowledgements

The authors are grateful to M. Bennett, N. Bruin, A. Duncan, S. Gaulhiac, D. Ghioca and B. Poonen for stimulating discussions.

Some of the results of this paper were announced during the Clifford Lectures at Tulane University in March 2015. The second author would like to thank M. Brion and M. Can for inviting him to take part in the Clifford Lectures and to submit a paper for the proceedings.

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Department of Mathematics, University of British Columbia, Vancouver, CANADA

E-mail address: mbrassil@math.ubc.ca, reichst@math.ubc.ca