JOUBERT’S THEOREM FAILS IN CHARACTERISTIC 2

ZINOVY REICHSTEIN

Abstract. Let L/K be a separable field extension of degree 6. An 1867 theorem of P. Joubert asserts that if char(K) ≠ 2 then L is generated over K by an element whose minimal polynomial is of the form t^{6} + at^{4} + bt^{2} + ct + d for some a, b, c, d ∈ K. We show that this theorem fails in characteristic 2.

I. Introduction. The starting point for this note is the following classical theorem.

Theorem 1. (P. Joubert, 1867) Let L/K be a separable field extension of degree 6. Assume char(K) ≠ 2. Then there is a generator y ∈ L for L/K (i.e., L = K(y)) whose minimal polynomial is of the form

(2)  t^{6} + at^{4} + bt^{2} + ct + d

for some a, b, c, d ∈ K.

Joubert [Jo] gave a formula, which associates to an arbitrary generator x for L/K another generator y ∈ L whose minimal polynomial is of the form (2). He did not state Theorem 1 in the above form, did not investigate under what assumptions on L, K and x his formula applies, and, most likely, only considered fields of characteristic zero. A proof of Theorem 1 based on an enhanced version of Joubert’s argument has been given by H. Kraft [Kr, Main Theorem (b)]. A different (earlier) modern proof of Theorem 1, based on arithmetic properties of cubic hypersurfaces, is due to D. Coray [Co, Theorem 3.1]. (Coray assumed that char(K) ≠ 2, 3.) Since both of these proofs break down in characteristic 2, Kraft [Kr, Remark 6] asked if Theorem 1 remains valid when char(K) = 2. In this paper we will show that the answer is “no" in general but “yes" under some additional assumptions on L/K.

II. Notational conventions. Suppose L/K is a field extension of of degree n. Every y ∈ L, defines a K-linear transformation L → L given by z ↦ yz. We will denote the characteristic polynomial of this linear transformation by t^{n} − σ_{1}(y)t^{n−1} + ⋯ + (−1)^{n}σ_{n}(y). It is common to write tr(y) in place of σ_{1}(y). The minimal and the characteristic polynomial of y coincide if and only if y is a generator for L/K.

If L/K is separable, then σ_{i}(y) = s_{i}(y_{1}, …, y_{n}), where y_{1}, …, y_{n} are the Galois conjugates of y and s_{i} is the ith elementary symmetric polynomial. Furthermore, if [L : K] = 6, then condition (2) of Theorem 1 is equivalent to σ_{1}(y) = σ_{3}(y) = 0.

We will be particularly interested in the “general" field extension L_{n}/K_{n} of degree n constructed as follows. Let F be a field and x_{1}, …, x_{n} be independent variables over F. The symmetric group S_{n} acts on F (x_{1}, …, x_{n}) by permuting x_{1}, …, x_{n}. Set K_{n} := F (x_{1}, …, x_{n})^{S^{n}} = F (a_{1}, …, a_{n}), where a_{i} = s_{i}(x_{1}, …, x_{n}), and L_{n} := F (x_{1}, …, x_{n})^{S^{n−1}} = K_{n}(x_{1}), where S_{n−1} permutes x_{2}, …, x_{n}. Note that by construction L_{n}/K_{n} is a separable extension of degree n.

2010 Mathematics Subject Classification. 12E10, 12F10.

The author was partially supported by National Sciences and Engineering Research Council of Canada Discovery grant 250217-2012.

We remark that since S_{n} has no proper subgroups containing S_{n−1}, there are no proper subextensions between K_{n} and L_{n}. Thus for n ⩾ 2, y ∈ L_{n} generates L_{n}/K_{n} if and only if y ∉ K_{n}.

III. Main results.

Theorem 3. Let F be a field of characteristic 2, m ≥ 1 be an integer, and n := 2 · 3^{m}. Then there is no y ∈ L_{n}−K_{n} such that σ_{1}(y) = σ_{3}(y) = 0.

In particular, setting m = 1, we see that Theorem 1 fails in characteristic 2. We will deduce Theorem 3 from the following more general

Theorem 4. Let F be a field of characteristic 2, m ⩾ 1 be an integer, p be an odd prime, and n := 2p^{m}. Then there is no y ∈ L_{n}−K_{n} such that tr(y) = tr(y^{2}) = ⋯ = tr(y^{p}) = 0.

By Newton’s formulas, tr(y^{3}) = tr(y)^{3} − 3 tr(y)σ_{2}(y) + 3σ_{3}(y). Thus in characteristic ≠ 3,

σ_{1}(y) = σ_{3}(y) = 0 ⟺ tr(y) = tr(y^{3}) = 0 .

Moreover, in characteristic 2, tr(z^{2}) = tr(z)^{2} for any z ∈ L_{n} and thus

tr(y) = tr(y^{2}) = ⋯ = tr(y^{p}) = 0 ⟺ tr(y) = tr(y^{3}) = ⋯ = tr(y^{p−2}) = tr(y^{p}) = 0 .

In particular, for p = 3, Theorem 4 reduces to Theorem 3.

Theorem 5. Let L/K be a separable field extension of degree 6. Assume char(K) = 2 and one of the following conditions holds:

(a) there exists an intermediate extension K ⊂ L′ ⊂ L such that [L′ : K] = 3,

(b) K is a C_{1}-field. Then there is a generator y ∈ L for L/K satisfying σ_{1}(y) = σ_{3}(y) = 0.

For background material on C_{1}-fields, see [Se_{1}, Sections II.3].

IV. Proof of Theorem 4: the overall strategy. It is easy to see that if Theorem 4 fails for a field F, it will also fail for the algebraic closure \overline{F}. We will thus assume throughout that F is algebraically closed.

Our proof of Theorem 4 will use the fixed point method, in the spirit of the arguments in [RY_{2}, Section 6]. The idea is as follows. Assume the contrary: tr(y) = ⋯ = tr(y^{p}) = 0 for some y ∈ L_{n}−K_{n}. Based on this assumption we will construct a projective F -variety \overline{X} with an S_{n}-action and an S_{n}-equivariant rational map φ_{y}: 𝔸^{n} −−→ \overline{X} defined over F. Here S_{n} acts on 𝔸^{n} by permuting coordinates in the usual way. The Going Down Theorem of J. Kollár and E. Szabó [RY_{1}, Proposition A.2] tells us that every abelian subgroup G ⊂ S_{n} of odd order has a fixed F -point in \overline{X}. On the other hand, we will construct an abelian p-subgroup G of S_{n} with no fixed F -points in \overline{X}. This leads to a contradiction, showing that y cannot exist. We will now supply the details of the proof, following this outline.

V. Construction of \overline{X}, φ_{y}, and the abelian subgroup G ⊂ S_{n}. Every y ∈ L_{n} gives rise to an S_{n}-equivariant rational map (i.e., a rational covariant)

\begin{matrix}f_{y}: 𝔸^{n} −−→ 𝔸^{n} \\ f_{y}(α) = (h_{1}(y)(α), …, h_{n}(y)(α)) , \end{matrix}

where 𝔸^{n} is the n-dimensional affine space defined over F, α = (a_{1}, …, a_{n}) ∈ 𝔸^{n}, elements of F (x_{1}, …, x_{n}) are viewed as rational functions on 𝔸^{n}, and h_{1}, …, h_{n} are representatives of the left cosets of S_{n−1} in S_{n}, such that h_{i}(1) = i. Note that h_{1}(y) = y, h_{2}(y), …, h_{n}(y) are the conjugates of y in F (x_{1}, …, x_{n}). Since y ∈ L_{n} := F (x_{1}, …, x_{n})^{S^{n−1}}, h_{i}(y) ∈ F (x_{1}, …, x_{n}) 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.

Recall that we are assuming that tr(y) = ⋯ = tr(y^{p}) = 0. Thus the image of f_{y} is contained in the S_{n}-invariant subvariety X ⊂ 𝔸^{n} given by

(6)  a_{1} + ⋯ + a_{n} = a_{1}^{2} + ⋯ + a_{n}^{2} = ⋯ = a_{1}^{p} + ⋯ + a_{n}^{p} = 0 .

Because n is even and we are working in characteristic 2, if X contains α ∈ 𝔸^{n} then X contains the linear span of α and α_{0} := (1, …, 1). Using this observation, we define an S_{n}-equivariant

rational map φ_{y}: 𝔸^{n} −−→ \overline{X} as a composition φ_{y}: 𝔸^{n} −−→^{f_{y}} X −−→^{π} \overline{X}, where π denotes the linear projection 𝔸^{n} −−→ ℙ(F^{n}/D), D := Span_{F} (α_{0}) is a 1-dimensional S_{n}-invariant subspace in F^{n}, and \overline{X} ⊂ ℙ(F^{n}/D) is the image of X under π. Points in the projective space ℙ(F^{n}/D) ≃ ℙ^{n−2} correspond to 2-dimensional linear subspaces L ⊂ F^{n} containing D. Points in \overline{X} correspond to 2-dimensional linear subspaces L ⊂ F^{n}, such that D ⊂ L ⊂ X. In particular, \overline{X} is closed in ℙ(F^{n}/D). The rational map π associates to a point α ∈ 𝔸^{n} the 2-dimensional subspace spanned by α and α_{0}. Note that π(α) is well defined if and only if α ∉ D. Since we are assuming that y ∉ K_{n}, the image of f_{y} is not contained in D. Thus the composition φ_{y} = π ∘ f_{y}: 𝔸^{n} −−→ \overline{X} is a well-defined S_{n}-equivariant rational map.

Finally, the abelian p-subgroup G ⊂ S_{n} we will be working with, is defined as follows. Recall that n = 2p^{m}. The regular action of (ℤ/pℤ)^{m} on itself allows us to view (ℤ/pℤ)^{m} as a subgroup of S_{p^{m}}. Here we denote the elements of (ℤ/pℤ)^{m} by g_{1}, …, g_{p^{m}} and identify {1, …, p^{m}} with {g_{1}, …, g_{p^{m}}}. We now set G := (ℤ/pℤ)^{m} × (ℤ/pℤ)^{m} ↪ S_{p^{m}} × S_{p^{m}} ↪ S_{n}.

VI. Conclusion of the proof of Theorem 4. It remains to show that G has no fixed F -points in \overline{X}. A fixed F -point for G in \overline{X} corresponds to a 2-dimensional G-invariant subspace L of F^{n} such that D ⊂ L ⊂ X. It will thus suffice to prove the following claim: no G-invariant 2-dimensional subspace of F^{n} is contained in X.

Since F is an algebraically closed field of characteristic 2 and G is an abelian p-group, where p ≠ 2, the G-representation on F^{n} is completely reducible. More precisely, F^{n} decomposes as F_{reg}^{p^{m}}[1] ⊕ F_{reg}^{p^{m}}[2], the direct sum of the regular representations of the two factors of G = (ℤ/pℤ)^{m} × (ℤ/pℤ)^{m}. Each F_{reg}^{p^{m}}[i] further decomposes as the direct sum of p^{m} one-dimensional invariant spaces

V_{χ}[i] := Span_{F} (χ(g_{1}), …, χ(g_{p^{m}})) ,

where χ: (ℤ/pℤ)^{m} → F^{∗} is a multiplicative character. Thus F^{n} = F_{reg}^{p^{m}}[1] ⊕ F_{reg}^{p^{m}}[2] is the direct sum of the two-dimensional subspace

(F^{n})^{G} = V_{0}[1] ⊕ V_{0}[2] = \{(\underbrace{a, …, a}_{p^{m} times}, \underbrace{b, …, b}_{p^{m} times}) | a, b ∈ F \} ,

where 0 denotes the trivial character of (ℤ/pℤ)^{m}, and 2p^{m} − 2 distinct non-trivial 1-dimensional representations V_{χ}[i], where i = 1, 2, and χ ranges over the non-trivial characters (ℤ/pℤ)^{m} → F^{∗}. Note that χ(g)^{p} = χ(g^{p}) = 1 for any character χ: (ℤ/pℤ)^{m} → F^{∗}, and thus

χ_{1}(g_{1})^{p} + ⋯ + χ(g_{p^{m}})^{p} = \underbrace{1 + ⋯ + 1}_{p^{m} times} = p^{m} = 1 in F.

(Recall that char(F) = 2 and p is odd.) Since one of the defining equations (6) for X is x_{1}^{p} + ⋯ + x_{n}^{p} = 0, we conclude that none of the 2p^{m} G-invariant 1-dimensional subspaces V_{χ}[i] is contained in X, and the claim follows. □

VII. Proof of Theorem 5. Let L_{0} be the 5-dimensional K-linear subspace of L given by tr(y) = 0. Let Y be the cubic threefold in ℙ_{K}^{4} = ℙ(L_{0}) given by σ_{3}(y) = 0 (or equivalently, tr(y^{3}) = 0). It is easy to see that Y is a cone, with vertex 1 ∈ L_{0}, over a cubic surface \overline{Y} in ℙ_{K}^{3} := ℙ(L_{0}/K), defined over K. Note that \overline{Y} is a K-form of the variety \overline{X} we considered in the proof of Theorem 4. Applying the Jacobian criterion to the defining equations (6) of \overline{X}, (with p = 3 and n = 6), we see that \overline{X} is a smooth surface, and hence, so is \overline{Y}. Either condition (a) or (b) implies that there exists a y ∈ L−K such that tr(y) = tr(y^{3}) = 0. Equivalently, \overline{Y} (K) ≠ ∅. It remains to show that we can choose a generator y ∈ L with tr(y) = tr(y^{3}) = 0 or equivalently, that \overline{Y} has a rational point away from of the “diagonal" hyperplanes x_{i} = x_{j} in ℙ^{3}, 1 ⩽ i < j ⩽ 6. (Note that the individual diagonal hyperplanes are defined over \overline{K}, but their union is defined over K.)

Suppose K is an infinite field. Since \overline{Y} (K) ≠ ∅, \overline{Y} is unirational; see [Ko]. Hence, K-points are dense in \overline{Y}, so that one (and in fact, infinitely many) of them lie away from the diagonal hyperplanes. Thus we may assume without loss of generality that K = 𝔽_{q} is a finite field of order q, where q is a power of 2, and L = 𝔽_{q^{6}}. (Note that in this case both conditions (a) and (b) are satisfied.) If y ∈ L is not a generator, it will lie in 𝔽_{q^{2}} or 𝔽_{q^{3}}. Clearly tr(y) ≠ 0 for any y ∈ 𝔽_{q^{2}}−𝔽_{q} and tr(y) = tr(y^{3}) = 0 for any y ∈ 𝔽_{q^{3}}. Thus a non-generator y ∈ L satisfies tr(y) = tr(y^{3}) = 0 if and only if y ∈ 𝔽_{q^{3}}. In geometric language, elements of 𝔽_{q^{3}} are the K-points of a line in \overline{Y}, defined over K = 𝔽_{q}. We will denote this line by Z. It suffices to show that \overline{Y} contains a K-point away from Z.

By [Ma, Corollary 27.1.1], |\overline{Y} (K)| ≥ q^{2} − 7q + 1. On the other hand, since Z ≃ ℙ^{1} over K, |Z(K)| = q + 1. Thus for q > 8, \overline{Y} has a K-point away from Z. In the remaining cases, where q = 2, 4 and 8, we will exhibit an explicit irreducible polynomial over 𝔽_{q} of the form (2):

t^{6} + t + 1 is irreducible over 𝔽_{2} see [Ch, p. 199],

t^{6} + t^{2} + t + α is irreducible, over 𝔽_{4}, where α ∈ 𝔽_{4}−F_{2}, and

t^{6} + t + β is irreducible over 𝔽_{8}, for some β ∈ 𝔽_{8}−𝔽_{2}; see [GT, Table 5]. □

VIII. Concluding remarks. (1) Theorem 1 extends a 1861 result of C. Hermite [He], which asserts that every separable extensions L/K of degree 5 has a generator y ∈ L with σ_{1}(y) = σ_{3}(y) = 0. Surprisingly, Hermite’s theorem is valid in any characteristic; see [Kr, Main Theorem (a)] or [Co, Theorem 2.2].

(2) It is natural to ask if results analogous to Theorem 1 are true for separable field extensions L/K of degree n, other than 5 and 6: does L/K always have a generator y ∈ L with σ_{1}(y) = σ_{3}(y) = 0? If n can be written in the form 3_{1}^{m} + 3^{m_{2}} for some integers m_{1} > m_{2} ⩾ 0, then the answer is “no" in any characteristic (other than 3); see [Re, Theorem 1.3(c)], [RY_{2}, Theorem 1.8]. Some partial results for other n can be found in [Co, §4].

(3) Using the Going Up Theorem for G-fixed points [RY_{1}, Proposition A.4], our proof of Theorem 4 can be modified, to yield the following stronger statement. Suppose K′/K_{n} is a finite field extension of degree prime to p. Set L′ := L_{n} ⊗_{K_{n}} K′. Then there is no y ∈ L′−K′ such that tr(y) = tr(y^{2}) = ⋯ = tr(y^{p}) = 0. In particular, under the assumptions of Theorem 3, there is no y ∈ L′−K′ with σ_{1}(y) = σ_{3}(y) = 0 for any finite field extension K′/K_{n} of degree prime to 3.

(4) Our argument shows that the G-action on \overline{X} is not versal in the sense of [Se_{2}, Section I.5] or [DR]. Otherwise \overline{X} would have a G-fixed point; see [DR, Remark 2.7]. Moreover, in view of Remark (3) above, the G-action on \overline{X} is not even p-versal. Since G ⊂ A_{n} ⊂ S_{n}, the same is true of the A_{n}- and S_{n}-actions on \overline{X}. This answers a question raised by J.-P. Serre in a letter to the author in 2005.

(5) Theorem 3 corrects an inaccuracy in the statement of Joubert’s theorem in [Re, Theorem 1.1], where the assumption that char(K) ≠ 2 was inadvertently left out.

(6) In the case where K = 𝔽_{q} is a finite field, Theorem 5 was communicated to the author by F. Voloch, along with an alternative proof, which is reproduced below with his permission.

“As in your comment after Theorem 4, it is enough to find y in 𝔽_{q^{6}}, not in a smaller field, with tr(y) = tr(y^{3}) = 0, where the trace is to 𝔽_{q}. These conditions are equivalent to the existence of x, z ∈ 𝔽_{q^{6}} with y = x^{q} − x, y^{3} = z^{q} − z, so z^{q} − z = (x^{q} − x)^{3}. Letting u = z + x^{3}, we get an affine plane curve u^{q} −u = x^{2q+1} +x^{q+2} over 𝔽_{q^{6}} (here q is a power of 2). It is a general fact that any affine plane curve of the form u^{q} − u = f(x), where f(x) polynomial of degree d prime to q, has genus (q − 1)(d − 1)/2, and its smooth projective model has exactly one point at infinity. In particular, our curve has genus q(q − 1). By the Weil bound the number of points on the smooth projective model of this curve is at least q^{6} + 1 − 2q(q − 1)q^{3}. There is one point at infinity and at most q^{5} points with y = x^{q} − x ∈ 𝔽_{q^{3}}; these are the bad points. If q > 2, our curve has a good point, one that gives rise to a generator of 𝔽_{q^{6}} over 𝔽_{q}, because q^{6}+1−2q(q−1)q^{3} > 1+q^{5} for any q > 2. For q = 2, I can exhibit an explicit ‘Joubert polynomial’, as in your formula (2). In fact, there are exactly two irreducible Joubert polynomials over 𝔽_{2}, t^{6} + t + 1 and t^{6} + t^{4} + t^{2} + t + 1."

Acknowledgement. The author is grateful to M. Florence, H. Kraft, D. Lorenzini, J.-P. Serre, F. Voloch, and the anonymous referee for helpful comments.

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

   E-mail address: reichst at math.ubc.ca, web page:  www.math.ubc.ca/˜ reichst