VARIETIES WITHOUT EXTRA AUTOMORPHISMS III: HYPERSURFACES

BJORN POONEN

Abstract. For any field k and integers n ≥ 1, d ≥ 3, with (n, d) not equal to (1, 3) or (2, 4), we exhibit a smooth hypersurface X over k of degree d in P^{n+1} such that X has no nontrivial automorphisms over \overline{k}. For (n, d) = (2, 4), we find a smooth hypersurface X with the weaker property of having no nontrivial automorphism induced by an automorphism of the ambient P^{n+1}.

1. Introduction

Let k be a field, and let p be its characteristic, which may be 0. Fix an algebraic closure \overline{k} of k. Let X in P^{n+1} be a smooth hypersurface of degree d over k. Let \overline{X} = X ×_{k} \overline{k}. Let Aut \overline{X} be the group of automorphisms of \overline{X} over \overline{k}. Call γ ∈ Aut \overline{X} linear with respect to the embedding X ↪ P^{n+1} if γ is induced by an automorphism of P^{n+1} over \overline{k}, i.e., by a linear transformation of the homogeneous coordinates. The linear automorphisms form a subgroup Lin \overline{X} of Aut \overline{X}.

We will study Lin \overline{X} primarily. Before stating our main result, Theorem 1.6, let us briefly survey known related results. First, it is known that for most (n, d), there is no difference between Aut \overline{X} and Lin \overline{X}:

Theorem 1.1. If X is a smooth hypersurface in P^{n+1} of degree d, where n ≥ 1, d ≥ 3, and (n, d) does not equal (1, 3) or (2, 4), then Aut \overline{X} = Lin \overline{X}.

Proof. The case n = 1 is Theorem 1 of [Cha78]. The case n ≥ 2 is Theorem 2 of [MM63]. □

Remark 1.2. The exclusion of (1, 3) and (2, 4) in Theorem 1.1 is necessary. When (n, d) = (1, 3), a choice of flex in X(\overline{k}) makes \overline{X} an elliptic curve, and if P ∈ X(\overline{k}) satisfies 3P ≠ 0, then translation by P is a nonlinear automorphism of \overline{X}. (See the proof of Theorem 1.3 below.) For (n, d) = (2, 4), the equality fails only for certain X; an example due to Fano and Severi is described in the proof of Theorem 4 in [MM63], for instance. What makes the proofs fail for (n, d) = (2, 4) is that the canonical bundle is trivial, and that Pic \overline{X} can be larger than Z. In fact, the Tate conjecture predicts that the latter is automatic for X over \overline{F}_{p} with (n, d) = (2, 4).

Theorem 1.3. If n ≥ 1 and d ≥ 3, then Lin \overline{X} is finite.

Proof. See the “Historical Remarks” section at the end of [OS77]. The result has apparently been known for at least one hundred years, at least when p = 0. Matsumura and Monsky [MM63] give a proof in arbitrary characteristic, at least when n ≥ 2. If n = 1 and d ≥ 4, then X is a curve of genus g = (d − 1)(d − 2)/2 ≥ 2, so Aut \overline{X} is finite [Sch38].

We are left with the easiest case, in which n = 1 and d = 3. Without loss of generality, k = \overline{k}. We can make X an elliptic curve by choosing a flex P as origin. The automorphism group Aut(X, P) of the elliptic curve is finite and of order dividing 24 [Sil92, Theorem III.10.1]. Also, Aut(X, P) ⊆ Lin X, since O_{X}(1) for the embedding X ↪ P^{2} is the line sheaf L(3P) on X. The orbit of P under Lin X is contained in the set of points P′ ∈ X(k) such that L(3P′) ≅ L(3P); this is the set of 3-torsion points of the elliptic curve (X, P), which is of size at most 9. Hence (Lin X : Aut(X, P)) ≤ 9, so #(Lin X) ≤ 216 (with equality if and only if X is supersingular and p = 2). □

Date: December 25, 2004.

2000 Mathematics Subject Classification. Primary 14J50; Secondary 14J70.

Key words and phrases. Hypersurface, automorphism, automorphism group scheme.

This research was initiated at Princeton University where the author was supported in part by an NSF Mathematical Sciences Postdoctoral Research Fellowship, and was continued at the University of California at Berkeley, supported in part by NSF grants DMS-9801104 and DMS-0301280, a Sloan Fellowship, and a Packard Fellowship.

This article is to appear in Finite fields and their applications.

Remark 1.4. Suppose that p = 0 and d ≥ 3. In unpublished work, Bott and Tate [BT61] used homological methods to show that there exists an upper bound for #(Lin \overline{X}) depending only on n and d. For n = 1 and d ≥ 4, one can use Hurwitz’s theorem that #(Aut \overline{X}) ≤ 84(g − 1) for any curve X of genus g. For n ≥ 2, Howard and Sommese [HS81] prove that there is a constant c_{n} depending only on n such that #(Lin \overline{X}) ≤ c_{n}d^{n}.

Let N = \binom{d+n+1}{d} be the number of monomials of degree d in variables x_{0}, …, x_{n+1}. Over any field k, smooth hypersurfaces of degree d in P^{n+1} correspond to the points of a dense open subset H_{n,d} of P^{N−1}, on which the homogeneous coordinates are the coefficients of the polynomial defining the hypersurface. For n ≥ 1, d ≥ 3, and (n, d) ≠ (1, 3), Katz and Sarnak [KS99, Lemma 11.8.5] show that there is an open subset U_{n,d} ⊂ H_{n,d} whose points correspond to the smooth hypersurfaces X with Lin \overline{X} = \{1\}.

Theorem 1.5. Suppose that n ≥ 1, d ≥ 3, and (n, d) ≠ (1, 3). Then U_{n,d} is nonempty. In other words, the generic hypersurface X of degree d in P^{n+1} has Lin \overline{X} = \{1\}.

Proof. Matsumura and Monsky [MM63] prove this for n ≥ 2, d ≥ 3, and their methods can be adapted to the case n = 1, d ≥ 4. A proof for n = 1, d ≥ 4 written out in full can be found in [Cha78] for p = 0, and in [KS99, 10.6.18] for arbitrary p using an alternative method. □

Combining Theorem 1.5 with the Lang-Weil method as in Corollary 11.8.7 of [KS99], one can show that for these (n, d), there exists N_{n,d} > 0 such that for any field k with #k > N_{n,d} (in particular, any infinite field), there exists a smooth hypersurface X of degree d in P^{n+1} over k with Lin \overline{X} = \{1\}. Our main result is that the same conclusion holds for all k:

Theorem 1.6. For any field k and integers n ≥ 1, d ≥ 3 with (n, d) ≠ (1, 3), there exists a smooth hypersurface X over k of degree d in P^{n+1} such that Lin \overline{X} = \{1\}.

Remark 1.7. The exclusion of (1, 3) is necessary. If (n, d) = (1, 3), then we may choose a flex to make X an elliptic curve, and then multiplication by −1 on the elliptic curve is a nontrivial linear automorphism.

Remark 1.8. There is a small overlap between Theorem 1.6 and the main result of [Poo00a], since a smooth hypersurface X of degree 4 in P^{2} with Lin \overline{X} = \{1\} is the same thing as a genus 3 curve X with Aut \overline{X} = \{1\}.

Our proof of Theorem 1.6 does not use Theorem 1.5, so it gives a new proof of Theorem 1.5. We can also combine Theorems 1.1 and 1.6 to obtain the following:

Corollary 1.9. For any field k and integers n ≥ 1, d ≥ 3 with (n, d) not equal to (1, 3) or (2, 4), there exists a smooth hypersurface X over k of degree d in P^{n+1} such that Aut \overline{X} = \{1\}.

Remark 1.10. Remark 1.7 shows that the exclusion of (1, 3) in Corollary 1.9 is necessary. But it may be that Corollary 1.9 holds for (n, d) = (2, 4).

Section 2 gives the definition of X for Theorem 1.6, which will depend on n, d, and p. Section 3 proves that X is smooth. Most of the rest of the paper is devoted to proving that Lin \overline{X} is trivial in the various cases. Finally, in Section 11, we mention a few consequences for the automorphism group scheme Aut \overline{X}.

2. Construction of X

The hypersurface X in Theorem 1.6 will be the subvariety of P^{n+1} defined by a homogeneous polynomial f(x_{0}, x_{1}, …, x_{n+1}). In order to help us control the automorphisms, we will choose an f that “endows the variables with an ordering.” As a first attempt, we could try

x_{0}x_{1}^{d−1} + x_{1}x_{2}^{d−1} + ⋯ + x_{n}x_{n+1}^{d−1},

but this fails for two reasons: first, the resulting hypersurface is singular at (1 : 0 : 0 : ⋯: 0); and second, it has nontrivial automorphisms if d − 1 is not a power of p, since one can multiply x_{n+1} by a nontrivial (d − 1)-th root of unity. In fact, if we choose any form with n + 1 or fewer monomials, there will be a nontrivial diagonal action of G_{m} on X in which λ ∈ \overline{k}^{∗} acts as

(x_{0} : x_{1} : ⋯: x_{n+1}) ↦ (λ^{a_{0}}x_{0} : λ^{a_{1}}x_{1} : ⋯: λ^{a_{n+1}}x_{n+1}),

for some integers a_{i} not all equal.

These problems can be fixed for most triples (n, d, p) by adding a few terms to the “ends” of f. In particular, we will show that adding cx_{0}^{d} and x_{n+1}^{d} to f will work when d ≢ 0, 1 (mod p), if we choose c ∈ k \ {0, c_{bad}}, where

c_{bad} := −d^{(1−d)^{n+1}−1}(1 − d)^{\frac{(1−d)^{n+2}−(1−d)}{d}} ∈ k^{∗},

except that we must also avoid c = 2^{4}3^{−9} if (n, d) = (2, 3). The hypersurface in Case I becomes singular for c = 0 or c = c_{bad}. If (n, d, c) = (2, 3, 2^{4}3^{−9}), then the resulting cubic surface in any characteristic not 2 or 3 has a nontrivial linear automorphism given by

[\begin{matrix}324 & 6561 & 1458 & 4374 \\ 16 & 324 & −72 & −216 \\ 0 & 0 & 648 & 0 \\ 48 & −972 & −216 & 0 \end{matrix}] ∈ P GL_{4}(k).

When d ≡ 0 (mod p), we need to add a term to rule out automorphisms mapping x_{0} ↦ x_{0} + λx_{1} and fixing all other x_{i}. (Actually such automorphisms create a problem only when d is a power of p.) When d ≡ 1 (mod p), we add a few terms in order that some of the second partial derivatives of f be nonvanishing, because our method for controlling the automorphisms relies on the fact that most, but not all, of the second partial derivatives of f vanish.

The definition of f in all cases is given in Table 1. The congruence conditions on d defining the cases are congruences modulo p. Note that in Cases I and II, we have p ≠ 2, and if (n, d) = (2, 3) in Case I, then p ≠ 3 also, so there is always at least one choice for c ∈ k. The reader who prefers to have c prescribed explicitly may take c = 2c_{bad} in Case I and c = 2(−2)^{d−2} in Case II.

3. Smoothness of X

This section proves that X is smooth in each case. This is not especially difficult. The hard part was finding the f for which this would be easy, and for which our methods for controlling the automorphisms would apply.

Case I: d ≢ 0, 1 (mod p)

Case  f  Conditions
I d ≢ 0, 1  cx_{0}^{d} + (∑_{i=0}^{n} x_{i}x_{i+1}^{d−1}) + x_{n+1}^{d}  \begin{matrix}c ≠ 0, c_{bad} \\ (n, d, c) ≠ (2, 3, 2^{4}3^{−9}) \end{matrix}

II \begin{matrix}d ≡ 0 \\ p ≠ 2 \end{matrix}  cx_{0}^{d} + x_{0}^{2}x_{1}^{d−2} + (∑_{i=0}^{n} x_{i}x_{i+1}^{d−1}) + x_{n+1}^{d}  c ≠ 0, (−2)^{d−2}

III \begin{matrix}d ≡ 0 \\ p = 2 \end{matrix}  x_{0}^{d−1}x_{1} + c(x_{1}^{d} + x_{2}^{d}) + (∑_{i=0}^{n} x_{i}x_{i+1}^{d−1}) + x_{n+1}^{d}  c = \begin{cases}0, if n = 1 \\ 1, if n ≥ 2 \end{cases}

IV \begin{matrix}d ≡ 1 \\ p ≠ 2 \end{matrix}  x_{0}^{d} + ( ∑_{i=0}^{⌊ \frac{n−1}{2} ⌋} x_{2i}^{2}x_{2i+1}^{d−2}) + (∑_{i=0}^{n} x_{i}x_{i+1}^{d−1}) + x_{n+1}x_{0}^{d−1}

V \begin{matrix}d ≡ 1 \\ p = 2 \\ n = 1 \end{matrix}  x_{0}x_{1}^{d−2}x_{2} + x_{0}x_{1}^{d−1} + x_{1}x_{2}^{d−1} + x_{2}x_{0}^{d−1} + x_{1}^{2}x_{2}^{d−2}

VI \begin{matrix}d ≡ 1 \\ p = 2 \\ n > 1 \end{matrix} x_{1}^{d} + ( ∑_{i=0}^{⌊ \frac{n−1}{3} ⌋} x_{3i}x_{3i+1}^{d−2}x_{3i+2}) + (∑_{i=1}^{n} x_{i}x_{i+1}^{d−1}) + x_{n+1}x_{0}^{d−1}
Table 1. Definition of f(x_{0}, x_{1}, …, x_{n+1}).

Suppose P is a singular point. At P the derivative ∂f/∂x_{i} must vanish for each i:

0 = cdx_{0}^{d−1} + x_{1}^{d−1}
0 = (d − 1)x_{0}x_{1}^{d−2} + x_{2}^{d−1}
(1)  0 = (d − 1)x_{1}x_{2}^{d−2} + x_{3}^{d−1}
⋮
0 = (d − 1)x_{n−1}x_{n}^{d−2} + x_{n+1}^{d−1}
0 = (d − 1)x_{n}x_{n+1}^{d−2} + dx_{n+1}^{d−1}.

Note that if 0 ≤ i ≤ n, and x_{i} = 0 at P, then x_{i+1} = 0 by the equation

0 = (d − 1)x_{i−1}x_{i}^{d−2} + x_{i+1}^{d−1}

(or the first equation, if i = 0), so that by induction x_{j} = 0 for all j ≥ i. On the other hand, if 2 ≤ i ≤ n + 1, and x_{i} = 0, we find from

0 = (d − 1)x_{i−2}x_{i−1}^{d−2} + x_{i}^{d−1}

that either x_{i−1} = 0 or x_{i−2} = 0, and the latter also implies x_{i−1} = 0 by what we just proved, so that x_{i−1} = 0 in any case. Also, if x_{1} = 0, then x_{0} = 0 by the first equation in (1).

Thus if any x_{i} is zero at P, all are zero at P. Hence if there is a singular point P, all its projective coordinates are nonzero. Without loss of generality, assume x_{n+1} = 1. Then from the last equation in (1) we find

x_{n} = d(1 − d)^{−1}.

Substituting into the penultimate equation in (1), we find

x_{n−1} = d^{2−d}(1 − d)^{d−3}.

Working our way up the list of equations, using all of them up to but not including the first, we prove by induction on i that

x_{n+1−i} = d^{\frac{1−(1−d)^{i}}{d}} (1 − d)^{\frac{1−(i+1)d−(1−d)^{i+1}}{d^{2}}}

for i = 1, 2, …, n + 1. The values of x_{0} and x_{1} so computed contradict the first equation in (1), provided that c ≠ c_{bad}.

Case II: d ≡ 0 (mod p), p ≠ 2

This time, the vanishing of the derivatives gives rise to the system

0 = 2x_{0}x_{1}^{d−2} + x_{1}^{d−1}
0 = −x_{0}x_{1}^{d−2} − 2x_{0}^{2}x_{1}^{d−3} + x_{2}^{d−1}
(2)  0 = −x_{1}x_{2}^{d−2} + x_{3}^{d−1}
⋮
0 = −x_{n−1}x_{n}^{d−2} + x_{n+1}^{d−1}
0 = −x_{n}x_{n+1}^{d−2}.

As in Case I, if 2 ≤ i ≤ n and x_{i} = 0, then x_{i+1} = 0 by the equation

0 = −x_{i−1}x_{i}^{d−2} + x_{i+1}^{d−1}.

On the other hand, if 4 ≤ i ≤ n + 1 and x_{i} = 0, then from

0 = −x_{i−2}x_{i−1}^{d−2} + x_{i}^{d−1}

we obtain x_{i−1} = 0 or x_{i−2} = 0, and the latter also implies x_{i−1} = 0 by what we just proved, so that x_{i−1} = 0 in any case.

From the last equation in (2), we obtain x_{n} = 0 or x_{n+1} = 0, so we immediately deduce x_{i} = 0 for 3 ≤ i ≤ n + 1. If x_{1} = 0, then we obtain x_{0} = 0 from the original equation f = 0, and x_{2} = 0 from the second equation in (2), which is a contradiction, as desired. Thus we may assume x_{1} = 1, and then the first and third equations in (2) yield x_{0} = −1/2 and x_{2} = 0. For (−\frac{1}{2} : 1 : 0 : 0 : ⋯: 0) to be a point on X we must have

c (−\frac{1}{2})^{d} + \frac{1}{4} − \frac{1}{2} = 0,

so we obtain the desired contradiction, provided that c ≠ (−2)^{d−2}.

Case III: d ≡ 0 (mod p), p = 2

The vanishing of the derivatives gives rise to the system

0 = x_{0}^{d−2}x_{1} + x_{1}^{d−1}
0 = x_{0}^{d−1} + x_{0}x_{1}^{d−2} + x_{2}^{d−1}
(3)  0 = x_{1}x_{2}^{d−2} + x_{3}^{d−1}

⋮

0 = x_{n−1}x_{n}^{d−2} + x_{n+1}^{d−1}
0 = x_{n}x_{n+1}^{d−2}.

We deduce as in Case II that x_{3} = x_{4} = ⋯ = x_{n+1} = 0. If x_{1} = 0, then we obtain x_{2} = 0 from the original equation f = 0, and x_{0} = 0 from the second equation in (3), so all x_{i} are zero, a contradiction. Thus we may assume x_{1} = 1, and the third and first equations in (3) yield x_{2} = 0 and x_{0}^{d−2} = 1. The original equation f = 0 becomes

x_{0} + 1(1 + 0) + x_{0} + 0 + 0 + ⋯ + 0 = 0,

a contradiction in characteristic 2.

Case IV: d ≡ 1 (mod p), p ≠ 2

The vanishing of the derivatives gives rise to the system

0 = x_{1}^{d−1} + 2x_{0}x_{1}^{d−2} + x_{0}^{d−1}
0 = x_{2}^{d−1} − x_{0}^{2}x_{1}^{d−3}
0 = x_{3}^{d−1} + 2x_{2}x_{3}^{d−2}
(4)  0 = x_{4}^{d−1} − x_{2}^{2}x_{3}^{d−3}
⋮
0 = x_{n+1}^{d−1}  (−x_{n−1}^{2}x_{n}^{d−3} if n is odd)
0 = x_{0}^{d−1}.

The last equation implies x_{0} = 0. The first then implies x_{1} = 0, and going down the list of equations we show by induction that x_{i} = 0 for all i. (Note that the conditions defining this case imply d ≥ 4, so the exponent d − 3 and anything larger will be positive.)

Case V: d ≡ 1 (mod p), p = 2, n = 1

The vanishing of the derivatives gives rise to the system

0 = x_{1}^{d−2}x_{2} + x_{1}^{d−1}
(5)  0 = x_{0}x_{1}^{d−3}x_{2} + x_{2}^{d−1}
0 = x_{0}x_{1}^{d−2} + x_{0}^{d−1} + x_{1}^{2}x_{2}^{d−3}.

Working from the bottom up, we find

x_{1} = 0 ⟹ x_{0} = 0 ⟹ x_{2} = 0 ⟹ x_{1} = 0.

Thus if any x_{i} is zero, all the x_{i} are zero, a contradiction. Hence all the x_{i} are nonzero. Then the first equation in (5) implies x_{1} = x_{2}. Substituting x_{2} = x_{1} in the second equation yields 0 = x_{0}x_{1}^{d−2} + x_{1}^{d−1}, so x_{0} = x_{1} = x_{2}. Substituting these into the third equation, we find x_{0}^{d−1} = 0, so x_{0} = 0, a contradiction.

Case VI: d ≡ 1 (mod p), p = 2, n > 1 The vanishing of the derivatives gives rise to the system

0 =  x_{1}^{d−2}x_{2}
0 = x_{2}^{d−1} + x_{0}x_{1}^{d−3}x_{2} + x_{1}^{d−1}
0 = x_{3}^{d−1} + x_{0}x_{1}^{d−2}
0 = x_{4}^{d−1} + x_{4}^{d−2}x_{5}
(6)  0 = x_{5}^{d−1} + x_{3}x_{4}^{d−3}x_{5}
0 = x_{6}^{d−1} + x_{3}x_{4}^{d−2}
⋮
0 = x_{0}^{d−1}  (+x_{n−1}x_{n}^{d−2} if n ≡ 1 (mod 3))

(To see the pattern, pretend that the exceptional term x_{1}^{d−1} in the second equation were actually in the first, and group the equations in threes.) Let m = 3⌊\frac{n+2}{3}⌋. There are zero, one, or two equations past the first m equations, and these final ones are those that are missing a “second term,” i.e., that are simply of the form 0 = x_{i}^{d−1} for some i.

By the first equation, either x_{1} or x_{2} is zero. If x_{2} = 0, then x_{1} = 0 by the second equation, so x_{1} = 0 in any case. The third equation then yields x_{3} = 0. For i = 3, 6, …, m − 3, the (i + 2)-th, (i + 1)-th, and (i + 3)-th equations show that

x_{i} = 0 ⟹ x_{i+2} = 0 ⟹ x_{i+1} = 0 ⟹ x_{i+3} = 0,

where we should interpret x_{n+2} as x_{0} if necessary. Thus we deduce x_{i} = 0 for 3 ≤ i ≤ m.

If m = n, we have x_{m+1} = 0 and x_{0} = 0 automatically from the last two equations in (6). If m = n + 1, we have x_{0} = 0 automatically from the last equation. If m = n + 2, we have already shown x_{0} = x_{n+2} = 0. Thus in every case we have x_{i} = 0 for all i except possibly i = 2. Finally, we obtain x_{2} = 0 from the second equation in (6).

4. Controlling the automorphisms: the idea

The remainder of the paper is devoted to proving that Lin \overline{X} is trivial in each case. In this section, we explain the main tool to be used, and introduce some notation.

Suppose we are in Case I. Then ∂f/∂x_{0} is killed by ∂/∂x_{i} for all i ≥ 2. If we have a linear automorphism of X given by the matrix L = (ℓ_{ij}) ∈ GL_{n+2}(\overline{k}), and if we set y_{i} = ∑_{j=0}^{n+1} ℓ_{ij}x_{j}, then

(7)  f(x_{0}, x_{1}, …, x_{n+1}) = αf(y_{0}, y_{1}, …, y_{n+1})

for some nonzero scalar α ∈ \overline{k}^{∗}, and

\frac{∂}{∂x_{0}} f(x_{0}, x_{1}, …, x_{n+1}) = α ∑_{i=0}^{n+1} ℓ_{i0} \frac{∂f(y_{0}, y_{1}, …, y_{n+1})}{∂y_{i}}

is killed by at least an (n − 2)-dimensional subspace of the span of the operators ∂/∂x_{j}, which is also the span of the ∂/∂y_{i}. Such considerations will severely constrain the possibilities for the entries of the matrix L.

In general, let A denote the Hessian matrix of f, with entries a_{ij} := \frac{∂^{2}f}{∂x_{i}∂x_{j}}. For (column) vectors

v = (v_{0}, v_{1}, …, v_{n+1}) and w = (w_{0}, w_{1}, …, w_{n+1}) in \overline{k}^{n+2}, we define a symmetric \overline{k}-linear pairing

⟨v, w⟩ := v^{t}Aw = ∑_{i=0}^{n+1} ∑_{j=0}^{n+1} v_{i}w_{j} \frac{∂^{2}f}{∂x_{i}∂x_{j}}

taking values in \overline{k}[x_{0}, x_{1}, …, x_{n+1}]. If L = (ℓ_{ij}) ∈ GL_{n+2}(\overline{k}) gives an automorphism of X, then ⟨v, w⟩ is a scalar multiple of the result of replacing each x_{i} by ∑_{j=0}^{n+1} ℓ_{ij}x_{j} in ⟨Lv, Lw⟩. In particular, and this is mainly what we will use,

⟨Lv, Lw⟩ = 0 ⟺ ⟨v, w⟩ = 0.

For a vector v ∈ \overline{k}^{n+2}, define a subspace

v^{⊥} := \{ w ∈ \overline{k}^{n+2} : ⟨v, w⟩ = 0 \}.

For any subspace V, let codim V denote the codimension of V as a subspace of \overline{k}^{n+2}. In the subsequent sections we will repeatedly use the following (trivial) observation.

Lemma 4.1. The number codim v^{⊥} equals the dimension of the \overline{k}-vector space spanned by the (polynomial) entries of the column vector Av.

Proof. Both numbers equal the dimension of the image of (Av)^{t}, considered as a linear function on \overline{k}^{n+2}. □

For a subspace V ⊆ \overline{k}^{n+2}, define a subspace

V^{⊥} := \{ w ∈ \overline{k}^{n+2} : ⟨v, w⟩ = 0 for all v ∈ V \}.

If L gives an automorphism of X, then for all vectors v and subspaces V,

(8)  (Lv)^{⊥} = L (v^{⊥})  and  (LV)^{⊥} = L (V^{⊥}) ,

so in particular

(9)  codim(Lv)^{⊥} = codim v^{⊥}  and  codim(LV)^{⊥} = dim V^{⊥}.

We let \{e_{0}, e_{1}, …, e_{n+1}\} denote the standard basis for \overline{k}^{n+2}.

For 0 ≤ m ≤ n + 1, define subspaces

S_{m} := ∑_{i=0}^{m} \overline{k} · e_{i}, T_{m} := ∑_{i=m}^{n+1} \overline{k} · e_{i}.

Also set S_{m} = 0 if m < 0, and T_{m} = 0 if m > n + 1.

Once we have taken full advantage of the fact that L respects the pairing, we can usually complete the proof that L is a scalar multiple of the identity simply by equating various coefficients in (7).

5. Controlling the automorphisms: Case I

In this case we have

A = [\begin{matrix}h_{0} & g_{1} & 0 & 0 & ⋯ & 0 & 0 \\ g_{1} & h_{1} & g_{2} & 0 & ⋯ & 0 & 0 \\ 0 & g_{2} & h_{2} & g_{3} & ⋯ & 0 & 0 \\ 0 & 0 & g_{3} & h_{3} & ⋯ & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ \\ 0 & 0 & 0 & 0 & ⋯ & h_{n} & g_{n+1} \\ 0 & 0 & 0 & 0 & ⋯ & g_{n+1} & h_{n+1} \end{matrix}]

where g_{i} := (d − 1)x_{i}^{d−2} and

h_{i} := \begin{cases}cd(d − 1)x_{0}^{d−2} & if i = 0 \\ (d − 1)(d − 2)x_{i−1}x_{i}^{d−3} & if 1 ≤ i ≤ n \\ d(d − 1)x_{n+1}^{d−2} + (d − 1)(d − 2)x_{n}x_{n+1}^{d−3} & if i = n + 1. \end{cases}

We will subdivide Case I as follows (recall that p ≠ 2 throughout this case):

• Case I.1: d ≢ 0, 1, 2 (mod p) and d ≠ 3

• Case I.2: d ≡ 2 (mod p) and p ≠ 2

• Case I.3: d = 3; p ≠ 2, 3; and n ≥ 2.

Case I.1: d ≢ 0, 1, 2 (mod p) and d ≠ 3

In this subcase, g_{1}, …, g_{n+1}, h_{0}, h_{1}, …, h_{n+1} are linearly independent over \overline{k}. In particular, note that e_{m}^{⊥} equals S_{m−2} + T_{m+2}, which is the \overline{k}-vector space spanned by all the e_{i} except e_{m−1}, e_{m}, and e_{m+1}.

Lemma 5.1. For any v = (v_{0}, v_{1}, …, v_{n+1}) ∈ \overline{k}^{n+2},

v^{⊥} = ⋂_{i:v_{i}≠0} e_{i}^{⊥}.

Proof. Suppose w = (w_{0}, w_{1}, …, w_{n+1}) ∈ \overline{k}^{n+2}. If the i-th coordinate of Aw is nonzero, then at least one of w_{i−1}, w_{i}, w_{i+1} is nonzero. If w_{i} is nonzero, then h_{i} occurs in the i-th coordinate of Aw and in no other coordinates. If w_{i} = 0 but w_{i−1} ≠ 0, then then g_{i} occurs in the i-th coordinate of Aw and in no other coordinates. If w_{i} = 0 but w_{i+1} ≠ 0, then then g_{i+1} occurs in the i-th coordinate of Aw and in no other coordinates. The nonzero coordinates of Aw are thus linearly independent over \overline{k}, since each involves a g or h not present in the other coordinates. In other words, the polynomials ⟨e_{i}, w⟩ for i = 0, 1, …, n + 1 that are nonzero are linearly independent. Thus if ⟨v, w⟩ = 0 and v_{i} ≠ 0, then ⟨e_{i}, w⟩ = 0. □

Lemma 5.1 and the remark preceding it let us immediately calculate v^{⊥} for any vector v, and also V^{⊥} for any subspace V, since V^{⊥} = ⋂_{v∈V} v^{⊥}. In particular, we obtain the following corollaries.

Corollary 5.2. If v ∈ \overline{k}^{n+2} is nonzero, then codim v^{⊥} ≥ 2, with equality if and only if v is a multiple of e_{0} or e_{n+1}.

Note that for 0 ≤ m ≤ n, the (m + 1)-dimensional subspace S_{m} ⊂ \overline{k}^{n+2} has S_{m}^{⊥} = T_{m+2}, and codim S_{m}^{⊥} = m + 2.

Corollary 5.3. Suppose 0 ≤ m ≤ n − 2. Let V be an (m + 2)-dimensional subspace of \overline{k}^{n+2} containing S_{m}. Then codim V^{⊥} ≥ m + 3, with equality if and only if V = S_{m+1}.

Proof. Write V = S_{m}+\overline{k}·v, so V^{⊥} = S_{m}^{⊥}∩v^{⊥}. If v has any nonzero coordinate v_{i} with m+2 ≤ i ≤ n, then the condition that an element w of S_{m}^{⊥} be in v^{⊥} places at least two linear conditions on w, namely w_{i} = 0 and w_{i+1} = 0, so codim V^{⊥} ≥ codim S_{m}^{⊥} + 2 = m + 4 in this case. Similarly, if v_{n+1} ≠ 0, then the condition that an element w of S_{m}^{⊥} be in v^{⊥} places the new conditions w_{n} = 0 and w_{n+1} = 0 on w, so that codim V^{⊥} ≥ m + 4 again. The only remaining possibility is that v_{i} = 0 for all i ≥ m + 2, in which case we must have V = S_{m+1} and codim V^{⊥} = codim T_{m+3} = m + 3. □

Corollary 5.4. Suppose 3 ≤ m ≤ n + 1. Let V be an (n − m + 3)-dimensional subspace of \overline{k}^{n+2} containing T_{m}. Then codim V^{⊥} ≥ n − m + 4, with equality if and only if V = T_{m−1}.

Proof. The proof is completely analogous to that of Corollary 5.3. □

Corollary 5.5. The vectors Le_{0} and Le_{n+1} are multiples of e_{0} and e_{n+1} in some order.

Proof. This follows from (9) and Corollary 5.2. □

We may now subdivide Case I.1 further into two subcases.

Case I.1.a: Le_{0} is a multiple of e_{0}

Corollary 5.3 gives a characterization of the flag S_{1} ⊂ S_{2} ⊂ ⋯ ⊂ S_{n−1} of vector spaces containing S_{0} that involves only dimensions and the ⊥-operation. Since L preserves S_{0} by assumption, we have L(S_{m}) = S_{m} for 0 ≤ m ≤ n − 1. Similarly by Corollary 5.4, L(T_{m}) = T_{m} for 2 ≤ m ≤ n + 1. Together, these imply that L is of the form^{1}

L = [\begin{matrix}∗ & ∗ & 0 & ⋯ & 0 & 0 & 0 \\ 0 & ∗ & 0 & ⋯ & 0 & 0 & 0 \\ 0 & 0 & ∗ & ⋯ & 0 & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ & ⋮ \\ 0 & 0 & 0 & ⋯ & ∗ & 0 & 0 \\ 0 & 0 & 0 & ⋯ & 0 & ∗ & 0 \\ 0 & 0 & 0 & ⋯ & 0 & ∗ & ∗ \end{matrix}] ,

with zeros off the diagonal except possibly in positions ℓ_{01} and ℓ_{n+1,n}. Since L is nonsingular, ℓ_{ii} ≠ 0 for all i, and by scaling L, we may assume ℓ_{n+1,n+1} = 1. By equating coefficients of x_{n+1}^{d} in (7), we see that α = 1. Equating coefficients of x_{n}^{d} and of x_{n}^{d−1}x_{n+1} in

f(x_{0}, x_{1}, …, x_{n}, x_{n+1}) = f(ℓ_{00}x_{0} + ℓ_{01}x_{1}, ℓ_{11}x_{1}, …, ℓ_{nn}x_{n}, ℓ_{n+1,n}x_{n} + x_{n+1}),

we obtain

0 = ℓ_{nn}ℓ_{n+1,n}^{d−1} + ℓ_{n+1,n}^{d}
0 = (d − 1)ℓ_{nn}ℓ_{n+1,n}^{d−2} + dℓ_{n+1,n}^{d−1}.

Multiply the first by (d − 1) and the second by ℓ_{n+1,n}, and subtract to deduce ℓ_{n+1,n} = 0. For i = n, n − 1, …, 1 in turn, we equate coefficients of x_{i}x_{i+1}^{d−1} to find ℓ_{ii} = 1. Equate coefficients of x_{0}^{d−1}x_{1} and use ℓ_{00} ≠ 0 and d ≢ 0 (mod p) to deduce ℓ_{01} = 0. Finally equate coefficients of x_{0}x_{1}^{d−1} to deduce ℓ_{00} = 1. Thus L is the identity, as desired.

Case I.1.b: Le_{0} is a multiple of e_{n+1}

This time Corollaries 5.3 and 5.4 imply that L(S_{m}) = T_{n+1−m} for 0 ≤ m ≤ n − 1 and L(T_{m}) = S_{n+1−m} for 2 ≤ m ≤ n + 1, so that L is of the form

L = [\begin{matrix}0 & 0 & 0 & ⋯ & 0 & ∗ & ∗ \\ 0 & 0 & 0 & ⋯ & 0 & ∗ & 0 \\ 0 & 0 & 0 & ⋯ & ∗ & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ & ⋮ \\ 0 & 0 & ∗ & ⋯ & 0 & 0 & 0 \\ 0 & ∗ & 0 & ⋯ & 0 & 0 & 0 \\ ∗ & ∗ & 0 & ⋯ & 0 & 0 & 0 \end{matrix}] ,

with nonzero entries on the reverse diagonal, and zero entries off it, except possibly at ℓ_{0n} and ℓ_{n+1,1}. We may assume ℓ_{0,n+1} = 1. Equating coefficients of x_{n}^{d} and of x_{n}^{d−1}x_{n+1} in

f(x_{0}, x_{1}, …, x_{n}, x_{n+1}) = f(ℓ_{0n}x_{n} + x_{n+1}, ℓ_{1n}x_{n}, …, ℓ_{n1}x_{1}, ℓ_{n+1,0}x_{0} + ℓ_{n+1,1}x_{1})

1Each asterisk in a matrix denotes an element of \overline{k} which may or may not be zero.

we find

0 = cℓ_{0n}^{d} + ℓ_{0n}ℓ_{1n}^{d−1}
0 = cdℓ_{0n}^{d−1} + ℓ_{1n}^{d−1}.

Subtracting ℓ_{0n} times the second from the first, we find c(1 − d)ℓ_{0n}^{d} = 0, so ℓ_{0n} = 0. Substituting back into the second, we find ℓ_{1n} = 0 as well. But this contradicts the nonsingularity of L.

Case I.2: d ≡ 2 (mod p) and p ≠ 2

We have

A = [\begin{matrix}2cg_{0} & g_{1} & 0 & 0 & ⋯ & 0 & 0 & 0 \\ g_{1} & 0 & g_{2} & 0 & ⋯ & 0 & 0 & 0 \\ 0 & g_{2} & 0 & g_{3} & ⋯ & 0 & 0 & 0 \\ 0 & 0 & g_{3} & 0 & ⋯ & 0 & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ & ⋮ \\ 0 & 0 & 0 & 0 & ⋯ & 0 & g_{n} & 0 \\ 0 & 0 & 0 & 0 & ⋯ & g_{n} & 0 & g_{n+1} \\ 0 & 0 & 0 & 0 & ⋯ & 0 & g_{n+1} & 2g_{n+1} \end{matrix}] ,

and g_{0}, g_{1}, …, g_{n+1} are linearly independent over \overline{k}.

Lemma 5.6. If v ∈ \overline{k}^{n+2} is nonzero, then codim v^{⊥} ≥ 1, with equality if and only if v is a multiple of e_{n+1}.

Proof. By Lemma 4.1, codim e_{n+1}^{⊥} = 1, and the same is true for any multiple of e_{n+1}.

Now assume instead that the first nonzero coordinate v_{i} in v occurs for i ≤ n. Then g_{i} appears exactly once in the coordinates of Av, namely in the (i−1)-th coordinate (or in the 0-th coordinate if i = 0). If i < n, then g_{i+1} appears in the (i + 1)-th coordinate of Av, since a_{i+1,i} is independent of the other entries of its row in A, so the span of the coordinates of Av has dimension at least 2. If i = n, then g_{n+1} appears in either the n-th or the (n + 1)-th coordinate of Av, so again the span of the coordinates of Av has dimension at least 2. Thus codim v^{⊥} ≥ 2 by Lemma 4.1. □

Corollary 5.7. We have L(T_{n+1}) = T_{n+1}.

The (n − m + 2)-dimensional space T_{m} satisfies T_{m}^{⊥} = S_{m−2} if 2 ≤ m ≤ n.

Lemma 5.8. For nonzero v ∈ \overline{k}^{n+2}, we have codim v^{⊥} ≤ 2 if and only if v is a multiple of some e_{i} or is a linear combination of e_{n} and e_{n+1}.

Proof. The “if” direction is clear from Lemma 4.1. Now suppose codim v^{⊥} ≤ 2 and that the first nonzero v_{i} in v occurs for i < n. We must show that v is a multiple of e_{i}. As in the proof of Lemma 5.6, g_{i} appears exactly once in the coordinates of Av, namely in the (i − 1)-th coordinate (or the 0-th coordinate if i = 0), and g_{i+1} appears in the (i + 1)-th coordinate, so at least these two coordinates are linearly independent. Suppose for sake of contradiction that v_{j} ≠ 0 for some j > i, and choose the largest such j. If j ≤ n, then g_{j+1} appears in the (j + 1)-th coordinate of Av but not before, so it is independent of the (i − 1)-th and (i + 1)-th coordinates, and the span is of dimension at least 3, as desired. If j = n + 1, then g_{n+1} appears in the n-th coordinate of Av and not before, so we are again done, unless i + 1 = n.

To handle the remaining case i = n − 1, j = n + 1 we break into cases according as v_{n} = 0 or not. If v_{n} = 0, then g_{n−1} appears only in the (n − 2)-th coordinate of Av, g_{n} appears only in the n-th coordinate of Av, and g_{n+1} appears in the (n + 1)-th coordinate of Av, so these three coordinates are independent. If v_{n} ≠ 0, then g_{n−1} appears only in the (n − 2)-th coordinate of Av, a pure multiple of g_{n} occurs in the (n − 1)-th coordinate, and a non-pure combination of g_{n} and g_{n+1} occurs in the n-th coordinate, so again the span of the coordinates is at least 3-dimensional. Hence codim v^{⊥} ≥ 3. □

Corollary 5.9. We have L(T_{n}) = T_{n}.

Proof. Lemma 5.8 shows that T_{n} is the only 2-dimensional subspace consisting entirely of vectors v for which codim v^{⊥} ≤ 2. □

Lemma 5.10. For 0 ≤ i ≤ n − 1, Le_{i} is a multiple of e_{i}.

Proof. By Lemma 5.8 and Corollary 5.9, L acts on e_{0}, e_{1}, …, e_{n−1} by scaling them independently and then permuting them. Equating coefficients of x_{0}^{d} in (7) we see that Le_{0} must be a multiple of e_{0}. By induction on i for 1 ≤ i ≤ n − 1, equating coefficients of x_{i−1}x_{i}^{d−1} shows that Le_{i} must be a multiple of e_{i}. □

Lemma 5.10 and Corollaries 5.7 and 5.9 together imply that L is of the form

L = [\begin{matrix}∗ & 0 & 0 & ⋯ & 0 & 0 & 0 \\ 0 & ∗ & 0 & ⋯ & 0 & 0 & 0 \\ 0 & 0 & ∗ & ⋯ & 0 & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ & ⋮ \\ 0 & 0 & 0 & ⋯ & ∗ & 0 & 0 \\ 0 & 0 & 0 & ⋯ & 0 & ∗ & 0 \\ 0 & 0 & 0 & ⋯ & 0 & ∗ & ∗ \end{matrix}] .

The argument at the end of Case I.1.a now implies that L is (a scalar multiple of) the identity.

Case I.3: d = 3; p ≠ 2, 3; and n ≥ 2

We have

A = 2 [\begin{matrix}3cx_{0} & x_{1} & 0 & 0 & ⋯ & 0 & 0 & 0 & 0 \\ x_{1} & x_{0} & x_{2} & 0 & ⋯ & 0 & 0 & 0 & 0 \\ 0 & x_{2} & x_{1} & x_{3} & ⋯ & 0 & 0 & 0 & 0 \\ 0 & 0 & x_{3} & x_{2} & ⋯ & 0 & 0 & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ & ⋮ & ⋮ \\ 0 & 0 & 0 & 0 & ⋯ & x_{n−3} & x_{n−1} & 0 & 0 \\ 0 & 0 & 0 & 0 & ⋯ & x_{n−1} & x_{n−2} & x_{n} & 0 \\ 0 & 0 & 0 & 0 & ⋯ & 0 & x_{n} & x_{n−1} & x_{n+1} \\ 0 & 0 & 0 & 0 & ⋯ & 0 & 0 & x_{n+1} & x_{n} + 3x_{n+1} \end{matrix}] .

Lemma 5.11. If n > 2, then codim v^{⊥} ≤ 2 if and only if v is a multiple of one of the following:

e_{0},  e_{0} ± \sqrt{3c}e_{1},  3e_{n} − e_{n+1},  e_{n+1}.

If n = 2, the same is true, except that multiples of e_{0} ± \sqrt{3c}(e_{1} + 3e_{3}) are also possible.

Proof. We may assume v ≠ 0. Let j be the largest integer such that v_{j} is nonzero.

If j = 0, then v is a multiple of e_{0}, and codim v^{⊥} = 2.

If j = 1, then to have codim v^{⊥} ≤ 2, we must have v_{0} ≠ 0, since codim e_{1}^{⊥} = 3. Assume v_{0} = 1. Then

Av = 2(3cx_{0} + v_{1}x_{1}, v_{1}x_{0} + x_{1}, v_{1}x_{2}, 0, 0, …, 0).

In order for the span of the coordinates to have dimension at most 2, the first two coordinates must be dependent. By looking at the coefficients of x_{1}, we see that this would imply

3cx_{0} + v_{1}x_{1} = v_{1}(v_{1}x_{0} + x_{1}),

which holds if and only if v_{1} = ±\sqrt{3c}.

If 2 ≤ j ≤ n, then x_{j+1} appears in the (j + 1)-th coordinate of Av but not before, and the j-th coordinate is a nonzero combination of x_{j−1} and x_{j}, so if codim v^{⊥} ≤ 2, then the 0-th, 1-st, …, (j − 1)-th coordinates of Av are all multiples of the j-th coordinate. In particular, x_{j−2} does not appear in the (j − 1)-th coordinate of Av, so v_{j−1} = 0. Thus the j-th coordinate of Av is a multiple of x_{j−1}. But the (j − 1)-th coordinate involves x_{j}, so it cannot be a multiple of the j-th coordinate, a contradiction.

Finally we have the case j = n + 1. Suppose codim v^{⊥} ≤ 2. If v_{n} ≠ 0, then x_{n−1} appears in the n-th coordinate of Av and not afterwards, and the (n + 1)-th coordinate is nonzero, so these coordinates already span a 2-dimensional space, and all others must be dependent on them. In this case, all coordinates must be combinations of x_{n−1}, x_{n}, and x_{n+1} only. If furthermore 0 ≤ i < n and v_{i} ≠ 0, we get a contradiction by observing that x_{i−1} (x_{0} if i = 0) appears in the i-th coordinate of Av. Thus, from our assumption v_{n} ≠ 0 we deduce that v is a combination of e_{n} and e_{n+1} in which both appear. The (n − 1)-th coordinate of Av is a nonzero multiple of x_{n}, and this can be in the span of the n-th and (n + 1)-th coordinates only if the (n + 1)-th coordinate also is a multiple of x_{n}, which happens if and only if v is a multiple of 3e_{n} − e_{n+1}.

Thus from now on, we may assume j = n + 1 and v_{n} = 0. If v_{n−1} ≠ v_{n+1}/3, then the last two coordinates of Av are independent, so in order to have codim v^{⊥} ≤ 2, all other coordinates must be combinations of these last two. In particular, they would all be combinations of x_{n} and x_{n+1} only. For 0 ≤ i ≤ n − 1, the non-appearance of x_{i−1} (of x_{0} if i = 0) in the i-th coordinate of Av then forces v_{i} = 0, so that v is a multiple of e_{n+1}, and in this case codim v^{⊥} = 2.

Finally we have the case v_{n} = 0, v_{n−1} = v_{n+1}/3 ≠ 0. The (n − 1)-th coordinate of Av is a combination of x_{n−1} and x_{n−2} in which the latter appears, and the (n + 1)-th coordinate is a multiple of x_{n} + 3x_{n+1}. These already span a 2-dimensional space, so if codim v^{⊥} ≤ 2, all other coordinates must be combinations of x_{n−2}, x_{n−1}, x_{n}, and x_{n+1}. Suppose that n > 2. Then for 0 ≤ i ≤ n − 2, the non-appearance of x_{i−1} (of x_{0} if i = 0) in the i-th coordinate of Av forces v_{i} = 0. The (n − 2)-th, (n − 1)-th, and (n + 1)-th coordinates of Av are now nonzero multiples of x_{n−1}, x_{n−2}, and x_{n} + 3x_{n+1}, respectively, so there are independent, and codim v^{⊥} ≥ 3.

We are left with the case n = 2, v_{2} = 0, v_{1} = v_{3}/3. If v_{1} ≠ ±\sqrt{3c}v_{0}, then the 0-th and 1-st coordinates of Av are independent, and neither involves x_{3}, so the last coordinate is independent of both of them, yielding codim v^{⊥} ≥ 3. Otherwise, if v_{1} = ±\sqrt{3c}v_{0}, then v is a nonzero multiple of e_{0} ± \sqrt{3c}(e_{1} + 3e_{3}), and we check that in this case codim v^{⊥} = 2. □

We next subdivide Case I.3 according as n = 2 or n > 2.

Case I.3.a: n > 2

Corollary 5.12. We have L(S_{1}) = S_{1} and L(T_{n}) = T_{n}.

Proof. By (8), L must permute the five lines generated by the vectors listed in Lemma 5.11. The only 2-dimensional subspace of \overline{k}^{n+2} containing three of these five lines in S_{1}, so L(S_{1}) = S_{1}. The subspace spanned by the other two lines is T_{n}, so L(T_{n}) = T_{n}. □

Lemma 5.13. The vectors Le_{n} and Le_{n+1} are nonzero multiples of e_{n} and e_{n+1}, respectively.

Proof. By Corollary 5.12, we know L(T_{n}) = T_{n}. Hence y_{0}, y_{1}, …, y_{n−1} are linear combinations of x_{0}, x_{1}, …, x_{n−1} only.

Substituting x_{0} = x_{1} = ⋯ = x_{n−1} = 0 in (7), we find

(10)  (x_{n} + x_{n+1})x_{n+1}^{2} = α(z_{n} + z_{n+1})z_{n+1}^{2},

where z_{i} denotes the part of the linear form y_{i} involving x_{n} and x_{n+1}. By unique factorization, this implies that z_{n+1} is a nonzero scalar multiple of x_{n+1}. Without loss of generality, we may assume z_{n+1} = x_{n+1}; i.e. ℓ_{n+1,n+1} = 1. Equating coefficients of x_{n+1}^{3} in (10), we obtain α = 1. Now (10) implies z_{n} = x_{n}. This gives the desired result. □

Corollary 5.14. We have L(S_{n−2}) = S_{n−2} and L(S_{n−1}) = S_{n−1}.

Proof. This follows from Lemma 5.13, since e_{n}^{⊥} = S_{n−2} and e_{n+1}^{⊥} = S_{n−1}. □

Lemma 5.15. For 1 ≤ m ≤ n + 1, T_{m}^{⊥} = S_{m−2}.

Proof. We use backwards induction on m. Clearly T_{n+1}^{⊥} = S_{n−1}. For 1 ≤ m ≤ n,

T_{m}^{⊥} = T_{m+1}^{⊥} ∩ e_{m}^{⊥} = S_{m−1} ∩ (S_{m−2} + T_{m+2}) = S_{m−2}.
□

Lemma 5.16. For 2 ≤ m ≤ n + 1, Le_{m} is a multiple of e_{m}.

Proof. We know it already for m = n + 1 and m = n. We use backwards induction on m. Suppose 2 ≤ m ≤ n − 1, and that Le_{m′} is a multiple of e_{m′} for m′ > m. Then T_{m+1} and T_{m+2} are each preserved by L, and so are S_{m−1} = T_{m+1}^{⊥} and S_{m} = T_{m+2}^{⊥} by Lemma 5.15. Hence if v = Le_{m}, then v ∈ S_{m}, since e_{m} ∈ S_{m}. Also v ∉ S_{m−1}, since otherwise L(S_{m}) ⊂ S_{m−1}, and L would not be invertible. Moreover v^{⊥} ∩ S_{m−1} has codimension 1 in S_{m−1}, since e_{m}^{⊥} ∩ S_{m−1} has codimension 1 in S_{m−1}. In other words, the span of the 0-th, 1-st, …, (m−1)-th coordinates of Av is 1-dimensional. But x_{m} appears in the (m − 1)-th coordinate of Av (since v ∈ S_{m} \ S_{m−1}), and not before, so the 0-th, 1-st, …, (m − 2)-th coordinates must all be zero. This forces v_{0} = v_{1} = ⋯ = v_{m−1} = 0, so v = Le_{m} is a multiple of e_{m}. □

We have S_{0} = S_{1} ∩ e_{2}^{⊥}, so S_{0} also is fixed by L. Putting this together with Corollary 5.12 and Lemma 5.16, we see that L is of the form

L = [\begin{matrix}∗ & ∗ & 0 & ⋯ & 0 & 0 & 0 \\ 0 & ∗ & 0 & ⋯ & 0 & 0 & 0 \\ 0 & 0 & ∗ & ⋯ & 0 & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ & ⋮ \\ 0 & 0 & 0 & ⋯ & ∗ & 0 & 0 \\ 0 & 0 & 0 & ⋯ & 0 & ∗ & 0 \\ 0 & 0 & 0 & ⋯ & 0 & 0 & ∗ \end{matrix}] .

The argument at the end of Case I.1.a now implies that L is (a scalar multiple of) the identity.

Case I.3.b: n = 2

The form defining X is

f := cx_{0}^{3} + x_{0}x_{1}^{2} + x_{1}x_{2}^{2} + x_{2}x_{3}^{2} + x_{3}^{3}.

For future convenience, we will make the change of coordinates

(x_{0}, x_{1}, x_{2}, x_{3}) ↦ (x_{0}, x_{1}, x_{3}, −x_{2} − x_{3}/3),

and for the rest of Case I.3.b, we will work with the new f, which is

f := cx_{0}^{3} + x_{0}x_{1}^{2} − x_{2}^{3} + (\frac{3x_{1} + x_{2}}{3} ) x_{3}^{2} + \frac{2x_{3}^{3}}{27} .

The new A is

A = 2 [\begin{matrix}3cx_{0} & x_{1} & 0 & 0 \\ x_{1} & x_{0} & 0 & x_{3} \\ 0 & 0 & −3x_{2} & \frac{1}{3}x_{3} \\ 0 & x_{3} & \frac{1}{3}x_{3} & x_{1} + \frac{1}{3}x_{2} + \frac{2}{9}x_{3} \end{matrix}] .

The set W of vectors v ∈ \overline{k}^{4} such that codim v^{⊥} ≤ 2 is the union of seven lines, the transform of those generated by the seven vectors in Lemma 5.11. They are the lines E_{1}, E_{2}, …, E_{7} generated by e_{0}, e_{0} + \sqrt{3c}e_{1}, e_{0} − \sqrt{3c}e_{1}, e_{3}, e_{2}, e_{0} + \sqrt{3c}(e_{1} − 3e_{2}), and e_{0} − \sqrt{3c}(e_{1} − 3e_{2}), respectively. By (8), L must permute the E_{i}.

There are four 2-dimensional subspaces of \overline{k}^{4} containing exactly three of these lines, namely

W_{1} := S_{1} ⊃ E_{1}, E_{2}, E_{3}
W_{2} := \overline{k}e_{0} + \overline{k}(e_{1} − 3e_{2}) ⊃ E_{1}, E_{6}, E_{7}
W_{3} := \overline{k}(e_{0} + \sqrt{3c}e_{1}) + \overline{k}e_{2} ⊃ E_{2}, E_{5}, E_{6}
W_{4} := \overline{k}(e_{0} − \sqrt{3c}e_{1}) + \overline{k}e_{2} ⊃ E_{3}, E_{5}, E_{7}.

The only E_{i} not contained in any W_{j} is E_{4}, so L(E_{4}) = E_{4}. The span of the other six E_{i} is S_{2}, so L(S_{2}) = S_{2}.

We now know that L has the form

L = [\begin{matrix}∗ & ∗ & ∗ & 0 \\ ∗ & ∗ & ∗ & 0 \\ ∗ & ∗ & ∗ & 0 \\ 0 & 0 & 0 & ∗ \end{matrix}] .

Without loss of generality we may assume ℓ_{33} = 1. Equating coefficients of x_{3}^{3} in (7), we find α = 1. By viewing both sides of (7) as polynomials in x_{3}, we see that the forms cx_{0}^{3} + x_{0}x_{1}^{2} − x_{2}^{3} and 3x_{1} + x_{2} are each preserved by L.

If γ ∈ \overline{k}^{∗} and the plane cubic

(11)  (cx_{0}^{3} + x_{0}x_{1}^{2} − x_{2}^{3}) + γ (3x_{1} + x_{2})^{3} = 0

has a unique singularity^{2}, then that singularity is preserved by the automorphism induced by L. A short calculation shows that the singularities on these curves are at the points (x_{0} : x_{1} : x_{2}) =

(−9s^{2} : 2 : 2s), where s ∈ \overline{k} \ \{0, −3\} satisfies c = −\frac{4}{243s^{4}} and γ = (\frac{s}{3+s})^{2}.

For c ≠ −2^{2}3^{−9}, 2^{4}3^{−9}, we find that there are four possibilities for s, giving rise to four distinct values of γ for which the curve has a unique singularity. The four distinct points so obtained are in general position in P^{2}, since they lie on the conic 2x_{0}x_{1} + 9x_{2}^{2} = 0. Hence an automorphism of P^{2} that fixes them is trivial. Together with the fact that L preserves the form 3x_{1} + x_{2} (and not just up to scalar multiple), this implies that the upper left 3 × 3 block of L is the identity, so L is the identity.

If c = −2^{2}3^{−9}, then we may assume p ≠ 5 in addition to p ≠ 2, 3, since c_{bad} = −2^{6}3^{−9} coincides with this c in characteristic 5. We dehomogenize the cubic

−2^{2}3^{−9}x_{0}^{3} + x_{0}x_{1}^{2} − x_{2}^{3} = 0

by setting x = x_{2}/x_{0} and y = x_{1}/x_{0}, to obtain the elliptic curve in Weierstrass form

E : y^{2} = x^{3} + 2^{2}3^{−9}.

2There is automatically at most one singularity if the cubic is irreducible.

(As usual, we choose the point at infinity as origin O on E, to make E an algebraic group.) Then L induces an automorphism σ of P^{2} preserving E. The automorphism σ also preserves the line 3y + x = 0, which is tangent to E at P := (2/27, −2/81) and meets E again at [−2]P = (−1/27, 1/81). Hence σ(P) = P. The action of σ on E is the composition of an automorphism η of E as an elliptic curve (i.e. fixing O), and a translation on E. Since σ preserves the class of a line section, which is the class of the divisor 3 · O, the translation must be a translation by a 3-torsion point T. It follows that η fixes [3]P = (−2/81, 10/729). The six automorphisms of E have the form (x, y) ↦ (±x, ωy), where ω^{3} = 1, but x([3]P) and y([3]P) are finite and nonzero in k, so η must be the identity. Since σ(P) = P, it then follows that T = O. Thus σ fixes E pointwise, and hence is the identity. Since L does not scale 3x_{1} + x_{2}, this implies that L is the identity.

The last case c = 2^{4}3^{−9} (in which two of the four s-values, namely −3/2 + 3i/2 and −3/2 − 3i/2 give rise to the same γ) was ruled out by assumption at the very beginning, so we are done with Case I.3.b, and indeed we are done with all of Case I.

6. Controlling the automorphisms: Case II

We will subdivide Case II as follows (recall that p ≠ 2 throughout this case):

• Case II.1: d ≡ 0 (mod p); d ≠ 3; and p ≠ 2, 3

• Case II.2: d ≡ 0 (mod p); d ≠ 3; and p = 3

• Case II.3: d = 3, p = 3, and n ≥ 2.

Case II.1: d ≡ 0 (mod p); d ≠ 3; and p ≠ 2, 3

We have

A = [\begin{matrix}h_{0} & g_{1} & 0 & 0 & ⋯ & 0 & 0 \\ g_{1} & h_{1} & g_{2} & 0 & ⋯ & 0 & 0 \\ 0 & g_{2} & h_{2} & g_{3} & ⋯ & 0 & 0 \\ 0 & 0 & g_{3} & h_{3} & ⋯ & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ \\ 0 & 0 & 0 & 0 & ⋯ & h_{n} & g_{n+1} \\ 0 & 0 & 0 & 0 & ⋯ & g_{n+1} & h_{n+1} \end{matrix}]

where

g_{i} := \begin{cases}−x_{1}^{d−2} − 4x_{0}x_{1}^{d−3} & if i = 1 \\ −x_{i}^{d−2} & if 2 ≤ i ≤ n + 1 \end{cases}

and

h_{i} := \begin{cases}2x_{1}^{d−2} & if i = 0 \\ 2x_{0}x_{1}^{d−3} + 6x_{0}^{2}x_{1}^{d−4} & if i = 1 \\ 2x_{i−1}x_{i}^{d−3} & if 2 ≤ i ≤ n + 1. \end{cases}

The polynomials g_{1}, …, g_{n+1}, h_{0}, …, h_{n+1} are linearly independent, and hence the same proof as in Case I.1 shows that L must be of the form

L = [\begin{matrix}∗ & ∗ & 0 & ⋯ & 0 & 0 & 0 \\ 0 & ∗ & 0 & ⋯ & 0 & 0 & 0 \\ 0 & 0 & ∗ & ⋯ & 0 & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ & ⋮ \\ 0 & 0 & 0 & ⋯ & ∗ & 0 & 0 \\ 0 & 0 & 0 & ⋯ & 0 & ∗ & 0 \\ 0 & 0 & 0 & ⋯ & 0 & ∗ & ∗ \end{matrix}] ,

or of the form

L = [\begin{matrix}0 & 0 & 0 & ⋯ & 0 & ∗ & ∗ \\ 0 & 0 & 0 & ⋯ & 0 & ∗ & 0 \\ 0 & 0 & 0 & ⋯ & ∗ & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ & ⋮ \\ 0 & 0 & ∗ & ⋯ & 0 & 0 & 0 \\ 0 & ∗ & 0 & ⋯ & 0 & 0 & 0 \\ ∗ & ∗ & 0 & ⋯ & 0 & 0 & 0 \end{matrix}] .

The second case is easily ruled out, since equating coefficients of x_{n}^{d−2}x_{n+1}^{2} in

f(x_{0}, x_{1}, …, x_{n}, x_{n+1}) = f(ℓ_{0n}x_{n} + ℓ_{0,n+1}x_{n+1}, ℓ_{1n}x_{n}, …, ℓ_{n1}x_{1}, ℓ_{n+1,0}x_{0} + ℓ_{n+1,1}x_{1})

yields 0 = ℓ_{1n}^{d−2}ℓ_{0,n+1}^{2}, which contradicts the nonsingularity of L.

In the first case, since L is nonsingular, ℓ_{ii} ≠ 0 for all i. By scaling L, we may assume ℓ_{n+1,n+1} = 1. By equating coefficients of x_{n+1}^{d} in (7), we see that α = 1. Equating coefficients of x_{n}^{d−1}x_{n+1} in

f(x_{0}, x_{1}, …, x_{n}, x_{n+1}) = f(ℓ_{00}x_{0} + ℓ_{01}x_{1}, ℓ_{11}x_{1}, …, ℓ_{nn}x_{n}, ℓ_{n+1,n}x_{n} + x_{n+1}),

we obtain

0 = (d − 1)ℓ_{nn}ℓ_{n+1,n}^{d−2}.

Since d − 1 and ℓ_{nn} are nonzero in \overline{k}, we have ℓ_{n+1,n} = 0. For i = n, n − 1, …, 1 in turn, we equate coefficients of x_{i}x_{i+1}^{d−1} to find ℓ_{ii} = 1. Equate coefficients of x_{0}^{2}x_{1}^{d−2}, of x_{0}x_{1}^{d−1} and of x_{1}^{d} to obtain

ℓ_{00}^{2} = 1
2ℓ_{00}ℓ_{01} + ℓ_{00} = 1
cℓ_{01}^{d} + ℓ_{01}^{2} + ℓ_{01} = 0.

The first two equations yield the possibilities (1, 0) and (−1, −1) for (ℓ_{00}, ℓ_{01}), but only (1, 0) is consistent with the third equation. Thus L is the identity.

Case II.2: d ≡ 0 (mod p); d ≠ 3; and p = 3

When p = 3, there is a single linear relation between the g’s and h’s (as defined in Case II.1), namely g_{1} = h_{0} + h_{1}.

Lemma 6.1. For nonzero v ∈ \overline{k}^{n+2}, we have codim v^{⊥} ≥ 2, with equality if and only if v is a multiple of e_{0}, e_{0} + e_{1}, or e_{n+1}.

Proof. The values of codim v^{⊥} will be exactly the same as in Case I.1 except possibly for v’s for which the appearances of g_{1}, h_{0}, h_{1} in the coordinates of Av are dependent due to the new relation between them. This happens when v_{0}h_{0} + v_{1}g_{1} is a scalar multiple of v_{0}g_{1} + v_{1}h_{1} and both are nonzero. Using g_{1} = h_{0} + h_{1}, we see that this holds exactly when v_{0} = v_{1} ≠ 0. We may assume this from now on, since otherwise the inequality and the equality cases are the same as in Case I.1.

Let j be the largest integer such that v_{j} ≠ 0. If j ≤ 1, then v is a multiple of e_{0} + e_{1} and we are done. If j > 1, then g_{j} appears in the (j − 1)-th coordinate of Av but not before, and h_{j} appears in the j-th coordinate of Av but not before, and the 0-th coordinate of Av is nonzero, so these three coordinates are linearly independent, and codim v^{⊥} ≥ 3. □

Corollary 6.2. The vector Le_{n+1} is a multiple of e_{n+1}, and L(S_{1}) = S_{1}.

Proof. We have

⟨e_{0}, e_{0} + e_{1}⟩ = h_{0} + g_{1} = x_{1}^{d−2} − x_{0}x_{1}^{d−3},
⟨e_{0}, e_{n+1}⟩ = 0,
⟨e_{0} + e_{1}, e_{n+1}⟩ = 0,

unless n = 1, in which case ⟨e_{0} + e_{1}, e_{n+1}⟩ = −x_{n+1}^{d−2} instead. If n > 1, the multiples of e_{n+1} are distinguished from the multiples of e_{0} and e_{0} + e_{1} by the fact that they pair to give zero with the latter two, so L maps e_{n+1} to itself, and fixes the subspace S_{1} generated by the multiples of the other two. If n = 1, then the multiples of e_{n+1} are distinguished by the fact that they pair with multiples of e_{0} or e_{0} + e_{1} to give perfect (d − 2)-th powers always, so the result again follows. □

Any easy induction on m proves that for 0 ≤ m ≤ n, S_{m}^{⊥} = T_{m+2}, which is of codimension m + 2.

Lemma 6.3. Suppose 1 ≤ m ≤ n−2. Let V be an (m+2)-dimensional subspace of \overline{k}^{n+2} containing S_{m}. Then codim V^{⊥} ≥ m + 3, with equality if and only if V = S_{m+1}.

Proof. Write V = S_{m} + \overline{k} · v, so

V^{⊥} = S_{m}^{⊥} ∩ v^{⊥} = T_{m+2} ∩ v^{⊥}.

If v has any nonzero coordinate v_{i} with m + 2 ≤ i ≤ n, then the condition that an element w of T_{m+2} be in v^{⊥} places at least two linear conditions on w, namely w_{i} = 0 and w_{i+1} = 0, so codim V^{⊥} ≥ codim T_{m+2} + 2 = m + 4 in this case. Similarly, if v_{n+1} ≠ 0, then the condition that an element w of S_{m}^{⊥} be in v^{⊥} places the new conditions w_{n} = 0 and w_{n+1} = 0 on w, so that codim V^{⊥} ≥ m + 4 again. The only remaining possibility is that v_{i} = 0 for all i ≥ m + 2, in which case we must have V = S_{m+1} and codim V^{⊥} = codim T_{m+3} = m + 3. □

Corollary 6.4. We have L(S_{m}) = S_{m} for 1 ≤ m ≤ n − 1.

Lemma 6.5. We have L(T_{m}) = T_{m} for 2 ≤ m ≤ n + 1.

Proof. Suppose n = 1. Then the needed fact L(T_{2}) = T_{2} follows from the first half of Corollary 6.2.

Suppose n ≥ 2. Using S_{m}^{⊥} = T_{m+2} and Corollary 6.4 proves the result for all the required m except m = 2. We know that L(T_{2}) is an n-dimensional subspace of \overline{k}^{n+2} containing L(T_{3}) = T_{3} such that codim L(T_{2})^{⊥} = n + 1. Write L(T_{2}) = T_{3} + \overline{k} · v, where v_{i} = 0 for i ≥ 3. Since T_{3}^{⊥} = S_{1}, which has codimension n, in order to have codim L(T_{2})^{⊥} = n + 1, the first two coordinates of Av must be linearly dependent. This is possible only if v is a multiple of e_{0} + e_{1} or a multiple of e_{2}. But L(T_{2}) ∩ S_{1} = L(T_{2} ∩ S_{1}) = \{0\}, so e_{0} + e_{1} ∉ L(T_{2}). Thus L(T_{2}) = T_{3} + \overline{k} · e_{2} = T_{2}. □

Corollary 6.6. We have L(S_{m}) = S_{m} for 0 ≤ m ≤ n − 1.

Proof. The new result, L(S_{0}) = S_{0}, follows from L(T_{2}) = T_{2} and T_{2}^{⊥} = S_{0}. □

By Lemma 6.5 and Corollary 6.6, L is of the form

L = [\begin{matrix}∗ & ∗ & 0 & ⋯ & 0 & 0 & 0 \\ 0 & ∗ & 0 & ⋯ & 0 & 0 & 0 \\ 0 & 0 & ∗ & ⋯ & 0 & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ & ⋮ \\ 0 & 0 & 0 & ⋯ & ∗ & 0 & 0 \\ 0 & 0 & 0 & ⋯ & 0 & ∗ & 0 \\ 0 & 0 & 0 & ⋯ & 0 & ∗ & ∗ \end{matrix}] .

Repeating the argument at the end of Case II.1 completes the proof in this case.

Case II.3: d = 3, p = 3, n ≥ 2

In this case we have

A = − [\begin{matrix}x_{1} & x_{0} + x_{1} & 0 & 0 & ⋯ & 0 & 0 & 0 \\ x_{0} + x_{1} & x_{0} & x_{2} & 0 & ⋯ & 0 & 0 & 0 \\ 0 & x_{2} & x_{1} & x_{3} & ⋯ & 0 & 0 & 0 \\ 0 & 0 & x_{3} & x_{2} & ⋯ & 0 & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ & ⋮ \\ 0 & 0 & 0 & 0 & ⋯ & x_{n−2} & x_{n} & 0 \\ 0 & 0 & 0 & 0 & ⋯ & x_{n} & x_{n−1} & x_{n+1} \\ 0 & 0 & 0 & 0 & ⋯ & 0 & x_{n+1} & x_{n} \end{matrix}] .

Lemma 6.7. For nonzero v ∈ \overline{k}^{n+2}, we have codim v^{⊥} ≥ 2, with equality if and only if v is a multiple of e_{0}, e_{0} + e_{1}, or e_{n+1}.

Proof. Let i be the smallest integer such that v_{i} is nonzero. Let j be the largest integer such that v_{j} is nonzero.

If i = 0 and j = 0, then v is a multiple of e_{0}, and codim v^{⊥} = 2.

If i = 0 and j = 1, then we may assume v = e_{0} + γe_{1} for some γ ∈ \overline{k}^{∗}. If codim v^{⊥} ≤ 2, then the first two coordinates of Av must be linearly dependent, which implies γ^{2} + γ + 1 = 0, which yields γ = 1 (since we are in characteristic 3). Hence v is a multiple of e_{0} + e_{1}.

If i = 0 and 2 ≤ j ≤ n, then x_{j+1} appears only in the (j + 1)-th coordinate of Av, x_{j} appears in the (j − 1)-th coordinate of Av and not before, and the 0-th coordinate of Av is nonzero, so these three coordinates are independent, and codim v^{⊥} ≥ 3.

If i = 0 and j = n + 1, then we branch according as v_{n} is zero or not. If v_{n} = 0, then the (n + 1)-th coordinate of Av is a nonzero multiple of x_{n}, the n-th coordinate of Av is a combination of x_{n} and x_{n+1} in which x_{n+1} appears, and the 0-th coordinate is a nonzero combination of x_{0} and x_{1}, so these three coordinates are independent, and codim v^{⊥} ≥ 3. If v_{n} ≠ 0 and n > 2, then the (n + 1)-th coordinate is a nonzero combination of x_{n} and x_{n+1}, the 0-th coordinate is a nonzero combination of x_{0} and x_{1}, and the n-th coordinate involves x_{n−1}, which appears in neither the 0-th nor the (n + 1)-th coordinate, so these three coordinates are independent, and codim v^{⊥} ≥ 3. Finally suppose v_{n} ≠ 0 and n = 2. The 0-th coordinate of Av is a nonzero combination of x_{0} and x_{1}, and the 3-rd coordinate of Av is a nonzero combination of x_{2} and x_{3}, so these two coordinates are independent. If moreover codim v^{⊥} ≤ 2, then the 2-nd coordinate must be a linear combination of the 0-th and 3-rd. The 0-th coordinate must appear in this combination since x_{1} appears in the 2-nd coordinate of Av. But x_{0} does not appear in the 2-nd coordinate, so x_{0} cannot appear in the 0-th coordinate, and this implies v_{1} = 0. Then x_{3} appears while x_{2} does not appear in the 2-nd coordinate, making it impossible for the 2-nd coordinate to be a combination of the 0-th and 3-rd coordinates.

If i ≥ 1 and j ≤ n, then the (i − 1)-th coordinate of Av is nonzero, and the i-th coordinate is not a multiple of it, so these two coordinates are independent. Also, the (j + 1)-th coordinate of Av is a multiple of x_{j+1}, which does not appear anywhere else in Av, so the (i − 1)-th, i-th, and (j + 1)-th coordinates are independent, and codim v^{⊥} ≥ 3.

If 1 ≤ i ≤ n − 1 and j = n + 1, then the (i − 1)-th coordinate of Av is a nonzero multiple of x_{i} (or of x_{0} + x_{1} if i = 1), the i-th coordinate of Av is a nonzero combination of x_{i−1} and x_{i+1}, and the n-th coordinate of Av involves x_{n+1}, which does not appear earlier, so these three coordinates are independent, and codim v^{⊥} ≥ 3.

If i = n and j = n + 1, then x_{n−1} appears only in the n-th coordinate of Av, the (n − 1)-th coordinate is a nonzero multiple of x_{n}, and the (n + 1)-th coordinate is a combination of x_{n} and x_{n+1} in which both appear, so these three coordinates are independent, and codim v^{⊥} ≥ 3.

If i = n + 1 and j = n + 1, then v is a multiple of e_{n+1}, and codim v^{⊥} = 2. □

Corollary 6.8. The vector Le_{n+1} is a multiple of e_{n+1}, and L(S_{1}) = S_{1}.

Proof. Since n ≥ 2, we have

⟨e_{0}, e_{0} + e_{1}⟩ = −x_{0} − 2x_{1},
⟨e_{0}, e_{n+1}⟩ = 0,
⟨e_{0} + e_{1}, e_{n+1}⟩ = 0,

so the multiples of e_{n+1} are distinguished from the multiples of e_{0} and e_{0} + e_{1} by the fact that they pair to give zero with the latter two. Thus L maps e_{n+1} to itself, and fixes the subspace S_{1} generated by the multiples of the other two. □

Lemma 6.9. Suppose 1 ≤ m ≤ n−2. Let V be an (m+2)-dimensional subspace of \overline{k}^{n+2} containing S_{m}. Then codim V^{⊥} ≥ m + 3, with equality if and only if V = S_{m+1}.

Proof. Write V = S_{m} + \overline{k} · v, so

V^{⊥} = S_{m}^{⊥} ∩ v^{⊥} = T_{m+2} ∩ v^{⊥}.

We may assume v_{i} = 0 for i ≤ m. We must show that the codimension of T_{m+2} ∩ v^{⊥} in T_{m+2} is at least 1, with equality if and only if v is a nonzero multiple of e_{m+1}. This is the same as showing that the span of the (m + 2)-th, …, (n + 1)-th coordinates of Av is of dimension at least 1, with equality if and only if v is a nonzero multiple of e_{m+1}.

Let j be the largest integer such that v_{j} is nonzero. If j = m + 1, then v is a nonzero multiple of e_{m+1}, the (m + 2)-th coordinate of Av is a nonzero multiple of x_{m+2}, and all later coordinates are zero, so we have equality, as desired.

If m + 2 ≤ j ≤ n, then the (j + 1)-th coordinate of Av is a nonzero multiple of x_{j+1}, but the j-th coordinate of Av involves x_{j−1}, so the span is of dimension at least 2.

If j = n + 1 and v_{n} = 0, then the (n + 1)-th coordinate of Av is a nonzero multiple of x_{n}, but the n-th coordinate involves x_{n+1}, so the span is of dimension at least 2.

If j = n + 1 and v_{n} ≠ 0, then x_{n−1} appears in the n-th coordinate of Av, and the (n + 1)- th coordinate of Av is nonzero but does not involve x_{n−1}, so again the span is of dimension at least 2. □

The rest of the proof of this case is exactly analogous to the corresponding final section of the proof in Case II.2, from Corollary 6.4 on.

7. Controlling the automorphisms: Case III

We have

A = [\begin{matrix}0 & g_{1} & 0 & 0 & ⋯ & 0 & 0 & 0 \\ g_{1} & 0 & g_{2} & 0 & ⋯ & 0 & 0 & 0 \\ 0 & g_{2} & 0 & g_{3} & ⋯ & 0 & 0 & 0 \\ 0 & 0 & g_{3} & 0 & ⋯ & 0 & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ & ⋮ \\ 0 & 0 & 0 & 0 & ⋯ & 0 & g_{n} & 0 \\ 0 & 0 & 0 & 0 & ⋯ & g_{n} & 0 & g_{n+1} \\ 0 & 0 & 0 & 0 & ⋯ & 0 & g_{n+1} & 0 \end{matrix}] ,

where

g_{i} := \begin{cases}x_{0}^{d−2} + x_{1}^{d−2} & if i = 1 \\ x_{i}^{d−2} & if 2 ≤ i ≤ n + 1. \end{cases}

Note that g_{1}, …, g_{n+1} are linearly independent over \overline{k}.

Lemma 7.1. Suppose n ≥ 4. If v ∈ \overline{k}^{n+2} is nonzero and codim v^{⊥} ≤ 2 then v is a multiple of some e_{i}, or v is a combination of e_{0} and e_{1}, or a combination of e_{0} and e_{2}, or a combination of e_{n} and e_{n+1}, or a combination of e_{n−1} and e_{n+1}, or a combination of e_{0} and e_{n+1}.

If n = 2 or n = 3, then the same result holds, except that combinations of e_{0}, e_{n−1}, e_{n+1} and combinations of e_{0}, e_{2}, e_{n+1} are also possible.

For all n ≥ 2, only the multiples of e_{0} and the multiples of e_{n+1} satisfy codim v^{⊥} = 1.

Proof. It is clear that the listed v’s satisfy codim v^{⊥} ≤ 2. Now suppose codim v^{⊥} ≤ 2.

Let i be the smallest integer such that v_{i} is nonzero. Let j be the largest integer such that v_{j} is nonzero.

If i = 0 and j ≤ 1, then v is a combination of e_{0} and e_{1}. If v is a multiple of e_{0}, then codim v^{⊥} = 1; otherwise, the 0-th and 2-nd coordinates of Av are independent and codim v^{⊥} = 2.

If i = 0 and j = 2, then v_{1} = 0, since otherwise, the 0-th coordinate of Av is a multiple of g_{1}, the 2-nd coordinate of Av is a multiple of g_{2}, and the 3-rd coordinate of Av is a multiple of g_{3}, which makes codim v^{⊥} ≥ 3. Hence v is a combination of e_{0} and e_{2}. The 0-th and 2-nd coordinates of Av are independent unless v is a multiple of e_{0}, in which case codim v^{⊥} = 1.

If i = 0 and 3 ≤ j ≤ n, then g_{j+1} appears in the (j + 1)-th coordinate of Av but not before, g_{j} appears in the (j − 1)-th coordinate of Av but not before, and the 1-st coordinate of Av is nonzero, so these three coordinates are independent, and codim v^{⊥} ≥ 3.

If i = 0 and j = n + 1, then the 1-st and n-th coordinates of Av are nonzero and independent because x_{n+1} appears only in the latter. Thus codim v^{⊥} ≥ 2. Hence if codim v^{⊥} ≤ 2, then every other coordinate of Av must be a combination of the 1-st and n-th. In particular, each nonzero coordinate of Av involves g_{1} or g_{n+1}, so the 2-nd, 3-rd, …, (n − 1)-th coordinates of Av must be zero. If n ≥ 4 this forces v_{1} = v_{2} = ⋯ = v_{n} = 0, as desired. If n = 3, then the vanishing of the 2-nd coordinate of Av forces only v_{1} = v_{3} = 0, so that v is a combination of e_{0}, e_{2}, and e_{4}, as desired. Finally, if n = 2, then either v_{1} = 0 or v_{2} = 0, since if all v_{i} were nonzero, then for m = 1, 2, 3, the term g_{m} occurs in the m-th coordinate of Av but not afterwards, making codim v^{⊥} ≥ 3. Thus v is a combination of e_{0}, e_{1}, and e_{3}, or a combination of e_{0}, e_{2}, and e_{3}, as desired.

We have now completely finished the case i = 0, and symmetrical considerations prove all cases in which j = n + 1. Therefore, from now on, we assume 1 ≤ i ≤ j ≤ n. If i = j, then v is a multiple of e_{i}, and codim v^{⊥} = 2, as desired. Otherwise, if 1 ≤ i < j ≤ n, then g_{i} appears only in the (i − 1)-th coordinate of Av, g_{j+1} appears only in the (j + 1)-th coordinate of Av, and g_{i+1} appears in the (i + 1)-th coordinate of Av, so these three coordinates are independent, and codim v^{⊥} ≥ 3. □

Lemma 7.2. The matrix L is of the form

L = [\begin{matrix}∗ & ∗ & ∗ & 0 & ⋯ & 0 & 0 & 0 & 0 \\ 0 & ∗ & 0 & 0 & ⋯ & 0 & 0 & 0 & 0 \\ 0 & 0 & ∗ & 0 & ⋯ & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & ∗ & ⋯ & 0 & 0 & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ & ⋮ & ⋮ \\ 0 & 0 & 0 & 0 & ⋯ & ∗ & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & ⋯ & 0 & ∗ & 0 & 0 \\ 0 & 0 & 0 & 0 & ⋯ & 0 & 0 & ∗ & 0 \\ 0 & 0 & 0 & 0 & ⋯ & 0 & ∗ & ∗ & ∗ \end{matrix}]

or of the form

L = [\begin{matrix}0 & 0 & 0 & 0 & ⋯ & 0 & ∗ & ∗ & ∗ \\ 0 & 0 & 0 & 0 & ⋯ & 0 & 0 & ∗ & 0 \\ 0 & 0 & 0 & 0 & ⋯ & 0 & ∗ & 0 & 0 \\ 0 & 0 & 0 & 0 & ⋯ & ∗ & 0 & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ & ⋮ & ⋮ \\ 0 & 0 & 0 & ∗ & ⋯ & 0 & 0 & 0 & 0 \\ 0 & 0 & ∗ & 0 & ⋯ & 0 & 0 & 0 & 0 \\ 0 & ∗ & 0 & 0 & ⋯ & 0 & 0 & 0 & 0 \\ ∗ & ∗ & ∗ & 0 & ⋯ & 0 & 0 & 0 & 0 \end{matrix}] .

In other words, either L has nonzero entries on the diagonal and zeros elsewhere except possibly at ℓ_{01}, ℓ_{02}, ℓ_{n+1,n−1}, ℓ_{n+1,n}, or L has nonzero entries on the reverse diagonal and zeros elsewhere except possibly at ℓ_{0,n−1}, ℓ_{0,n}, ℓ_{n+1,1}, ℓ_{n+1,2}.

Proof. If n = 1, then the nonzero v ∈ \overline{k}^{n+2} for which codim v^{⊥} = 1 are exactly the combinations of e_{0} and e_{2}, so L must preserve the subspace S_{0} + T_{2}; i.e., L must have the form

L = [\begin{matrix}∗ & ∗ & ∗ \\ 0 & ∗ & 0 \\ ∗ & ∗ & ∗ \end{matrix}] ,

and this is what Lemma 7.2 is claiming in this case.

For n ≥ 2, Lemma 7.1 implies that L maps e_{0} and e_{n+1} to themselves or interchanges them, up to scalar multiple. By symmetry, we may assume that Le_{0} is a multiple of e_{0}, and that Le_{n+1} is a multiple of e_{n+1}. (The possibilities where Le_{0} is a multiple of e_{n+1} will give rise to the mirror reflections of the possibilities for L in the first case.) The subset W := \{v : codim v^{⊥} ≤ 2\} of \overline{k}^{n+2} is preserved by L, and Lemma 7.1 gives an explicit description of W.

If n = 2, then e_{1} ∈ e_{3}^{⊥}, so Le_{1} ∈ (Le_{3})^{⊥} = e_{3}^{⊥} = S_{1} + T_{3}. Similarly Le_{2} ∈ e_{0}^{⊥} = S_{0} + T_{2}. This completes the proof in the case n = 2.

If n = 3, then the subspace V of \overline{k}^{n+2} generated by e_{0}, e_{2}, and e_{4} is preserved by L, since by Lemma 7.1 it is the only 3-dimensional subspace contained in W. Also, L preserves S_{1} (resp. T_{3}), since by Lemma 7.1 this is the only 2-dimensional subspace that contains \overline{k}e_{0} (resp. \overline{k}e_{4}), that is not contained in V, and that is contained in W. These restrictions together imply that L has the desired shape.

From now on, we assume n ≥ 4. By Lemma 7.1, the 2-dimensional subspaces containing \overline{k}e_{0} and contained in W are S_{1} = \overline{k}e_{0} + \overline{k}e_{1} and R := \overline{k}e_{0} + \overline{k}e_{2}. Hence L preserves \{S_{1}, R\}, and preserves their sum, which is S_{2}. Similarly L preserves T_{n−1}. It then follows from Lemma 7.1 that L permutes e_{3}, e_{4}, …, e_{n−2} up to scalar multiple, since these (and their multiples) are the only vectors of W outside S_{2} + T_{n−1}.

We next prove by induction on i that L(S_{i}) = S_{i} for 2 ≤ i ≤ n − 2, and that Le_{i} is a multiple of e_{i} for 3 ≤ i ≤ n − 2. The base case L(S_{2}) = S_{2} is already known. Suppose 3 ≤ i ≤ n − 2, and L(S_{i−1}) = S_{i−1}, and L(e_{j}) is a multiple of e_{j} for 3 ≤ j ≤ i − 1. We know already that Le_{i} is a multiple of some e_{k}, k ≥ i, but the only such e_{k} that can pair with some vector in S_{i−1} to give something nonzero is e_{i}, so Le_{i} must be a multiple of e_{i}. Hence also L(S_{i}) = S_{i}, which completes the induction step.

In particular, we now know that L(T_{3}) = T_{3}. The subspaces S_{1} and R can be distinguished using the fact that only the latter contains elements that can pair with some vector in T_{3} to give something nonzero, so L(S_{1}) = S_{1} and L(R) = R. Similarly we deduce that L(T_{n}) = T_{n} and that L preserves the subspace R′ := \overline{k}e_{n−1} + \overline{k}e_{n+1}. The restrictions we have deduced, taken together, imply that L has the desired shape. □

Case III.1: n = 1

Equation (7) becomes

(12) x_{0}^{d−1}x_{1} + x_{0}x_{1}^{d−1} + x_{1}x_{2}^{d−1} + x_{2}^{d}

= α [(ℓ_{00}x_{0} + ℓ_{01}x_{1} + ℓ_{02}x_{2})^{d−1}ℓ_{11}x_{1} + (ℓ_{00}x_{0} + ℓ_{01}x_{1} + ℓ_{02}x_{2})(ℓ_{11}x_{1})^{d−1}

+ ℓ_{11}x_{1}(ℓ_{20}x_{0} + ℓ_{21}x_{1} + ℓ_{22}x_{2})^{d−1} + (ℓ_{20}x_{0} + ℓ_{21}x_{1} + ℓ_{22}x_{2})^{d}] .

Equating coefficients of x_{0}^{d} yields 0 = αℓ_{20}^{d}, so ℓ_{20} = 0. Since L is nonsingular, ℓ_{00} ≠ 0. Equating coefficients of x_{0}^{d−2}x_{1}^{2} yields 0 = αℓ_{00}^{d−2}ℓ_{01}ℓ_{11}, but α, ℓ_{00}, ℓ_{11} must all be nonzero, so ℓ_{01} = 0. Equating coefficients of x_{0}^{d−2}x_{1}x_{2} yields 0 = αℓ_{00}^{d−2}ℓ_{02}ℓ_{11}, so ℓ_{02} = 0. Equating coefficients of x_{1}^{d−1}x_{2} yields 0 = αℓ_{11}ℓ_{21}^{d−2}ℓ_{22}, and ℓ_{11}, ℓ_{22} ≠ 0 by nonsingularity, so ℓ_{21} = 0. We now know that L is diagonal. Without generality assume ℓ_{22} = 1. Equating coefficients of x_{2}^{d} in (12) shows α = 1. Equating coefficients of x_{1}x_{2}^{d−1} shows ℓ_{11} = 1. Equating coefficients of x_{0}x_{1}^{d−1} shows ℓ_{00} = 1. Thus L is the identity.

Case III.2: n ≥ 2

Equating coefficients of x_{0}^{d} in (7) rules out the second possibility in Lemma 7.2, so L is of the form

L = [\begin{matrix}∗ & ∗ & ∗ & 0 & ⋯ & 0 & 0 & 0 & 0 \\ 0 & ∗ & 0 & 0 & ⋯ & 0 & 0 & 0 & 0 \\ 0 & 0 & ∗ & 0 & ⋯ & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & ∗ & ⋯ & 0 & 0 & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ & ⋮ & ⋮ \\ 0 & 0 & 0 & 0 & ⋯ & ∗ & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & ⋯ & 0 & ∗ & 0 & 0 \\ 0 & 0 & 0 & 0 & ⋯ & 0 & 0 & ∗ & 0 \\ 0 & 0 & 0 & 0 & ⋯ & 0 & ∗ & ∗ & ∗ \end{matrix}] ,

and (7) becomes

(13) f(x_{0}, x_{1}, …, x_{n+1})

= αf(ℓ_{00}x_{0} + ℓ_{01}x_{1} + ℓ_{02}x_{2}, ℓ_{11}x_{1}, …, ℓ_{nn}x_{n}, ℓ_{n+1,n−1}x_{n−1} + ℓ_{n+1,n}x_{n} + ℓ_{n+1,n+1}x_{n+1}).

Note that ℓ_{ii} ≠ 0 for all i, since L is nonsingular. Equating coefficients of x_{0}^{d−2}x_{1}^{2} in (13) yields 0 = αℓ_{00}^{d−2}ℓ_{01}ℓ_{11} so ℓ_{01} = 0. Equating coefficients of x_{0}^{d−2}x_{1}x_{2} yields 0 = αℓ_{00}^{d−2}ℓ_{02}ℓ_{11} so ℓ_{02} = 0.

Equating coefficients of x_{n}^{d−1}x_{n+1} yields 0 = αℓ_{nn}ℓ_{n+1,n}^{d−2}ℓ_{n+1,n+1} so ℓ_{n+1,n} = 0. Equating coefficients of x_{n−1}^{d−2}x_{n}x_{n+1} yields 0 = αℓ_{nn}ℓ_{n+1,n−1}^{d−2}ℓ_{n+1,n+1} so ℓ_{n+1,n−1} = 0. We now know that L is diagonal. Without loss of generality assume ℓ_{n+1,n+1} = 1. Equating coefficients of x_{n+1}^{d} in (13) shows α = 1. We now prove ℓ_{ii} = 1 for all i by backwards induction, by equating coefficients of x_{i}x_{i+1}^{d−1}. Thus L is the identity.

8. Controlling the automorphisms: Case IV

The matrix A will have ⌊\frac{n+1}{2}⌋ nonzero 2 × 2 blocks along the diagonal, and zeros elsewhere. For odd i, define g_{i} := −2x_{i−1}x_{i}^{d−3}. Also define

h_{i} := \begin{cases}2x_{i+1}^{d−2} & if i is even \\ 2x_{i−1}^{2}x_{i}^{d−4} & if i is odd. \end{cases}

(Note that d ≥ 4 in Case IV.) The g’s and h’s are linearly independent over \overline{k}.

Case IV.1: n is odd

We have

A = [\begin{matrix}h_{0} & g_{1} & 0 & 0 & ⋯ & 0 & 0 & 0 \\ g_{1} & h_{1} & 0 & 0 & ⋯ & 0 & 0 & 0 \\ 0 & 0 & h_{2} & g_{3} & ⋯ & 0 & 0 & 0 \\ 0 & 0 & g_{3} & h_{3} & ⋯ & 0 & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ & ⋮ \\ 0 & 0 & 0 & 0 & ⋯ & h_{n−1} & g_{n} & 0 \\ 0 & 0 & 0 & 0 & ⋯ & g_{n} & h_{n} & 0 \\ 0 & 0 & 0 & 0 & ⋯ & 0 & 0 & 0 \end{matrix}] .

Clearly Le_{n+1} is a multiple of e_{n+1}, because only the multiples of e_{n+1} pair under ⟨ , ⟩ with all vectors to give zero. For i = 0, 2, 4, …, n − 1, let V_{i} be the \overline{k}-vector space spanned by e_{i}, e_{i+1}, and e_{n+1}. It is clear that V_{0}, V_{2}, V_{4}, …, V_{n−1} are the only 3-dimensional subspaces V of \overline{k}^{n+2} such that codim V^{⊥} = 2. Thus L(V_{i}) = V_{π(i)} for some permutation π of {0, 2, 4, …, n − 1}. In other words, L has the form of a permutation matrix, except with 2×2 blocks, and with an added row at the bottom with potentially nonzero entries, and with zeros in an added final column on the right (except for the lower right corner, which must be nonzero).

If we view both sides of (7) as polynomials in x_{n+1} and equate coefficients of x_{n+1}^{d−1}, we find that y_{n} is a nonzero multiple of x_{n}. Thus L(V_{n−1}) = V_{n−1}. If we instead equate coefficients of x_{n+1}, we find that y_{0} is a nonzero multiple of x_{0}. Thus L(V_{0}) = V_{0}.

We now prove by backwards induction that y_{i} is a nonzero multiple of x_{i} for i = n−1, n−2, …, 1. (We already know it for i = n and i = 0.) First suppose i is even. By assumption, y_{i+1} is a multiple of x_{i+1}, so π(i) = i. It follows that y_{i} is a linear combination of x_{i} and x_{i+1}. Moreover, x_{i} occurs in this combination, since otherwise L would be singular. Suppose ℓ_{i,i+1} ≠ 0. Then for each j < i, equating coefficients of x_{j}x_{i+1}^{d−1} in (7) yields

0 = αℓ_{i−1,j}ℓ_{i,i+1}^{d−1}

so ℓ_{i−1,j} = 0. The block form of L implies ℓ_{i−1,j} = 0 for j ≥ i as well, so L has a row of zeros, which is a contradiction. Thus ℓ_{i,i+1} must have been zero, and hence y_{i} is a (nonzero) multiple of x_{i}.

Next suppose i is odd, 1 ≤ i ≤ n − 2. For j < i, equating coefficients of x_{j}x_{i+1}^{d−1} in (7) yields 0 = αℓ_{i,j}ℓ_{i+1,i+1}^{d−1}, and ℓ_{i+1,i+1} is nonzero (since y_{i+1} is a nonzero multiple of x_{i+1}), so ℓ_{i,j} = 0 for j < i. On the other hand, the block form of L implies ℓ_{i,j} = 0 for j > i also, so y_{i} is a (nonzero) multiple of x_{i}.

Equating coefficients of x_{n}^{d} in (7) yields 0 = αℓ_{nn}ℓ_{n+1,n}^{d−1}, so ℓ_{n+1,n} = 0. For each j < n, equating coefficients of x_{j}^{d−1}x_{n} shows that 0 = αℓ_{nn}ℓ_{n+1,j}^{d−1}, so ℓ_{n+1,j} = 0. Thus y_{n+1} is a (nonzero) multiple of x_{n+1}.

We now know that L is diagonal. We may assume ℓ_{00} = 1. Equating coefficients of x_{0}^{d} in (7) shows α = 1. Equating coefficients of x_{n+1}x_{0}^{d−1} in (7) shows ℓ_{n+1,n+1} = 1. We can now show ℓ_{ii} = 0 for i = n, n − 1, …, 1 as well, by backwards induction: equating coefficients of x_{i}x_{i+1}^{d−1} in (7) yields ℓ_{ii}ℓ_{i+1,i+1}^{d−1} = 1, so if ℓ_{i+1,i+1} = 1, then ℓ_{ii} = 1. Thus L is the identity.

Case IV.2: n is even

The matrix A has the same form as in Case IV.1 except that it ends with two final rows of zeros and two final columns of zeros, instead of only one of each.

The subspace of v in \overline{k}^{n+2} such that ⟨v, w⟩ = 0 for all w in \overline{k}^{n+2} is T_{n}, so L(T_{n}) = T_{n}. For i = 0, 2, 4, …, n − 1, let V_{i} be the \overline{k}-vector space spanned by e_{i}, e_{i+1}, e_{n}, and e_{n+1}. It is clear that V_{0}, V_{2}, V_{4}, …, V_{n−2} are the only 4-dimensional subspaces V of \overline{k}^{n+2} such that codim V^{⊥} = 2. Thus L(V_{i}) = V_{π(i)} for some permutation π of {0, 2, 4, …, n − 2}. In other words, L has the form of a permutation matrix, except with 2 × 2 blocks, and with two added rows at the bottom with potentially nonzero entries, and with zeros in two added final columns on the right (except for the lower right 2 × 2 block, which may have nonzero entries).

If we substitute x_{0} = x_{1} = ⋯ = x_{n−1} = 0 in (7), we obtain

x_{n}x_{n+1}^{d−1} = α(ℓ_{nn}x_{n} + ℓ_{n,n+1}x_{n+1})(ℓ_{n+1,n}x_{n} + ℓ_{n+1,n+1}x_{n+1})^{d−1}.

By unique factorization, ℓ_{nn}x_{n}+ℓ_{n,n+1}x_{n+1} is a nonzero multiple of x_{n}, and ℓ_{n+1,n}x_{n}+ℓ_{n+1,n+1}x_{n+1} is a nonzero multiple of x_{n+1}. Hence the lower right 2 × 2 block of L is diagonal, with nonzero entries on the diagonal.

View both sides of (7) as polynomials in x_{n+1}. Equating coefficients of x_{n+1}^{d−1} shows that y_{n} is a nonzero multiple of x_{n}. Equating coefficients of x_{n+1} shows that y_{0}^{d−1} is a nonzero multiple of x_{0}^{d−1}, so y_{0} is a nonzero multiple of x_{0}.

Now view both sides of (7) as polynomials in x_{n}. Equating coefficients of x_{n} shows that y_{n+1}^{d−1} is a nonzero multiple of x_{n+1}^{d−1}, so y_{n+1} is a nonzero multiple of x_{n+1}. Equating coefficients of x_{n}^{d−1} shows that y_{n−1} is a nonzero multiple of x_{n−1}.

The same backwards induction on i as in Case IV.1 now shows that y_{i} is a nonzero multiple of x_{i} for all i. (We already know it for i = 0, n − 1, n, n + 1.) Thus L is diagonal. We deduce that L is (a scalar multiple of) the identity as in the end of Case IV.1.

9. Controlling the automorphisms: Case V

Note that d ≥ 5 in Case V. We have

A = [\begin{matrix}0 & x_{1}^{d−3}x_{2} & x_{1}^{d−2} \\ x_{1}^{d−3}x_{2} & 0 & x_{0}x_{1}^{d−3} \\ x_{1}^{d−2} & x_{0}x_{1}^{d−3} & 0 \end{matrix}] .

The greatest common divisor of the entries of A is x_{1}^{d−3}, so y_{1}^{d−3} must be a nonzero multiple of x_{1}^{d−3}. Hence y_{1} is a nonzero multiple of x_{1}. Without loss of generality we may assume y_{1} = x_{1}.

Then (7) becomes

(14)  f(x_{0}, x_{1}, x_{2}) = αf(ℓ_{00}x_{0} + ℓ_{01}x_{1} + ℓ_{02}x_{2}, x_{1}, ℓ_{20}x_{0} + ℓ_{21}x_{1} + ℓ_{22}x_{2}).

If we set x_{1} = 0 and use the definition of f, we obtain

x_{2}x_{0}^{d−1} = α(ℓ_{20}x_{0} + ℓ_{22}x_{2})(ℓ_{00}x_{0} + ℓ_{02}x_{2})^{d−1}.

By unique factorization, we deduce that ℓ_{20} = 0 and ℓ_{02} = 0. Now (14) becomes

(15) α^{−1}(x_{0}x_{1}^{d−2}x_{2} + x_{0}x_{1}^{d−1} + x_{1}x_{2}^{d−1} + x_{2}x_{0}^{d−1} + x_{1}^{2}x_{2}^{d−2})

= (ℓ_{00}x_{0} + ℓ_{01}x_{1})x_{1}^{d−2}(ℓ_{21}x_{1} + ℓ_{22}x_{2}) + (ℓ_{00}x_{0} + ℓ_{01}x_{1})x_{1}^{d−1} + x_{1}(ℓ_{21}x_{1} + ℓ_{22}x_{2})^{d−1}

+ (ℓ_{21}x_{1} + ℓ_{22}x_{2})(ℓ_{00}x_{0} + ℓ_{01}x_{1})^{d−1} + x_{1}^{2}(ℓ_{21}x_{1} + ℓ_{22}x_{2})^{d−2}

Equating coefficients of x_{0}^{d−1}x_{1} yields 0 = ℓ_{21}ℓ_{00}^{d−1}. The nonsingularity of L guarantees ℓ_{00} ≠ 0, so ℓ_{21} = 0. Equating coefficients of x_{1}^{d} yields 0 = ℓ_{01}, so L is diagonal.

Equating coefficients of x_{1}x_{2}^{d−1} and of x_{1}^{2}x_{2}^{d−2} yields

α^{−1} = ℓ_{22}^{d−1},
α^{−1} = ℓ_{22}^{d−2}.

Dividing, we find ℓ_{22} = 1, and then α = 1. Equating coefficients of x_{0}x_{1}^{d−1} now shows ℓ_{00} = 1. Thus L is the identity.

10. Controlling the automorphisms: Case VI

Let m = 3⌊\frac{n+2}{3}⌋. For i = 0, 3, 6, …, m − 3, define f_{i} := x_{i+1}^{d−3}x_{i+2}, g_{i} := x_{i+1}^{d−2}, and h_{i} := x_{i}x_{i+1}^{d−3}. We have

A = [\begin{matrix}0 & f_{0} & g_{0} & 0 & 0 & 0 & ⋯ \\ f_{0} & 0 & h_{0} & 0 & 0 & 0 & ⋯ \\ g_{0} & h_{0} & 0 & 0 & 0 & 0 & ⋯ \\ 0 & 0 & 0 & 0 & f_{3} & g_{3} & ⋯ \\ 0 & 0 & 0 & f_{3} & 0 & h_{3} & ⋯ \\ 0 & 0 & 0 & g_{3} & h_{3} & 0 & ⋯ \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix}] ,

in which there are ⌊\frac{n+2}{3}⌋ 3 × 3 blocks along the diagonal, and zeros elsewhere. (There will be (n + 2 − m) rows of zeros at the bottom, and also (n + 2 − m) columns of zeros at the right.) Note that f_{0}, g_{0}, h_{0}, f_{3}, g_{3}, h_{3}, … are linearly independent over \overline{k}.

The set of v in \overline{k}^{n+2} such that ⟨v, w⟩ = 0 for all w in \overline{k}^{n+2} is T_{m}, so L(T_{m}) = T_{m}. Note that dim T_{m} = (n + 2) mod 3 = n + 2 − m. For i = 0, 3, 6, …, m − 3, let V_{i} be the (n + 2 − m) + 3dimensional vector space spanned by T_{m}, e_{i}, e_{i+1}, and e_{i+2}.

Lemma 10.1. If codim v^{⊥} ≤ 3, then v ∈ V_{i} for some i.

Proof. If v is not contained in any V_{i}, then there are at least two distinct i, j ∈ {0, 3, 6, …, m − 3} such that v equals a nonzero combination w of e_{i}, e_{i+1}, e_{i+2}, plus a nonzero combination w′ of e_{j}, e_{j+1}, e_{j+2}, plus an element of T_{m}. Any nonzero combination of the three columns A_{i}, A_{i+1}, A_{i+2} will have entries spanning a vector space of dimension at least 2, because there will be at least two nonzero entries, and there will be one form f_{i}, g_{i}, or h_{i} that appears in some but not all of these nonzero entries. The span of the nonzero entries of this combination does not intersect the span of the entries of a nonzero combination of A_{j}, A_{j+1}, A_{j+2}, so we see that codim v^{⊥} ≥ 2 + 2 = 4. □

Corollary 10.2. We have L(V_{i}) = V_{π(i)} for some permutation π of {0, 3, 6, …, m − 3}.

Proof. By Lemma 10.1, the V_{i} are the only ((n + 2 − m) + 3)-dimensional subspaces V such that codim V^{⊥} = 3. □

Corollary 10.3. For i = 0, 3, 6, …, m − 3, each of y_{i}, y_{i+1}, y_{i+2} is a linear combination of x_{j}, x_{j+1}, x_{j+2}, where j = π^{−1}(i).

Proof. This is a direct consequence of Corollary 10.2 and the fact L(T_{m}) = T_{m}. □

Before proceeding further, we subdivide Case VI as follows.

• Case VI.1: n ≡ 0 (mod 3)

• Case VI.2: n ≡ 1 (mod 3) and n ≥ 4

• Case VI.3: n ≡ 2 (mod 3). (Also, remember that throughout Case VI, p = 2, d is odd, and n > 1.)

Case VI.1: n ≡ 0 (mod 3)

We have m = n. Each of y_{0}, y_{1}, …, y_{n−1} is a linear combination of x_{0}, x_{1}, …, x_{n−1}, by Corollary 10.3. Thus if we substitute x_{0} = x_{1} = ⋯ = x_{n−1} = 0 in (7), we obtain

x_{n}x_{n+1}^{d−1} = α(ℓ_{n,n}x_{n} + ℓ_{n,n+1}x_{n+1})(ℓ_{n+1,n}x_{n} + ℓ_{n+1,n+1}x_{n+1})^{d−1}.

By unique factorization, ℓ_{n,n+1} = ℓ_{n+1,n} = 0.

If we consider both sides of (7) as polynomials in x_{n+1} and equate coefficients of x_{n+1}^{d−1}, we deduce that y_{n} is a multiple of x_{n}. In particular, the only y_{i} in which x_{n} appears is y_{n}. If we consider both sides of (7) as polynomials in x_{n} and equate coefficients of x_{n}, we deduce that y_{n+1}^{d−1} is a multiple of x_{n+1}^{d−1}, so y_{n+1} is a multiple of x_{n+1}.

If we again consider both sides of (7) as polynomials in x_{n+1}, but this time equate coefficients of x_{n+1}, we deduce that y_{0}^{d−1} is a multiple of x_{0}^{d−1}, so y_{0} is a multiple of x_{0}. Corollary 10.3 implies π(0) = 0.

Similarly if we consider both sides of (7) as polynomials in x_{n} and equate coefficients of x_{n}^{d−1}, we deduce that y_{n−1} is a multiple of x_{n−1}, and π(m − 3) = m − 3.

We now prove π(i) for all i = 0, 3, 6, …, m − 3 by induction. Suppose i ≥ 3, and we know π(j) = j for j < i. By Corollary 10.3, the only y-monomial of f(y_{0}, y_{1}, …, y_{n+1}) whose expansion can contain x_{i−1}x_{i}^{d−1} is y_{i−1}y_{i}^{d−1}. It follows that y_{i} must involve x_{i}, so π(i) = i by Corollary 10.3.

Fix i ∈ {3, 6, 9, …, m−6}. If we expand f(y_{0}, y_{1}, …, y_{n+1}) and discard all monomials unless they involve both one of x_{i−3}, x_{i−2}, x_{i−1} and one of x_{i}, x_{i+1}, x_{i+2}, then what remains, by Corollary 10.3, is exactly the expansion of y_{i−1}y_{i}^{d−1}. Hence y_{i−1}y_{i}^{d−1} is a multiple of x_{i−1}x_{i}^{d−1}, and by unique factorization, we see that y_{i−1} is a multiple of x_{i−1} and y_{i} is a multiple of x_{i}.

We now know that L is of the form

L = [\begin{matrix}∗ & 0 & 0 & 0 & 0 & 0 & ⋯ & 0 & 0 & 0 & 0 & 0 \\ ∗ & ∗ & ∗ & 0 & 0 & 0 & ⋯ & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & ∗ & 0 & 0 & 0 & ⋯ & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & ∗ & 0 & 0 & ⋯ & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & ∗ & ∗ & ∗ & ⋯ & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & ∗ & ⋯ & 0 & 0 & 0 & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ \\ 0 & 0 & 0 & 0 & 0 & 0 & ⋯ & ∗ & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & ⋯ & ∗ & ∗ & ∗ & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & ⋯ & 0 & 0 & ∗ & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & ⋯ & 0 & 0 & 0 & ∗ & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & ⋯ & 0 & 0 & 0 & 0 & ∗ \end{matrix}] ,

and the diagonal entries must be nonzero, since L is nonsingular.

Equating coefficients of x_{0}x_{1}^{d−1} in (7) yields

0 = αℓ_{10}ℓ_{11}^{d−1}

so ℓ_{10} = 0. Equating coefficients of x_{1}^{d−1}x_{2} yields

0 = αℓ_{11}^{d−1}ℓ_{12}

so ℓ_{12} = 0.

Let i be a positive multiple of 3. Equating coefficients of x_{i}^{d} yields

0 = αℓ_{i,i}ℓ_{i+1,i}^{d−1},

so ℓ_{i+1,i} = 0. Similarly, equating coefficients of x_{i+2}^{d} yields

0 = αℓ_{i+1,i+2}ℓ_{i+2,i+2}^{d−1}

so ℓ_{i+1,i+2} = 0.

We now know that L is diagonal. Without loss of generality suppose ℓ_{11} = 1. Equating coefficients of x_{1}^{d} in (7) shows α = 1. Equating coefficients of x_{n+1}x_{0}^{d−1} shows ℓ_{n+1,n+1} = ℓ_{00}^{1−d}. Equating coefficients of x_{n}x_{n+1}^{d−1} shows

ℓ_{n,n} = ℓ_{n+1,n+1}^{1−d} = ℓ_{00}^{(1−d)^{2}}.

By backwards induction on i, we show

(16)  ℓ_{i,i} = ℓ_{00}^{(1−d)^{n+2−i}}

for all i ≥ 1. In particular,

1 = ℓ_{11} = ℓ_{00}^{(1−d)^{n+1}}.

On the other hand, equating coefficients of x_{0}x_{1}^{d−3}x_{2}, we find

1 = ℓ_{00}ℓ_{22} = ℓ_{00}^{1+(1−d)^{n}}.

Since the exponents (1 − d)^{n+1} and 1 + (1 − d)^{n} are relatively prime, it follows that ℓ_{00} = 1, and then by (16), ℓ_{i,i} = 1 for all i. Thus L is the identity.

Case VI.2: n ≡ 1 (mod 3) and n ≥ 4

We have m = n + 2. In what follows, subscripts are to be considered modulo m. Suppose i ∈ {0, 3, 6, …, m − 3}. Because of Corollary 10.3, the only y-monomials in f(y_{0}, y_{1}, …, y_{n+1}) whose expansions could possibly contain x_{i−1}x_{i}^{d−1} are those of the form y_{j−1}y_{j}^{d−1} for some j ∈ {0, 3, 6, …, m − 3}. Moreover, for fixed i, at most one of these y-monomials can contribute an x_{i−1}x_{i}^{d−1} term. On the other hand, by (7), x_{i−1}x_{i}^{d−1} must appear in one of them, since it appears in f(x_{0}, x_{1}, …, x_{n+1}). Suppose it appears in y_{j−1}y_{j}^{d−1}. Then, again by Corollary 10.3, the monomials in the expansion of y_{j−1}y_{j}^{d−1} are exactly those monomials in the expansion of f(y_{0}, y_{1}, …, y_{n+1}) involving both one of x_{i−3}, x_{i−2}, x_{i−1} and one of x_{i}, x_{i+1}, x_{i+2}. By (7) it then follows that y_{j−1}y_{j}^{d−1} is a multiple of x_{i−1}x_{i}^{d−1}. By unique factorization, we deduce that y_{j−1} is a multiple of x_{j−1} and y_{j} is a multiple of x_{j}. By Corollary 10.3, it follows that π(i) = j and π(i − 3) = j − 3. (We should identify −3 with m − 3 when necessary.) Thus π acts as a rotation of {0, 3, 6, …, m − 3}, and there exists an integer r divisible by 3, determined up to a multiple of m, such that if j ≢ 1 (mod 3), then y_{j} is a multiple of x_{j+r}. If j ≡ 1 (mod 3), then by Corollary 10.3, y_{j} is a combination of x_{j+r−1}, x_{j+r}, and x_{j+r+1}.

It follows that the only y-monomial on the right hand side of (7) whose expansion could contain x_{1}^{d} is y_{1}^{d}. Thus r ≡ 0 (mod m).

Equating coefficients of x_{0}^{d} in (7), we find ℓ_{10} = 0. Equating coefficients of x_{0}x_{2}^{d−1}, we find ℓ_{12} = 0. Thus y_{1} is a multiple of x_{1}.

Now suppose j ∈ {4, 7, 10, …, m − 2}. Equating coefficients of x_{j−1}^{d} in (7), we deduce that ℓ_{j,j−1} = 0. Equating coefficients of x_{j+1}^{d}, we deduce that ℓ_{j,j+1} = 0.

We now know that L is diagonal. The same proof as at the end of Case VI.1 shows that L is (a scalar multiple of) the identity.

Case VI.3: n ≡ 2 (mod 3)

We have m = n + 1. Since T_{m} is the one-dimensional vector space generated by e_{n+1}, we know that Le_{n+1} is a multiple of e_{n+1}. In other words, the only y_{i} that involves x_{n+1} is y_{n+1}.

If we view both sides of (7) as polynomials in x_{n+1}, and equate coefficients of x_{n+1}^{d−1} in (7), we deduce that y_{n} is a multiple of x_{n}. Similarly, equating coefficients of x_{n+1} shows that y_{0} is a multiple of x_{0}. In particular, we have π(0) = 0 and π(m − 3) = m − 3.

We now show π(i) = i for all i ∈ {0, 3, 6, …, m − 3} by induction on i. Suppose i ≥ 3, and we know π(j) = j for j < i. By Corollary 10.3, the only y-monomial in the right hand side of (7) whose expansion can contain x_{i−1}x_{i}^{d−1} is y_{i−1}y_{i}^{d−1}. It follows that y_{i} involves x_{i}, and π(i) = i, as desired. In fact, the monomials in the expansion of y_{i−1}y_{i}^{d−1} are exactly those monomials in the expansion of f(y_{0}, y_{1}, …, y_{n+1}) involving both one of x_{i−3}, x_{i−2}, x_{i−1} and one of x_{i}, x_{i+1}, x_{i+2}, and in which the exponents of x_{i}, x_{i+1}, x_{i+2} are even. By (7) it then follows that y_{i−1}y_{i}^{d−1} is a multiple of x_{i−1}x_{i}^{d−1}. By unique factorization, we deduce that y_{i−1} is a multiple of x_{i−1} and y_{i} is a multiple of x_{i}.

We now know that L is of the form

L = [\begin{matrix}∗ & 0 & 0 & 0 & 0 & 0 & ⋯ & 0 & 0 & 0 & 0 \\ ∗ & ∗ & ∗ & 0 & 0 & 0 & ⋯ & 0 & 0 & 0 & 0 \\ 0 & 0 & ∗ & 0 & 0 & 0 & ⋯ & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & ∗ & 0 & 0 & ⋯ & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & ∗ & ∗ & ∗ & ⋯ & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & ∗ & ⋯ & 0 & 0 & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ & ⋮ & ⋮ \\ 0 & 0 & 0 & 0 & 0 & 0 & ⋯ & ∗ & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & ⋯ & ∗ & ∗ & ∗ & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & ⋯ & 0 & 0 & ∗ & 0 \\ ∗ & ∗ & ∗ & ∗ & ∗ & ∗ & ⋯ & ∗ & ∗ & ∗ & ∗ \end{matrix}] ,

and the diagonal entries must be nonzero, since L is nonsingular.

Equating coefficients of x_{0}x_{1}^{d−1} in (7), we find 0 = αℓ_{10}ℓ_{11}^{d−1}, so ℓ_{10} = 0. Equating coefficients of x_{0}^{d}, we find 0 = αℓ_{n+1,0}ℓ_{00}^{d−1}, so ℓ_{n+1,0} = 0. Thus y_{0} is the only y_{i} that involves x_{0}.

If we view both sides of (7) as polynomials in x_{0}, and equate coefficients of x_{0}^{d−1}, we deduce that y_{n+1} is a multiple of x_{n+1}.

Equating coefficients of x_{1}^{d−1}x_{2}, we find 0 = αℓ_{11}^{d−1}ℓ_{12}, so ℓ_{12} = 0.

Now suppose j ∈ {4, 7, 10, …, m − 2}. Equating coefficients of x_{j−1}^{d} in (7), we deduce that ℓ_{j,j−1} = 0. Equating coefficients of x_{j+1}^{d}, we deduce that ℓ_{j,j+1} = 0.

We now know that L is diagonal. The same proof as at the end of Case VI.1 shows that L is (a scalar multiple of) the identity.

11. The automorphism group scheme

Finally, we consider the automorphism group scheme Aut \overline{X} of a smooth hypersurface \overline{X} over \overline{k}. One can recover Aut \overline{X} as the group of \overline{k}-points of Aut \overline{X}, but the triviality of Aut \overline{X} cannot be deduced immediately from the triviality of Aut \overline{X}, because a priori Aut \overline{X} could be non-reduced. Fortunately, it is usually reduced:

Theorem 11.1. If X is a smooth hypersurface in P^{n+1} of degree d, where n ≥ 1, d ≥ 3, and (n, d) does not equal (1, 3), then the connected component of the identity of Aut \overline{X} is trivial.

Proof. Let T\overline{X} denote the tangent sheaf of \overline{X} over \overline{k}. Under the hypotheses on (n, d), we have H^{0}(\overline{X}, T\overline{X}) = 0 by [KS99, Theorem 11.5.2]. Thus the tangent space at the identity of Aut \overline{X} is trivial, so the connected component of the identity of Aut \overline{X} is trivial. □

Combining Corollary 1.9 and Theorem 11.1, we obtain:

Corollary 11.2. For any field k and integers n ≥ 1, d ≥ 3 with (n, d) not equal to (1, 3) or (2, 4), there exists a smooth hypersurface X over k of degree d in P^{n+1} such that Aut \overline{X} is trivial.

Acknowledgements

I thank Nick Katz for bringing the problem considered in this paper to my attention, and for referring me to [MM63]. I thank Nick Shepherd-Barron for several discussions, and in particular for the suggestion to discuss the automorphism group scheme in addition to the automorphism group.

References

[BT61] Raoul Bott and John Tate, 1961 (approximately). Colloquium lecture, University of Pennsylvania.

[Cha78] H. C. Chang, On plane algebraic curves, Chinese J. Math. 6 (1978), no. 2, 185–189.MR 80f:14012

[HS81] Alan Howard and Andrew John Sommese, On the orders of the automorphism groups of certain projective manifolds, Manifolds and Lie groups (Notre Dame, Ind., 1980), 1981, pp. 145–158.MR 84e:14011

[KS99] Nicholas M. Katz and Peter Sarnak, Random matrices, Frobenius eigenvalues, and monodromy, American Mathematical Society Colloquium Publications, vol. 45, American Mathematical Society, Providence, RI, 1999.MR 2000b:11070

[MM63] Hideyuki Matsumura and Paul Monsky, On the automorphisms of hypersurfaces, J. Math. Kyoto Univ. 3 (1963/1964), 347–361.MR 29 #5819

[OS77] Peter Orlik and Louis Solomon, Singularities. II. Automorphisms of forms, Math. Ann. 231 (1977/78), no. 3, 229–240.MR 57 #16292b

[Poo00a] Bjorn Poonen, Varieties without extra automorphisms. I. Curves, Math. Res. Lett. 7 (2000), no. 1, 67– 76.MR 2001g:14052a

[Poo00b] , Varieties without extra automorphisms. II. Hyperelliptic curves, Math. Res. Lett. 7 (2000), no. 1, 77–82.MR 2001g:14052b

[Sch38] Hermann Ludwig Schmid, Über die Automorphismen eines algebraischen Funktionenkörpers von Primzahlcharakteristik, J. Reine Angew. Math. 179 (1938), 5–15.

[Sil92] Joseph H. Silverman, The arithmetic of elliptic curves, Graduate Texts in Mathematics, vol. 106, Springer-Verlag, New York, 1992. Corrected reprint of the 1986 original.MR 95m:11054

Department of Mathematics, University of California, Berkeley, CA 94720-3840, USA

   E-mail address: poonen@math.berkeley.edu
   URL: http://math.berkeley.edu/~poonen