Pairs of homogeneous additive equations

Michael P. Knapp

Abstract

In 1966, Davenport & Lewis published their paper Notes on congruences III, in which they proved that under some mild conditions a system of two additive forms of equal degrees must have a nonsingular simultaneous zero modulo any prime number. In their paper, they ask whether the theorem is true in general finite fields, and point out that one of their key lemmas is no longer true in this situation. In this paper we answer their question in the affirmative, proving that under the same conditions a system of two additive forms over any finite field must have a nonsingular simultaneous zero. We then apply this result to obtain an upper bound on the number of variables required to ensure that a system of two additive forms of equal degrees has a nontrivial zero in a 𝔭-adic field.

1. Introduction

A special case of a well-known theorem due to Chevalley [Che35] states that if f_{1}(x_{1}, …, x_{s}) and f_{2}(x_{1}, …, x_{s}) are polynomials defined over a finite field 𝔽_{q} which have degree k and no constant terms, and if s > 2k, then the system of equations

f_{1}(x) = f_{2}(x) = 0

has a nontrivial solution in 𝔽_{q}. That is, there exists a solution with at least one variable nonzero. Sometimes it is necessary, however, to know that a system of equations over a finite field has a nonsingular solution. For example, to show that a system over a 𝔭-adic field has a solution, one

2000 Mathematics Subject Classification 11D72, (11T06, 11E95)

Work supported by NSF grant DMS-0344082.

often reduces modulo the maximal ideal to a finite field. If the resulting system has a nonsingular solution, then Hensel’s Lemma guarantees that there exists a solution over the 𝔭-adic field.

A theorem of Davenport & Lewis [DL66] gives conditions under which a system of two diagonal forms has a nonsingular zero over a finite field 𝔽_{p}, where p is prime. Specifically, they consider the system

\begin{aligned}a_{1}x_{1}^{k} + ⋯ + a_{s}x_{s}^{k} & = & 0 \\ b_{1}x_{1}^{k} + ⋯ + b_{s}x_{s}^{k} & = & 0 \end{aligned}  (1)

and show that under certain conditions the system possesses a nonsingular solution. That is, there is a solution such that the expression

k^{2}x_{i}^{k−1}x_{j}^{k−1}(a_{i}b_{j} − a_{j}b_{i})  (2)

is nonzero (in 𝔽_{p}) for some variables x_{i} and x_{j}.

In [DL66], Davenport & Lewis ask whether their theorem holds over general finite fields. They point out that one of their lemmas is no longer true in this case, and say that they were unable to determine whether the theorem should be true. The main purpose of the present paper is to answer this 40-year-old question in the affirmative. In particular, we prove the following theorem.

Theorem 1. Let p be a prime number and let q = p^{f} for some positive integer f. Consider the system (1) where the coefficients are in the finite field 𝔽_{q} and the coefficients a_{i} and b_{i} are never both zero. Suppose that p |̸ k, that s ⩾ 2k + 1, and that any nontrivial linear combination of the two forms contains at least k + 1 variables with nonzero coefficients. Then the system (1) has a nonsingular solution.

If f = 1, so that q = p, then Theorem 1 reduces to the Davenport-Lewis theorem. The condition p |̸ k is necessary, as otherwise the expression (2) would be zero for any pair x_{i}, x_{j} since 𝔽_{q} has characteristic p. The condition s ⩾ 2k + 1 is included to guarantee (by Chevalley’s theorem) that (1) has a nontrivial solution. The final condition guarantees that every nontrivial linear combination of the forms has a nontrivial zero. As Davenport & Lewis show in [DL66], if any nontrivial linear combination of the forms has only the trivial zero then the system (1) has only singular solutions.

It is worth mentioning that these conditions on numbers of variables cannot be replaced by the corresponding conditions on zeros of forms, even over fields with a prime number of elements. For example, consider the system

\begin{aligned}F_{1}(x) = & x_{1}^{6} − x_{2}^{6} + x_{3}^{6} + x_{4}^{6} + x_{5}^{6} & = & 0 \\ F_{2}(x) = & x_{3}^{6} + x_{4}^{6} − 2x_{5}^{6} & = & 0 \end{aligned}

over 𝔽_{7}. Since the only values of x^{6} in 𝔽_{7} are 0 and 1, it is not too hard to check that all of the nontrivial solutions of this system have the form x = (x_{1}, x_{2}, 0, 0, 0) where x_{1} and x_{2} are both nonzero, and that these solutions are all singular. Moreover, if one considers any nontrivial linear combination mF_{1} + nF_{2}, one finds that if m ≠ 0 then x = (1, 1, 0, 0, 0) is a nontrivial zero of this form and if m = 0 then x = (0, 0, 1, 1, 1) is a nontrivial zero.

As an application of Theorem 1, standard methods allow us to prove the following theorem about zeros of forms over 𝔭-adic fields.

Theorem 2. Let p be a prime number and let 𝕃 be any (possibly infinite) algebraic extension of the field ℚ_{p}. Suppose that p|̸ k and consider the system (1) where the coefficients may be any elements of 𝕃. If s ⩾ 2k^{2} + 1, then the system has a nontrivial solution.

While the condition that p |̸ k is annoying, it should be noted that obtaining bounds in the case where p|k is a quite difficult problem. One suspects that the bound s ⩾ 2k^{2} + 1 will also hold in this situation, but this is not even known when 𝕃 is ℚ_{p} itself. When 𝕃 is an extension of ℚ_{p}, the problem is even harder due to the possibility of ramification.

The proof of Theorem 1 is split into two cases depending on the field generated by the k^{th} powers in 𝔽_{q} under addition. If the k^{th} powers generate 𝔽_{q} under addition, then the ideas of Davenport & Lewis suffice to prove the theorem, although one of their key lemmas seems to require a much different proof over general finite fields. If the k^{th} powers generate a proper subfield of 𝔽_{q}, then we replace the system (1) by another system of equations with coefficients in this subfield. This increases the number of equations we need to solve, but it turns out that we are able to simultaneously lower the degrees of the equations. Then the aforementioned result of Chevalley suffices to prove the theorem. Although it is not needed in our proof, we point out that Bhaskaran [Bha66, Theorem G] (see [Bha71] for a corrected proof) has characterized the exponents k which fall into each case.

Once Theorem 1 is proven, Theorem 2 can be proved using techniques which are now fairly standard. First we apply a normalization procedure which allows us to restrict to forms which have certain desirable properties. Then, as indicated above, we reduce the forms modulo the maximal ideal of the field 𝕃 and find a nonsingular solution of the system in the residue field. Finally, we use a version of Hensel’s Lemma to lift this solution to a solution in 𝕃.

2. Notation and Preliminaries

First we fix some notation for the proof of Theorem 1. We define p, q, f, k and s as in the statement of the theorem. Note that since the set of k^{th} powers in 𝔽_{q} is the same as the set of (k, q − 1)-st powers, we may assume without loss of generality that k|(q −1), and we do so throughout the proof. Let F be the set of nonzero k^{th} powers in 𝔽_{q}, and let 𝕂 be the subfield of 𝔽_{q} generated by F under addition. Define r so that 𝕂 contains p^{r} elements, and note that r ⩾ 1 and that 𝔽_{q} is a vector space over 𝕂 of dimension d = f/r.

In both parts of the proof, we use the notion of colored variables developed by Brüdern & Godinho in [BG02]. Consider the vectors

e_{∞} = ( \begin{matrix}1 \\ 0 \end{matrix} )  and  e_{v} = ( \begin{matrix}v \\ 1 \end{matrix} ) , v ∈ 𝔽_{q}.

Note that for any variable x_{i}, the vector of coefficients of x_{i} can be written as

( \begin{matrix}a_{i} \\ b_{i} \end{matrix} ) = ae_{v}

for some a ∈ 𝔽_{q} and v ∈ 𝔽_{q} ∪ {∞}. In this case we call v the color of the variable x_{i}. We say that a color v is zero-representing if the system (1) has a nontrivial solution in which only variables of color v are nonzero.

The utility of this definition is clear upon considering the nonsingularity condition. One can see that there are variables x_{i} and x_{j} such that (2) is nonzero if and only if p |̸ k and the matrix

( \begin{matrix}a_{1}x_{1} & ⋯ & a_{s}x_{s} \\ b_{1}x_{1} & ⋯ & b_{s}x_{s} \end{matrix} )

has rank 2. Equivalently, a solution of (1) is nonsingular if and only if it contains at least two nonzero variables of different colors. This observation leads immediately to the following lemma, which (in the context of congruences modulo powers of p) is Lemma 4.1 of [BG02].

Lemma 1. If two colors are zero-representing, then the system (1) has a nonsingular solution.

Another consequence of this observation implicit in the work of Brüdern & Godinho is that we may assume that exactly one color is zero-representing. If two or more colors represent zero, then Lemma 1 guarantees that the system (1) has a nonsingular solution. If no colors represent zero, then Chevalley’s theorem implies that (1) has a nontrivial solution. But since no color represents zero, this solution must have nonzero variables from at least two different colors and so the solution is nonsingular. Therefore we may assume that precisely one color represents zero. Moreover, by taking linear combinations of the forms if necessary we may assume if we wish that ∞ is the zero-representing color.

3. The Proof of Theorem 1 When 𝕂 = 𝔽_{q}

If the k^{th} powers of 𝔽_{q} actually generate 𝔽_{q} under addition, then our proof proceeds along the lines of the one given by Davenport & Lewis in [DL66]. In fact, our two preliminary lemmata are direct extensions of theirs, although our proof of Lemma 2 is completely different since theirs does not seem to extend to general finite fields. Our proof of Theorem 1 is also slightly different than theirs in order to highlight the use of colored variables, although it retains the same flavor. We begin with the lemmata. Lemma 2. Consider the additive form

G(x) = a_{1}x_{1}^{k} + ⋯ + a_{n}x_{n}^{k}

over 𝔽_{q}, where k|(q − 1) and all of the coefficients are nonzero. Suppose that the k^{th} powers in 𝔽_{q} generate 𝔽_{q} under addition. Then the number of distinct nonzero elements of 𝔽_{q} represented by G is at least min\{\frac{n(q−1)}{k}, q − 1\}. Proof. This proof was inspired by the proof of Theorem 1 of [Tor38]. Note that since the elements of the set of k^{th} powers generate 𝔽_{q} under addition, they are a spanning set for 𝔽_{q} when 𝔽_{q} is considered as a vector space over 𝔽_{p}. Similarly, if a ∈ 𝔽_{q} and a ≠ 0, then the set {ax^{k} : x ∈ 𝔽_{q}} spans 𝔽_{q}. Hence each such set contains a basis for 𝔽_{q} as a vector space over 𝔽_{p}.

Now, if 1 ⩽ i ⩽ n, let

K_{i} = {a_{1}x_{1}^{k} + ⋯ + a_{i}x_{i}^{k} : x_{1}, …, x_{i} ∈ 𝔽_{q}}

and note that K_{i} ⊆ K_{j} whenever i ⩽ j. Our first goal is to show that if K_{i} = K_{i+1} for some i, then we must have K_{i} = 𝔽_{q}. Since the set {a_{i+1}x_{i+1}^{k} : x_{i+1} ∈ 𝔽_{q}} contains an 𝔽_{p}-basis for 𝔽_{q}, the condition K_{i} = K_{i+1} implies that adding any basis element to a member of K_{i} yields another member of K_{i}. Therefore adding any finite sum of basis elements to a member of K_{i} yields another member of K_{i}. Since any element of 𝔽_{q} is a finite sum of basis elements, we see that if one takes any element of K_{i} and adds any element of 𝔽_{q} to it, the sum is still in K_{i}. But this can only happen if K_{i} = 𝔽_{q}.

Now, if K_{i} = K_{i+1} for any i then since K_{i} = 𝔽_{q} the form G represents every nonzero element of 𝔽_{q}, and so the lemma is true. In light of this, the lemma will follow if we can show first that K_{1} contains both zero and at least (q − 1)/k nonzero elements and also that if K_{i} ≠ K_{i+1}, then we have

|K_{i+1}| ⩾ |K_{i}| + \frac{q − 1}{k}.

The first statement is trivial since there are (q − 1)/k nonzero k^{th} powers in 𝔽_{q}, each giving rise to a different nonzero element of K_{1} and since setting x_{1} = 0 gives 0 ∈ K_{1}.

To prove the second statement, suppose that x ∈ K_{i+1} − K_{i}. Since 0 ∈ K_{1}, we know that x ≠ 0. We now show that if y ∈ 𝔽_{q} and y ≠ 0 then y^{k}x ∈ K_{i+1} − K_{i}. Indeed, since x ∈ K_{i+1}, we may write x = a_{1}x_{1}^{k}+⋯+a_{i+1}x_{i+1}^{k} for some x_{1}, …, x_{i+1}, whence we have y^{k}x = a_{1}(yx_{1})^{k}+⋯+a_{i+1}(yx_{i+1})^{k} ∈ K_{i+1}. Now suppose that we had y^{k}x ∈ K_{i}. Then we could write y^{k}x = a_{1}x_{1}^{k} + ⋯ + a_{i}x_{i}^{k} for some (different) x_{1}, …, x_{i}. But since y ≠ 0, we would then have x = a_{1}(y^{−1}x_{1})^{k} + ⋯ + a_{i}(y^{−1}x_{i})^{k} ∈ K_{i}, a contradiction. Again since there are (q − 1)/k nonzero values of y^{k} in 𝔽_{q}, this provides (q − 1)/k elements of K_{i+1} which are not in K_{i}. The second statement, and hence the lemma, follow. □

Lemma 3. Let q = p^{f} and suppose that k|(q − 1). Let G, H be forms defined over 𝔽_{q} of degree k in s variables, where s > k. Let γ_{1}, …, γ_{μ} be distinct nonzero elements of 𝔽_{q}, where

μ > (q − 1)(2k − s)/k.

Then there exist y_{1}, …, y_{s} not all zero such that

\begin{aligned}G(y) & = & 0 or γ_{i} \\ H(y) & = & 0 \end{aligned}

for some i.

This is the generalization to arbitrary finite fields of Lemma 2 of [DL66]. This time, the proof given by Davenport & Lewis generalizes almost without change, and so we will not give a proof here. The only nontrivial alteration involves their appeal to a “well-known principle” (see page 56 of [Dav82]) to guarantee that there is a choice of variables such that a particular auxiliary polynomial is not equal to zero. In the more general setting, this principle should be replaced by a theorem of Alon (Theorem 1.2 of [Alo99]) which generalizes the principle to polynomials over arbitrary fields.

Now we can prove the theorem when 𝕂 = 𝔽_{q}. As mentioned earlier, we may assume that exactly one color of variables represents zero, and without loss of generality we may assume that this color is ∞. Suppose that there are n variables with this color. Then the system (1) looks like

a_{1}x_{1}^{k} + ⋯ + a_{n}x_{n}^{k} + a_{n+1}x_{n+1}^{k} + ⋯ + a_{s}x_{s}^{k} = 0
b_{n+1}x_{n+1}^{k} + ⋯ + b_{s}x_{s}^{k} = 0,

where b_{n+1}, …, b_{s} ≠ 0. The hypotheses of the theorem imply that s ⩾ 2k + 1 and s − n ⩾ k + 1.

Since we are assuming that k|(q − 1), Lemma 2 tells us that the expression a_{1}x_{1}^{k} + ⋯ + a_{n}x_{n}^{k} takes on at least min{n(q − 1)/k, q − 1} nonzero values in 𝔽_{q}. Suppose first that n ⩾ k, so we know that the expression takes on q − 1 nonzero values. Since the color ∞ is zero-representing, this expression nontrivially represents every element of 𝔽_{q}. Since s−n ⩾ k +1, there exist by Chevalley’s theorem elements y_{n+1}, …, y_{s} ∈ 𝔽_{q} not all zero such that b_{n+1}y_{n+1}^{k} + ⋯ + b_{s}y_{s}^{k} = 0. Suppose that a_{n+1}y_{n+1}^{k}+⋯+a_{s}y_{s}^{k} = A. Then we can find y_{1}, …, y_{n} not all zero such that a_{1}y_{1}^{k} +⋯+a_{n}y_{n}^{k} = −A. Then x = y yields a solution of the system (1). Since this solution involves nonzero variables of different colors (one variable from x_{1}, …, x_{n} and at least one from x_{n+1}, …, x_{s}), this solution is nonsingular.

Now suppose that n < k, so that we can only guarantee that the expression a_{1}x_{1}^{k} + ⋯ + a_{n}x_{n}^{k} takes on at least n(q − 1)/k nonzero values. Call these values −γ_{1}, …, −γ_{μ}, where μ ⩾ n(q − 1)/k. Then we want to find y_{n+1}, …, y_{s} not all zero such that we have

a_{n+1}y_{n+1}^{k} + ⋯ + a_{s}y_{s}^{k} = 0 or γ_{i} for some i
b_{n+1}y_{n+1}^{k} + ⋯ + b_{s}y_{s}^{k} = 0.

Now, note that the number of variables here is s − n > k and that since s > 2k we have

μ ⩾ \frac{n(q − 1)}{k}
= \frac{(q − 1)}{k}(2k − (s − n)) + \frac{(q − 1)}{k}(s − 2k)
> \frac{(q − 1)}{k}(2k − (s − n)).

Hence such values of y_{n+1}, …, y_{s} exist by Lemma 3. Suppose that these values give a_{n+1}y_{n+1}^{k} + ⋯ + a_{s}y_{s}^{k} = γ. Then since either γ = 0 or γ = γ_{i} for some i, we can choose y_{1}, …, y_{n} not all zero such that a_{1}y_{1}^{k} + ⋯ + a_{n}y_{n}^{k} = −γ. Thus x = y is a nontrivial solution of the system (1). As before, this solution contains nonzero variables of different colors (one from x_{1}, …, x_{n} and at least one from x_{n+1}, …, x_{s}), and so this solution is nonsingular. This completes the proof of Theorem 1 in the case where 𝕂 = 𝔽_{q}. □

4. The Proof of Theorem 1 When 𝕂 ≠ 𝔽_{q}

If the k^{th} powers in 𝔽_{q} only generate a proper subfield, then the strategy used in the previous section no longer works. In particular, the assumption 𝕂 = 𝔽_{q} is critical in the proof of Lemma 2, and Davenport & Lewis [DL66] give an example showing that the lemma is false without this assumption. Thus we need a different strategy to deal with this situation. We begin with a lemma.

Lemma 4. Suppose that k|(q − 1) and the k^{th} powers in 𝔽_{q} generate the subfield 𝕂 = 𝔽_{p^{r}} under addition. Then the set of k^{th} powers in 𝔽_{q} is the same as the set of m^{th} powers in 𝔽_{p^{r}}, where

m = \frac{k(p^{r} − 1)}{q − 1}.

Proof. Note that 0 is both a k^{th} power in 𝔽_{q} and an m^{th} power in 𝔽_{p^{r}}. Since k|(q − 1), the set F of nonzero k^{th} powers in 𝔽_{q} forms a multiplicative subgroup of 𝔽_{q}^{∗} of cardinality (q − 1)/k. This set is also a multiplicative subgroup of 𝕂^{∗}, and so (q − 1)/k must evenly divide |𝕂^{∗}| = p^{r} − 1. In other words, we have

\frac{q − 1}{k} = \frac{p^{r} − 1}{m}

for some integer m. Now, since 𝕂^{∗} is a cyclic group of order p^{r} − 1 it has a unique subgroup of order (p^{r} − 1)/m, and we have just seen that the set F forms a subgroup of 𝕂^{∗} of this order. Also, since m|(p^{r} − 1) we know that there are (p^{r} − 1)/m nonzero m^{th} powers in 𝕂^{∗}, and these form a subgroup. Hence these subgroups must be the same. The lemma follows. □

Our next lemma provides the meat of the proof. It shows that when 𝕂 ≠ 𝔽_{q}, only a small number of variables are necessary to guarantee that a system such as (1) has a nontrivial (not a nonsingular) solution.

Lemma 5. Suppose that k|(q − 1), that the k^{th} powers in 𝔽_{q} generate under addition the subfield 𝕂 containing p^{r} elements and that q ≠ p^{r}. If q ≠ 4, then the system (1) has a nontrivial solution provided only that s ⩾ k + 1. If q = 4, then the system (1) has a nontrivial solution provided only that s ⩾ k + 2. Proof. Write d = [𝔽_{q} : 𝕂] = f/r. Then we have 𝔽_{q} = 𝕂(α) for some α ∈ 𝔽_{q} having degree d over 𝕂. So any element a_{i} ∈ 𝔽_{q} can be written as

a_{i} = a_{i,0} + a_{i,1}α + ⋯ + a_{i,d−1}α^{d−1},  (3)

where a_{i,0}, …, a_{i,d−1} ∈ 𝕂. Then we can replace an equation such as

a_{1}x_{1}^{k} + ⋯ + a_{s}x_{s}^{k} = 0  (4)

with coefficients in 𝔽_{q} by a system of equations

\begin{aligned}a_{1,0}x_{1}^{k} + ⋯ + a_{s,0}x_{s}^{k} & = & 0 \\ ⋮ & & ⋮ \\ a_{1,d−1}x_{1}^{k} + ⋯ + a_{s,d−1}x_{s}^{k} & = & 0, \end{aligned}  (5)

where the coefficients are now in 𝕂, noting that if the system (5) has a nontrivial solution then so does the equation (4).

Similarly, the system (1) can be replaced by a system

\begin{matrix}a_{1,i}x_{1}^{k} + ⋯ + a_{s,i}x_{s}^{k} = 0 \\ b_{1,i}x_{1}^{k} + ⋯ + b_{s,i}x_{s}^{k} = 0 \end{matrix}  (0 ⩽ i ⩽ d − 1)  (6)

of 2d diagonal equations of degree k with coefficients in 𝕂. As above, we note that if the system (6) has a nontrivial solution then so does the system (1). Since the set of k^{th} powers in 𝔽_{q} equals the set of m^{th} powers in 𝕂 by Lemma 4, replacing all occurrences of x_{i}^{k} in (6) by x_{i}^{m} yields a new system of 2d equations of degree m which has nontrivial solutions over 𝕂 if and only if the system (6) does. Thus, in order to solve (1) over 𝔽_{q}, it suffices to solve this system of degree m equations with all of the variables in 𝕂.

By Chevalley’s theorem, we may do this whenever we have

s > 2dm = 2d\frac{k(p^{r} − 1)}{q − 1}.

So we need to show that

k + 1 > 2d\frac{k(p^{r} − 1)}{q − 1}  (7)

when q ≠ 4 and that k + 2 is greater than the right-hand side of (7) when q = 4. Suppose first that 2d ⩽ p^{d−1}. Then we have

2d\frac{k(p^{r} − 1)}{q − 1} ⩽ kp^{d−1} \frac{p^{r} − 1}{p^{f} − 1}
= kp^{d−1} \frac{1}{p^{f−r} + p^{f−2r} + ⋯ + p^{r} + 1}
⩽ \frac{kp^{d−1}}{p^{f−r}}
⩽ k
< k + 1,

as desired, where the second to last inequality holds because f = dr. So the lemma holds whenever 2d ⩽ p^{d−1}.

Now, since 𝕂 is a proper subfield of 𝔽_{q}, we must have d ⩾ 2. It is not too hard to check that when d ⩾ 2 the only pairs (p, d) for which the condition 2d ⩽ p^{d−1} is false are (2, 3), (3, 2) and (2, 2). If (p, d) = (2, 3), then we can calculate that

2d\frac{k(p^{r} − 1)}{q − 1} = \frac{6k}{2^{2r} + 2^{r} + 1},

which is less than k for all r ⩾ 1. If (p, d) = (3, 2), then we calculate that

2d\frac{k(p^{r} − 1)}{q − 1} = \frac{4k}{3^{r} + 1},

which is at most k for all r ⩾ 1. In either case, we have k + 1 > 2dk(p^{r} − 1)/(q − 1) for all r ⩾ 1.

Finally, if (p, d) = (2, 2), then we have

2d\frac{k(p^{r} − 1)}{q − 1} = \frac{4k}{2^{r} + 1}.

If r ⩾ 2, then 4/(2^{r} + 1) < 1 and we obtain k + 1 > 2dk(p^{r} − 1)/(q − 1). If r = 1 then this inequality does not hold, but in this case we have f = dr = 2 and so q = p^{f} = 4. In this case, the assumptions that k|(q − 1) and 𝕂 ≠ 𝔽_{q} imply that k = 3. Then we can easily see that

k + 2 = 5 > 4 = 2d\frac{k(p^{r} − 1)}{q − 1},

and so Chevalley’s theorem again implies that the system has a nontrivial solution. This concludes the proof of the lemma. □

Now we can prove the theorem when 𝕂 ≠ 𝔽_{q}. As mentioned in Section 2, we can assume that there is precisely one color – call it v – of variables which is zero-representing. Also, the condition in the theorem about linear combinations of the forms implies that there are at least k + 1 variables which are not of this color. Set all but one of the variables of color v equal to zero. Then we are left with one variable of color v and at least k + 1 variables of other colors. By Lemma 5, there exists a nontrivial zero of the system (1) involving only these variables. Note that any nontrivial solution of the system must have at least two nonzero variables. If the variable having color v is nonzero, then this variable and any other nonzero variable are two nonzero variables of different colors, and so the solution is nonsingular. If the variable of color v is equal to zero, then all of the nonzero variables in this solution are from colors which do not represent zero. Thus it is impossible for all the nonzero variables to have the same color, and so there must again be two nonzero variables of different colors. So in this case our solution is also nonsingular. This completes the proof of the theorem. □

5. The Proof of Theorem 2

In order to prove Theorem 2, we first note that it suffices to prove the theorem for finite extensions of ℚ_{p}. Suppose that the theorem is true for finite extensions, and that 𝕃 is an infinite algebraic extension of ℚ_{p}. If we have a system

\begin{aligned}a_{1}x_{1}^{k} + ⋯ + a_{s}x_{s}^{k} & = & 0 \\ b_{1}x_{1}^{k} + ⋯ + b_{s}x_{s}^{k} & = & 0 \end{aligned}  (8)

defined over 𝕃, then we can also think of this system as being defined over the subfield 𝕂 = ℚ_{p}(a_{1}, …, a_{s}, b_{1}, …, b_{s}), which is a finite extension of ℚ_{p}. If s ⩾ 2k^{2} + 1, then the system has a nontrivial solution in 𝕂, which will also be a nontrivial solution in 𝕃. Thus the truth of the theorem for finite extensions implies that it is true for any algebraic extension of ℚ_{p}.

We begin the proof for finite extensions by applying a normalization procedure to the forms. Suppose that 𝕃 is a finite extension of ℚ_{p}, and let π be a generator of the maximal ideal of 𝕃. First note that by clearing denominators if necessary, we may assume that all of the coefficients in our system lie in 𝔒_{𝕃}, the ring of integers of 𝕃. For a system of equations F = (F_{1}(x), F_{2}(x)) = 0 as in (8), we define

Θ(F) = ∏_{1⩽i<j⩽s}(a_{i}b_{j} − a_{j}b_{i}).

By a standard argument involving the compactness of 𝔒_{𝕃}, we may assume that Θ(F) ≠ 0. (One may see [DL69, pp. 572-573] for an example of this argument. As with [DL69, Lemma 11], which will be quoted shortly, although this fact is written down only for the case 𝕃 = ℚ_{p}, it is not hard to see that its proof extends to general 𝔭-adic fields by merely replacing occurrences of ℚ_{p} and p by 𝕃 and π respectively.) Next, we say that two systems of additive equations with coefficients in 𝔒_{𝕃} are equivalent if one can be obtained from the other through a combination of the following three operations:

(i) replacing a variable x_{i} by π^{α}x_{i} for some integer α (ii) dividing one or more equations by an integral power of π (iii) taking nonsingular 𝔒_{𝕃}-linear combinations of the equations.

A system F is said to be π-normalized if both Θ(F) ≠ 0 and the power of π dividing Θ(F) is less than or equal to the power of π dividing Θ(G) for all systems G equivalent to F. Since any system F with Θ(F) ≠ 0 is equivalent to a π-normalized system, it suffices to prove the theorem for πnormalized systems. The next lemma shows that π-normalized systems have some useful properties. After making the slight changes mentioned above, this lemma is [DL69, Lemma 11], specialized to two forms. Lemma 6. A π-normalized system of two additive forms as in (8) can be written (after renumbering the variables) as

F_{1} = f_{1}(x_{1}, …, x_{r}) + πg_{1}(x_{r+1}, …, x_{s})
F_{2} = f_{2}(x_{1}, …, x_{r}) + πg_{2}(x_{r+1}, …, x_{s}),

where r ⩾ s/k and if 1 ⩽ i ⩽ r, then the coefficient of x_{i} in either f_{1} or f_{2} is not divisible by π. Moreover, if we form any nontrivial linear combination of F_{1} and F_{2}, then at least s/2k variables will appear in this linear combination with a coefficient not divisible by π.

Now for i = 1, 2, let F_{i}^{∗} be the form obtained by reducing F_{i} modulo π, and consider the system of congruences

\begin{matrix}F_{1}^{∗} ≡ 0 (mod π) \\ F_{2}^{∗} ≡ 0 (mod π). \end{matrix}  (9)

Since 𝕃 is a finite extension of ℚ_{p}, the residue field 𝔒_{𝕃}/(π) is a finite field containing q = p^{f} elements, where f is the residue degree of 𝕃 over ℚ_{p}. Also, since we are assuming that s ⩾ 2k^{2} + 1, Lemma 6 tells us that (9) is explicit in at least 2k + 1 variables, and that any nontrivial linear combination of F_{1}^{∗} and F_{2}^{∗} is explicit in at least k + 1 variables. Since we are assuming that p|̸ k, the conditions of Theorem 1 are satisfied, and we know that (9) has a nonsingular solution. That is, (9) has a solution x = t such that the expression

k^{2}t_{i}^{k−1}t_{j}^{k−1}(a_{i}b_{j} − a_{j}b_{i})

is nonzero modulo π for some variables t_{i} and t_{j}.

Finally, we must lift this solution of congruences modulo π to a solution in 𝕃 of our original system. For this we use the following version of Hensel’s lemma. This is a special case of [Gre69, Lemma 5.21]. Lemma 7. Let 𝕃 be a finite extension of ℚ_{p} with maximal ideal generated by π. Consider the system F = 0 as in (8) in s ⩾ 2 variables defined over 𝕃. Suppose that there exists t ∈ 𝕃^{s} such that

F(t) ≡ 0 (mod π)

and that the Jacobian matrix

[ \begin{matrix}ka_{1}t_{1}^{k−1} & ⋯ & ka_{s}t_{s}^{k−1} \\ kb_{1}t_{1}^{k−1} & ⋯ & kb_{s}t_{s}^{k−1} \end{matrix} ]

has maximal rank modulo π. Then there exists u ∈ 𝕃^{s} such that F(u) = 0 and u ≡ t (mod π).

Now, we know that there exists t ∈ 𝕃^{s} which satisfies (9). Since t is a nonsingular solution, the Jacobian matrix has a 2×2 submatrix with nonzero determinant modulo π, and hence has maximal rank modulo π. Therefore Lemma 7 may be applied to our nonsingular solution of (9), and this solution lifts to a solution u of (8). Finally, we note that since t ≢ 0 (mod π) and u ≡ t (mod π), we know that u is a nontrivial solution of (8), as desired.□

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Michael P. Knapp mpknapp@loyola.edu Mathematical Sciences Department, Loyola College, 4501 North Charles Street, Baltimore, MD 21210-2699