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Solubility of Additive Forms of Twice Odd Degree over Totally Ramified Extensions of 2\mathbb{Q}_{2}

Drew Duncan
Abstract

We prove that an additive form of degree d=2md=2m, mm odd over any totally ramified extension of 2\mathbb{Q}_{2} has a nontrivial zero if the number of variables ss satisifies sd24+3d+1s\geq\frac{d^{2}}{4}+3d+1.

1 Introduction

Consider additive forms of degree dd over a p-adic field KK in ss variables:

a1x1d+a2x2d++asxsd.a_{1}x_{1}^{d}+a_{2}x_{2}^{d}+\ldots+a_{s}x_{s}^{d}. (1)

Let Γ(d,K)\Gamma^{*}(d,K) represent the minimum number such that any additive form of degree dd over KK in at least that many variables has a nontrivial zero. A refinement of a famous conjecture by Artin holds that Γ(d,K)d2+1\Gamma^{*}(d,K)\leq d^{2}+1 for any pp-adic field KK. The seminal paper in this direction by Davenport and Lewis [1] showed that the bound holds for all fields of pp-adic numbers p\mathbb{Q}_{p}, and that equality holds when d=p1d=p-1. In fact, a similar argument gives an additive form in d2d^{2} variables with no nontrivial zero for any totally ramified pp-adic field and d=p1d=p-1.

Aside from combinations of degrees and fields which make the problem relatively trivial, this bound of d2+1d^{2}+1 has been extended to proper extensions of p\mathbb{Q}_{p} in only a few restricted cases: In [4] it was shown that Γ(d,K)d2+1\Gamma^{*}(d,K)\leq d^{2}+1 for KK any unramified extension of p\mathbb{Q}_{p} with pp odd. It was demonstrated in [3] that Γ(4,K)d2+1\Gamma^{*}(4,K)\leq d^{2}+1 for any KK of the four (out of the six total) ramified quadratic extensions 2(2)\mathbb{Q}_{2}(\sqrt{2}), 2(2)\mathbb{Q}_{2}(\sqrt{-2}), 2(10)\mathbb{Q}_{2}(\sqrt{10}), and 2(10)\mathbb{Q}_{2}(\sqrt{-10}). And, most recently it was proven in [2] that Γ(d,K)d2+1\Gamma^{*}(d,K)\leq d^{2}+1 for KK any of the seven quadratic extensions of 2\mathbb{Q}_{2} and dd not a power of 2. Note that in all of these cases the degree of ramification is at most two. In this paper, I will show that the bound holds for an infinite family of degrees over any totally ramified extension of 2\mathbb{Q}_{2} of arbitrarily large degree. In fact (aside from the case d=2d=2, which is known by other means) I will show that a bound strictly lower than d2+1d^{2}+1 holds in all of these cases.

Theorem 1.

Let d=2md=2m, with mm any odd number at least 3, and KK be any totally ramified extension of 2\mathbb{Q}_{2}. Then,

Γ(d,K)d24+3d+1.\Gamma^{*}(d,K)\leq\frac{d^{2}}{4}+3d+1.

2 Preliminaries

Let KK denote a totally ramified extension of 2\mathbb{Q}_{2}, 𝒪\mathcal{O} denote its ring of integers, and ee denote its degree of ramification, so that 2=uπe2=u\pi^{e} where uu is a unit in 𝒪\mathcal{O}. Without loss of generality, assume ai𝒪\{0}a_{i}\in\mathcal{O}\backslash\{0\}. Let π\pi be a uniformizer (generator of the unique maximal ideal) of 𝒪\mathcal{O}. Any a𝒪a\in\mathcal{O} can be written a=π(a0+a1π+a2π2+a3π3+)a=\pi^{\ell}(a_{0}+a_{1}\pi+a_{2}\pi^{2}+a_{3}\pi^{3}+\ldots), with ai{0,1}a_{i}\in\{0,1\} and a00a_{0}\neq 0. I will refer to \ell as the level of the variable.

By the change of variables πrxd=πrid(πix)d=πridyd\pi^{r}x^{d}=\pi^{r-id}(\pi^{i}x)^{d}=\pi^{r-id}y^{d} for ii\in\mathbb{Z}, we will consider the level of a variable modulo dd for the remainder of the paper. Multiplying a form by π\pi increases the level of each variable by one, and does not affect the existence of a nontrivial zero. Considering the levels of variables modulo dd, applying this transformation any number of times effects a cyclic permutation of the levels. This is useful for arranging the variables in an order which is more convenient, a process to which we will refer as normalization. See Lemma 3 of [1] for a proof of the following Lemma.

Lemma 2.

Given an additive form of degree dd in an arbitrary local field KK, let ss be the total number of variables, and sis_{i} be the number of variables in level i(modd)i\pmod{d}. By a change of variables, the form may be transformed to one with:

s0sd,s0+s12sd,,s0++sd1=s.\displaystyle\begin{split}s_{0}\geq\frac{s}{d},\end{split}\begin{split}s_{0}+s_{1}\geq\frac{2s}{d},\end{split}\begin{split}\ldots,\end{split}\begin{split}s_{0}+\ldots+s_{d-1}=s.\end{split} (2)

Consider a collection of terms ai1xi1d+ai2xi2d++ainxinda_{i_{1}}x_{i_{1}}^{d}+a_{i_{2}}x_{i_{2}}^{d}+\ldots+a_{i_{n}}x_{i_{n}}^{d}. Some assignment xij=bijx_{i_{j}}=b_{i_{j}} of values to the variables yields some value bb. If instead we assign xij=bijyx_{i_{j}}=b_{i_{j}}y, then the terms are replaced with the new term bydby^{d}. I call this replacement a contraction of the original variables. The level of the resulting variable is at least as high as the lowest level of the variables used in the contraction. I proceed by showing that given a configuration of variables enough contractions can be performed so that a variable is produced which is at a sufficiently high level relative to the levels of the variables used to produce it. A nontrivial zero then follows from the following version of Hensel’s Lemma specialized to additive forms of degree d=2md=2m over totally ramified extensions of 2\mathbb{Q}_{2}. (For a proof, see [4].)

Lemma 3 (Hensel’s Lemma).

Let xix_{i} be a variable of (1) at level kk. Suppose that xix_{i} can be used in a contraction of variables (or one in a series of contractions) which produces a new variable at level at least 2e+12e+1. Then (1) has a nontrivial zero.

As a result, we will henceforth assume whenever possible that a contraction or series of contractions produces a variable at most 2e2e levels higher than the variables used in the contractions.

Consider two variables in the same level with coefficients a=π(a0+a1π+)a=\pi^{\ell}(a_{0}+a_{1}\pi+\ldots) and b=π(b0+b1π+)b=\pi^{\ell}(b_{0}+b_{1}\pi+\ldots) with ai,bi{0,1}a_{i},b_{i}\in\{0,1\}. Because a0a_{0} and b0b_{0} are nonzero, they must both be 11, so a+b=2+(a1+b1)π+=(a1+b1)π++πe+a+b=2+(a_{1}+b_{1})\pi+\ldots=(a_{1}+b_{1})\pi+\ldots+\pi^{e}+\ldots, and their sum is divisible by π\pi. Thus, any two variables can be contracted at least one level higher.

Note that (1+πk)d=(1+2πk+π2k)m=(1+uπk+e+π2k)m(1+\pi^{k})^{d}=(1+2\pi^{k}+\pi^{2k})^{m}=(1+u\pi^{k+e}+\pi^{2k})^{m}. If k<ek<e, then π2k\pi^{2k} is a lower order term than every term of uπk+eu\pi^{k+e}, and so (1+πk)d=(1+π2k+)m=1+mπ2k+1+π2k(modπ2k+1)(1+\pi^{k})^{d}=(1+\pi^{2k}+\ldots)^{m}=1+m\pi^{2k}+\ldots\equiv 1+\pi^{2k}\pmod{\pi^{2k+1}}. Suppose a contraction involving a term aixia_{i}x_{i} in level \ell results in a variable in level +2k\ell+2k with k<ek<e. That is, there is some substitution for the variables such that the resulting sum is c=π+2k(1+c1π+)c=\pi^{\ell+2k}(1+c_{1}\pi+\ldots). Replacing the substitution βi\beta_{i} of xix_{i} with βi(1+πk)\beta_{i}(1+\pi^{k}), the resulting sum is c+aiπ2k=π+2k(1+c1π+)+π+2k(1+ai1π+)0(modπ+2k+1)c+a_{i}\pi^{2k}=\pi^{\ell+2k}(1+c_{1}\pi+\ldots)+\pi^{\ell+2k}(1+a_{i_{1}}\pi+\ldots)\equiv 0\pmod{\pi^{\ell+2k+1}}. Thus the contraction which would have a produced a variable 2k2k levels higher can be arranged so that the new variable is instead produced at least 2k+12k+1 levels higher. Similarly, if the contraction would have gone up at least 2k+12k+1 levels, it can be arranged to go up exactly 2k2k levels, for any k<ek<e. I will refer to this ability to make a contraction involving a variable from level \ell bypass or stop at any of the levels +2,+4,,+2e2\ell+2,\ell+4,\ldots,\ell+2e-2 as the resulting variable being free in those levels. Note that if a variable which is free in some levels is used in a subsequent contraction, then the new resulting variable is also free in those levels (as well as new levels arising from this contraction) as the earlier contraction can be arranged to produce an additional term of the appropriate order without affecting any terms of lower order.

3 Proof

Lemma 4.

Let every unordered pair among nn objects be assigned to one of mm bins. If nm+3n\geq m+3, then two disjoint pairs get assigned to the same bin.

Proof.

Suppose there is an assignment of pairs of m+3m+3 objects to mm bins which does not assign two disjoint pairs to the same bin. If a bin contains two pairs, then by hypothesis they must have an object in common, say (a,b)(a,b) and (a,c)(a,c). If that bin contains an addition pair, by hypothesis it is either of the form (b,c)(b,c) or (a,d)(a,d). If it is of the form (b,c)(b,c), then any addition pair would be disjoint to one of the pairs already assigned to that bin. If it is of the form (a,d)(a,d), then any additional pair which does not contain the common object aa would be disjoint to one of the pairs already assigned to that bin. In the former case, the bin contains at most 3 pairs, and in the latter case it contains at most m+2m+2 pairs.

Consider now the number of pairs in each bin. Start with one of these bins with the maximum number of pairs; it contains at most m+2m+2 pairs and has some common object, say aa. Next, take a bin with the maximum number of pairs among the remaining bins; it has some common object bb. If the previous bin contained the pair (a,b)(a,b), then this bin does not, and contains at most m+1m+1 pairs. If the previous bin did not contain (a,b)(a,b), then it contained at most m+1m+1 pairs and thus the present bin also contains at most m+1m+1 pairs. Continuing in this way, note that the next bin contains at most mm pairs, and so on, and so the kth bin considered contains at most m+3km+3-k pairs. Thus we can conclude that the number of pairs contained in all bins is no more than (m+2)+(m+1)++3=(m+32)3(m+2)+(m+1)+\ldots+3=\binom{m+3}{2}-3. ∎

Lemma 5.

Suppose any level contains at least m+7m+7 variables. Then (1) has a nontrivial zero.

Proof.

Without loss of generality, assume the level containing the m+7m+7 variables is level 0. Contracting a pair of variables from level 0 results in a variable that is free in levels 2,4,,2e22,4,\ldots,2e-2. Thus we may assume the new variable is produced in an odd numbered level or in level 2e2e. If there are any pairs which contract to level 2e2e, first perform these contractions. If there are at least two such pairs, then the two resulting variables can be contracted to at least level 2e+12e+1 and a zero follows from Hensel’s Lemma. Now, consider all the pairs which contract to odd numbered levels. There are mm such levels and at least m+3m+3 variables with which to form pairs. By Lemma 4, two pairs contract to the same odd level kk. Contracting this pair results in a variable which is free in levels 2,4,,2e22,4,\ldots,2e-2 and k+2,k+4,,2e1k+2,k+4,\ldots,2e-1, and so can be contracted to level 2e2e. If a variable in level 2e2e was produced in the initial phase, it can be contracted with this new variable to level at least 2e+12e+1 and a zero follows from Hensel’s Lemma. Otherwise, there remain at least m+3m+3 variables in level 0, and so by Lemma 4, another variable can be produced in level 2e2e. A zero follows as before. ∎

Lemma 6.

Suppose level kk contains at least m+3m+3 variables and level k+1k+1 contains at least 22 variables. Then (1) has a nontrivial zero.

Proof.

Without loss of generality, assume k=0k=0. As in Lemma 5, either a pair from level 0 can be contracted to level 2e2e, or two pairs can be contracted to the same odd level, and from there to level 2e2e by lemma 4. Now, contract a pair from level 0 and level 1 to variables y0y_{0} and y1y_{1} at levels ii and jj respectively. We may assume by the free levels of the variables that ii is odd or 2e2e and jj is even or 2e+12e+1. Further, if i=2ei=2e, then y0y_{0} can be contract at least one level higher. Thus assume ii is odd. If ii is less than jj, then because y1y_{1} is free at level jj, we can arrange the contraction so that it goes to level jj instead. The two variables in level jj can be contracted, resulting in a variable which is free at every level lower than 2e2e, the variable is then produced in 2e2e, and a zero follows as before. Thus assume j<ij<i. Then y0y_{0} is free in level jj and a zero follows as before. ∎

Lemma 7.

Suppose level kk contains at least 44 variables and level k+1k+1 contains at least 44 variables. Then (1) has a nontrivial zero.

Proof.

Contract a pair from level 0 and level 1 to variables y0y_{0} and y1y_{1} at levels ii and jj respectively. We may assume by the free levels of the variables that ii is odd or 2e2e and jj is even or 2e+12e+1. If i=2ei=2e, proceed to the next step. Otherwise, assume ii is odd. If ii is less than jj, then because y1y_{1} is free at level jj, we can arrange the contraction so that it goes to level jj instead. The two variables in level jj can be contracted, resulting in a variable which is free at every level lower than 2e2e, the variable is then produced in 2e2e, and a zero follows as before. Thus assume j<ij<i. Then y0y_{0} is free in level jj and a zero follows as before.

Repeat this process again, producing a second variable at level 2e2e. A zero follows. ∎

If the form contains at least d24+3d+1\frac{d^{2}}{4}+3d+1, then by normalization it falls under one of the three lemmas above, completing the proof of Theorem 1.

References

  • [1] H. Davenport and D. J. Lewis. Homogeneous additive equations. Proc. Roy. Soc. London Ser. A, 274:443–460, 1963.
  • [2] Bruno de Paula Miranda, Hemar Godinho, and Michael P. Knapp. Diagonal forms over quadratic extensions of 2\mathbb{Q}_{2}. Publ. Math. Debrecen, to appear.
  • [3] Drew Duncan and David B Leep. Solubility of additive quartic forms over ramified quadratic extensions of 2\mathbb{Q}_{2}. International Journal of Number Theory, 2022.
  • [4] David B. Leep and Luis Sordo Vieira. Diagonal equations over unramified extensions of p\mathbb{Q}_{p}. Bull. Lond. Math. Soc., 50(4):619–634, 2018.