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### An extension to the Hermite-Joubert problem Brassil, Matthew 2016

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`An extension to the Hermite-Joubert problembyMatthew BrassilA THESIS SUBMITTED IN PARTIAL FULFILLMENTOF THE REQUIREMENTS FOR THE DEGREE OFMaster of ScienceinTHE FACULTY OF GRADUATE AND POSTDOCTORAL STUDIES(Mathematics)The University of British Columbia(Vancouver)April 2016c©Matthew Brassil, 2016AbstractLet E/F be a field extension of degree n. A classical problem is to find a generating element in Ewhose characteristic polynomial over F is as simple is possible. An 1861 theorem of Ch. Hermite[5] asserts that for every separable field E/F of degree n there exists an element a ∈ E whosecharacteristic polynomial is of the formf (x) = x5+b2x3+b4x+b5or equivalently, trE/F(a) = trE/F(a3) = 0. A similar result for extensions of degree 6 was provenby P. Joubert in 1867; see [6].In this thesis we ask if these results can be extended to field extensions of larger degree. Specif-ically, we give a necessary and sufficient condition for a field F , a prime p and an integer n≥ 3 tohave the following property: Every separable field extension E/F of degree n contains an elementa ∈ E such that a generates E over F , and trE/F(a) = trE/F(ap) = 0.As a corollary we show for infinitely many new values of n that the theorems of Hermite andJoubert do not extend to field extensions of degree n. We conjecture the same for more values of nand provide computational evidence for a large number of these.iiPrefaceMuch of this thesis is based on an earlier joint paper [1] by Reichstein and myself. Theorem 1.3 iscovered in [1] in a higher generality and some of the introduction is borrowed from [1]. Chapters 2and 3 of this thesis are taken and modified from [1]. Theorem 4.1 appears in [1]. The rest of Chapter4 is originial to this thesis. Some of the results of Chapter 5 appear in [1], others are original tothis thesis and expand upon these results. Further results in Chapter 6 to support Conjecture 1.5 areoriginal to this thesis, as well as the code appearing in Appendix A,iiiTable of ContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Proof of Theorem 1.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.1 Proof of Theorem 1.3: (1) =⇒ (2) . . . . . . . . . . . . . . . . . . . . . . . . . . 62.2 Proof of Theorem 1.3: (2) =⇒ (3) . . . . . . . . . . . . . . . . . . . . . . . . . . 82.3 Proof of Theorem 1.3: (3) =⇒ (1) . . . . . . . . . . . . . . . . . . . . . . . . . . 112.4 Density of F-points on hypersurfaces . . . . . . . . . . . . . . . . . . . . . . . . 132.5 Geometry of the hypersurface Xn,p . . . . . . . . . . . . . . . . . . . . . . . . . . 162.6 Proof of Theorem 1.3: (∗) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 When are there solutions to (1.1)? . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 The Hermite-Joubert problem for p = 2 . . . . . . . . . . . . . . . . . . . . . . . . . 224.1 m≤ 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244.2 m = 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244.3 m = 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.4 m = 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264.5 m≥ 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274.6 Proof of Theorem 4.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275 The Hermite-Joubert problem for p = 3 . . . . . . . . . . . . . . . . . . . . . . . . . 29iv6 Numerical evidence in support of Conjecture 1.5 . . . . . . . . . . . . . . . . . . . . 337 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39A Maple code . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41vChapter 1IntroductionLet E/F be a field extension of degree n. A classical problem is to find a generating element in Ewhose characteristic polynomialf (x) = xn+b1xn−1+ · · ·+bn−1x+nnover F is as simple as possible. For example, any separable quadratic extension F [x]/(x2+bx+ c)over a field F of characteristic 6= 2 is generated by √b2−4c, which has characteristic polynomialx2+(4c−b2). Similarly, any separable cubic extension of a field of characteristic 6= 3 is generatedby an element whose characteristic polynomial is of the form x3+bx+ c for some b,c ∈ F .An 1861 theorem of Ch. Hermite [5] asserts that for every separable field extension E/F ofdegree 5 there exists an element a ∈ E whose characteristic polynomial is of the formf (x) = x5+b2x3+b4x+b5 .A similar result for extensions of degree 6 was proven by P. Joubert in 1867; see [6]. For modernproofs of these results, see [3, 8].Let a ∈ E and a1, . . . ,an be the conjugates of a (with possible repetitions if the degree of a isless than n). Let si denote the elementary symmetric polynomials in n variables, that is, the sum ofmonomials of degree i. Then the characteristic polynomial of a over F is given byxn− s1(a1, . . . ,an)+ · · ·+(−1)nsn(a1, . . . ,an) .The above results of Hermite and Joubert can then be phrased as the existence of an element a in1E of degree n over F such that its conjugates satisfy s1(a1, . . . ,an) = s3(a1, . . . ,an) = 0. Denote byti the ith power sum in n variables, that is,ti(a1, . . .an) = ai1+ · · ·+ain .Newton’s identities yield the following:t1 = s1t2 = s21−2s2t3 = s31−3s1s2+3s3 .This implies we have s1 = s3 = 0 if and only if t1 = t3 = 0. If E = F [α], for each a ∈ E weconsider the linear transformation x 7→ ax with basis 1,α,α2, . . . ,αn−1. We have ti(a1, . . . ,an) =ai1+ · · ·+ain = trE/F(ai), where the trace here is the trace of the linear transformation x 7→ ax. Notethat the trace depends only on the field extension E/F and not on the choice of basis. Coray’sapproach to the Hermite-Joubert problem studied the existence of F-points in the projective varietyin Pn−1 defined by trE/F(x) = trE/F(x3) = 0, with coordinates being a basis 1,α, . . . ,αn−1 of Eover F .In this thesis we will consider a generalization of the Hermite-Joubert problem: For a fixedn ≥ 7, does every separable field extension E/F of degree n contain an element a ∈ E whichgenerates E over F and satisfies trE/F(a) = trE/F(a3) = 0? The answer is “no” if n is of the form 3kor 3k1 +3k2 where k1 > k2 ≥ 0, see [11, Theorem 1.3]. We will see that the answer remains “no” forn of the form 3k1 +3k2 +3k3 where k1 > k2 > k3 ≥ 0, under additional assumptions on the integersk1,k2,k3, and we conjecture that the additional assumptions are unneeded. For other values of n(in particular, for n = 7), the question remains open. One can also ask a similar question for anarbitrary prime p.Question 1.1. Let n≥ 2 be an integer, p be a prime, and F0 be a base field. Which triples (F0, p,n)have the property that for every field extension F/F0 and every e´tale algebra E/F of degree n, thereexists an element a ∈ E such that a generates E as an F-algebra and trE/F(a) = trE/F(ap) = 0?Here an e´tale algebra E/F is defined to be a finite product of separable field extensions overF . The degree of an e´tale algebra E/F is the dimension of E as an F-algebra. For example, anyseparable polynomial f in x gives rise to an e´tale algebra F [x]/ f . If f = f1 · · · fn, with each fiirreducible and separable, then F [x]/ f ∼= F [x]/ f1×·· ·×F [x]/ fn.2In this paper we will show that this question becomes tractable if we restrict our attention to thecase where F is a p-field. A field F is called a p-field if the degree of every finite field extensionof F is a power of p. If F is a p-field we say that F is p-closed. Then our question becomes:Question 1.2. Let n≥ 2 be an integer, p be a prime, and F0 be a base field. Which triples (F0, p,n)have the property that for every p-field F containing F0, and every e´tale algebra E/F of degree n,there exists an element a∈E such that a generates E as an F-algebra and trE/F(a) = trE/F(ap) = 0?Before stating our main results, we recall the definition of the “general field extension” En/Fnof degree n. Let F0 be a base field and x1, . . . ,xn be independent variables. Set Ln := F0(x1, . . . ,xn),Fn := LSnn and En := LSn−1n = Fn(x1), where Sn acts on Ln by permuting x1, . . . ,xn and Sn−1 bypermuting x2, . . . ,xn.Let p be a prime. We will say that n = pk1 + pk2 + · · ·+ pkm is the base p expansion of n if eachpower of p appears in the sum at most p−1 times. It is well known that the base p expansion of nis unique.Theorem 1.3. Let p be a prime, F0 be a field of characteristic 6= 2,3 or p containing a primitivepth root of unity ζp, and n = pk1 + · · ·+ pkm be the base p expansion of an integer n≥ 3. Then thefollowing conditions are equivalent.(1) For every p-field F containing F0 and every n-dimensional e´tale algebra E/F, there exists anelement 0 6= a ∈ E such that trE/F(a) = trE/F(ap) = 0.(2) There exists a finite field extension F ′/Fn of degree prime to p and an element 0 6= a ∈ E ′ :=En⊗Fn F ′ such that trE ′/F ′(a) = trE ′/F ′(ap) = 0.(3) The system of equationspk1y1+ · · ·+ pkmym = 0pk1yp1 + · · ·+ pkmypm = 0(1.1)has a solution y = (y1 : · · · : ym) ∈ Pm−1(F0).(∗) Moreover, assume that condition (3) holds and one of the following additional conditions ismet:(i) There is a solution y = (y1 : · · · : ym) ∈ Pm−1(F0) to (1.1) such that y 6= (1 : . . . : 1), or3(ii) p > 2.Then the element a in parts (1) and (2) can be chosen so that E = F [a] and E ′ = F ′[a], respectively.Note that conditions (i) and (ii) of assertion (∗) are very mild. In particular, condition (i) issatisfied whenever charF0 does not divide n, since in this case (1 : · · · : 1) is not a solution to (1.1).We will prove Theorem 1.3 in Chapter 2. Sections 2.1, 2.2 and 2.3 will cover the main statement ofthe theorem. The assertion (∗) will be proven in section 2.6, after establishing preliminary resultsin sections 2.4 and 2.5.It is natural to ask for which n, p and F0 there exist solutions to the system (1.1). We will givesome partial results to this question in Chapter 3. In particular, we show that for p≥ 3 the system(1.1) has a solution over any field F0, if when we write n = [ndnd−1 . . .n0]p in base p, one of thedigits ni is ≥ 2, or if the number of non zero digits is ≥ p+3. (Here ni is the number of times pioccurs in the base p expansion n= pk1 + · · ·+ pkm . If each ni is 0 or 1, then the number of non zerodigits is m.)We will also see that the system (1.1) has no solutions when n = pk for any k ≥ 1, or whenn = pk1 + pk2 and charF0 does not divide p(k1−k2)(p−1)+(−1)p, for any k1 > k2 ≥ 0.We will give more detailed results for particular cases of n when p = 2 and p = 3 in Chapters4 and 5 respectively. Due to a theorem of Springer theorem, proven in [9, VII. Theorem 2.7], wecan remove the requirement of F being a 2-field in condition (1) in the case of p = 2. This leads toa nearly complete solution of Question 1.1 when p = 2.If we assume an analogous conjecture of Cassels and Swinnerton-Dyer [2] then we could alsoremove the requirement of F being a 3-field in the case of p = 3. We obtain the following resultfor p = 3 when n has 3 digits base 3.Theorem 1.4. Let En/Fn be the general field extension of degree n, over the base field F0 = Q.Suppose n = 3k1 + 3k2 + 3k3 , where k1 > k2 > k3 ≥ 0. If k1 + k2 + k3 6≡ 2 (mod 3) or k1 6≡ k2(mod 3), then for any finite field extension F ′/Fn of degree prime to 3 there does not exist anelement 0 6= a ∈ E ′n := En⊗Fn F ′ such that trE ′/F ′(a) = trE ′/F ′(a3) = 0.We provide a proof of this theorem in Chapter 5. This yields new examples where the Hermite-Joubert problem has a negative solution, the smallest being n= 13= 32+31+30. In Chapter 6 weprovide some evidence in support of the following conjecture.Conjecture 1.5. Theorem 1.4 remains true for all triples k1 > k2 > k3 ≥ 0.The maple code used to establish this evidence is provided in Appendix A.4Much of this thesis is based on an earlier joint paper of Reichstein and myself. Theorem 1.3is covered in [1] in a higher generality, and Chapters 2 and 3 of this thesis are taken from [1].Chapters 4 and 5 expand upon the paper, and further results to support Conjecture 1.5 are originalto this thesis.5Chapter 2Proof of Theorem 1.32.1 Proof of Theorem 1.3: (1) =⇒ (2)For every field F there exists an algebraic extension F(p)/F , such that F(p) is p-closed, and forevery finite subextension F ⊆ F ′ ⊆ F(p), the degree [F ′ : F ] is prime to p. We call this field F(p)the p-closure of F . This field is unique with respect to these conditions up to F-isomorphism. Fora proof of its existence, see [4, Proposition 101.16].Lemma 2.1.1. Let E/F be an e´tale algebra of degree n.(a) Every element a ∈ E⊗F F(p) lies in the image of the natural mapφ : E⊗F F ′ ↪→ E⊗F F(p)for some intermediate field F ⊂ F ′ ⊂ F(p) (depending on a), such that [F ′ : F ] is finite (and thusprime to p).(b) a ∈ E ′ := E⊗F F ′ generates E ′ over F ′ if and only if φ(a) generates E⊗F F(p) over F(p).Proof. (a) Let e1, . . . ,en be a basis of E, viewed as an F-vector space. Thena = f1(e1⊗1)+ · · ·+ fn(en⊗1)for some f1, . . . , fn ∈ F(p), and we can take F ′ = F( f1, . . . , fn).(b) Let e1, . . . ,en be a basis for E over F . Then e1⊗ 1, . . . ,en⊗ 1 forms a basis for E ⊗F F ′over F ′ as well as for E⊗F F(p) over F(p). Working in these bases we then have 1,a, . . . ,an−1 arelinearly dependent over F ′ if and only if 1,φ(a), . . . ,φ(a)n−1 are linearly dependent over F(p).6Lemma 2.1.2. The condition (1) of Theorem 1.3 is equivalent to the following:(1′) For every field F containing F0 and every n-dimensional e´tale algebra E/F, there exists afinite field extension F ′/F of degree prime to p and an element 0 6= a ∈ E ′ := E ⊗F F ′ such thattrE ′/F ′(a) = trE ′/F ′(ap) = 0.Proof. For (1′) =⇒ (1) the result follows immediately, as F ′ = F is the only finite field extensionof a p-field F such that [F ′ : F ] is prime to p.For the other direction, apply (1) to E ′′ := E⊗F F(p) over F(p) to obtain an element 0 6= a ∈ E ′′satisfying trE ′′/F(p)(a) = trE ′′/F(p)(ap) = 0. Then by Lemma 2.1.1 there exists a subfield F ⊆ F ′ ⊆F(p) such that [F ′ : F ] is prime to p and a lies in the natural mapE ′ := E⊗F F ′ ↪→ E ′′ = E⊗F F(p) .This a as an element of E ′ then satisfies trE ′/F ′(a) = trE ′/F ′(ap) = 0.The implication (1) =⇒ (2) of Theorem 1.3 follows from Lemma 2.1.2 by setting F = Fn andE = En.72.2 Proof of Theorem 1.3: (2) =⇒ (3)Let Z be an F0-variety whose F0-algebra of rational functions is isomorphic to L′ := Ln⊗Fn F ′.Define an Sn-action on Z to be the action induced by the Sn-action on L′, so that σ( f )(z) = f (σ(z))for all z ∈ Z, f ∈ L′. The Sn-equivariant inclusionLn ↪→ L′ = Ln⊗Fn F ′then gives rise to a dominant Sn-equivariant rational map Z 99K An of degree d := [F ′ : Fn].Let σi ∈ Sn be such that σi(1) = i. These represent the n cosets of Sn−1 in Sn. As a ∈ E ′ =(L′)Sn−1 , σi(a) are the conjugates of a in L′ and depend only on the coset hi Sn−1 and not on theparticular choice of hi in this coset. Now we define the rational mapfa : Z 99K Pn−1z 7→ (σ1(a)(z) : · · · : σn(a)(z)) .This map is well-defined, as σ1(a) = a 6= 0.Let Sn act on Pn−1 in the natural way. For any τ ∈ Sn and z ∈ Z we haveτ( fa(z)) = τ(σ1(a)(z) : · · · : σn(a)(z))= τ(a(σ1(z)) : · · · : a(σn(z)))= (a(στ−1(1)(z)) : · · · : a(στ−1(n)(z)))= fa(τ(z))and so fa is Sn-equivariant.Let Xn,p ⊆ Pn−1 be the variety defined byy1+ · · ·+ yn = 0yp1 + · · ·+ ypn = 0 .As trE ′/F ′(a) =σ1(a)+ · · ·+σn(a) = 0 and trE ′/F ′(ap) = 0, the image of fa lies in Xn,p. In summary,we have the following diagram of Sn-equivariant rational maps:8Zfa''generically d : 1An Xn,p  // Pn−1Let H =(Z/pZ)k1×·· ·×(Z/pZ)km , viewed as a subgroup of Sn by embedding into Spk1 ×·· ·×Spkm via regular representations on each factor.We note that An contains a smooth H-fixed point (e.g. (0, . . . ,0)) and apply the going uptheorem of J. Kolla´r and E. Szabo´ [12, Proposition A.2]. The assumptions of the theorem aresatisfied, being that X is complete, fa is rational and that every linear representation of H has a1-dimensional invariant subspace (as H is abelian and F0 contains all pth roots of unity). HenceXn,p contains a H-fixed point defined over F0(ζp).Let z ∈ Pn−1 be a H-fixed point. Then z corresponds to a 1-dimensional subspace of An fixedby H. As H ⊆ Spk1 ×·· ·×Spkm , its representation on An = F0[H] decomposes as F0[H] = W1⊕·· ·⊕Wm, where Wi = Apki and we have H acts invariantly on each Wi.The action of H on each Wi is given by the regular representation of (Z/pZ)ki . As (Z/pZ)ki isabelian, its action on Wi is decomposable as a direct sum of irreducible representations of dimension1 so that Wi =V1⊕·· ·⊕Vpki with (Z/pZ)ki acting invariantly on each Vj.Denote an element of (Z/pZ)ki by (γ1, . . . ,γki). Then the characters of (Z/pZ)ki can be de-scribed asχγ1...γki : (Z/pZ)ki → F0(β1, . . . ,βki) 7→ ζ γ1β1p · · ·ζγkiβkipwhere we have γ j,β j ∈ {0, . . . , p−1}. Then the irreducible subrepresentation of Wi correspondingto (γ1, . . .γk) is spanned by (χγ1...γki (g1), . . . ,χγ1...γki (gpki )), where g j are the elements of (Z/pZ)ki .In particular, the coordinates are all pth roots of unity.Let Vh be the subspace spanned by (χh(g1), . . . ,χh(gpki )). We check that these Vh are the only1-dimensional invariant subspaces of Wi. Let w ∈Wi and write w = ∑h∈(Z/pZ)ki vh, with vh ∈ Vhfor each h. For each g ∈ (Z/pZ)ki we have g.(∑vh) = ∑χh(g−1)vh. If w lies in a 1-dimensionalsubspace of Wi and vh,v′h are nonzero, we must have χh(g) = χh′(g) for all g ∈ (Z/pZ)ki and soh = h′ and w ∈Vh. Thus the 1-dimensional invariant subspaces of Wi are exactly these Vh.Write z = (z1, . . . ,zm) ∈W1⊕·· ·⊕Wm ∼= An. Then for each i, j such that zi,z j 6= 0, there existhi ∈ (Z/pZ)ki and h j ∈ (Z/pZ)k j corresponding to characters χi of Wi and χ j of Wj such that9χi(gi) = χ j(g j) for all gi ∈ (Z/pZ)ki and g j ∈ (Z/pZ)k j . If zi 6= 0 for at least 2 different i, then thisrequires χi being the trivial character for all i such that zi 6= 0. Otherwise zi = 0 for all i 6= j for aparticular j.Fix j ∈ {1, . . . ,m} and suppose zi = 0 for all i 6= j. Then the coordinates of z j are all λζ `p forsome pth roots of unity ζ `p and for some fixed scalar λ 6= 0. As z ∈ X we thus have (λζ `1p )p+ · · ·+(λζ`pkip )p = λ p pki = 0, which is impossible in characteristic 6= p.Thus z is acted on by H by the trivial character. Writing z = (z1 : · · · : zn) as an elementof Pn−1 we then have zi = z j when i and j lie in the same H-orbit and so there are at most mdifferent coordinates of z. Using these coordinates, and as z ∈ X , we have thus obtained a solutiony = (y1 : · · · : yn) ∈ Pm−1(F0) to the system of equationspk1y1+ · · ·+ pkmym = 0pk1yp1 + · · ·+ pkmypm = 0 ,as required.102.3 Proof of Theorem 1.3: (3) =⇒ (1)For the implication (3) =⇒ (1) it suffices to prove the following:Lemma 2.3.1. Let p be a prime and n ≥ 1 be a positive integer with base p expansion n = pk1 +· · ·+ pkm . Suppose F is a p-field and E is an e´tale algebra of degree n over F. Then we can writeE ∼= E1×·· ·×Em, where each Ei is an e´tale algebra of degree pki over F.Indeed, suppose this lemma holds, and let y = (y1 : · · · : ym) ∈ Pm−1(F0) be a solution to (1.1)as in condition (3). Let F be a p-closed field and E/F be an n-dimensional e´tale algebra. Fromthe lemma, we can write E ∼= E1×·· ·×Em with each Ei an e´tale algebra of degree pki over F . Weview a := (y11E1 , . . . ,ym1Em) as a nonzero element of E1×·· ·×Em. As each yi lies in F0 ⊆ F wehave trEi/F(yi) = pkiyi and trEi/F(ypi ) = pkiypi . Then we havetrE/F(a) = trE1/F(yi)+ · · ·+ trEm/F(yi) = pk1y1+ · · ·+ pkmym = 0trE/F(ap) = trE1/F(ypi )+ · · ·+ trEm/F(ypi ) = pk1yp1 + · · ·+ pkmypm = 0and soa = (y11E1 , . . . ,ym1Em) (2.3.1)gives us a solution to (1).We now consider Lemma 2.3.1. Let F be a p-field and E/F be an e´tale algebra of degree n overF . As the only finite field extensions of F have degree a power of p, we can write E ∼= F1×·· ·×Ftwith each Fi being a field extension of F of degree a power of p, and the sum of the degrees beingn.To prove the lemma, it suffices to arrange these fields into m distinct groups G1, . . . ,Gm, suchthat Ei :=∏Fj∈Gi Fj is an e´tale algebra of degree pki over F , for each i. This is satisfied if and onlyif ∑Fj∈Gi [Fj : F ] = pki . Thus, it suffices to rearrange the integers [Fj : F ], which partition n, intodistinct groups such that the sum of the integers in each group i is pki .A partition λ = [λ1, . . . ,λr] of n is an unordered collection of (possibly repeated) positive inte-gers λ1, . . . ,λr such that λ1+ · · ·+λr = n. For two partitions λ = [λ1, . . . ,λr] and µ = [µ1, . . . ,µs]of n we say that λ is a refinement of µ , and write µ  λ , if µ can be obtained from λ by par-titioning each of the numbers λ1,λr. Equivalently, we have µ  λ if µ1, . . . ,µs can be arrangedinto r distinct collections such that the sum of the numbers in each group i is λi. For example, wehave [3,2,1] ≺ [3,3] ≺ [6] and [3,2,1] ≺ [5,1] ≺ [6]. This operation defines a partial order on thepartitions of n.11Let p be a prime. We call a partition λ = [λ1, . . . ,λr] a p-partition of n if each λi is a powerof p. Every integer n ≥ 1 has a unique p-partition [pk1 , . . . , pkm ] corresponding to the base p ex-pansion n = pk1 + · · · = pkm . We denote this partition by [n]p. As the partition arising from thedecomposition E ∼= F1×·· ·×Fr is a p-partition of n, we reduce Lemma 2.3.1 to the following:Lemma 2.3.2. Let λ = [λ1, . . . ,λr] be a p-partition of n. Then λ  [n]p.LetP be the set of all p-partitions of n. As there are finitely many p-partitions of n, to provethis lemma it suffices to show that [n]p is the unique maximal element of P with respect to therefinement ordering. Again, asP is finite, we know that there exists at least one maximal partition,so it suffices to show that if λ 6= [n]p, then λ is not maximal.We assume λ 6= [n]p. The base p expansion n = pk1 + . . . pkm is unique with the property thateach power of p appears in the sum at most p− 1 times, so the partition [n]p = [pk1 , . . . , pkm ] isunique with the same property. Thus some power of p appears in λ at least p times, say λ1 = · · ·=λp = p`. Then we haveλ = [p`, . . . , p`︸ ︷︷ ︸p times,λp+1, . . . ,λr]≺ [p`+1,λp+1, . . . ,λr]with the latter partition again lying inP . Thus λ is not maximal and so [n]p is the unique maximalelement ofP , proving Lemma 2.3.2, and hence the implication (3) =⇒ (1).122.4 Density of F-points on hypersurfacesRecall that a closed subvariety X of projective space is called a cone over a point c∈X if X containsthe line through c and c′ for every c 6= c′ ∈ X . We will say that X is a cone if it is a cone over oneof its points.Lemma 2.4.1. Let X be a degree d hypersurface in P` defined over a field F, which is not a coneover c ∈ X. If H ' P` is a hyperplane defined over F, then the projection pi : X \ {c} → H isdominant.Proof. Without loss of generality, after a change of basis we may assume c=(1 : 0 : · · · : 0) and H isdefined by x0 = 0. Let X be defined by f of degree d. Let U = {(0 : a1 : . . .a`) | ai 6= 0 for each ai},which is dense and open in H.For a contradiction we suppose there exists a = (0 : a1 : · · · : a`) ∈U such that a /∈ pi(X \{c}).Then f (x0 : a1 : · · · : a`) is nonzero of degree zero as a polynomial in x. As each of the ai arenonzero we thus have x0 does not appear in f . But then for all b = (b0 : · · · : b`) ∈ X we havef (λ1+λ2b0 : λ2b1 : · · · : λ2b`) = f (λ2b0 : λ2b1 : · · · : λ2b`) = f (b) = 0 .Hence X contains the line through c and b. As b ∈ X was arbitrary we have a contradiction. ThusU ⊆ pi(X \{c}) and hence pi is dominant.An effective zero cycle in X is a subvariety (possibly reducible) of X of dimension 0. Thedegree of a zero cycle Z in X is defined as dimF(F [Z]). Our main lemma for this section is thefollowing.Lemma 2.4.2. Let F be a p-closed field of characteristic 6= p. Suppose X ⊂ P` is a closed hyper-surface of degree ≤ p defined over a p-closed field F of characteristic 6= p. Assume that X has anF-point c such that X is not a cone over c. Then F-points are dense in X.Proof. Case 1: X is a hypersurface of degree d < p. Note that effective zero cycles of degree dare dense in X (these can be obtained by intersecting X with lines defined over F in P`). Since Fis a p-closed field, every effective zero cycle of degree d < p splits over F , that is, is a sum of dF-points. Consequently, F-points are dense in X .Case 2: X is reducible over F , that is, its irreducible components, X1, . . . ,Xr are defined over Fand r ≥ 2. Here each Xi is a hypersurface of degree < p. By Case 1, F-points are dense in each Xi;hence, they are dense in X .13Case 3: X is irreducible over F but reducible over F . Note that since F is a p-closed field,and charF 6= p, F is perfect. Hence, the irreducible components X1, . . . ,Xr of X are transitivelypermuted by the Galois group Gal(F/F), which is a pro-p group. Thus r ≥ 2 is a power of p.Moreover, since deg(X) = deg(X1)+ · · ·+ deg(Xr) ≤ p, we conclude that r = p and deg(X1) =· · · = deg(Xp) = 1. In other words, X is a union of the hyperplanes X1, . . . ,Xp. Now observe thatc ∈ X(F) is fixed by Gal(F/F). After relabeling the components, we may assume that c ∈ X1.Translating X1 by Gal(F/F), we see that c lies on every translate of X1, that is, on every Xi fori = 1, . . . ,r. Since each Xi is a hyperplane, we conclude that X is a cone over c, contradicting ourassumption.Case 4: X is absolutely irreducible. Choose a hyperplane H ' P`−1 in P` such that H is definedover F and c 6∈H. Let pi : X \{c}→H be projection from c. Since X is not a cone over c, this mapis dominant, by Lemma 2.4.1. In particular, there is a dense open subset U ⊂H such that pi is finiteover U . The preimage pi−1(u) of any F-point u ∈U(F) is then an effective zero cycle of degree≤ p− 1. Once again, every such zero cycle splits over F , that is, pi−1(u) is a union of F-points.Taking the union of pi−1(u), as u varies over U(F), we obtain a dense set of F-points in X .If p= 2 or 3, then Lemma 2.4.2 remains true for all infinite fields F (not necessarily p-closed),under mild additional assumptions. We say X is rational over F if X is birationally isomorphic tosome projective space Pk over F .Lemma 2.4.3. (a) Let X ⊂ P` be a hypersurface of degree 2 defined over a field F of characteristic6= 2. Assume X has an F-point and X is not a cone. Then X is rational over F. In particular,F-points are dense in X.(b) (J. Kolla´r) Let X ⊂ P` be an absolutely irreducible cubic hypersurface of dimension ≥ 2defined over a field F. Assume X has an F-point, and X is not a cone. If X is singular, assume alsothat charF 6= 2 or 3. Then X is unirational over F. In particular, if F is infinite, then F-points aredense in X.Proof. (a) Suppose X is given by q = 0, where q is a quadratic form on F`+1. If X were reducible,then q would be a product of two linear forms, say q = l1l2. Then X would be a cone over theintersection of the lines defined by l1 and l2. This is not the case, hence X is irreducible. As q isirreducible, X is also smooth. The stereographic projection from a point c ∈ X(F) to a hyperplaneH ⊂ P` defined over F and not passing through c, gives rise to a birational isomorphism betweenX and H.14(b) If X is smooth, see [7, Theorem 1.1]. If X is singular and F is perfect, see [7, Theorem 1.2].Finally, if X is singular and F is an imperfect field of characteristic 6= 2,3, see the remark after thestatement of Theorem 1.2 on [7, p. 468].152.5 Geometry of the hypersurface Xn,pIn this section we will prove some geometric properties of the hypersurfaceXn,p := {(x1 : · · · : xn) | x1+ · · ·+ xn = xp1 + · · ·+ xpn = 0} ⊂ Pn−1defined over the base field F0. These will be used in the next section to prove assertion (∗).Let ∆n be the union of the “diagonal” hyperplanes xi = x j, over all 1≤ i < j ≤ n.Lemma 2.5.1. Assume charF0 6= 2,3 or p. Then(a) The singular locus of Xn,p is Xn,p∩{(x1 : · · · : xn) | xp−11 = · · ·= xp−1n }.(b) Xn,p is absolutely irreducible if n≥ 5.(c) Xn,p is not contained in ∆n if n≥ 3.(d) Let (1 : · · · : 1) 6= c ∈ Xn,p. Then Xn,p is not a cone over c.(e) Xn,p is not a cone if p > 2.Proof. In order to prove the lemma we may, without loss of generality, pass to the algebraic closureof F0, that is, assume that F0 is algebraically closed.(a) follows from the Jacobian criterion.(b) Assume the contrary: Xn,p has at least two irreducible components, X1 and X2. Since Xn,pis a hypersurface in Pn−1, dim(X1) = dim(X2) = n− 3, and dim(X1 ∩X2) = n− 4 with X1 ∩X2contained in the singular locus of X . By part (a), the singular locus of X is 0-dimensional. Thusn−4≤ 0, a contradiction, and the result follows.(c) First assume n ≥ 5 and assume the contrary. We have Xn,p is irreducible by part (b) andhence lies in one the hyperplanes defined by xi = x j. Since Xn,p is invariant under the action of Sn,it is contained in every hyperplane Hi j. That is,Xn,p ⊂⋂1≤i< j≤nHi j = {(1 : · · · : 1)} ,contradicting dim(Xn,p) = n−3≥ 2.In the remaining cases, where n = 3 or 4, we will exhibit a point y ∈ Xn,p which does not lie in∆n.If n = 4 we can take y := (1 : ζ4 : ζ 24 : ζ 34 ), where ζ4 ∈ F0 is a primitive 4th root of unity (recallthat we are assuming F0 to be algebraically closed and charF0 6= 2).16Now suppose n = 3. If p 6= 2, then we can take y := (1 : −1 : 0). Otherwise we can takey := (1 : ζ3 : ζ 23 ). This completes the proof of (c).(d) Assume the contrary. Since Sn acts on Xn,p ⊂ Pn−1 by permuting coordinates, Xn,p is a coneover g · c for every g ∈ Sn. From this it follows that Xn,p contains the linear span of {g · c | g ∈ Sn}.Denote this linear span by L. Let H ' Pn−2 be defined byH := {(x1 : · · · : xn) | x1+ · · ·+ xn = 0} .Then L is an Sn-invariant linear subspace of H. If charF0 does not divide n, then the Sn-representationon H is irreducible. Hence, the only Sn-invariant subspace of H is H itself. Thus H = L ⊂ Xn,p,a contradiction. If charF0 divides n, then the only other possibility is L = {(1 : 1 · · · : 1)}. This isruled out by our assumption that c 6= (1 : · · · : 1).(e) Assume the contrary, that Xn,p is a cone over some point c∈Xn,p. By part (d), c=(1 : · · · : 1).Note here that this implies charF0 divides n. Whenever Xn,p contains a point y = (y1 : · · · : yn), itcontains the entire line through c and y. That is,(1+ ty1)p+ · · ·+(1+ tyn)p = 0as a polynomial in t. In particular, p(yp−11 + · · ·+ yp−1n ), which is the coefficient of t p−1 in thispolynomial, should vanish for every y= (y1 : · · · : yn) ∈ Xn,p. Set y := (−1 : 1 : 0 : · · · : 0). Note thaty ∈ Xn,p, because p > 2. For this y, p(yp−11 + · · ·+ yp−1n ) = 0 reduces to 2p = 0, contradicting ourassumptions that charF0 6= 2 or p.172.6 Proof of Theorem 1.3: (∗)In this section we complete the proof of Theorem 1.3 by proving assertion (∗).Given an e´tale algebra E/F of degree n, we define XE/F,p as the degree p hypersurface inP(E) = Pn−1F , given by trE/F(x) = trE/F(xp) = 0. Let ∆E/F be the discriminant locus in P(E), thatis, the closed subvariety of P(E) defined by the condition 1,a, . . . ,an−1 are linearly dependent overF (here a ∈ E).Lemma 2.6.1. XE/F,p and ∆E/F are F-forms of the varieties Xn,p and ∆n defined above.Proof. It suffices to show that XE/F,p ∼= Xn,p and ∆E/F ∼= ∆n when we pass to the algebraic clo-sure of F . That is, it suffices to show that the varieties are isomorphic when we assume F to bealgebraically closed.When F is algebraically closed, there exist no finite field extensions and hence E ∼= Fn. Thenwe have trE/F(x1, . . . ,xn) = x1 + · · ·+ xn and trE/F(xp1 , . . . ,xpn) = xp1 + · · ·+ xpn and hence XE/F,p ∼=Xn,p. In the standard basis for Fn over F , the discriminant locus becomes the determinant of theVandermonde matrix of an element (a1, . . . ,an). This is nonzero if and only if the coordinates aredistinct. That is, we have ∆E/F is isomorphic to ∆n, being the union of the hyperplanes xi = x j.We now prove the assertion (∗) of Theorem 1.3, which we do by appealing to Lemma 2.4.2.Assume (3) holds. Our goal is to show that a can be chosen so that (†) E = F [a] in part (1), and(††) E ′ = F ′[a] in part (2). We will see that it suffices to prove (†) in part (1).Suppose (†) holds in part (1), and in particular that a generates En⊗Fn F(p)n over F(p)n . Thenby Lemma 2.1.1(a), a lies in the image of the natural map φ : E⊗F F ′→ E⊗F F(p) for some fieldextension F ′/Fn such that [F ′ : F ] is prime to p. By Lemma 2.1.1(b) we have φ−1(a) generatesEn⊗Fn F ′ over F ′. Thus (††) follows from (†) and so it suffices to prove (†) for part (1), which canbe restated as follows:If XE/F,p has an F-point, then XE/F,p has an F-point away from ∆E/F .As XE/F,p is an F-form of Xn,p and n≥ 3, Lemma 2.5.1(c) tells us that XE/F,p is not containedin ∆E/F . Assuming that condition (3) of Theorem 1.3 holds, we have constructed an F-point a ofXE/F,p(F), as in (2.3.1). Thus it suffices to prove that F-points are dense in XE/F,p.We claim that under the assumptions of Theorem 1.3(∗), XE/F,p is not a cone over a.Indeed, if condition (i) of Theorem 1.3(∗) holds, that is, a = (y1 : . . . : ym) 6= (1 : · · · : 1), thenformula (2.3.1) tells us that a is not a scalar (that is, a 6∈ F ·1E). If XE/F,p were a cone over a, then it18would remain a cone over a after passing to the algebraic closure F of F . When we pass from F toF , E becomes split, and so isomorphic to Fn; so P(E) reduces to P(Fn)' Pn−1, XE/F,p reduces toXn,p, and the condition that a is not a scalar in E translates into a 6= (1 : · · · : 1) in Pn−1. By Lemma2.5.1(d), Xn,p cannot be a cone over a, a contradiction. We conclude that XE/F,p is not a cone overa, as claimed.On the other hand, if condition (ii) of Theorem 1.3(∗) holds, that is if p > 2 then by Lemma2.5.1(e), Xn,p is not a cone over any of its points, and hence, neither is XE/F,p. This proves theclaim.Thus in order to prove assertion (∗), it suffices to establish the following lemma.Lemma 2.6.2. Suppose p is a prime, F is a p-closed field, and E/F is an e´tale algebra of degreen ≥ 3. If XE/F,p has an F-point, and XE/F,p is not a cone over this point, then F-points are densein XE/F,p.Lemma 2.6.2 is a special case of Lemma 2.4.2. This completes the proof of assertion (∗) ofTheorem 1.3, and the proof of the theorem.19Chapter 3When are there solutions to (1.1)?Before exploring the existence of solutions to (1.1) in the cases of p = 2 and p = 3 in more detailwe will prove some more general results.Lemma 3.1. Let F0 be a field such that charF0 6= p. Then the system (1.1) has solution in Pm−1(F0)if any of the following conditions hold:(1) ki ≡ k j (mod p) and ki′ ≡ k j′ (mod p) for distinct i, i′, j, j′, and either p is odd, or√−1 ∈ F0.(2) m≥ p+3 and either p is odd or√−1 ∈ F0.(3) charF0 6= 0 and m≥ p+2Proof. (1) Let y = (y1 : · · · : ym), where yi = 1, y j = p√−pki−k j and yh = 0 for every other h 6= i, jand y′ = (y′1 : . . .y′j), where yi′ = 1, y j′ =p√−pki′−k j′ and yh′ = 0 for every other h′ 6= i′, j′. Theny and y′ lie in Pm−1(F0), as p√−1 ∈ F0 and ki− k j ≡ ki′ − k j′ ≡ 0 (mod p). These points, andalso the line between them, lie in the hypersurface defined by pk1yp1 + · · ·+ ykmypm. Intersectingthis line with the hyperplane defined by pk1y1+ · · ·+ pkmym, we obtain a solution to (1.1).(2) This follows from (1) by two applications of the pigeonhole principle.(3) Let q = charF0 > 0. Then F0 contains a copy of Z/qZ. By Chevalley’s theorem, finite fieldsare C1 (see [10, Theorem 5.2.1]), which implies that any quadratic form of dimension m− 1is isotropic when m− 1 > 2, that is, when m ≥ 4. Thus (1.3) has a nontrivial solution inZ/qZ⊆ F0 when m≥ 4.20Lemma 3.2. Let m = 2 and F0 be a field such that charF0 6= p. The system (1.1) has a solution inP1(F0) if and only if p(k1−k2)(p−1)+(−1)p = 0 in F0.Proof. A solution to the system when m = 2 is equivalent to a nonzero solution in F0 to pk1−k2yp1 +(−pk1−k2y1)p = 0. This occurs precisely when pk1−k2(1+(−p)(k1−k2)(p−1)) = 0 in F0.21Chapter 4The Hermite-Joubert problem for p = 2For p = 2 we can strengthen Theorem 1.3 to the following:Theorem 4.1. Let F0 be a field of characteristic 6= 2 and n = 2k1 + · · ·+2km ≥ 3 where k1 > · · ·>km ≥ 0.(a) Conditions (1), (2) and (3) of Theorem 1.3 (with p = 2) are equivalent to:(4) For every field F containing F0 and every n-dimensional e´tale algebra E/F, there exists anelement 0 6= a ∈ E such that trE/F(a) = trE/F(a2) = 0.(b) Moreover, if (4) holds, charF0 does not divide n, and F is an infinite field, then the elementa ∈ E in (4) can be chosen so that E = F [a].Proof. To prove (a), by applying Lemma 2.1.2 it suffices to prove the equivalence of conditions(1′) and (4).Let F ′ be a field of finite odd degree over F . By a theorem of Springer, a quadratic form definedover a field F of characteristic 6= 2 has a solution in F if and only if it has a solution in F ′, see [9,Theorem VII.2.7].Let E ′ = E ⊗F F ′. The equations trE ′/F ′(x) = trE ′/F ′(x2) = 0 and trE/F(x) = trE/F(x2) = 0define the same quadratic form over F ′ and F respectively. This and Springer’s theorem give (1′)is equivalent to (4)To prove (b), we wish to show that F-points are dense in XE/F,2, by applying Lemma 2.4.3.As we are assuming charF0 6= 2, we have Xn,2 is a smooth hypersurface of degree 2, by Lemma2.5.1(a). As XE/F,2 is an F-form of Xn,2 we then have XE/F,2 is a smooth hypersurface of degree 2,and hence not a cone. As we are assuming (4) holds, XE/F,2 has an F-point and hence by Lemma2.4.3 we have F-points are dense in XE/F,2.22By Lemma 2.5.1(c) we have Xn,2 is not contained in ∆n and hence XE/F,2 is not contained in∆E/F . Thus XE/F,2 contains an F-point away from ∆E/F . This point generates E over F .In the the rest of the chapter we will describe and prove necessary and sufficient conditions onthe field F0 to satisfy condition (4) of Theorem 4.1 for a given n, by appealing to condition (3) ofTheorem 1.3. We provide a summary here, and go in to more detail for particular cases of m in thefollowing subsections.Theorem 4.2. Let k1 > · · ·> km ≥ 0 be integers and F0 be a field of characteristic 6= 2. Considerthe system of equations:2k1y1+ · · ·+2kmym = 02k1y21+ · · ·+2kmy2m = 0 .(4.1)The following statements hold:• If m = 1 then (4.1) has no solution in P0(F0).• If m = 2 then (4.1) has a solution in P1(F0) if and only if charF0 | 2k1−k2 +1.• If m = 3 then (4.1) has a solution in P2(F0) if and only if −2k1+k2+k3n is a square in F0.• If m≥ 4,√−1 ∈ F0 and√2 ∈ F0, then (4.1) has a solution in Pm−1(F0).• If m≥ 4 and charF0 6= 0, then (4.1) has a solution in Pm−1(F0).• If m≥ 5 and √−1 ∈ F0, then (4.1) has a solution in Pm−1(F0).• If m≥ 6 and F0 is a nonreal number field, then (4.1) has a solution in Pm−1(F0).• In all cases of m, it is necessary for F0 to be nonreal for (4.1) to have a solution in Pm−1(F0).In the statements throughout the rest of this chapter we assume k1 > k2 > · · · > km ≥ 0 andcharF0 6= 2. The subsections are sorted by m, being the number of digits in the base 2 expansion ofn.234.1 m≤ 2Proposition 4.1.1.If m = 1, then (4.1) has no solution in P0(F0).If m = 2, then (4.1) has a solution in P1(F0) if and only if charF0 | 2k1−k2 +1.Proof. If m= 1 then P0(F0) consists of a point, which is not a solution to (4.1). If m= 2, the resultfollows from Lemma 3.2.4.2 m = 3Proposition 4.2.1. Suppose m = 3. Then (4.1) has a solution in P2(F0) if and only if −2k1+k2+k3nis a square in F0.Proof. Let `1 = k1− k3 and `2 = k2− k3. Then (4.1) has a solution in P2(F0) if and only if thequadratic formQ(y1,y2) = 2`1y21+2`2y22+(2`1y1+2`2y2)2is isotropic. We haveQ(y1,y2) = 2`1(1+2`1)y21+2`2(1+2`2)y22+2`1+`2+1y1y2= 2`1(1+2`1)(y1+2`1+`22`1(1+2`1)y2)2+(2`2(1+2`2)− 22`1+2`22`1(1+2`1))y22and soQ∼= 〈2`1(1+2`1), 2`1+`2(1+2`1)(1+2`2)−22`1+2`22`1(1+2`1)〉 .This gives us2`1(1+2`1)Q∼= 〈1,2`1+`2((1+2`1)(1+2`2)−2`1+`2)〉= 〈1,2`1+`2(1+2`1 +2`2)〉= 〈1,2k1+k2−2k3 ·2−k3(2k1 +2k2 +2k3)〉= 〈1,2k1+k2−3k3n〉∼= 〈1,2k1+k2+k3n〉24which is isotropic if and only if Q is isotropic, if and only if −2k1+k2+k3n is a square in F0.4.3 m = 4Proposition 4.3.1. Suppose m = 4. If either charF0 6= 0, or both√−1 ∈ F0 and√2 ∈ F0, then(4.1) has a solution in P3(F0).Here a nonreal field is one in which −1 is a sum of squares.We will prove a more general lemma here, giving more detailed necessary and sufficient conditionsfor the existence of solutions to (4.1) by considering different classes of quadratic forms.In the following lemma we will use the following result: If the quadratic form 〈2k1 ,2k2 , . . . ,2km〉is equivalent to a scalar multiple of a form with Schur index 2 (a form with a 2-dimensional totallyisotropic subspace), then (4.1) has a solution. This is implied by the 2-dimensional subspacehaving nontrivial intersection with the line defined by 2k1y1+ · · ·+2kmym = 0.We will also use the result that in any field, the smallest r such that −1 is a sum of r squares isa power of 2 (see [10, Theorem 3.1.3]).Lemma 4.3.2. Let m = 4. The quadratic form 〈2k1 ,2k2 ,2k3 ,2k4〉 is equivalent to a scalar multipleof one of:• 〈1,1,1,1〉. In this case (4.1) has a solution in P3(F0) if and only if −1 is a sum of 2 squaresin F0.• 〈1,1,1,2〉 or 〈1,1,2,2〉. In these cases, if √2 ∈ F0 and −1 is a sum of 2 squares in F0, then(4.1) has a solution in P3(F0).If (4.1) has a solution in P3(F0), then −1 is a sum of 4 squares in F0.Proof. By removing squares, reordering and possibly multiplying by 2, the quadratic form 〈2k1 ,2k2 ,2k3 ,2k4〉is equivalent to a scalar multiple of either 〈1,1,1,1〉, 〈1,1,1,2〉 or 〈1,1,2,2〉. We split these in totwo cases.– Suppose 〈2k1 ,2k2 ,2k3 ,2k4〉 is equivalent to a multiple of 〈1,1,1,1〉, which is a Pfister form〈1,1,1,1〉 = 〈1,1〉 ⊗ 〈1,1〉. Suppose −1 is a sum of 2 squares in F0. Then 〈1,1,1,1〉 isisotropic. By [9, X. Theorem 1.7], an isotropic Pfister form is hyperbolic and hence hasSchur index 2, thus (4.1) has a solution in P3(F0).25Conversely, suppose (4.1) has a solution in P3(F0). Then 〈1,1,1,1〉 is isotropic, and so −1is a sum of 3, and hence 2 squares in F0.– Suppose 〈2k1 ,2k2 ,2k3 ,2k4〉 is equivalent to a multiple of 〈1,1,1,2〉 or 〈1,1,2,2〉. Suppose√2 ∈ F0 and −1 is a sum of 2 squares in F0. Then 〈2k1 ,2k2 ,2k3 ,2k4〉 ∼= 〈1,1,1,1〉. By theabove case, this then has Schur index 2 and so (4.1) has a solution in P3(F0).Conversely, suppose (4.1) has a solution in P3(F0). Then either 〈1,1,1,2〉 or 〈1,1,2,2〉 isisotropic. Both cases imply that −1 is a sum of 5, and hence 4 squares in F0.We now consider Proposition 4.3.1. If both√−1 ∈ F0 and√2 ∈ F0 then (4.1) has a solutionin P3(F0) as a special case of Lemma 4.3.2. If charF0 6= 0, then (4.1) has a solution by Lemma3.1(3).4.4 m = 5Proposition 4.4.1. Suppose m = 5. If√−1 ∈ F0, then (4.1) has a solution in P4(F0).Again, we will prove a more general lemma here, giving more detailed necessary and sufficientconditions for the existence of solutions to (4.1).Lemma 4.4.2. Let m = 5. The quadratic form 〈2k1 ,2k2 ,2k3 ,2k4 ,2k5〉 is equivalent to a scalar mul-tiple of one of:• 〈1,1,1,2,2〉. In this case, if√−2 ∈ F0 or√−1 ∈ F0, then (4.1) has a solution in P4(F0).• 〈1,1,1,1,1〉 or 〈1,1,1,1,2〉. In these cases, if√−1∈ F0, then (4.1) has a solution in P4(F0).If (4.1) has a solution in P3(F0), then −1 is a sum of 4 squares in F0.Proof. By removing squares, reordering and possibly multiplying by 2, the quadratic form 〈2k1 , . . . ,2k5〉is equivalent to a scalar multiple of either 〈1,1,1,1,1〉, 〈1,1,1,1,2〉 or 〈1,1,1,2,2〉. We considereach case in turn.– Suppose 〈2k1 , . . . ,2k5〉 is equivalent to a scalar multiple of 〈1,1,1,2,2〉. If √−2 ∈ F0, then〈1,1,1,2,2〉 ∼= 〈1,−1,1,−1,1〉 which has Schur index 2 and so there exists a solution to(4.1). If√−1 ∈ F0, then by Lemma 3.1(2) there exists a solution to (4.1).26– Otherwise 〈2k1 , . . . ,2k5〉 is equivalent to a multiple of 〈1,1,1,1,c〉 for some c ∈ {1,2}. If√−1∈ F0, then 〈1,1,1,1,c〉 ∼= 〈1,−1,1,−1,c〉, which has Schur index 2 and so has nontriv-ial intersection with the line defined by 2k1y1+2k2y2+2k3y3+2k4y4+2k5y5 = 0. Thus thereexists a solution to (4.1).Conversely, suppose (4.1) has a solution in P4(F0). Then one of 〈1,1,1,2,2〉, 〈1,1,1,1,2〉 or〈1,1,1,1,1〉 is isotropic. In all cases we must have −1 is a sum of 6, and hence 4 squares in F0.4.5 m≥ 6Proposition 4.5.1. Suppose m ≥ 6. If F0 is a nonreal number field, then (4.1) has a solution inP4(F0).Proof. Suppose m≥ 6 and F0 is a nonreal number field. After dividing out by 2km we can substitutethe first equation of (4.1) into the second to obtain a quadratic form of degree m−1 representingthe system (4.1). Over a nonreal number field, any quadratic form of dimension ≥ 5 is isotropic(see [9, XI, Corollary 1.5, 379]), and thus we have a solution to (4.1).4.6 Proof of Theorem 4.2The proof of Theorem 4.2 has mostly been summarized in propositions of the earlier subsections.We need to extend Propositions 4.3.1 and 4.4.1 to arbitrary m≥ 4 and m≥ 5 respectively. We willneed the following:For a fixed field F0, if (4.1) has a solution in Pm−1(F0) for all n with m digits in its base 2expansion, then (4.1) will have a solution for all n with ≥ m digits in its base 2 expansion.Suppose n = 2`1 + · · ·+2`m′ +2k1 + · · ·+2km with `1 > · · · > `m′ > k1 > · · · > km and let (a1 :· · · : am) be a solution to (4.1) in Pm−1(F0) for 2k1 + · · ·+2km . Then we obtain a solution to (4.1)in P`+m−1(F0) for 2`1 + · · ·+2`m′ +2k1 + · · ·+2km being (0 : · · · : 0 : a1 : · · · : am).Lastly we need to show that if (4.1) has a solution in Pm−1(F0) then F0 must be nonreal.Suppose (4.1) has a solution in Pm−1(F0). Then as each of the 2ki are positive integers, the secondequation gives us 0 as the sum of n squares, not all zero:y21+ · · ·+ y21+ · · ·+ y2m+ · · ·+ y2m = 0 .27By dividing out by a nonzero yi we get −1 as a sum of squares. This completes the Theorem.28Chapter 5The Hermite-Joubert problem for p = 3Theorem 5.1. Let n= 3k1 + · · ·+3km ≥ 0 be the base 3 representation of n, with k1 ≥ ·· · ≥ km, andF0 be a field with characteristic 6= 3. Consider the system of equations:3k1y1+ · · ·+3kmym = 03k1y31+ · · ·+3kmy3m = 0 .(5.1)The following statements hold:• If m = 1 then (5.1) has no solution in P0(F0).• If m = 2 then (5.1) has a solution in P1(F0) if and only if charF0 divides 32(k1−k2)− 1 ork1 = k2.• If m= 3, F0 =Q, the ki are distinct and k1+k2+k3 6≡ 2 (mod 3), then (5.1) has no solutionin P2(F0).• If m = 3, F0 = Q, the ki are distinct and k1 6≡ k2 (mod 3), then (5.1) has no solution inP2(F0).• If m≥ 5 and charF0 6= 0 then (5.1) has a solution in Pm−1(F0).• If m≥ 6 then (5.1) has a solution in P1(F0).• If the ki are not distinct then (5.1) has a solution in P1(F0).29We will first prove the statements in the theorem for when m= 3. Define an equivalence relationon m-tuples of integers by (k1, . . . ,km) ∼ (`1, . . . `m) if and only if there exist σ ∈ Sm and λ ∈ Zsuch that ki ≡ `σ(i)+λ (mod 3) for each i.Remark 5.2. If (k1, . . . ,km)∼ (`1, . . . `m) then the solutions to 3k1y31+ · · ·+3kmy3m = 0 in Pm−1(F0)are in bijective correspondence with the solutions to 3`1z31 + · · ·+ 3`mz3m = 0 in Pm−1(F0). If `i =kσ(i)+λ +3si, then this correspondence is given by zi = 3−siyσ(i). In particular, if all solutions to3k1y31+ · · ·+3kmy3m = 0 have a zero coordinate yi, then all solutions to 3`1z31+ · · ·+3`mz3m = 0 havea zero coordinate z j.Lemma 5.3. Let k1 > k2 > k3 ≥ 0 be integers such that k1+k2+k3 6≡ 2 (mod 3). Then the systemof equations3k1y1+3k2y2+3k3y3 = 03k1y31+3k2y32+3k3y33 = 0(5.2)has no solution in P2(Q).Proof. Suppose we have a solution (y1 : y2 : y3) to (5.2) with a zero coordinate, say yi = 0. Thenthis gives rise to a solution in P1(Q) to (5.1) for n′ = n−3ki , which has only two terms in its base3 expansion. By Lemma 3.2, this is impossible in characteristic 0. Thus it suffices to show that allsolutions 3k1y31+3k2y32+3k3y33 = 0 have a zero coordinate.By Remark 5.2, it suffices to show that for some (`1, . . . `m)∼ (k1,k2,k3) the only solutions to3`1y31+3`2y32+3`3y33 = 0 have a zero coordinate.Under the equivalence relation described above, (k1,k2,k3) is equivalent to one of (0,0,0),(0,0,1), (0,1,2), or (0,0,2). For this lemma we need only consider the first three, noting that if(k1,k2,k3)∼ (`1, . . . `m), then k1+ k2+ k3 ≡ `1+ `2+ `3 (mod 3).Thus to prove the lemma it suffices to show that the only solutions in P2(Q) to 3`1y31+3`2y32+3`3y33 = 0 have a zero coordinate, for each case of (`1, `2, `3) ∈ {(0,0,0,),(0,0,1),(0,1,2)}. Weconsider each case in turn.• The only solutions to y31 + y32 + y33 = 0 have a zero coordinate, as a result of Fermat’s LastTheorem.• By [13], the only solution to y31+ y32+3y33 = 0 is (1 :−1 : 0).30• Suppose (y1 : y2 : y3) is a solution to y31 + 3y32 + 9y33 = 0. We may choose a representationsatisfying gcd(y1,y2,y3)= 1. Then we have 3 | y1, and hence 3 | y2 and 3 | y3. This contradictsthe minimality of (y1,y2,y3).Lemma 5.4. Let k1 > k2 > k3 ≥ 0 be integers such that k1 6≡ k2 (mod 3). Then the system ofequations3k1y1+3k2y2+3k3y3 = 03k1y31+3k2y32+3k3y33 = 0(5.3)has no solution in P2(Q).Proof. Assume the contrary and suppose we have a nontrivial solution (y1,y2,y3) 6= (0,0,0) to(5.3) with y1,y2,y3 ∈ Q. By dividing each equation by 3k3 and replacing k1 and k2 by k1− k3and k2− k3 respectively, we may assume without loss of generality that k3 = 0. Now the secondequation is y3 =−3k1y1−3k2y2, and by substituting this into the first, we gain a nontrivial solutionto3k1y31+3k2y32−33k1y31−32k1+k2+1y21y2−3k1+2k2+1y1y22−33k2y32 = 0 . (5.4)As (y1,y2,y3) is nontrivial we must have one of y1 and y2 are nonzero, and hence by (5.4) both arenonzero. DefineM1 := ν3(3k1y31) = k1+3ν3(y1),M2 := ν3(3k2y32) = k2+3ν3(y2), andM := min(M1,M2).Here ν3 denotes the 3-adic valuation. Since k1 6≡ k2 (mod 3), we have M1 6=M2, so that ν3(3k1y31+3k2y32) = M. We now show that the 3-adic valuation of the rest of the terms in the left hand sideof (5.4) is larger than M, which would imply that the 3-adic valuation of the entire equation is M,contradicting it being 0. We consider each term separately:• ν3(33k1y31) = 3k1+3ν3(y1)> M1 ≥M• ν3(32k1+k2+1y21y2) = 2k1+ k2+2ν3(y1)+ν3(y2)+1 > 23 M1+ 13 M2 > 23 M+ 13 M = M• ν3(3k1+2k2+1y1y22) = k1+2k2+ν3(y1)+2ν3(y2)+1 > 13 M1+ 23 M2 > 23 M+ 13 M = M31• ν3(33k2y32) = 3k2+3ν3(y2)> M2 ≥MThus we have obtained a contradiction in the 3-adic valuation of the equation and so there are nosolutions to (5.3) when k1 6≡ k2 (mod 3).Proof of Theorem 5.1. If m = 1 then P0(F0) consists of a point, which is not a solution to (5.1).If m = 2, our result follows from Lemma 3.2.If m = 3, our result follows from Lemmas 5.3 and 5.4.If the ki are not distinct, say ki = k j, then we obtain a solution y = (y1 : · · · : ym) by settingyi = 1, y j =−1 and yh = 0 for h 6= i, j.The other cases are covered by Lemma 3.1(2) and (3).Note that Lemma 3.1(1) shows that in characteristic 6= 3 there exist solutions to (5.1)when m=4 and (k1,k2,k3,k4)∼ (0,0,1,1) or when m= 5 and (k1,k2,k3,k4,k5) 6∼ (0,0,0,1,2). Except for theremaining cases of m= 3,4 or 5, being when (k1, . . . ,km) is equivalent to one of (0,0,2),(0,0,1,2)or (0,0,0,1,2), we have a complete description of when there exist solutions to (5.1) in Pm−1(Q).Proof of Theorem 1.4. Theorem 1.4 readily follows from Theorem 1.3 and Theorem 5.132Chapter 6Numerical evidence in support ofConjecture 1.5We now consider Conjecture 1.5. By Theorem 1.3, it is equivalent to the followingConjecture 6.1. Let k1 > k2 > 0. Then the system of equations3k1y1+3k2y2+ y3 = 03k1y31+3k2y32+ y33 = 0(6.1)has no solution in P2(Q).Note in this statement we have assumed k3 = 0. This is without loss of generality, as in theproof of Lemma 5.4.The remaining case left unsolved by Lemmas 5.3 and 5.4 is when k1 ≡ k2 (mod 3) and k1 +k2 ≡ 2 (mod 3). That is, when k1 ≡ k2 ≡ 1 (mod 3). Indeed in this case we have (k1,k2,0) ∼(0,0,2), but there do exist nonzero solutions toy31+ y32+9y33as illustrated by (1 : 2 : −1). However the set of solutions to this equation (after a change ofvariables) does not necessarily intersect with 3k1y1 + 3k2y2 + y3 = 0 to give a solution to (6.1), aswe have already seen when k1 6≡ k2 (mod 3).The smallest n of this form, not covered by Theorem 1.4, is 85 = 34+31+30. In this case wehave shown computationally that (6.1) does not have a solution in P2(Q). In fact we have covered33a lot more.Proposition 6.2. Theorem 1.4 remains true for all k1 > k2 > k3 ≥ 0 such that k2− k3 < k1− k3 <30000.We introduce some notation before giving proofs of our computational evidence. For a,b non-negative integers, definefa,b := (36a+2−1)x3+34a+2b+3x2+32a+4b+3x+(36b+2−1) .Here fa,b ∈Z[x] is primitive. For abbreviation, we write la,b to denote the leading coefficient of fa,b(that is la,b := 36a+2−1).Lemma 6.3. Let k1 ≥ k2 > 0 with k1 ≡ k2 ≡ 1 (mod 3). Write k1 = 3a+1 and k2 = 3b+1. Thenthe system (6.1) has a rational solution in P2(Q) if and only if fa,b has a rational root.Moreover, suppose there exists an integer m such that (la,b,m) = 1 and fa,b(z) 6≡ 0 (mod m)for all integers 0≤ z < m. Then fa,b does not have a rational root.Proof. The system (6.1) is equivalent to3k1y31+3k2y32−33k1y31−32k1+k2+1y21y2−3k1+2k2+1y1y22−33k2y32 = 0 .We may consider this as a cubic polynomial in one variable y = y1/y2:3k1(1−32k1)y3−32k1+k2+1y2−3k1+2k2+1y+3k2(1−32k1) = 0 .After taking out a factor of −3k2 and writing k1 = 3a+1, k2 = 3b+1 for integers a > b ≥ 0, thisbecomes33a−3b(36a+2−1)y3+36a+3y2+33a+3b+3y+(36b+2−1) = 0 .After the substitution y = 3b−ax, we then have that (6.1) has a rational solution in P2(Q) if andonly iffa,b = (36a+2−1)x3+34a+2b+3x2+32a+4b+3x+(36b+2−1)has a rational root.Now suppose there exists an integer m such that (la,b,m) = 1 and fa,b(z) 6≡ 0 (mod m) for allintegers 0≤ z < m.34As fa,b is primitive, by Gauss’s lemma it suffices to show that fa,b is irreducible over the inte-gers. By assumption we have fa,b is irreducible modulo m, and as (la,b,m) = 1 we thus have fa,b isirreducible over the integers.The most efficient algorithm we have found to prove that fa,b has no rational root is to findsome small prime q such that q does not divide la,b, and fa,b is irreducible modulo q, as in Lemma6.3. For the b < a < 10000 of which we checked, we found primes q (dependent on a and b)smaller than 500 for which q does not divide la,b and fa,b is irreducible modulo q.Remark 6.4. Note that there is no such absolute bound on primes independent of a and b forwhich this algorithm will work. The equation fa,b does indeed have solutions if a = b. Thus if allthe primes less than some integer B appear in the prime factorization of both a and b, then fa,b willhave solutions modulo q for all primes less than B.To aid in computation, we first establish some preliminary results.For tuples of integers we say (a,b)≡ (x,y) (mod m) if both a≡ x (mod m) and b≡ y (mod m),so there are m2 classes of tuples of integers modulo m.Lemma 6.5. Let m be a positive integer with (3,m) = 1.(a) Suppose (a′,b′) ≡ (a,b) (mod ordm(3)), fa,b is irreducible modulo m, and (la,b,m) = 1.Then fa′,b′ has no rational root.(b) Let o be a positive integer multiple of ordm(3). Suppose fa,b is irreducible modulo m and(la,b,m) = 1. Then fa′,b′ has no rational solutions if (a′,b′)∼= (a+ iordm(3),b+ j ordm(3)) for anyi, j ∈ Z.(c) Let r be a positive integer divisor of ordm(3) and write ordm(3) = rs. Suppose fa,b isirreducible modulo m and (la,b,m) = 1. If fri+a,r j+b is irreducible modulo m and (lri+a,r j+b,m) = 1for all tuples (ri+a,r j+b) with 0≤ i, j < s, then fa′,b′ has no rational solutions for any (a′,b′)∼=(a,b) (mod r).Proof. (a) As a and b only appear in fa,b in exponents of 3, and (a′,b′) ≡ (a,b) (mod ordm(3)),we have fa′,b′ ≡ fa,b (mod m). In particular we have la′,b′ ≡ la,b (mod m) and fa′,b′ is irreduciblemodulo m. Thus fa′,b′ is irreducible over the integers and hence as fa′,b′ is primitive, it is irreducibleover the rationals.Part (b) follows from (a) by the projection Z/(oZ)→ Z/(ordm(3)Z). Part (c) follows from (a)by the natural isomorphism Z/(sZ)→ (rZ)/(ordm(3)Z) of additive groups.35For a fixed modulus o, parts (b) and (c) of Lemma 6.5 give us a way of searching for classes oftuples (a,b) modulo o for which fa′,b′ has no rational root, for any (a′,b′)≡ (a,b) (mod o).Proposition 6.6. Suppose (a,b) is not congruent to one of(0,0),(1,1),(2,2),(3,3),(4,4),(1,3),(2,3),(3,1),(3,2),(3,4),(4,3)modulo 5. Then fa,b has no rational solution.Proof. By Lemma 6.5 (c), it suffices to find an integer m where ordm(3) = 5s for some integer s,such that f5i+a,5 j+b is irreducible modulo m and (lri+a,r j+b,m) = 1 for all 0≤ i, j < s.Suppose (a,b) is congruent to one of(0,1),(0,3),(0,4),(1,0),(1,2),(2,1),(2,4),(3,0),(4,0),(4,2)modulo 5. Let m = 31, where we have ord31(3) = 30. By checking computationally, we prove thatthe assumptions of Lemma 6.5 (c) are satisfied. For more detail, see Appendix A.Otherwise, (a,b) is congruent to one of(0,1),(0,2),(1,0),(1,4),(2,0),(4,1)modulo 5. Now let m = 61, where we have ord61(3) = 10. Again, by checking computationally,we prove that the assumptions of Lemma 6.5 (c) are satisfied.This proposition allows the algorithm based on Lemma 6.3 to skip many values of a and b.We return to Proposition 6.2. Without loss of generality we may assume k3 = 0 and k2 <k1 < 30000. By Theorem 1.4 we may then assume k1 ≡ k2 ≡ 1 (mod 3). Let a = (k1−1)/3 andb = (k2−1)/3. Then b < a < 10000. By Lemma 6.3, it suffices to prove that fa,b has no rationalroots. By Proposition 6.6, we may assume (a,b) is congruent to one of(0,0),(1,1),(2,2),(3,3),(4,4),(1,3),(2,3),(3,1),(3,2),(3,4),(4,3)modulo 5. We ensured computationally (see Appendix A) that there exists a prime p such thatfa,b has no roots modulo p and (la,b, p) = 1. Thus by Lemma 6.3 fa,b has no rational root. Thiscompletes the proof of the proposition.36Remark 6.7. Testing with other orders o of 3 (as opposed to 5) has ruled out a larger fraction oftuples in (Z/oZ)2. By checking moduli up to 300, we have ruled out 14/25 of the tuples (a,b)modulo 5. The same approach has ruled out 42548/44100 tuples modulo 210. Unfortunately, dueto the large scale, this fact can not be easily used to optimize the algorithm proving no rationalsolutions to fa,b.Remark 6.8. An alternate approach to proving the conjecture would be to consider x3+y3+9z3 =0 as an elliptic curve. When k1 + k2 + k3 ≡ 2 (mod 3), after substitution the system of equationsbecomes equivalent tox3+ y3+9z3 = 03ax+3by+ z = 0for some integers a> b> 0. The group of rational points lying on the elliptic curve x3+y3+9z3 = 0is cyclic, generated by (1 : 2 :−1), see [13].37Chapter 7ConclusionIn this thesis we addressed a generalization of the classical Hermite-Joubert problem, as in Question1.1. By restricting to the case of p-fields, we gave a necessary and sufficient condition in Theorem1.3 to answer the analogous Question 1.2. This condition is phrased in terms of the existence ofan F-point on a particular variety, depending on n and p. In Theorems 4.1 and 4.2 we obtain analmost complete description of the Hermite-Joubert problem in the case of p = 2.Theorem 1.4 yields us new examples in the case of p = 3 where the Hermite-Joubert problemhas a negative solution, the smallest being where n = 13. In Chapter 6 we give numerical evidencein support of Conjecture 1.5, showing more examples where the Hermite-Joubert problem has anegative solution.A further potential research direction would be to give a proof of Conjecture 1.5. Theorem1.3 considers the existence of an element a ∈ E such that trE/F(a) = trE/F(ap) = 0. In the caseof p = 2 or p = 3 this is analagous to the classical Hermite-Joubert problem, where this conditionis equivalent to the 1st and pth coefficients of the characteristic polynomial of a being zero. An-other possible direction would be to study the existence of an element a such that the 1st and pthcoefficients of the characteristic polynomial of a are zero, for a more general p.In the case of p = 2, for particular separable field extensions E/F , we showed that there existsan element a∈ E such that trE/F(a) = trE/F(a2) = 0. It would be of interest to see how this elementmight be constructed.38Bibliography[1] M. Brassil and Z. Reichstein. 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Acta Math., 85:203–362 (1plate), 1951.40Appendix AMaple codewi th ( L i s t T o o l s )# The following function tests for solutions for specific k1 = 3a+1, k2 = 3b+1 and a specificmodulus m. It returns an empty list if a solution is found, and [(a,b)] otherwise.f i n d r a t i o n a l r o o t t e s t m o d u l u s := proc ( a , b , m) l o c a l i , g , l c ,f , d , n , found ;l c := 3 ˆ ( 6∗ a +2)−1;g := l c ∗x ˆ 3 + 3 ˆ ( 4∗ a +2∗b +3) ∗x ˆ 2 + 3 ˆ ( 2∗ a +4∗b +3) ∗x +3ˆ (6∗ b +2)−1;i f i g c d ( l c , m) = 1 thenf := modp1 ( C o n v e r t I n ( g , x ) , m) ;found := f a l s e ;f o r i from 0 to m−1 whi le not found doi f modp1 ( Eva l ( f , i ) , m) = 0 thenfound := t r u eend i fend do ;i f not found thenreturn [ [ a , b ] ]end i fend i f ;re turn [ ]end proc41# The following function returns tuples of integers (t1, t2) such that for all (a,b) satisfyinga ≡ t2 (mod ordm(3)) and b ≡ t2 (mod ordm(3)) there are provably no solutions to our system.Here o is the order of 3 modulo mf i n d r a t i o n a l r o o t t e s t m o d u l u s c o m p l e t e := proc (m, o ) l o c a l i ,j , l i s t , r e v l i s t , ord , n ;n := 3 ;l i s t := [ ] ;f o r i from o to 2∗o−1 dof o r j from 0 to o−1 dol i s t := [ op ( l i s t ) , op ( f i n d r a t i o n a l r o o t t e s t m o d u l u s ( i , j ,m) ) ]end doend do ;l i s t := map ( x → map ( y → y mod o , x ) , l i s t ) ;re turn s o r t ( l i s t )end proc# Assumes (3,m) = 1.ord 3 modulo m := proc (m) l o c a l ord , n ;n := 3 ;o rd := 1 ;whi le n mod m <> 1 do n := 3∗n ;o rd := ord +1end do ;re turn ordend proc# The following function returns integers for which 3 has order a multiple of o, and their orders.o r d 3 m u l t i p l e o := proc ( l , h , o ) l o c a l m, ord , n , l i s t ;l i s t := [ ] ;m := max ( l , 4 ) ;whi le m <= h don := 3 ;o rd := ord 3 modulo m (m) ;42i f ord mod o = 0 thenl i s t := [ op ( l i s t ) , [m, ord ] ]end i f ;m := m+1;i f m mod 3 = 0 thenm := m+1end i fend do ;re turn l i s tend proc# The following function takes as input a modulus and a list of tuples. It is intended for modulithat are a multiple of o. It outputs the set of tuples modulo o which occur completely in the list.For example, if m = 10, o = 5 and all of (1,4), (1,9), (6,4), (6,9) occur in the list, then (1,4) willbe included in the output.r e d u c e t o m o d o := proc (m, l i s t , o ) l o c a l i , j , i2 , j2 , l2 ,found ;l 2 := [ ] ;f o r i from 0 to o−1 dof o r j from 0 to o−1 dofound := t r u e ;f o r i 2 from 0 to i quo (m, o )−1 whi le found dof o r j 2 from 0 to i quo (m, o )−1 whi le found doi f not member ( [ i +o∗ i2 , j +o∗ j 2 ] , l i s t ) thenfound := f a l s eend i fend doend do ;i f found = t r u e thenl 2 := [ op ( l 2 ) , [ i , j ] ]end i fend doend do ;re turn l 243end proc# The following function takes as input a modulus and a list of tuples. It is intended for modulithat divide o. It outputs the set of tuples modulo m which reduce mod o to something occuring inthe list. For example, if m = 5, o = 10 and (1,4) occurs in the list, then (1,4), (1,9), (6,4), (6,9)will be included in the output.i n c r e a s e t o m o d o := proc (m, l i s t , o ) l o c a l i , j , d , t , l 2 ;l 2 := [ ] ;d := o /m;f o r t i n l i s t dof o r i from 0 to d−1 dof o r j from 0 to d−1 dol 2 := [ op ( l 2 ) , [ t [ 1 ] +m∗ i , t [ 2 ] +m∗ j ] ]end doend doend do ;re turn l 2end proc# The following function returns tuples of integers (t1, t2) such that for all (k1,k2) satisfyingk1≡ t2 (mod o) and k2≡ t2 (mod o) there are provably no solutions to our system.f i n d a l l m o d o := proc ( l , h , o ) l o c a l x , l i s t ;l i s t := [ ] ;f o r x i n o r d 3 m u l t i p l e o ( l , h , o ) dol i s t := [ op ( l i s t ) , op ( r e d u c e t o m o d o ( x [ 2 ] ,f i n d r a t i o n a l r o o t t e s t m o d u l u s c o m p l e t e ( x [ 1 ] , x [ 2 ] ) , o ) ) ]end do ;f o r x from 2 to o−1 doi f o mod x = 0 thenl i s t := [ op ( l i s t ) , op ( i n c r e a s e t o m o d o ( x , f i n d a l l m o d o ( l, h , x ) , o ) ) ]end i fend do ;re turn MakeUnique ( s o r t ( l i s t ) )end proc44# The result of the following command shows that the only possible values of (a (mod 5),b(mod 5)) are (0,0), (1,1), (2,2), (3,3), (4,4), (1,3), (2,3), (3,1), (3,2), (3,4), (4,3). Note in oursystem we are proving for k1 = 3a+1, k2 = 3b+1.MakeUnique ( s o r t ( [ op ( f i n d a l l m o d o ( 3 1 , 31 , 5 ) ) , op ( f i n d a l l m o d o( 6 1 , 61 , 5 ) ) ] ) )# The above command outputs[[0,1], [0,2], [0,3], [0,4], [1,0], [1,2], [1,4], [2,0], [2,1], [2,4], [3,0], [4,0], [4,1], [4,2]]# The following function attempts to prove that our system has no solution for a fixed k1, k2.It does this by checking modulo primes up to a bound B until it finds no solution.f i n d r a t i o n a l r o o t t e s t p r i m e := proc ( a , b , B) l o c a l i , g , f , d ,l c , found , p ;l c := 3 ˆ ( 6∗ a +2)−1;g := l c ∗x ˆ 3 + 3 ˆ ( 4∗ a +2∗b +3) ∗x ˆ 2 + 3 ˆ ( 2∗ a +4∗b +3) ∗x +3ˆ (6∗ b +2)−1;d := d i f f ( g , x ) ;p := 2 ;whi le p <= B doi f i r em ( lc , p ) <> 0 and ‘mod ‘ ( Gcd ( g , d ) , p ) = 1 thenf := modp1 ( C o n v e r t I n ( g , x ) , p ) ;found := f a l s e ;f o r i from 0 to p−1 whi le not found doi f modp1 ( Eva l ( f , i ) , p ) = 0 thenfound := t r u eend i fend do ;i f not found thenreturn pend i fend i f ;p := n e x t p r i m e ( p )end do ;re turn f a l s e45end proc# This is the main function. It checks for k1 > k2 for a > b between l and h. Here we sayk1 = 3a+ 1 and k2 = 3b+ 1. B is an upper bound on the primes used in the subfunction. It isoptimized to check only when (a (mod 5),b (mod 5)) is one of (0,0), (1,1), (2,2), (3,3), (4,4),(1,3), (2,3), (3,1), (3,2), (3,4), (4,3), which is proven sufficient above. It returns false after nosolutions are found.f i n d r a t i o n a l r o o t t e s t p r i m e m a i n o p t i m i z e d := proc ( l , h , B)l o c a l i , j ;f o r i from i quo ( l , 5 ) to i quo ( h , 5 ) dof o r j from 0 to i−1 doi f f i n d r a t i o n a l r o o t t e s t p r i m e (5∗ i , 5∗ j , B) = f a l s e thenreturn t r u e , 5∗ i , 5∗ je l i f f i n d r a t i o n a l r o o t t e s t p r i m e (5∗ i +1 , 5∗ j +1 , B) = f a l s ethenreturn t r u e , 5∗ i +1 , 5∗ j +1e l i f f i n d r a t i o n a l r o o t t e s t p r i m e (5∗ i +1 , 5∗ j +3 , B) = f a l s ethenreturn t r u e , 5∗ i +1 , 5∗ j +3e l i f f i n d r a t i o n a l r o o t t e s t p r i m e (5∗ i +2 , 5∗ j +2 , B) = f a l s ethenreturn t r u e , 5∗ i +2 , 5∗ j +2e l i f f i n d r a t i o n a l r o o t t e s t p r i m e (5∗ i +2 , 5∗ j +3 , B) = f a l s ethenreturn t r u e , 5∗ i +2 , 5∗ j +3e l i f f i n d r a t i o n a l r o o t t e s t p r i m e (5∗ i +3 , 5∗ j +1 , B) = f a l s ethenreturn t r u e , 5∗ i +3 , 5∗ j +1e l i f f i n d r a t i o n a l r o o t t e s t p r i m e (5∗ i +3 , 5∗ j +2 , B) = f a l s ethenreturn t r u e , 5∗ i +3 , 5∗ j +2e l i f f i n d r a t i o n a l r o o t t e s t p r i m e (5∗ i +3 , 5∗ j +3 , B) = f a l s ethenreturn t r u e , 5∗ i +3 , 5∗ j +346e l i f f i n d r a t i o n a l r o o t t e s t p r i m e (5∗ i +3 , 5∗ j +4 , B) = f a l s ethenreturn t r u e , 5∗ i +3 , 5∗ j +4e l i f f i n d r a t i o n a l r o o t t e s t p r i m e (5∗ i +4 , 5∗ j +3 , B) = f a l s ethenreturn t r u e , 5∗ i +4 , 5∗ j +3e l i f f i n d r a t i o n a l r o o t t e s t p r i m e (5∗ i +4 , 5∗ j +4 , B) = f a l s ethenreturn t r u e , 5∗ i +4 , 5∗ j +4end i fend do ;i f f i n d r a t i o n a l r o o t t e s t p r i m e (5∗ i +3 , 5∗ i +1 , B) = f a l s ethenreturn t r u e , 5∗ i +3 , 5∗ i +1e l i f f i n d r a t i o n a l r o o t t e s t p r i m e (5∗ i +3 , 5∗ i +2 , B) = f a l s ethenreturn t r u e , 5∗ i +3 , 5∗ i +2e l i f f i n d r a t i o n a l r o o t t e s t p r i m e (5∗ i +4 , 5∗ i +3 , B) = f a l s ethenreturn t r u e , 5∗ i +4 , 5∗ i +3end i fend do ;re turn f a l s eend proc# The following command returns false.f i n d r a t i o n a l r o o t t e s t p r i m e m a i n o p t i m i z e d ( 0 , 10000 , 500)# The following command returns 42548.numelems ( f i n d a l l m o d o ( 0 , 300 , 210) )47`

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