ARTICLE IN JUNE 2014

<span class='text_page_counter'>(1)</span>GENERIC SINGULARITIES ON LINEAR COMBINATIONS OF PRESCRIBED MONOMIALS IN TWO VARIABLES DARREN J. FINNIGAN, ALAA HAJ ALI, KYUNGYONG LEE, CHRIS M. LOCRICCHIO, THE MINH TRAN, RAFAL URBANIUK. Abstract. The polynomial interpolation problem on the plane is to determine if general multiple points impose independent conditions on plane algebraic curves with given restriction(s). This restriction is classically given by a degree. In this paper we use a different restriction, namely we fix a set of monomials and consider their linear combinations. We study whether singular points in general position impose independent conditions on linear combinations of monomials satisfying certain parity properties.. 1. Introduction Let K be an infinite field. It is an important problem to determine if general multiple points on P2K impose independent conditions on plane curves of a given degree. This problem has a very long history (for a very small fraction of references, see [1, 2, 3, 4, 5, 6, 8, 9]). In this paper we continue the characteristic 2 reduction approach following the spirit of [7]. The following is Before we state the main theorem, we need some notation. For integers i, j, a, b, we say that (i, j) ≡ (a, b)(mod 2) if i ≡ a(mod 2) and j ≡ b(mod 2). For a, b ∈ {0, 1}, let Ma,b := {(i, j) ∈ (Z≥0 )2 | (i, j) ≡ (a, b)(mod 2)}. For L ⊂ (Z≥0 )2 , define L1 = M0,0 ∩ L, L2 = M0,1 ∩ L, L3 = M1,0 ∩ L and L4 = M1,1 ∩ L. Also, define xL = {xi y j | (i, j) ∈ L}. The following is our main theorem. Theorem 1.1. Fix a positive integer n. Let L ⊂ (Z≥0 )2 be a set of lattice points with |L| ≤ 3n. Suppose that |Li | ≤ n for all 1 ≤ i ≤ 4. Then no curve in Span(xL ) passes through n general points with multiplicity≥ 2. Applying this to plane curves with a given degree, we obtain several corollaries. Corollary 1.2. Fix positive integers d ≥ 2 and n. Let n0 = n + d 2d+5 e + 1, and let (mi )1≤i≤n0 3 be a sequence consisting of positive integers with mi = 2 for i ∈ {n + 1, ..., n0 }. Assume that no curve of degree d passes through general points p1 , ..., pn with multiplicity≥ mi at pi . Then no curve of degree d + 2 passes through general points p1 , ..., pn0 with multiplicity≥ mi at pi . Corollary 1.3. Let t, d and n be integers with t ≥ 3 and d ≥ t + 5. If (d + 1)(d + 2)/2 ≤ 3n + t(t + 1)/2 then there is no polynomial of degree d that passes through n + 1 general points with multiplicity ≥ 2 at the first n points and with multiplicity≥ t at the last point. 1.

<span class='text_page_counter'>(2)</span> 2. FINNIGAN ET AL. As another corollary, we give a new proof of the following theorem which is obtained more than a century ago by Campbell [4], Palatini [8] and Terracini [9]. Corollary 1.4. Fix two positive integers n and d with (n, d) 6= (2, 2) and (n, d) 6= (5, 4). Then (d + 1)(d + 2)/2 ≤ 3n if and only if no polynomial of degree d vanishes at n general points with multiplicity ≥ 2.. 2. Proof of the main theorem In this section we prove Theorem 1.1. Let L ⊂ (Z≥0 )2 be a set of lattice points and (x1 , y1 ), ..., (xn , yn ) be n general points on K . A nonzero element f in Span(xL ) is of the form 2. X. ci,j xi y j ,. (i,j)∈L. and f vanishes at a point (xk , yk ) with multiplicity ≥ 2 if and only if f (xk , yk ) = ∂f (xk , yk ) = 0. ∂y. ∂f (xk , yk ) ∂x. =. To prove Theorem 1.1, it is enough to prove it for |L| = 3n. If |L| < 3n we can always 0 find a set of lattice points S ⊂ (Z≥0 )2 \ L where |S| = 3n − |L| and L := L ∪ S satisfies 0 |Li | ≤ n for every i ∈ [1, 4]. For the rest of this section, we will assume that |L| = 3n and |Li | ≤ n for every i ∈ [1, 4]. Let L = {(i1 , j1 ), (i2 , j2 ), ..., (i3n , j3n )}. If the system of equations           . xi11 y1j1 xi12 y1j2 i1 xi11 −1 y1j1 i2 xi12 −1 y1j2 j1 xi11 y1j1 −1 j2 xi12 y1j2 −1 .. .. . . xin1 ynj1 xin2 ynj2 i1 xin1 −1 ynj1 i2 xin2 −1 ynj2 j1 xin1 ynj1 −1 j2 xin2 ynj2 −1. ··· xi13n y1j3n · · · i3n xi13n −1 y1j3n · · · j3n xi13n y1j3n −1 .. .. . . ··· ··· ···. xin3n ynj3n i3n xin3n −1 ynj3n j3n xin3n ynj3n −1.    ci1 ,j1    ci ,j  2 2  ..  .   ci3n ,j3n .   =0 . has no nontrivial solution, then no curve in Span(xL ) passes through n general points with multiplicity≥ 2. Give a total order T on the set of all lattice points as follows: (0, 0) < (0, 1) < (1, 0) < 0 0 (0, 2) < (1, 1) < (2, 0) < (0, 3) < (1, 2) < (2, 1) < (3, 0) < ... i.e. (i, j) < (i , j ) 0 0 if i + j < i + j or 0 0 0 if i + j = i + j and if i < i ..

<span class='text_page_counter'>(3)</span> GENERIC SINGULARITIES ON LINEAR COMBINATIONS OF PRESCRIBED MONOMIALS IN TWO VARIABLES 3. Notation. Let       L D = det     . xi12 y1j2 xi11 y1j1 i1 xi11 −1 y1j1 i2 xi12 −1 y1j2 j1 xi11 y1j1 −1 j2 xi12 y1j2 −1 .. .. . . xin2 ynj2 xin1 ynj1 i1 xin1 −1 ynj1 i2 xin2 −1 ynj2 j1 xin1 ynj1 −1 j2 xin2 ynj2 −1. ··· xi13n y1j3n · · · i3n xi13n −1 y1j3n · · · j3n xi13n y1j3n −1 .. .. . . ··· ··· ···. xin3n ynj3n i3n xin3n −1 ynj3n j3n xin3n ynj3n −1.       ,    . and for every W ⊂ L with |W | = 3 , if W = {(u1 , v1 ), (u2 , v2 ), (u3 , v3 )} and i ∈ {1, 2, .., n} we let   xui 1 yiv1 xui 2 yiv2 xui 3 yiv3 DiW = det  u1 xui 1 −1 yiv1 u2 xui 2 −1 yiv2 u3 xui 3 −1 yiv3  . v1 xui 1 yiv1 −1 v2 xui 2 yiv2 −1 v3 xui 3 yiv3 −1 Definition 2.1. A triangle decomposition of L is a sequence T = (W (1) , ..., W (n) ) such that ∪1≤i≤n W (i) = L, W (i) ∩ W (j) = ∅ for all i 6= j and |W (i) | = 3 for every i. Let TL be the collection of all triangle decompositions of L. For every T = {W (1) , ..., W (n) } ∈ TL , let n Q (i) P(T ) = DiW . i=1. Lemma 2.2. We can always obtain a triangular decomposition T0 = {W (1) , W (2) , ...W (n) } using the following recursive algorithm. Let S be a set of lattice points and j and k positive integers. Set S to L and j = 1; (1) (2) (3) (4). Step Step Step Step. 1: 2: 3: 4:. Choose k such that |Sk | = M in{|Si |, i ∈ [1, 4]}. Let W (j) = ∪i6=k,i∈[1,4] {smallest element in Si }. remove W (j) from S and increment j by 1 if S 6= ∅ go back to step 1; otherwise the construction is complete.. Proof. Prooving the Lemma by induction on n Base Case: For n = 1, |L| = 3 and |Li | ≤ 1 for every i ∈ [1, 4]. So we excute the steps 1,2,3,4 ones and we exit because S becomes empty. inductive case: Suppose that the algorithm works successfully for n = m. For n = m + 1, |L| = 3m + 3 and |Li | ≤ m + 1 for every i ∈ [1, 4]. But there exists i0 ∈ [1, 4] such that |Li0 | ≤ m, because, if not, we would have |L| = 4(m + 1) > 3(m + 1) which is a contradiction. Thus, after implementing the first three steps of the algorithm, we have |S| = 3m and |Si | ≤ m for every i ∈ [1, 4]; and by the induction assumption the rest of the algorithm works successfully.  We want to show that P(T0 ) 6= 0 and P(T0 ) cannot be cancelled by. P T ∈TL ,T 6=T0. can conclude that DL 6= 0.. P(T ) so we.

<span class='text_page_counter'>(4)</span> 4. FINNIGAN ET AL.  xa y b xc y d xe y f Lemma 2.3. Let D =detaxa−1 y b cxc−1 y d exe−1 y f  bxa y b−1 dxc y d−1 f xe y f −1 . Then D is of the form D = Cxs y t with C 6= 0 (mod 2) if and only if (a, b) 6≡ (c, d) 6≡ (e, f ). Proof. D = [(cf − de) − (af − be) + (ad − bc)]xa+c+e−1+ y b+d+f −1 . Then C = (cf − de) − (af − be) + (ad − bc) It is easy to show that for any (α, β), (γ, δ) ∈ (Z≥0 )2 , αδ − βγ = 1 (mod 2) iff (α, β) 6≡ (γ, δ) 6≡ (0, 0) (mod 2). If (a, b) ≡ (0, 0), C 6= 0 (mod 2) iff cf − de = 1(mod 2) i.e. (c, d) 6≡ (e, f ) 6≡ (0, 0). If not, and (c, d) ≡ (0, 0), C 6= 0 (mod 2) iff af − be ≡ 1 i.e. (a, b) 6≡ (e, f ) 6≡ (0, 0). Otherwise, that is (a, b) 6≡ (0, 0) and (c, d) 6≡ (0, 0), if (a, b) ≡ (c, d) then ad − bc = 0 (mod 2) and (cf − de) − (af − be) = 0 (mod 2). Thus C = 0 (mod 2). if (a, b) 6≡ (c, d) then ad − bc = 1 (mod 2). Thus C 6= 0 (mod 2) iff (cf − de) − (af − be) = 0 (mod 2) i.e. [(e, f ) ≡ (0, 0)] or [(cf − de) = 1 (mod 2) and(af − be) = 1 (mod 2)]. Thus, C 6= 0 (mod 2) iff (a, b) 6≡ (c, d) 6≡ (e, f )  0. 0. Lemma 2.4. Let W and W two subsets of L with |W | = |W | = 3 and let DiW = CiW xαi yiβ 0 0 0 and DiW = CiW xγi yiδ . If CiW 6= 0 (mod 2) and CiW 6= 0 (mod 2) then there exist (a, b) and 0 (c, d) ∈ (Z≥0 )2 such that W ∩ Ma,b = ∅ and W ∩ Mc,d = ∅ Proof. Clear by Lemma 2.4 Lemma 2.5. two subsets of that CiW 6= 0 W ∩ Ma,b = ∅.  0. 0. 0. 0. 0. 0. 0. Let W = {(u1 , v1 ), (u2 , v2 ), (u3 , v3 )} and W = {(u1 , v1 ), (u2 , v2 ), (u3 , v3 )} be 0 0 0 L with |W | = |W | = 3 and let DiW = CiW xαi yiβ and DiW = CiW xγi yiδ . Assume 0 (mod 2) and CiW 6= 0 (mod 2) and let (a, b) and (c, d) ∈ (Z≥0 )2 such that 0 and W ∩ Mc,d = ∅. (1) if (a, b) 6= (c, d) then (α 6= γ)or (β 6= δ). 0 0 (2) if (a, b) = (c, d) and (ui , vi ) ≤ (ui , vi ) for every i ∈ {1, 2, 3} then (α 6= γ)or (β 6= δ). Proof. for (1) 0 0 0 0 0 0 α = u1 + u2 + u3 − 1, β = v1 + v2 + v3 − 1, γ = u1 + u2 + u3 − 1, and δ = v1 + v2 + v3 − 1 Thus, α = a − 1 (mod 2), β = b − 1 (mod 2) and γ = c − 1 (mod 2), δ = d − 1 (mod 2) Therefore, if (a, b) 6= (c, d), (α 6= γ)or (β 6= δ) for (2), 0. 0. If ui + vi < ui + vi for some i then. 3 P. (ui + vi ) <. i=1. or. 3 P. i=1. vi 6=. 3 P i=1. 0. vi . Thus (α 6= γ)or (β 6= δ). 3 P. 0. 0. (ui + vi ) which implies that. i=1. 3 P i=1. ui 6=. 3 P i=1. 0. ui.

<span class='text_page_counter'>(5)</span> GENERIC SINGULARITIES ON LINEAR COMBINATIONS OF PRESCRIBED MONOMIALS IN TWO VARIABLES 5 0. 0. If ui + vi = ui + vi for all i then. 3 P. ui <. i=1. 3 P. 0. ui and (α 6= γ)or (β 6= δ). i=1.  Proof of theorem 1.1. Let T0 = {W (1) , W (2) , ...W (n) } the triangular decomposition found in lemma 2.3 It is sufficient to show that P(T0 ) 6= 0 and P(T0 ) cannot be cancelled by P P(T ) . T ∈TL ,T 6=T0. Let P(T0 ) =. n Q i=1. DiW. (i). =. n Q. (i). CiW xαi i yiβi. i=1. By construction, |W (i) ∩ Mα,β | ≤ 1 for any (α, β) ∈ {0, 1}2 and for any i ∈ {1, 2, .., n}.Thus, (i) by lemma 2.4, CiW 6= 0 (mod 2) for any i ∈ {1, 2, .., n} and the coefficient of P(T0 ) is odd. In particular, P(T0 ) 6= 0. n n Q Q (i) (i) CiV xγi i yiδi and For any T ∈ TL , T 6= T0 , if T = {V (1) , V (2) , ...V (n) } let P(T ) = DiV = i=1. i=1. j the first integer such that W (j) 6= V (j) . Also, let W (j) = {(u1 , v1 ), (u2 , v2 ), (u3 , v3 )} and 0 0 0 0 0 0 V (j) = {(u1 , v1 ), (u2 , v2 ), (u3 , v3 )}. (j) If CjV ≡ 0 (mode 2), then the coefficient of P(T ) is even and it cannot cancel P(T0 ) . (j) If CjV 6≡ 0 (mode 2), then by lemma 2.4 |V (j) ∩ Ms,t | ≤ 1 for any (s, t) ∈ {0, 1}2 . Then there exist (a, b), (c, d) ∈ {0, 1}2 such that W (j) ∩ Ma,b = ∅ and V (j) ∩ Mc,d = ∅ If (a, b) 6= (c, d) then by lemma 2.6 (1),(αj 6= γj )or (βj 6= δj ) . Thus P(T ) cannot cancel P(T0 ) . 0 0 If (a, b) = (c, d) then by the construction of P(T0 ) , (ui , vi ) ≤ (ui , vi ) for every i ∈ {1, 2, 3}. Thus by lemma 2.6 (2), (αj 6= γj )or (βj 6= δj ) and P(T ) cannot cancel P(T0 ) . P Therefore, P(T0 ) cannot be cancelled by P(T ) and DL 6= 0.  T ∈TL ,T 6=T0. 3. Proofs of Corollaries 0. Proof of Corollary 1.2. Let (x1 , y1 ),P ..., (xn , yn ), (xn+1 , yn+1 ), ..., (xn0 , yn0 ) be n general points. 00 2d+5 Let n = d 3 e + 1 and P (x, y) = i+j≤d+2 ci,j xi y j a polynomial of degree d + 2. We have
P 00 a system of ni=1 mi2+1 + 3n homogeneous linear equations with (d+3)(d+4)/2 unknowns. Let A be the matrix corresponding to this system of linear equations, N = {(i, j), i + j ≤ d} and L = {(i, j), d + 1 ≤ i + j ≤ d + 2}. Also let B be the matrix formed by choosing the rows corresponding to the first n points and the columns correspoonding to N ; 00 and C be the submatrix formed by choosing the rows corresponding to the last n points and the columns corresponding to L. To show that our system of equations has no trivial solution it is enough to show that both of the systems of homegeneous linear equations associated with B and C have no trivial solutions. The claim is true for B by assumption. For C, we try to use theorem 1.1. We have |L| = 2d + 5. On the other hand, 00 00 e + 3 = 2d + 5 − (2d + 5)(mod2) + 3 ≥ 2d + 5. Thus |L| ≤ 3n . To use theorem 3n = 3d 2d+5 3.

<span class='text_page_counter'>(6)</span> 6. FINNIGAN ET AL 00. 1.1 we only need to show that |Li | ≤ n for every i ∈ [1, 4]. It is easy to show that  (d + 4)/2 if d is even |L1 | = (d + 3)/2 if d is odd;  (d + 2)/2 if d is even |L2 | = |L3 | = (d + 3)/2 if d is odd;  |L4 | =. (d + 2)/2 if d is even (d + 1)/2 if d is odd; 00. Since by assumption we have (d + 4)/2 ≤ (2d + 5)/3; therefore, |Li | ≤ n for every i ∈ [1, 4].  Proof of Corollary 1.3. Let (x1 , y1 ), ..., (xn+1 , yn+1 ) P be n + 1 general points. As in the previous two corollaries we let P (x, y) = i+j≤d ci,j xi y j be a polynomial of degree d. We have a system of 3n + t(t + 1)/2 homogeneous linear equations with (d+1)(d+2)/2 unknowns. Let A be the matrix corresponding to this system of linear equations, N = {(i, j), i+j ≤ t−1} and L = {(i, j), t ≤ i + j ≤ d}. Also let B be the matrix formed by choosing the rows corresponding to the (n + 1)th point and the columns correspoonding to N ; and C be the matrix formed by choosing the rows corresponding to the first n points and the columns corresponding to L. To proove that our system of equations has no trivial solution it is enough to show that both of the systems of homogeneous linear equations corresponding to the matrices B and C have no trivial solutions. The claim is true for B because there is no curve of degree t − 1 that passes through a point with multiplicity ≥ t. Since |N | = t(t + 1)/2 and |L| = (d + 1)(d + 2)/2 − t(t + 1)/2 ≤ 3n, we try to use theorem 1.1 to proove our claim for the matrix C. We need to show that |Li | ≤ n for every i ∈ [1, 4]. As in the first corollary  (d + 2)(d + 4)/8 if d is even |L1 ∪ N1 | = (d − 1)(d + 1)/8 if dis odd;  (d(d + 2)/8 if d is even |L2 ∪ N2 | = |L3 ∪ N3 | = |L4 ∪ N4 | = (d + 1)(d + 3)/8 if dis odd; and  (t + 1)(t + 3)/8 if d is even |N1 | = (t − 2)t/8 if dis odd;  ((t − 1)(t + 1)/8 if d is even |N2 | = |N3 | = |N4 | = t(t + 2)/8 if dis odd; Since |Li | = |Li ∪ Ni | − |Ni | for every i ∈ [1, 4], it is suffices to show that the greatest possible value of |Li ∪ Li | mince the smallest possible values of |Li | is ≤ n. That is it is suffices to show that (d + 2)(d + 4)/8 − (t − 2)t/8 ≤ n p (d + 1)(d + 2)/2 ≤ t(t + 1)/2 + 3n is equivalent to d ≤ ( −3 + ( 14 + t2 + t + 6n)) . 2 p To show that (d + 2)(d + 4)/8 − (t − 2)t/8 ≤ n is equivalent to show that d ≤ (−3 + (1 +.

<span class='text_page_counter'>(7)</span> GENERIC SINGULARITIES ON LINEAR COMBINATIONS OF PRESCRIBED MONOMIALS IN TWO VARIABLES 7. t2 − 2t + 8n)). p p + ( 14 + t2 + t + 6n)) ≤ (−3 + (1 + t2 − 2t + 8n)). It can be shown that if n ≥ t + 5, d ≤ ( −3 2 But for d ≥ t + 4, (d + 1)(d + 2)/2 ≤ t(t + 1)/2 + 3n implies that t + 5 ≤ 53 n + 2. We notice that in this case, n is necessarly ≥ 5. Because otherwise we would have t + 5 ≤ 5 i.e. t ≤ 0. Thus n ≥ 5 and t + 5 ≤ 35 n + 2 ≤ n which prooves the claim.  P Proof of Corollary 1.4. A polynomial P (x, y) of degree d is of the form i+j≤d ci,j xi y j . Such a polynomial vanishes at n general points with multiplicity ≥ 2 iff the system of 3n homogeneous linear equations P (xk , yk ) = ∂P/∂x(xk , yk ) = ∂P/∂y(xk , yk ) = 0 for any 1 ≤ k ≤ n has at least one solution. The number of unknown in this system is the cardinality of the set {(i, j), i + j ≤ d}. This number is equal to (d + 1)(d + 2)/2. If (d + 1)(d + 2)/2 > 3n, the number of unknown is strictly greater than the number of equations; therefore, there are an infinitely many curves of degree d vanishing at n general points with multiplicity ≥ 2. If (d + 1)(d + 2)/2 ≤ 3n, Let L = {(i, j), i + j ≤ d}. Then |L| = (d + 1)(d + 2)/2 ≤ 3n. To apply theorem 1.1 it suffices to show that |Li | ≤ n for every i ∈ [1, 4]. It is easy to show that  (d + 2)(d + 4)/8 if d is even |L1 | = (d − 1)(d + 1)/8 if d is odd;  (d(d + 2)/8 if d is even |L2 | = |L3 | = |L4 | = (d + 1)(d + 3)/8 if d is odd; If d is even, |Li | ≤ |L1 | for every i ∈ [1, 4]. So it suffices p 1 to proove that |L1 | ≤ n. −3 (d + 1)(d + 2)/2 ≤ 3n is equivalent to d ≤ ( 2 + ( 4 + 6n)) . p (1 + 8n)). To show that (d + 2)(d + 4)/8 ≤ n is equivalent to show that d ≤ (−3 + p 1 p −3 It can be shown that if n ≥ 15 then ( 2 + ( 4 +6n)) ≤ (−3+ (1+8n)). therefore |L1 | ≤ n in this case. p 1 If n < 15, the only possible even values of d that satisfies d ≤ ( −3 + ( 4 + 6n)) are 2 and 2 4; so we still need to study these two cases. For d = 2 , (d + 1)(d + 2)/2 ≤ 3n with n 6= 2 implies that n ≥ 3. But (d + 2)(d + 4)/8 = 3; therefore |L1 | ≤ n in this case. For d = 4, (d + 1)(d + 2)/2 ≤ 3n with n 6= 5 implies that n ≥ 6. But (d + 2)(d + 4)/8 = 6; therefore |L1 | ≤ n in this case. For d is odd, |Li | ≤ |L2 | for every i ∈ [1, 4]. So it is sufficient to show that |L2 | ≤ n. This is true for n ≥ 15 because (d + 1)(d + 3)/8 < (d + 2)(d + 4)/8. So, we only need to show that the claim is true for n < 15. p + ( 14 + 6n)) are 1 and 3. If n < 15, the only possible odd values of d that satisfy d ≤ ( −3 2 For d = 3, (d + 1)(d + 2)/2 ≤ 3n implies that n ≥ 4. But (d + 1)(d + 3)/8 = 3 ≤ n; therefore |L2 | ≤ n in this case. For d = 1, (d + 1)(d + 3)/8 = 1 ≤ n for any n ≥ 1. therefore |L2 | ≤ n in this case. .

<span class='text_page_counter'>(8)</span> 8. FINNIGAN ET AL. References [1] J. Alexander, Singularités imposables en position générale à une hypersurface projective, Compositio Math. 68 (1988), no. 3, 305–354 [2] J. Alexander, A. Hirschowitz, Polynomial interpolation in several variables, J. Alg. Geom. 4 (1995), no.2, 201–222 [3] J. Alexander, A. Hirschowitz, Generic Hypersurface Singularities, Proc. Indian Acad. Sci. Math. Sci. 107 (1997), no. 2, 139- 154 [4] J.E. Campbell, Note on the maximum number of arbitrary points which can be double points on a curve, or surface, of any degree, The Messenger of Mathematics, XXI, 158–164 (1891–92) [5] K. Chandler, A brief proof of a maximal rank theorem for generic double points in projective space, Trans. Amer. Math. Soc. 353 (2001), no. 5, 1907–1920 [6] K. Chandler, Linear systems of cubics singular at general points of projective space, Compositio Mathematica 134 (2002), 269– 282 [7] K. Lee, Characteristic 2 approach to bivariate interpolation problems, [8] F. Palatini, Sulla rappresentazione delle forme ternarie mediante la soma di potenze di forme lineari, Rend. Accad. Lincei V, t.12 (1903), 378–384 [9] A. Terracini, Sulla rappresentazione delle coppie di forme ternarie mediante somme di potenze di forme lineari, Annali Mat., 24 (1915), Selecta vol. I Department of Mathematics, Wayne State University, Detroit, MI 48202.

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