504 ARITHMETIC 4.6.4 a) The first construction is based on the identity abc + ABC = (a + A)(b + B)(c + C) -(a + A)bC -A(b + B)c -aB(c + C). It follows that c xij y2 k zkz = c ( ii + Zi)(?, jk + ?&)(zki + ZjR) -cl -x2 -x3, 1 7711 hand side of the above identity and Ci, CZ, Cs correspond respectively to the next three groups of terms; the remaining terms (namely those corresponding to uBC + ABC + AbC) cancel out of the sum. The set S in this case is different from the S in part (a); it consists of all (i, j, k) E 0 X 0 X 0 such that i 5 j and i < k. It follows from this construction that M(n) 2 $(($)3 -(4)) + 6n2 when n is even. 61. [M23] Let (tijk) be a tensor over an arbitrary field. We define rankd(t,,k) as the minimum value of r such that there is a realization of the form where ail(u), bil(u), cki(U) are polynomials in u over the field. Thus ranks is the ordinary rank of a tensor. Prove that (a) rankd+l(tijk) 5 rankd(tijk); (b) rank(tijk) 2 (d$2)rankd(tt3k); (c) rankd((t%,k) @ (t&)) < rank&&k) + rankd(ti,,), in the sense of exercise 48; (d) rankd+,+jk) @ (t&k)) 2 rank&k) . rank&jk).
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