An Illuminating Counterexample

Sd+1PI1 occurs exactly as a product (aci . .* aliklaak ... aid)(/llaj), where the index j belongs to {1, 2, ..., n} \ {il, ..., id) and ik-1 < j < ik. (Of course if j < ii, then aj is the first factor, while if id < j, then aj is the last factor in the bracket.) Therefore each such summand ailai2 ... a d Occurs n d times. Precisely because (n d)sd consists of the monomials y-(n d)ail ... aid where 1 < i1 < i2 < ... < id < n, we see that B contains no positive monomial. The upshot: we see that in the sum A + B there are only negative monomials, whereas in C there are only positive monomials. Therefore no monomial in A + B can be cancelled out by a monomial in C. Thus A + B = 0 = C, and our proof is complete.