Non-vanishing of Alternants

Let p be prime, K a field of characteristic 0. Let (x1, . . . , xn) ∈ Kn such that xi 6= 0 for all i and xi/xj is not a root of unity for all i 6= j. We prove that there exist integers 0 < e1 < e2 < · · · < en such that det (x ei j ) 6= 0. The proof uses p-adic arguments. As corollaries, we derive the linear independence of certain Witt vectors and study the result of applying the Witt functor to a mod p representation of a finite group. Let B be a commutative ring, xi elements in B and fi : B → B arbitrary functions, i = 1, . . . , n. The n×n matrix (fi(xj)) is called an alternant matrix, and its determinant is called an alternant. Alternants have a long history. For basic facts about them, see [A]. The most common alternants, which are the topic of this note, occur when the fi are polynomial functions. Classically, one wants to evaluate alternants in terms of simpler functions. In this paper, we are concerned instead with the nonvanishing of certain alternants that come up in the study of rings of Witt vectors. The application to Witt vectors, presented in Theorem 6, will be of use (we hope) in work in progress that compares cohomology of arithmetic groups with coefficients in Z/p and in Q. We use bold face to denote n-dimensional vectors. For x = (x1, . . . , xn) and m = (m1, . . . ,mn), set A(x,m) = det(xi j ). For example, A(x, (0, 1, . . . , n−1)) := V (x) = ∏ i>j(xi − xj) is the Vandermonde determinant. It vanishes if and only if two of the xi’s are equal. What is the condition for vanishing of the alternant if the exponents are some other sequence of non-negative integers? In particular we ask this question when the exponents are all powers of a fixed prime p. Our main theorem states: 1991 Mathematics Subject Classification. Primary 15A15, Secondary 15A03, 13K05,20C11.