On differential independence of the Riemann zeta function and the Euler gamma function

Abstract

It is well-known ([2] and [4]) that Γ , as well as ζ, is not a solution of any algebraic differential equation with coefficients in C. In other words, if P (u0, u1, . . . , um) is any polynomial in u0, u1, . . . , um over C, and P (Γ, Γ ′, . . . , Γ )(z) ≡ 0 or P (ζ, ζ ′, . . . , ζ)(z) ≡ 0, for all z ∈ C, then the polynomial P is identically zero. This answers Hilbert’s conjecture [1] in his 18th problem. For a detailed discussion of this subject and other related topics, we refer the reader to [5]. Let h(z) = ζ(sin(2πz)). Recently, Markus [3] proved that if P (h, h′, . . . , h;Γ, Γ ′, . . . , Γ )(z) ≡ 0 for z ∈ C, then the polynomial P (u0, u1, . . . , um; v0, v1, . . . , vn) is identically zero. Thus, in the terminology of differential algebraic theory, Γ and h are differentially independent over C (hence, over C(z)). Furthermore, Markus [3] conjectured that if P (ζ, ζ ′, . . . , ζ;Γ, Γ ′, . . . , Γ )(z) ≡ 0 for z ∈ C, then the polynomial P (u0, u1, . . . , um; v0, v1, . . . , vn) is identically zero, i.e. Γ and ζ are differentially independent. In this short note, we prove that ζ and Γ cannot satisfy a class of algebraic differential equations. Let P (u0, u1, . . . , um; v0, v1, . . . , vn) be any polynomial with coefficients in C. For a non-negative integer μ, we let Λ = Λ(μ) = {(λ0, λ1, . . . , λμ) : λj is a non-negative integer and 0 ≤ j ≤ μ <∞}