Integrality of resolvents

This is from Stauduhar's "classic" article on resolvents:

Let p(x) in Z[x] be monic irreducible with roots r_1,...,r_n... Suppose H is transitive and that Gal(p) < H. Let G < H and F(x_1,...,x_n) a function belonging to G in H [i.e. stab_H(F)=G]. Let s_1,...,s_k be a transversal of H/G. Then Q_(H,G)(y) = prod_{i=1}^k (y - s_i F(r_1,...,r_n)) has integer coefficients.

Proof: For each i, s_i F(r_1,...,r_n) is an algebraic integer. Hence the coefficients of Q are algebraic integers. Now suppose t is in Gal(p). Then t is in H, and hence

tQ = prod (y-ts_i F(r_1,...,r_n)) = prod (y - (ts_i) F(r_1,...,r_n)). But the ts_i is also a transversal of H/G. Thus the application of t has merely permuted the roots of Q, leaving the coefficients fixed. The coefficients of Q are then algebraic integers left fixed by Gal(p) and are therefore rational integers. (end proof)

I posted the whole proof for entirety, but for now my question is about the very first assertion, that s_i F(r_1,...,r_n) is an algebraic integer. Umm... Why? If e_i is the i:th elementary symmetric polynomial and \Omega is the root tuple, this assertion means that [; e_i(s_1 F(\Omega),...,s_k F(\Omega)) ;] is an algebraic integer. I just don't see it, even if it's obvious to Stauduhar.

Re: Integrality of resolvents