(b) if then since is surjective, there exists such that thus: hence is surjective.
(c) let then thus: which proves that generates as module, because
to prove that the set is linearly independent, we assume that which after clearing denominators gives us: where since
is assumed to be a basis for M, it's a set of linearly independent elements of R. thus hence
(d) let which is obviously multiplicatively closed because R is a domain. now is the field of fractions of on the other hand we have:
now since is a field, the LHS and RHS of (1) are two vector spaces. since they are isomorphic, their dimensions must be equal. (recall from linear algebra!) thus