consider the module over the rational numbers (R = Q) generated by 1: that is the module Q itself (since Q being a ring, is a module over Q).

if Q was a finitely-generated module over Z (abelian groups are Z-modules), we would be able to express any rational number as a Z-linear combination of afiniteset of rationals:

B = {q_{1},...,q_{n}}.

write q_{j}= a_{j}/b_{j}, where a_{j},b_{j}≠ 0, and gcd(a_{j},b_{j}) = 1.

consider k = lcm(b_{1},...,b_{n}).

certainly any Z-linear combination of elements of B can be written as N/k, where N is some integer (which N we get depends on which Z-linear combination we have, but we can use the same k).

taking out common factors, we have N/k = N'/k', where gcd(N',k') = 1.

let p be a prime number that does not divide k (we can ALWAYS find such a prime number, since there are infinitely many primes, but k has only a finite number of prime factors).

i claim 1/p is not in span(B) (where this indicates the Z-linear span).

for if it were, we have 1/p = N'/k', and thus: k' = N'p, hence p divides k', and thus divides k, contradiction.