This definition remains valid when M is a left module over a noncommutative ring, but more often, faithful refers to modules over a commutative ring.
The tensor product of two faithful modules is faithful.
If one component of a direct sum is faithful, then the entire direct sum is faithful.
since RU = U, R is faithful.
Take the direct sum of copies of R, and every free module is faithful.
If R is a division ring then every R module is free, and faithful.
If R is a pid then every finitely generated torsion free module is free, and faithful.
We now have a round-about proof that Q is not a free Z module. If it were it would be faithful, but we have seen that Q×Zp = 0. So Q is not faithful or free.
Q is not faithful, but it is flat. Are there faithful modules that aren't flat?
Let M = Zp, which is an R module. In fact it happens to be isomorphic to R/J. Tensor M with the exact sequence 0 → J → R → M → 0. The middle module is easy; R tensor M is M. To evaluate M×M, use the quotient formula, giving M mod JM. Since JM is 0, the result is M. finally J×M becomes J mod JJ, which is J. The result is J → M → M → 0. These are all M vector spaces, so if J embeds in M then M has dimension 2 instead of 1. This is a contradiction, hence M is not flat.
Suppose f is not injective, with kernel K, and restrict N to the image of M under f. Tensor the following sequence with T.
0 → K → M → N → 0
Since T is flat, the resulting sequence is exact. Also, since T is faithful, K×T is nonzero. Thus K×T is a nontrivial kernel of M×T. This is impossible, hence f is injective.
Suppose f is not surjective, and let C be the cokernel. Tensor the following sequence with T.
M → N → C → 0
Again, C×T becomes a nonzero cokernel, and that is impossible. Therefore f is an isomorphism, and M and N are equivalent.