Projective Modules, Faithful Modules

Faithful Module

A module M is faithful if, for every module U, M×U = 0 implies U = 0.

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?

Faithful Not Flat

A faithful module need not be flat. Let R be the integers mod p2, with J equal to the ideal generated by p, i.e. the multiples of p. Note that R is a local ring, and its maximal ideal J is nilpotent. If the tensor product of two R modules is zero then one of the two modules is zero. Therefore every nonzero module over R is faithful. We only need find one that isn'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.

Faithful Homomorphism

A ring homomorphism from R into S is faithful if S is a faithful R module. This is similar to the definition of a flat homomorphism, as presented in the previous section. A homomorphism is faithfully flat if it is faithful and flat, i.e. if S is a faithful flat R module. For example, embed R into R[x], the constants in the ring of polynomials. Since R[x] is a free R module it is both flat and faithful.

Unwrapping the Isomorphism

Let T be a faithfully flat R module, and let M and N be R modules, with a module homomorphism f from M into N. Tensor M and N with T, and assume the resulting homomorphism, f cross the identity map on T, is an isomorphism. We can pull back and prove M and N are isomorphic.

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.