Projective Modules, Faithful Modules

Faithful Module

A module M is faithful if, for every module U, MU = 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 QZp = 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 MM, use the quotient formula, giving M mod JM.  Since JM is 0, the result is M.  finally JM 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, KT is nonzero.  Thus KT is a nontrivial kernel of MT.  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, CT becomes a nonzero cokernel, and that is impossible.  Therefore f is an isomorphism, and M and N are equivalent.