## 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**×**Z**p = 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 = **Z**p, 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.
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.