If R is not commutative, RP is both a left and a right R module. Tensor the left module M with the right module RP, and MP is a well defined abelian group. However, this does not come up very often. Most of the time we want MP to be an R module. Therefore I will assume, throughout this topic, that rings are commutative.
Since RP is itself a ring, tensoring with RP performs a base change, so that MP is an RP module. For example, if R is an integral domain and P = 0, MP becomes a vector space.
The new module is often interpreted as an RP module. Thus we can concisely say: if M is projective then MP is projective. This means M is a projective R module, and MP is a projective RP module. Now that we've said it, let's prove it.
A free R module becomes a free RP module of the same rank. The generators simply carry across. Let F = M*U, where M is projective, and tensor with RP. Remember that tensor and direct sum commute, thus MP is the summand of a free module, and is projective.
This applies to any base change S, not just localization. A projective R module becomes a projective S module. However, the result need not be a projective R module. Z×Q = Q, which is a free Q module, but not a free or projective Z module. (We'll explore this further in the next section.)
If R does not contain 1, use d/d2 instead of 1/d. Thus everything in M×S is some x in M crossed with some d/d2, where d is a denominator of S.
Before you procede, review the formal definition of a fraction ring. The characteristics of the tensor product are similar, and are in fact identical when M = R. This reaffirms R×S = S, as one would expect.
An element x/d ∈ M×S cannot be squashed down to zero by the relations of bilinearity unless something in R can be passed between the two components, driving one of the components to zero. Let's look at the possibilities.
Suppose a fraction of S, equivalent to 0, is multiplied by c to get 1/d. Then 1/d = 0, which is impossible. (Remember that 1/d is a unit in S.)
Perhaps 1/cd is nonzero, but cx becomes 0. This can happen only if c kills x in M, and some fraction of S, times c, = 1/d. Let a/b times c = 1/d in the fraction ring S. Write acdv = bv, where v is a denominator of S. Now acdv kills x, hence a denominator, namely bv, kills x. We'll see this condition again in the next paragraph.
Going the other way, we might factor c out of x and apply it to 1/d. As before, x/c cannot be 0, else x would be 0. Is c/d equal to 0? Only if c times a denominator of S equals 0. The same denominator kills x, which is the condition we saw earlier.
In summary, x/d becomes 0 in M×S iff some denominator of S kills x. Also, x1/d1 equals x2/d2 iff their difference is 0, iff (d2x1-d1x2)*v = 0 for some denominator v. If this looks familiar, it should; it is analogous to the equivalence relation originally defined for fraction rings.
Continuing the above, assume the denominators of S don't kill anything in M. Thus each x/d is nonzero, and M embeds into M×S via x/1. This is analogous to embedding an integral domain into its fraction field. In fact, if R is an integral domain and M is a direct sum/product of ideals of R, then M embeds into M×S, where S is any fraction ring of R. As a special case, M could be a free R module, or a submodule thereof.
All these theorems apply even if R does not contain 1. Use d/d2 instead of 1/d throughout. Note that S always contains 1, in the form of d/d, and M×S becomes a unitary S module. Multiply xd/d2 by d/d and demonstrate equality.