Projective Localization, Projective is a Local Property

Zero is a Local Property

If x = 0 then M tensored with any R algebra sets x,1 equal to 0. This holds for any fraction ring, and any localization. We only need consider the converse.

Assume each localization about a maximal ideal maps x to 0. Let J be the ideal in R that kills x. Since 1 does not kill x, J is a proper ideal, and embeds in a maximal ideal P. Suppose x/1 = 0 in M_P. A denominator v kills x, yet v lies in J, lies in P. This is a contradiction, hence the localization of x remains nonzero in M_P.

The entire module M is 0 iff each M_P is 0.

Projective is a Local Property

Let M be a finitely presented module such that each M_P is a projective R_P module, and suppose M is not projective. Recall that M is projective iff the natural map from hom(M,R)×W into hom(M,W) is an isomorphism for every W. Therefore there is some W such that the aforementioned map is not an isomorphism. In the following sequence, K is the kernel and C is the cokernel, and either K or C or both is nonzero.

0 → K → hom(M,R)×W → hom(M,W) → C → 0

If C is nonzero, concentrate on the last four modules. If C = 0, concentrate on the middle modules from K to C. In either case we have a sequence that is almost short exact, except for the leading 0.

Find a maximal ideal P such that the localization of the kernel, or cockernel, remains nonzero. Tensor the four term sequence with R_P, and the resulting map between the middle modules has a nontrivial kernel or cokernel.

Build another natural map from hom(M_P,R_P)×W_P into hom(M_P,W_P). Once we prove this map is not an isomorphism, M_P will not be projective, and that will finish the proof.

Consider the following diagram.

The down arrows are carefully selected isomorphisms. Let's look at the left vertical arrow. When viewed as an R_P module, the upper left module is isomorphic, in a natural way, to hom(M,R)×R_P×W×R_P, or hom(M,W)×R_P×W_P. Now apply the isomorphism described in the previous section, between hom(M,W)×R_P and hom(M_P,R_P). (This is where we need M to be finitely presented.) The composition of these two isomorphisms equates the upper left module with the lower left module.

The vertical arrow on the right is once again an application of the previous theorem, an interchange between dual and localization.

Next, show that the diagram commutes. Moving right and down is the same as moving down and right. Select an element f,x,a/b from hom(M,R) cross W cross R_P and follow this triple as we travel down either path. The result is f*x*a/b in the lower right. I'll leave the details to you.

If the kernel of the top row is nontrivial, let y be a nonzero element of the kernel. Move down to something nonzero in the lower left, and over to 0 in the lower right. Thus the bottom row has a nontrivial kernel.

If the kernel of the top row = 0, then the cokernel is nontrivial. Let y be an element in the upper right that is not in the image of the upper left. Apply the rightmost isomorphism, mapping y to z in the lower right. Suppose the cokernel of the bottom row is 0. Pull z back to something in the lower left module, and lift this up to x in the upper left. Starting at x, follow either path to the lower right module, and you have to reach z. Now z has only one preimage in the upper right, namely y. Thus x maps to y across the top, which is a contradiction. There is indeed a cokernel down below. The bottom row fails to be an isomorphism, and this particular localization is not projective.

In summary, a finitely presented module is projective iff all its localizations are projective. Projective is a local property.

Applying an earlier theorem, a finitely presented module is projective iff all its localizations are free.

Combining Duals

Let U be S×T, as an R module, and let U′ be the dual of U into R. Note that U is also finite flat, or finitely presented, as the case may be.

Build a map j from S′×T′ into U′ by applying the two functions of S and T on the two components of U respectively. Since both functions are R homomorphisms, R can be passed between the two components, and the image is the same. Thus the resulting function is well defined on U.

Since j is bilinear on S′ cross T′ into U′, it factors through the tensor product. In other words, j is an R module homomorphism. We will prove j is an isomorphism.

We only need demonstrate the isomorphism locally. Since base change distributes across tensor product, we can localize S′ and T′ separately, then take their tensor product. Since S and T are finite flat, or finitely presented, and R_P is flat, dual and tensor product commute. We're really talking about the functions from S_P into R_P. The same thing happens to T′, and U′.

To simplify notation, assume localization has already occurred. Thus S is free over R. The dual of S is also free, of the same rank. Tensor T with S and you are taking n copies of T, where n is the rank of S. Similarly, S′×T′ is n copies of T′. Evaluate U′ and get the same set of homomorphisms, from n copies of T into R. We are taking advantage of the fact that dual and direct sum commute. Thus j is an isomorphism, and that completes the proof.

Here is another way to combine duals. Let S be finite flat over R, and let T be finite flat over S, so that T is finite flat over R. We know that T′ is an S module, via postmultiplication by S. In the same way, S′ is an R module, via postmultiplication by R. However, S′ can also be viewed as an S module, via premultiplication by S. Let c be an element of S. Multiply S by c first, then apply the particular homomorphism from S into R to get a new homomorphism. Multiplying S by c is R linear, so the resulting homomorphism is still R linear. Verify the distributive properties, and S′ becomes an S module.

Since T′ and S′ are both S modules, we can consider the tensor product T′×S′. An element c from S can be multiplied on the right of T′, or on the left of S′, and the result is the same.

The map j is composition. Combine a function from T into S, with a function from S into R, and get a function from T into R. Since c is trapped in the middle, j is well defined on the tensor product. As before, j is bilinear on the two duals, and it becomes an S homomorphism on their tensor product. We will show this is an isomorphism by working locally. As we saw earlier, localization and dual commute, so there is no trouble.

Assume localization has already occured, so that T and S are free over R. This does not mean T is free over S, but we can localize about Q in S, lying over P, to make this hapen. This was described earlier. Now T is free over S, is free over R, and the homomorphisms from T into S, or into R, look just like T, as an S module or an R module. Just follow the images of the basis of T.

Let h be an R linear function from T into R. In practice, h is a vector of elements of R, and the function is the dot product of h with the vectors of T. Is h in the image of j?

Since j is an R module homomorphism, its image is an R module. It is sufficient to prove every projection from T onto R is in the image of j. These projections can then be scaled and combined to build h.

Remember that T is a direct sum of copies of S. Let f extract a particular copy of S, and let g extract a particular copy of R within S. Note that f is S linear and g is R linear. Their composition, f followed by g, produces the desired projection of T onto R. Therefore j is surjective.

Now for injective. Let f map 1 in the i^th component of T to some nonzero c in S. The image is the principal ideal generated by c. Let g be a function from S into R, such that the composition h = fg becomes 0. Within the tensor product, we can pass the action of S between f and g. Let f map 1 to 1, and premultiply g by c. Now g maps all of S to 0 in R, and is the 0 function. The combination f,g is 0 in the tensor product, and j is injective.

Since j is an isomorphism locally, it is an isomorphism on the original modules. The composition of hom(T,S) with hom(S,R) builds an isomorphism between their tensor product and hom(T,R).