Projective Localization, RP is a Flat R Module

RP is a Flat R Module

Naturally RP is a flat RP module, but it is also a flat R module. Start with the following short exact sequence and ask when the localization embeds KP into MP.

0 → K → M → M/K → 0

Everything in KP can be represented by an element in K over a denominator drawn from R-P. Let x/d be a nonzero member of KP that becomes 0 in MP. As shown in the introduction, Some denominator kills x in M, and the same thing happens in K, hence x/d was really 0 after all. Therefore KP embeds, and RP is flat.

There is nothing special about localization here; every fraction ring of R is flat. This holds even if R does not contain 1.

As a special case, Q is a flat Z module. Moving from the rationals to the reals, R is a Q vector space, with a basis, hence R is a direct sum of flat Z modules, and R is a flat Z module.

If M is flat, tensor M with a fraction ring S of R, which is also flat, and the resulting module M×S is a flat R module, and a flat S module. Localization preserves flatness.

Flat and Not Projective

If R is a nontrivial pid, its fraction field F is not projective (as an R module), hence there are flat modules that are not free or projective.

Since R is a pid, the summand of a free R module is free. Thus free and projective are synonymous. We will prove F is not free, hence it is not projective.

build a module R/P analogous to the integers mod P. When this is tensored with F the result is 0. The proof is similar to ZP tensor Q. Since R/P tensor F is 0, F cannot be a free R module.

Isomorphism is a Local Property

Let f be an R module homomorphism from U into V. If P is a prime ideal, tensor the exact sequence 0 → U → V → 0 with RP and find an isomorphism between UP and VP, as R modules, or as RP modules. In fact this generalizes to any base change. This makes sense, intuitively, because the points of V are merely the points of U relabeled.

In general, the tensor product preserves surjective, and injective if the module is flat.

Conversely, assume f is not injective, such that f(x) = 0. Find a prime ideal P such that xP is nonzero. Tensor the following sequence with the flat module RP.

0 → K → U → V → 0

The result remains exact, and KP is nonzero, containing xP, hence the localization is not injective.

If f is not surjective let C be the cokernel.

K → U → V → C → 0

Select P such that CP is nonzero, and tensor with RP. The new homomorphism from U to V is not surjective.

In summary, a function is injective, surjective, or bijective iff it is the same under maximal localizations.

Containment is a Local Property

If A is contained in B, write the exact sequence 0 → A → B → C → 0. The cokernel C is nonzero iff containment is proper. As shown above, the cokernel is 0 iff it is 0 locally. Thus module equality is a local property.

Let A and B be submodules of a module M. Note that B contains A iff (B+A)/B is zero. Tensor the sequence 0 → B → B+A → (B+A)/B → 0 with RP. Remember that B+A commutes with localization. The result embeds BP into BP+AP. As above, the cokernel is zero iff it is zero locally. Thus containment of A by B is a local property.