Projective Localization, Tensor Product and Dual Commute

Tensor Product and Dual Commute

Let M and W be R modules, so that hom(M,W), also known as the dual of M into W, is an R module.

Let S be an R algebra, so that tensoring with S produces an S module. Sometimes S is a fraction ring of R, so that tensoring with S is a form of localization. However, other algebras are possible.

We're going to build a map j from the tensor of the dual into the dual of the tensors. In other words, j maps hom(M,W)×S into hom(M×S,W×S). The domain and range are both S modules, and in some cases, j is an S module isomorphism. When it is, the domain and range are isomorphic, and the dual and the tensor product commute. If M is finitely presented, and S is a fraction ring of R, j becomes an isomorphism (shown below), and dual and localization commute.

Start with a function f from M in to W,crossed with an element c from S. This represents an element of hom(M,W)×S. If f(x) = y, let the corresponding homomorphism g map x,1 to y,c. In other words, g is the aplication of f to all of R, then the image is crossed with c. Thanks to f, g respects addition and the action of R. In other words, g is an R module homomorphism from M into WS.

Extend g by tensoring with the identity map on S. Now g is an R module homomorphism from MS into WSS. Let's see what this homomorphism looks like. Given x,b in MS, let g(x,b) = y,c,b. In other words, the contribution from S carries across. Given d in S, apply d to x,b and y,c,b, producing x,bd and y,c,bd. Thus g respects the action of S, and g is an S module homomorphism. View everything as S modules, and WSS becomes WS. The two S components are simply folded together. Thus y,c,b is simply y,cb. Now g is an S module homomorphism from MS into WS. This completes the definition of j, as a map on sets.

Note that j is bilinear in hom(M,W) cross S. It therefore becomes a well defined R homomorphism on the tensor product. Apply d in S, and g is multiplied by d, which is the action of d in hom(MS,WS). Thus j is an S module homomorphism.

Free of Finite Rank

Let M = R, and suppose j carries f,c to the trivial homomorphism. Of course f is defined by the image of 1, which I will call y. Thus g, as an S module homomorphism, is defined by g(1,1) = y,c. Therefore y,c is equivalent to 0 in WS. Some z in R passes between y and c, making one of the two components 0. Use the same z, passed between f and c, to kill f or c. Therefore f,c is equivalent to 0, and j is injective.

If g(1,1) = y,c, let f(1) = y, and j maps f,c onto g. Therefore j is surjective, and j is an isomorphism. This is a lot of algebra to prove something rather intuitive. The homomorphisms from R into W are equivalent to W, hence the domain is WS. At the same time, the homomorphisms from S into WS are equivalent to WS. Thus j implements an isomorphism from WS onto WS.

Let M be a free module of finite rank. In other words, M is the direct product of n copies of R. Since finite direct product commutes with dual and with tensor product, j respects direct product. Since j is an isomorphism for each copy of R, j is an isomorphism on M. The domain and range both look like n parallel copies of WS.

If M is projective it is the summand of a free module. Since j is an isomorphism on the aforementioned free module, it must be an isomorphism when restricted to M. Thus tensor and dual commute when M is projective. If M is not projective, we need to know that M is finitely presented, as described below.

Finitely Presented

If M is finitely presented, consider the following exact sequence.

0 → K → F1 → F0 → M → 0

Moving from right to left, M is the quotient of a free module F0, which is finitely generated. In other words, the generators of F0 map onto the generators of M. The kernel of F0 is also finitely generated. This is what we mean when we say M is finitely presented. Thus there is a finite set of words in F1 that generate the kernel, and map to 0 in M. Let these words span a free module F1. Now F1 maps into F0, and the kernel of this map is K. We don't know anything about K, and we don't care. We're only interested in the right half of the sequence: F1 → F0 → M → 0.

Tensor with S and the result remains exact. The first two modules become free S modules of finite rank, and the third is MS.

Then take the dual into WS. This gives an exact sequence running in the reverse direction.

0 → hom(MS,WS) → hom(F0S,WS) → hom(F1S,WS)

At the same time, take the dual, and then tensor with S.

? → hom(M,W)S → hom(F0,W)S → hom(F1,W)S

We lost the leading 0 when we tensored with S, but we really want it back again. Assume S is flat, so there is no trouble. As a special case, S could be a fraction ring or a localization of R, which is flat.

Place the two sequences above each other, like this.

0 → hom(M,W)S → hom(F0,W)S → hom(F1,W)S
↓ ↓ ↓ ↓
0 → hom(MS,WS) → hom(F0S,WS) → hom(F1S,WS)

The down arrows are the homomorphisms described at the top of this page, denoted by the letter j. Since F0 and F1 are free of finite rank, the last two homomorphisms are actually isomorphisms. The first homomorphism is also an isomorphism, mapping 0 to 0.

At this point we need some tedious algebra to show the diagram commutes. Moving right and down is the same as moving down and right. It's not as hard as it looks, because the definition of j is a natural combination of the dual of a module and S. I'll leave the details to you.

Ok, if you believe this is a commutative diagram, we're home free. Put an extra 0 at the left of each sequence and run another isomorphism between these two zeros. Than apply the 5 lemma, and you're done. The map j from hom(M,W)S into hom(MS,WS) is an isomorphism, and the two modules are equivalent.

If M is finitely presented, and if S is a fraction ring of R, or a flat R algebra, then dual and tensor product commute.