Projective Modules, Tensoring a Short Exact Sequence

Tensoring a Short Exact Sequence

Recall that a short exact sequence is an embedding of A into B, with quotient module C, and is denoted as follows.

0 → A → B → C → 0

If these are left modules, and M is a right module, consider the three tensor products: AM, BM, and CM. These are abelian groups, or R modules if R is commutative.

A homomorphism carries A into B, and another homomorphism carries M onto M, namely the identity map. In the previous section we described the tensor product of two homomorphisms. Thus there is an induced homomorphism from AM into BM, and another induced homomorphism from BM into CM. We will show that the derived sequence is almost short exact. The first 0 is lost; AM need not embed into BM.

AM → BM → CM → 0

Don't let the notation confuse you. We are really saying that there is a homomorphism from AM into BM, and the quotient is CM. That's all.

We know that AM maps into BM and BM maps into CM. To complete the proof we need to show that BM maps onto CM, and that the kernel of this map is the image of AM.

Let's start by mapping a pair generator xy in BM over to CM. Remember that the diagram commutes. We can go from B cross M down to BM and over to CM, or from B cross M over to C cross M and down to CM. The image of xy in BM is therefore g(x)y in CM, where g is the homomorphism from B onto C. Since g is onto, each pair generator in CM has a preimage in BM, and the map from BM to CM is onto.

Next show that AM maps into the kernel of BM. Let x and y in A cross B lead to the generator xy in AM. The image of x in B, and then in C, is 0. Travel across the top and down to CM and get 0. Since the diagram commutes, travel down to xy in AM, then through BM and into CM. The result must be 0. All the generators of AM map into the kernel of BM, hence the image of AM lies in the kernel of BM.

Finally show that the kernel of BM lies in the image of AM. Consider an element e in the kernel of BM. Remember that e is a linear combination of generating pairs from B cross M, and the image of any one of these generators xy becomes g(x)y in CM. This applies to all of e. for instance, if e = 5x2y3-x6y7, then the image of e in CM is 5g(x2)y3-g(x6)y7. This has to equal 0.

The image of e is spanned by the relations of bilinearity in CM. One of these relations might look like z3y-z2y-z1y. Perhaps the image of e is equal to this relation plus a handful of others. Pull the relation back to a relation in B cross M. Select any w1 in the preimage of z1, and so on for z2 and z3 and write w3y-w2y-w1y-w0y. The extra element w0 lies in the image of A. Since w3 = w1+w2 as cosets, we use w0 to attain proper equality. This maps forward to the relation that defines CM. Furthermore, it is a bilinear relation in B cross M, used to define BM. Every relation can be pulled back in this manner. Therefore e becomes the sum of relations in B cross M, plus some extra terms from A cross M. Looking to the tensor product, e is 0, plus some terms in A cross M. These come from AM, hence e is in the image of AM after all. The image and the kernel coincide.

As a generalization of the above, tensor A B and C with three isomorphic modules MA MB and MC. The proof is the same, and the following exact sequence appears.

AMA → BMB → CMC → 0

This makes intuitive sense; the elements of M are merely relabeled, as it is tensored with A B and C.

Tensor and Quotient Commute

Without realizing it, we have proved that tensor and quotient commute. In other words, C tensor M equals B tensor M mod A tensor M, where AM is interpreted in the context of BM.

An Inexact Example

Note that AM need not embed in BM. We lost the leading zero. Let the inclusion of the even integers into Z produce a short exact sequence as follows.

0 → 2Z → Z → Z2 → 0

Take the tensor product with Z2. Remember that Z2 is a quotient ring, so we can use the formula for tensoring with a quotient ring. In each case the tensor product is Z2. However, we cannot embed Z2 into Z2 and find a quotient group equal to Z2. The first tensor function maps Z2 to 0, and the second is an isomorphism.

Split Exact

Let our short exact sequence be split exact. In other words, B contains a copy of A and C, and B is the direct sum of A and C. Tensor with M, and BM is the direct sum of AM and CM. As we saw in the previous section, the induced homomorphism from BM onto CM is the direct sum of the induced homomorphisms from AM into CM and from CM into CM. In other words, the map from BM onto CM is a projection. Its kernel is AM, or more accurately, the image of AM, which has nothing to do with C, and is an embedding. The resulting sequence has its leading 0, and is split exact.

0 → AM → BM ⇔ CM → 0

Mathematical Simulation for Cooking the Perfect Rare or Medium Steak:
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