Let R = Z[p]/p2. This is the integers adjoin p, where p2 is 0. Let J be the principal ideal generated by p.
Let M be R adjoin q, with pq = 0. Thus M is an R module. Note that J kills off p and q in M, hence JM is isomorphic to Z. Is this the tensor product?
Let g map J cross M into Z2 by carrying px cross a+bp+cq to x×(a,c). Clearly g respects addition in J and in M. Assume d+ep is a factor of px. This means d is a divisor of x. Pull this out and move it to the other side. In other words, multiply d+ep by a+bp+cq. Extract the first and third components and multiply by x/d; the image under g is the same. Thus the action of R can be passed between J and M, and the map is bilinear. Since the tensor product is universal, a group homomorphism completes the diagram. It agrees with g, and is onto, hence there is a surjective group homomorphism from Z on to Z2, which is impossible. Even though J is principal, J×M does not equal JM.
Are there ideals and modules that satisfy JM = J×M? Let's start again, with J a right ideal and M a left module.
Let T be J×M, which exists, and is unique up to isomorphism. Construct T as we did on the previous page, with generators xy for every x in J and every y in M.
Let g(x,y) = xy in JM. Verify that g is a bilinear map. There is therefore a unique homomorphism h(T) into JM that makes the diagram commute. If h is an isomorphism, then JM is indeed the tensor product.
Since h is a homomorphism, we only need show injective and surjective. Let's start with the latter. Each generator xy in JM comes from g(x,y), and it also comes from h(xy) for xy in T. Linear combinations of generators in T map to linear combinations of generators in JM, and every element of JM has some preimage in T. Thus h is onto.
To show injective, let J be a principal right ideal with generator p. Suppose a linear cobination of generators lies in the kernel of h. Since J is principal we can "normalize" each generator. Instead of using px cross y, use p cross xy. Thus each generator is p combined with something in M. By linearity, the generators combine to form pq, where p generates J and q is an element of M. Thus pq is in the kernel of h. We saw this in our earlier example; pq was equal to 0. But what if p does not kill anything in M? Then pq is nonzero, the image of pq is nonzero in JM, and pq is not in the kernel of h after all. The homomorphism is 1-1, and JM is the tensor product.
similar reasoning holds when M is cyclic, generated by some element q, and nothing in J kills q. I'll leave the details to you.
Let JM be the subgroup of M spanned by the elements of J times the elements of M. If R is commutative JM is a submodule of M.
Let T consist of cosets of JM in M, i.e. the quotient group. Given x in R and y in M, xy represents a coset of JM in M, i.e. an element of T. If x changes by an element in J, the difference lies in JM, and represents the same element of T. Therefore we have a well defined map from R/J cross M into T. Verify that this is a bilinear map, and if R is commutative it respects the action of R.
Since R/J contains 1, and M is unitary (by assumption), the map is onto. The pair 1 cross c is a valid preimage of c in T.
Let g(R/J,M) be a bilinear map into U. Pull c back to a preimage in R/J cross M. A preimage xy can be normalized to 1(xy), or 1c. We don't have to worry about 1d, for some d ≠ c, because modules are assumed unitary. Hence 1c is the only "normalized" preimage of c. All the preimages of c map to the same thing under the belinear map g. Let h(c) = g(1,c). Thus h is well defined, and uniquely determined by g. Pull c+d back to R/J cross M, and forward to U to show h is a group homomorphism, or a module homomorphism if R is commutative. Therefore (R/J)×M equals M/JM.
As before, JM is shorthand for the finite sums of elements of J times elements of M. Note that KM is a subgroup of JM is a subgroup of M. These subgroups become submodules if R is commutative.
Let T be the cosets of KM in JM, and map x in J and y in M to xy, a representative of KM in JM. Verify that this map is well defined and bilinear.
We need J to be principal to show the map is onto. Given c in T, or in JM, remember that c is a linear combination of product pairs. However, each pair can be normalized to p times something in M. The components from M can then be added together, hence something in J times something in M yields c. Every element of T is accessible.
Given g(J/K,M) into U, build h as we did before. The normalized preimage of c is py, for some y in M. If some other pz leads to c then p(y-z) equals 0, yet p kills nothing in M. Thus all the preimages lead to the same thing in U, and this becomes h(c). Therefore J/K tensor M equals JM/KM.
Elements of M are cosets of K in R. Use coset representatives to denote the elements of M. These are multiplied by the elements of J, and the pairwise products combined, to build JM. But what is an element of J times a coset representative in M? Since J is a right ideal, it is simply another element of J. When these are combined, the sum is again in J. Therefore the elements of M that lie in JM are the elements of M whose coset representatives lie in J. Conversely, an element in M that happens to lie in J can be premultiplied by anything in J to produce the same coset of K, the same element of M. Therefore JM = M∩J.
As an abelian group, M = R/K. Take the image of J in this quotient group and find the elements that we call JM. We must mod out by the image of J to find the tensor product. Both J and K combine to span the kernel, and the result is R/(J+K). This is an R module if R is commutative. Multiplication in R implements the bilinear map.
As a corollary, Zm tensor Zn = Zgcd(m,n).
Let S and T be finitely generated modules over a pid. Each can be decomposed into a direct sum of cyclic modules, and a cyclic module is isomorphic to a quotient ring. Since tensor commutes with direct sum, and since the tensor product of two quotient rings is well characterized, S×T can be derived.
Z6Z15 tensor Z9Z35 = Z32Z5Z1
Of course Z1 is trivial, and drops out.