Problem: Let . Show that in .
If the notation is unfamiliar to you, read up on tensor products of modules.
What we are doing here is forming a -module comprised of sums of symbols of the form , where , subject to the condition that the map be -bilinear, that is, linear in each variable and satisfying for all .
It is possible to show that two different symbols represent the same element of . A basic example is . This illustrates an important thing about tensor products: in general, it is easy to prove that elements are equal and difficult to prove that they are not equal.
As in the previous article, the solution here is to approach this computation obliquely, through the universal property of tensor products, which we will describe briefly for commutative rings.
Suppose that is a commutative ring and and are -modules. Then the tensor product is universal in the following sense: let be any -module and an -bilinear map. Then there is a unique -module homomorphism such that the following diagram commutes:
(again, apologies for the poor diagram) Here is simply the map .
This suggests a general line of attack for our problem: to construct a bilinear map from to some -module that maps and to different elements.
One failure would be to define by . This map is -bilinear, but it doesn’t distinguish between these two elements. A better choice is , which sends to , and to .
By the universal property above, it follows that there exists a homomorphism such that and , and it follows that .
A mistake that I used to make when working with tensor products was to try to define homomorphisms like this directly. In a proof you might write “Define by .” But this is incorrect; we do not know a priori that this is a real definition, that is a well-defined map. So understanding the universal property is really fundamental when working with tensor products.
Related Problem: Show that is not a simple tensor; that is, it cannot be written in the form for .