**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 .