Universal Properties II: Tensor Products

Problem: Let I=(2,x)\subset \mathbb{Z}[x]. Show that 2\otimes x\neq x\otimes 2 in I\bigotimes_{\mathbb{Z}[x]} I.

If the notation is unfamiliar to you, read up on tensor products of modules.

What we are doing here is forming a \mathbb{Z}[x]-module comprised of sums of symbols of the form f\otimes g, where f,g\in I, subject to the condition that the map \Box \otimes \Box be R-bilinear, that is, linear in each variable and satisfying r(f\otimes g)=(rf)\otimes g = f\otimes (r g) for all r\in \mathbb{Z}[x].

It is possible to show that two different symbols represent the same element of I\bigotimes_{\mathbb{Z}[x]} I. A basic example is 2\otimes 4 = 2(2\otimes 2) = 4 \otimes 2. 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 R is a commutative ring and M and N are R-modules. Then the tensor product M\bigotimes_R N is universal in the following sense: let T be any R-module and \phi: M\times N \to T an R-bilinear map. Then there is a unique R-module homomorphism \Phi: M\bigotimes_R N \to T such that the following diagram commutes:

\begin{array}{rclc} M\bigotimes_R N \\ \iota \uparrow & \searrow \Phi \\ M\times N & \rightarrow & T \\ & \phi \end{array}

(again, apologies for the poor diagram) Here \iota: M\times N \to M\bigotimes_R N is simply the map \iota(m,n)=m\otimes n.

This suggests a general line of attack for our problem: to construct a bilinear map from I\times I to some \mathbb{Z}[x]-module that maps (2,x) and (x,2) to different elements.

One failure would be to define \phi: I\times I \to \mathbb{Z} by \phi(f, g)=f(0)g(0). This map is \mathbb{Z}[x]-bilinear, but it doesn’t distinguish between these two elements. A better choice is \phi(f,g)=f(0)g'(0), which sends (2,x) to 2, and (x,2) to {0}.

By the universal property above, it follows that there exists a homomorphism \Phi: I\bigotimes_{\mathbb{Z}[x]} I \to \mathbb{Z} such that \Phi(2\otimes x)=2 and \Phi(x\otimes 2)=0, and it follows that 2\otimes x\neq x\otimes 2.

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 \Phi by \Phi(\sum_i r_i (f_i\otimes g_i))= \sum_i r_i f(0)g'(0).” But this is incorrect; we do not know a priori that this is a real definition, that \Phi is a well-defined map. So understanding the universal property is really fundamental when working with tensor products.

Related Problem: Show that 2\otimes 2 + x\otimes x is not a simple tensor; that is, it cannot be written in the form f\otimes g for f,g\in I.

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2 Responses to “Universal Properties II: Tensor Products”

  1. Cotton Seed Says:
    September 1, 2008 at 12:43 am | Reply

    Since \phi is defined on I\times I, shouldn’t the sentence read: “A better choice is \phi(f,g)=f(0)g’(0), which sends (2,x) to 2, and (x,2) to 0.”?

    (P.S. Can I put latex in comments?)

  2. lydianrain Says:
    September 1, 2008 at 1:13 am | Reply

    That’s right, I had several typos like that. Fixed.

    I don’t know if you can put latex in comments. Let’s find out: x\otimes 2

    It works, you need to do it like this: D latex \phi D, where the D’s are dollar signs.

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