Tensor product does not distribute over direct product

Posted: February 21, 2011 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: ,

Problem 1. Let G be an abelian group. Prove that if G has no element of infinite order,  then \mathbb{Q} \otimes_{\mathbb{Z}}G=(0).

Solution.  If g \in G and o(g)=n, then for all r \in \mathbb{Q} we have

r \otimes_{\mathbb{Z}}g=(nr/n) \otimes_{\mathbb{Z}}g=(r/n) \otimes_{\mathbb{Z}}(ng)=(r/n) \otimes_{\mathbb{Z}}0=0. \Box

Problem 2. Let G be an abelian group. Prove that if G has an element of infinite order, then \mathbb{Q} \otimes_{\mathbb{Z}} G \neq (0).

Solution. Let g be an element of infinite order in G. Since \mathbb{Q} is a flat \mathbb{Z}-module, \mathbb{Q} \otimes_{\mathbb{Z}} \langle g \rangle is a subgroup of \mathbb{Q} \otimes_{\mathbb{Z}} G. So we only need to prove that \mathbb{Q} \otimes_{\mathbb{Z}} \langle g \rangle \neq (0). This is obvious because, since g has infinite order, we have \langle g \rangle \cong \mathbb{Z} and thus \mathbb{Q} \otimes_{\mathbb{Z}} \langle g \rangle \cong \mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{Z} \cong \mathbb{Q} \neq (0). \ \Box

Let R be a commutative ring with unity and let I be a set which may or may not be finite. Suppose that  Y_i, \ i\in I, are R-modules. Let X be an R-module. Recall that tensor product distributes over direct sum, i.e.

X \otimes_R \bigoplus_{i \in I} Y_i \cong \bigoplus_{i \in I}(X \otimes_R Y_i),

as R-modules. Next problem shows that the above is not true in general if \bigoplus is replaced with \prod.

Problem 3. Let G_i = \mathbb{Z}/2^i\mathbb{Z}, \ i\in \mathbb{N}. Prove that \mathbb{Q} \otimes_{\mathbb{Z}} \prod_{i=1}^{\infty} G_i \ncong \prod_{i=1}^{\infty} (\mathbb{Q} \otimes_{\mathbb{Z}}G_i).

Solution. Each G_i is a finite abelian group and thus \mathbb{Q} \otimes_{\mathbb{Z}} G_i = (0), by Problem 1. Thus

\prod_{i=1}^{\infty} (\mathbb{Q} \otimes_{\mathbb{Z}} G_i)=(0). \ \ \ \ \ \ \ \ \ \ \ \ \ (1)

Now let G = \prod_{i=1}^{\infty}G_i. Let g = (1 + 2^i \mathbb{Z})_{i \in \mathbb{N}} \in G. Suppose that the order of g is finite, say n. Then n + 2^i \mathbb{Z} = 0, for all i \in \mathbb{N}. That means  2^i divides n for all i \in \mathbb{N}, which is nonsense. So the order of g in G is infinite and hence, by Problem 2,

\mathbb{Q} \otimes_{\mathbb{Z}} G \neq (0). \ \ \ \ \ \ \ \ \ \ \ \ (2)

The result now follows from (1) and (2). \Box


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