Tensor product does not distribute over direct product

Posted: February 21, 2011 in Elementary Algebra; Problems & Solutions, Rings and Modules
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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). $\Box$