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The following problem was subject of examination that was taken place in June. The document is here. Problem 1 states:

The tensor product $\mathbb{Q}\otimes_{\mathbb Z}\mathbb{Q}$ is a vector space over $\mathbb{Q}$ by multiplication in the left factor, i.e. $\lambda(x\otimes y)=(\lambda x)\otimes y$ for $\lambda, x, y\in\mathbb{Q}$. What is the dimension of $\mathbb{Q}\otimes_{\mathbb{Z}}\mathbb{Q}$ as a vector space over $\mathbb{Q}$?

I only know the definition of tensor product for modules (via universal property). How does one go about calculating dimension of such a vector space?

Thanks!

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  • $\begingroup$ $\mathbb Z$ isn't a field so it doesn't make sense to ask about vector spaces over the integers. Perhaps what was meant was module? $\endgroup$
    – kahen
    Aug 2, 2013 at 20:21
  • $\begingroup$ @kahen: sorry it was a typo $\endgroup$
    – Prism
    Aug 2, 2013 at 20:21
  • $\begingroup$ As a vector space over $\mathbb{Z}$? Don't you mean as a free module? $\endgroup$
    – Dedalus
    Aug 2, 2013 at 20:22
  • $\begingroup$ @Dedalus: I meant $\mathbb{Q}$. Fixed. $\endgroup$
    – Prism
    Aug 2, 2013 at 20:22
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    $\begingroup$ I advocate my answer here as a good discussion of three important properties of tensor products, in particular #3 is relevant to this question. (BenjaLim generalizes it a bit.) $\endgroup$
    – anon
    Aug 3, 2013 at 0:17

3 Answers 3

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Begin by checking

\begin{align}\frac{a}{b}\otimes \frac{c}{d}&=\frac{ad}{bd}\otimes\frac{c}{d}\\ &=\frac{a}{bd}\otimes\frac{cd}{d}\\ &=\frac{a}{bd}\otimes c\\ &=\frac{ac}{bd}\otimes 1 \end{align}

So, $\mathbb{Q}\otimes_{\mathbb{Z}}\mathbb{Q}=\mathbb{Q}$.

Now you can end it.

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    $\begingroup$ Obviously, the more general things going on here is that $\Frac(R)\otimes_R\Frac(R)$ is isomorphic to $\Frac(R)$ as a $\Frac(R)$-module. $\endgroup$ Aug 2, 2013 at 21:26
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    $\begingroup$ There is a subtle point in this year which perhaps should be pointed out to the OP. The last step is not in general legit for general modules (which don't have $1$ in them, e.g. $\Bbb{2}\Bbb{Z}$ as a $\Bbb{Z} - module). $\endgroup$
    – user38268
    Aug 3, 2013 at 0:18
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    $\begingroup$ @AlexYoucis: Ah indeed! A very nice presentation is given in Keith Conrad's expository note on tensor products. (Theorem 4.20, and Example 4.21) $\endgroup$
    – Prism
    Aug 3, 2013 at 1:58
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I just want to make remark that RGB's answer more generally shows the following. Suppose $M,N$ are $S^{-1}A$ - modules. Then we can form both $M \otimes_{S^{-1}A} N$ and $M \otimes N$. The former is an $S^{-1}A$ - module and the latter an $A$ - module by restriction of scalars. The ultimate point now is this: $M \otimes_A N$ already has the structure of an $S^{-1}A$ - module built into it! Thus $M \otimes_{S^{-1}A} N$ and $M \otimes_A N$ are canonically isomorphic as $S^{-1}A$ modules; this is basically the content of RGB's answer.

If you want to show $\Bbb{Q} \otimes_{\Bbb{Z}} \Bbb{Q} \cong \Bbb{Q}$ using universal properties here is what we do: Let $f : \Bbb{Q} \times \Bbb{Q} \to M$ be a $\Bbb{Z}$ - bilinear map. Consider $\pi : \Bbb{Q} \times \Bbb{Q} \to \Bbb{Q}$ that sends $(a,b)$ to $ab$. Now we have linearity in the first variable because $$\begin{eqnarray*} \pi(na_1 + ma_2,b) &=& (na_1+ma_2)b \\ &=& n(a_1b) + m(a_2b) \\ &=& n\pi(a_1,b) + m\pi(a_2,b)\end{eqnarray*}$$ for any $a_1,a_2,b \in \Bbb{Q}$ and $n,m\in \Bbb{Z}$. By symmetry linearity in the second variable follows and so $\pi$ is bilinear. Let us now define a map $g : \Bbb{Q} \to M$ by $g(a) = f(a,1)$ for any $a \in \Bbb{Q}$. This map $g$ is linear because $f$ is linear in the first variable. Also $g$ is well - defined and furthermore is unique: Any linear map out of $\Bbb{Q}$ to $M$ $\tilde{g}$ such that $\tilde{g} \circ \pi = f$ must necessarily satisfy $$ \tilde{g}(a) = \tilde{g}(\pi(a,1)) = f(a,1).$$

Thus we have shown that $\Bbb{Q}$ satisfies the universal property of $\Bbb{Q} \otimes_{\Bbb{Z}} \Bbb{Q}$ and so the answer to your question comes easily: $\dim_\Bbb{Q} \Bbb{Q} \otimes_{\Bbb{Z}} \Bbb{Q} = 1$.

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    $\begingroup$ Thank you very much for taking your time to compose this detailed answer. I am drawing diagrams to convince myself :) $\endgroup$
    – Prism
    Aug 3, 2013 at 1:59
  • $\begingroup$ +1. why are they canonically isomorphic? is there a "functorial" proof? I know that $-\otimes_{A} S^{-1}A\cong S^{-1}$ (functorial isom.) $\endgroup$
    – user111072
    Jan 26, 2014 at 15:39
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    $\begingroup$ I have some concerns about the proof. 1) If you want to show the universal property is satisfied for the $\mathbb{Z}$ module $\mathbb{Q}$ then you would have to show that $g\circ \pi=f$ and I am not sure how to show this by using the $\mathbb{Z}$-module structure on $\mathbb{Q}\times \mathbb{Q}$. $\endgroup$
    – Babai
    Sep 24, 2020 at 8:46
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There is another solution using category theory : tensor product is the fiber coproduct in the category of commutative rings. $\mathbb{Z} \rightarrow \mathbb{Q}$ is an epimorphism since it's a localization.

If $A \rightarrow B$ is an epimorphism, and $B \rightarrow C, B \rightarrow D$ are any morphisms, then $C \coprod_A D \simeq C \coprod_B D$. Therefore, $\mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{Q} = \mathbb{Q} \otimes_{\mathbb{Q}} \mathbb{Q}= \mathbb{Q}$

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