# Dimension of $\mathbb{Q}\otimes_{\mathbb{Z}} \mathbb{Q}$ as a vector space over $\mathbb{Q}$

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!

• $\mathbb Z$ isn't a field so it doesn't make sense to ask about vector spaces over the integers. Perhaps what was meant was module? – kahen Aug 2 '13 at 20:21
• @kahen: sorry it was a typo – Prism Aug 2 '13 at 20:21
• As a vector space over $\mathbb{Z}$? Don't you mean as a free module? – Dedalus Aug 2 '13 at 20:22
• @Dedalus: I meant $\mathbb{Q}$. Fixed. – Prism Aug 2 '13 at 20:22
• 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.) – anon Aug 3 '13 at 0:17

So, $\mathbb{Q}\otimes_{\mathbb{Z}}\mathbb{Q}=\mathbb{Q}$.
• 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. – Alex Youcis Aug 2 '13 at 21:26
• 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 $\Bbb{Z} - module). – user38268 Aug 3 '13 at 0:18 • @AlexYoucis: Ah indeed! A very nice presentation is given in Keith Conrad's expository note on tensor products. (Theorem 4.20, and Example 4.21) – Prism Aug 3 '13 at 1:58 I just want to make remark that RGB's answer more generally shows the following. Suppose$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:$\dim_\Bbb{Q} \Bbb{Q} \otimes_{\Bbb{Z}} \Bbb{Q} = 1$. • Thank you very much for taking your time to compose this detailed answer. I am drawing diagrams to convince myself :) – Prism Aug 3 '13 at 1:59 • +1. why are they canonically isomorphic? is there a "functorial" proof? I know that$-\otimes_{A} S^{-1}A\cong S^{-1}\$ (functorial isom.) – user111072 Jan 26 '14 at 15:39