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I want to make sure that I understand this correctly: $V- \mathbb{Q}$ vector space. Then $V_\mathbb{R} := V \otimes \mathbb{R} $ is naturally an R-vector space (next to being a $\mathbb{Q}$-vector space).

If I am not mistaken this can be explained in the following way: $\mathbb{R}$ is a $\mathbb{Q}$ vector space, but the basis of $\mathbb{R}$ cannot be handwritten,so we construct $V \otimes_{\mathbb{Q}} \mathbb{R}$ s.t it is generated by $\{v_i\otimes r | v_i \in basis( V) ,r \in \mathbb{R}\}$ indeed if we take any $v\otimes r \in V \otimes_\mathbb{Q} \mathbb{R}$ then: $$v\otimes r= \sum \gamma _i v_i \otimes r = \sum (v_i \otimes \gamma_i r)=\sum v_i\otimes r'$$ where $v_i$ is basis vectors of $V$ and $\gamma_i \in \mathbb{Q}$. Now,consider $\mathbb{Q}$-linear mapping : $$V \otimes_\mathbb{Q} \mathbb{R} \rightarrow V_\mathbb{R}$$ defined as $$v_i \otimes r \rightarrow rv_i$$ then it induces isomorphism between $V_{\mathbb{R}}$ and $V \otimes_\mathbb{Q} \mathbb{R}$. Is there any flaw in my understanding, is there something here which is not obvious and I am skipping? Please help,any help would be very much appreciated!

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$V_\mathbb{R} = V \otimes_\mathbb{Q} \mathbb{R}$ is a $\mathbb{R}$-vector space in "the natural way". You do not need to talk about basis elements of neither $V$ nor $\mathbb{R}$ to define the structure.

Any element of of $V_\mathbb{R}$ looks like a sum of terms $v \otimes r$, where $v \in V$ and $r \in \mathbb{R}$. To define a $\mathbb{R}$-vector space structure, we need only define what it means to multiply a vector $v \otimes r$ with another $s \in \mathbb{R}$. But we simply define $$ s \cdot ( v \otimes r) := v \otimes (sr).$$

This makes $V_\mathbb{R}$ a $\mathbb{R}$-vector space. If $v_i \in V$ is a $\mathbb{Q}$-basis for $V$, then (a subset of) $\{v_i \otimes 1\}$ is a $\mathbb{R}$-basis of $V_\mathbb{R}$.

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  • $\begingroup$ This is not correct. Tensor products of free modules of finite rank are free of the same rank. Here $\dim_{\mathbf R}(V_\mathbf R) = 2$, since $1 \otimes 1$ and $\sqrt 2 \otimes 1$ are an $\mathbf R$-basis of $V_{\mathbf R}$. We don't have $\sqrt 2 (1 \otimes 1) = \sqrt 2 \otimes 1$, since $\sqrt 2 ( 1 \otimes 1) = 1 \otimes \sqrt 2$ and you cannot pull $\sqrt 2$ in the first ''coordinate''. $\endgroup$ – Hans Giebenrath Jul 30 '14 at 13:34
  • $\begingroup$ @HansGiebenrath Thank you for the comment. I've removed my flawed example. $\endgroup$ – Fredrik Meyer Aug 2 '14 at 9:28

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