TITLE: Tensor product of real numbers over the rationals
QUESTION [5 upvotes]: How do I show that $\mathbb{R}\otimes_{\mathbb{Q}}\mathbb{R}\not\cong\mathbb{R}$ as $\mathbb{R}$-vector spaces?
Possible approaches I can think of (but can't implement) are to show that this tensor product is not 1-dimensional as a real vector space, or to use some exactness property of the tensor product functor, or to take a further tensor product on both sides.

REPLY [5 votes]: Let $K$ be a field, $V$ a $K$-vectorspace and $\mathcal B\subset V$ a $K$-basis. If $L\supset K$ is a field extension, then $L\otimes_KV$ is an $L$-vectorspace and $1\otimes_K\mathcal B\subset L\otimes_KV$ is an $L$-basis. This shows that $\dim_L(L\otimes_KV)=\dim_KV$. 
In your case $\dim_{\mathbb R}(\mathbb R\otimes_{\mathbb Q}\mathbb R)=\dim_{\mathbb Q}\mathbb R>1$ and $\dim_{\mathbb R}\mathbb R=1$.