# Tensor product of real numbers over the rationals

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.

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Suggested approach- try finding an $\mathbb{R}$-linear transformation on $\mathbb{R}\otimes_\mathbb{Q}\mathbb{R}$ which is not given by multiplying by a scalar. – kneidell Feb 23 at 23:02
What is the action of $\mathbb{R}$ on $\mathbb{R}\otimes_\mathbb{Q}\mathbb{R}$, btw? – kneidell Feb 23 at 23:05
@kneidell: I'll take the action to be given by $\mathbb{R}$ acting on the first coordinate. – user44532 Feb 23 at 23:07
@CYC, what about starting with a $\mathbb{Q}$-basis of $\mathbb{R}$, say $\mathcal{B}$, and note that $1 \otimes \mathcal{B}$ gives you a $\mathbb{R}$-basis of $\mathbb{R} \otimes_{\mathbb{Q}} \mathbb{R}$? – Sanchez Feb 23 at 23:44

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 $\dim_{\mathbb R}\mathbb R=1$.