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I would like to know what would be tensor product of set of reals over reals would be?

That is, $\Bbb{R} \otimes_\Bbb{R} \Bbb{R}$

I think it should be $\Bbb{R}^{2}$ as tensor product combines two vector spaces to give another vector space. Therefore, it should be $\Bbb{R}^2$ but sometimes it doesn't.

Similarly, what would be $\Bbb{R} \otimes_\Bbb{Z} \Bbb{R}$ ?

madeel
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    I'm not sure that's a good enough reason; indeed, one always has $V \otimes_k k \simeq V$. – Hoot Dec 22 '14 at 06:58

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As stated in the comments, $V\otimes_kk\cong V$ for any vector space $V$ over a field $k$. Tensoring against another space does not always increase dimension. Indeed one purpose of tensoring is to extend scalars (this application, among others, is discussed here), and rather trivially if we extend scalars to the field extension $k/k$ ... we have introduced no new scalars at all, and so have done nothing really. Can you think of the isomorphism $V\otimes_kk\cong V$?

That takes care of $\Bbb R\otimes_{\Bbb R}\Bbb R$. What about $\Bbb R\otimes_{\Bbb Z}\Bbb R$? I assume you mean as a vector space over $\Bbb R$, as opposed to as a ring. Take any vector space basis of $\Bbb R/\Bbb Q$ (invoking such a thing requires some Choice) writing $\Bbb R=\bigoplus_B\Bbb Qc$ as $c$ ranges over the basis. Then I leave you to check that the set $\{1\otimes c:c\in B\}$ is an $\Bbb R$-basis for $\Bbb R \otimes_{\Bbb Z}\Bbb R$. It is important to really understand what elements of tensor products look like in order for this to be an intuitive fact. Some technology can also be used: writing $\Bbb R$ as an uncountable direct sum of copies of $\Bbb Q$ (using Choice to invoke a basis for $\Bbb R$ over $\Bbb Q$), using the fact that tensor products distribute over direct sums, and that tensoring over a domain (like $\Bbb Z$) against its fraction field (like $\Bbb Q$) does not change anything. (See the "sliding" and "distributivity" rules over here.)

Community
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anon
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  • One should point out that the easiest way to see $\mathbb R \otimes_{\mathbb R} \mathbb R \simeq \mathbb R$ is to first note that for any simple tensor we have $a \otimes b = 1 \otimes ab$ and then for any sum of those $1 \otimes a + 1 \otimes b = 1 \otimes (a + b)$. Thus every element of $\mathbb R \otimes_{\mathbb R} \mathbb R$ is a simple tensor of the form $1 \otimes a$. – Jim Dec 23 '14 at 05:55
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Consider $$\phi: \Bbb{R} \times \Bbb{R} \to \Bbb{R} $$ which acts on elementary tensors in the following way $$(a,b) \mapsto ab $$ This map is $\Bbb{R}$-bilinear and so by the universal property of tensor product there is a $\Bbb{R}$-linear map $$\phi' : \Bbb{R} \otimes_{\Bbb{R}} \Bbb{R} \to \Bbb{R} $$ $$ a \otimes_{\Bbb{R}} b \mapsto ab $$ The inverse of $\phi'$ is the $\Bbb{R}$-linear map $$\lambda : \Bbb{R} \to \Bbb{R} \otimes_{\Bbb{R}} \Bbb{R}$$ $$r \mapsto 1 \otimes_{\Bbb{R}} r$$ as you can check.

Thus $$\Bbb{R} \otimes_{\Bbb{R}} \Bbb{R} \cong \Bbb{R}$$ as $\Bbb{R}$-modules.

WLOG
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