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I know how to calculate some easy tensor products like $\mathbb{Z}/m\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}/n\mathbb{Z}\cong_{\mathbb{Z}} \mathbb{Z}/(m,n)\mathbb{Z} $ or $F[X] \otimes_{F} F[Y] \cong_F F[X,Y]$ or $\mathbb{C} \otimes_\mathbb{R} \mathbb{C} \cong_\mathbb{R} \mathbb{R}^4$ but the reason that I can do this is that the structure of these rings is kinda well-known or they're finitely generated modules I know that continuous real functions over the interval $[0,1]$ form an $\mathbb{R}$-module, so, as a challenge for myself, I tried to find $C([0,1]) \otimes_\mathbb{R} C([0,1])$, but nothing came to my mind. So I thought I would need help and hence that's why I'm writing it here now. I also like to know what $\mathbb{R} \otimes_\mathbb{Q} \mathbb{R}$ looks like, is it algebraically the same as $\mathbb{R}$?

Thanks in advance.

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Please clarify if you are interested in the structure of these tensor products as $R$-modules or as $R$-algebras (with $R$ being the ring you are tensoring over). – KConrad Mar 7 '13 at 14:15
By the way, I'm new here, I didn't know that here is filled with PhD students and professors that ask really sophisticated questions. I'm just a 2nd-year undergraduate student, for me, tensors, is the highest level of mathematics that I've ever encountered! So, please, go easy on me and explain things to me in the simplest way possible. – some1.new4u Mar 7 '13 at 15:15
@KConrad: How is ${\mathbb R}\otimes_{\mathbb Q}{\mathbb R}$ as a real vector space isomorphic to $\mathbb R$? To start with, how is the vector space structure defined? One can multiply on the first or the second slot. Let's say the first slot, then $\sqrt 2\otimes 1$ and $1\otimes 1$ are linearly dependent over $\mathbb R$, but $1\otimes \sqrt 2$ and $1\otimes 1$ are not. It rather seems to me that ${\mathbb R}\otimes_{\mathbb Q}\mathbb R$ has the same diemnsion over $\mathbb R$ as $\mathbb R$ has over $\mathbb Q$? – Anton Mar 7 '13 at 16:05
@some1.new4u: Then perhaps is the more appropriate site for you to ask your questions. MO is for research-level questions. – Johannes Hahn Mar 7 '13 at 18:16
To expand on Johannes' comment: at the answers and discussion might be at a level more suitable to your background. It is a natural and worthwhile question, though. – Yemon Choi Mar 7 '13 at 19:06

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