How to find the tensor product of modules that we don't know a basis for them?

Hi

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 $\mathbb{R}$?

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
@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