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$$ \mathbb{C} \otimes_{\mathbb{R}} \mathbb{R} = \;? $$

I guess this guy is just $\mathbb{C}$, is this correct?

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Remember that $\mathbb R$ is like an identity with respect to $\otimes_\mathbb R$. –  Asaf Karagila May 25 '12 at 10:30
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Yes, you are correct. That is why it is called complexifying. –  M.B. May 25 '12 at 10:31
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Yes this is correct. One way of seeing this is considering the map $$\mathbb C\otimes_\mathbb R \mathbb R\to \mathbb C$$ which maps $$z\otimes x\mapsto xz$$ and showing that it is an isomorphism. Another way is to see that $\mathbb C\otimes_\mathbb R \mathbb R$ is $2$-dimensional as a real vector space and that the induced product is the same as the product of $\mathbb C$. Hence they are isomorphic as algebras.

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Thx Benjamin, I made the edit. –  Simon Markett May 25 '12 at 11:06
    
Perhaps you may also want to emphasize that you are doing $\Bbb{C} \otimes_\Bbb{R} \Bbb{R}$. –  user38268 May 25 '12 at 11:07
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Yes. In fact, for any commutative ring $R$ with module $M$, you can say this about the tensor product with free modules: $$M\otimes_R(\oplus_{i=1}^n R)\cong\oplus_{i=1}^n M$$

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funny we have the exact same reputation...I wonder whether this question changes this ;) –  Simon Markett May 25 '12 at 10:42
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@SimonMarkett Haha, quite a coincidence! Now I see my speculation that these icons are generated directly from rep is incorrect (I really should ask around and look for the true source.) –  rschwieb May 25 '12 at 10:48
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Now guys I don't know which of your answer I should accept. I don't want to ruin this symmetry! ;-) –  Abramo May 25 '12 at 10:54
    
@rschwieb If you mean the badges by "icons" then you will find an explanation under "badges" ;) –  Simon Markett May 25 '12 at 10:58
    
@SimonMarkett No, I was referring to the colorful fractalish looking square picture which is everchanging. –  rschwieb May 25 '12 at 10:59
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Recall that given commutative rings $S$, $R$ together with a ring homomorphism $f:R \longrightarrow S$ and an $R$ - module $M$ we can consider the tensor product

$$S \otimes_R M.$$

Now $S$ can be considered as an $R$ - module by defining the action of $R$ on $S$ by $r\cdot s = f(r)s$ so that $S \otimes_R M$ has the structure of an $R$ - module. However what is even more interesting is that there is a unique $S$ - module structure on $S \otimes_R M$; we may define $S$ - multiplication by

$$s_1( s_2 \otimes m ) = (s_1s_2) \otimes m$$

for all $s_1,s_2 \in S$ and $m \in M$. One still needs to check that this is well defined and extends from elementary tensors to all tensors. This process of turning an $R$ - module like that into an $S$ - module is also known as extension of scalars.

Now back to your problem. We already know that $\Bbb{C} \otimes_\mathbb{R} \Bbb{R}$ has the structure of an $\Bbb{R}$ - module. By extension of scalars, we can consider this as a $\Bbb{C}$ - module. Now $\Bbb{R}$ as a module over itself is a free $\Bbb{R}$ - module with basis just the real number $1$ say. Then it is possible to show that the $\Bbb{C}$ - module

$$\Bbb{C} \otimes_\Bbb{R} \Bbb{R} $$

has basis $1 \otimes 1$. However a $\Bbb{C}$ - module is nothing more than a complex vector space and in our case a vector space that is of dimension 1 (over $\Bbb{C}$). Because two vector spaces over the same field of equal dimension are isomorphic we conclude that

$$\Bbb{C} \otimes_\Bbb{R} \Bbb{R} \cong \Bbb{C}$$

as $\Bbb{C}$ - modules.

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Do we need extension of scalars here? Isn't $M = \mathbb C$ an $R = \mathbb R$-module? So we could use the (natural) isomorphism $M \otimes R \cong M$ directly. –  Matt N. Jun 15 '12 at 21:47
    
@MattN. I suppose you could. However I just took a different approach by considering everything as complex vector spaces from the beginning. –  user38268 Jun 15 '12 at 21:51
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You could use that you have a (natural) isomorphism $M \cong M \otimes_R R$, see here for a proof.

Then $M = \mathbb C$ is an $R$-module for $R = \mathbb R$ and you get (as you suspected) that $\mathbb C \cong \mathbb R \otimes_{\mathbb R} \mathbb R$.

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