$\Bbb{C} \otimes_{\Bbb{R}} \Bbb{C}$ and $\Bbb{C} \otimes_{\Bbb{C}} \Bbb{C}$ are not isomorphic as $\Bbb{R}$-vector spaces 
$\Bbb{C} \otimes_{\Bbb{R}} \Bbb{C}$ and $\Bbb{C} \otimes_{\Bbb{C}} \Bbb{C}$ are not isomorphic as $\Bbb{R}$-vector spaces.

Clearly each tensor product is both a left and right $R$-module. But how do I show that they're not isomorphic?
I'm trying to argue as in other examples, so in $M = \Bbb{C} \otimes_{\Bbb{C}} \Bbb{C}$, there's the simple tensor $(-1)\otimes i = (i^2) \otimes i = i \otimes(-1)$.   But those two simple tensors are not equal in $N = \Bbb{C} \otimes_{\Bbb{R}} \Bbb{C}$.  I'm not sure how to show that.  Let $\phi : M \to N$ be an isomorphism.    Then $\phi((-1)\otimes i)$ must equal $\phi (i \otimes (-1))$.
 A: Recall that as $k$-vector spaces, we have $\dim(V\otimes_k W)=\dim(V)\dim(W)$. Now also recall that if $V$ is an $n$-dimensional $\mathbb{C}$-vector space, then it is also a $2n$-dimensional $\mathbb{R}$-vector space.
A: Let $z \otimes w$ be a simple tensor in $\mathbb{C} \otimes_\mathbb{C} \mathbb{C}$.
Certainly we have $z \otimes w = 1 \otimes zw$. Every simple tensor may be written in this form, so that certainly every element of $\mathbb{C} \otimes_\mathbb{C} \mathbb{C}$ has the form $1 \otimes (a+bi)$. Note that $1 \otimes (a+bi) = 1 \otimes a + 1 \otimes bi = a \otimes 1 + b \otimes i = a \cdot (1 \otimes 1) + b \cdot (1 \otimes i)$. In particular, every element of $\mathbb{C} \otimes_\mathbb{C} \mathbb{C}$ may be written as an $\mathbb{R}$-linear combination of $1 \otimes 1$ and $1 \otimes i$.
We claim that $1 \otimes 1$ and $1 \otimes i$ form a basis for $\mathbb{C} \otimes_\mathbb{C} \mathbb{C}$ over $\mathbb{R}$, so that this module is free with rank 2 over $\mathbb{R}$. 
by the universal property of free modules, the mapping $(1,0) \mapsto 1 \otimes 1$,$ (0,1) \mapsto 1 \otimes i$ extends to a surjective $\mathbb{R}$-module homomorphism $\varphi : \mathbb{R}^2 \rightarrow \mathbb{C} \otimes_\mathbb{C} \mathbb{C}$.
Now define $\psi$ : $\mathbb{C} \times \mathbb{C} \rightarrow \mathbb{R}^2$ by $\psi(a+bi,c+di) = (ac-bd, ad+bc)$. Evidently, $\psi$ is $\mathbb{C}$-balanced, and thus induces an $\mathbb{R}$-module homomorphism $\Psi$ : $\mathbb{C} \otimes_\mathbb{C} \mathbb{C} \rightarrow \mathbb{R}^2$. Moreover, $\Psi$ is surjective. Evidently, $\varphi$ and $\psi$ are mutual inverses. Thus $\mathbb{C} \otimes_\mathbb{C} \mathbb{C}$ is free of rank 2 as a left $\mathbb{R}$-module.
Now let $(a+bi) \otimes (c+di)$ be a simple tensor in $\mathbb{C} \otimes_\mathbb{R} \mathbb{C}$. Evidently, $(a+bi) \otimes (c+di) = a \otimes c + a \otimes di + bi \otimes c + bi \otimes di = ac(1 \otimes 1) + ad(1 \otimes i) + bc(i \otimes 1) + bd(i \otimes i)$. Certainly then every element of $\mathbb{C} \otimes_\mathbb{R} \mathbb{C}$ is an $\mathbb{R}$-linear combination of $1 \otimes 1, 1 \otimes i, i \otimes 1$, and $i \otimes i$.
We claim that these four elements form a basis for $\mathbb{C} \otimes_\mathbb{R} \mathbb{C}$ as a left $\mathbb{R}$-module. To see this, note that the universal property of free modules again induces a surjective $\mathbb{R}$-module homomorphism $\mathbb{R}^4 \rightarrow \mathbb{C} \otimes_\mathbb{R} \mathbb{C}$. 
Evidently, the mapping $\mathbb{C} \times   \mathbb{C} \rightarrow \mathbb{R}^4$ given by $(a+bi,c+di) \mapsto (ac,ad,bc,bd)$ is $\mathbb{R}$-balanced, and so induces a (clearly surjective) $\mathbb{R}$-module homomorphism $\mathbb{C} \otimes_\mathbb{R} \mathbb{C} \rightarrow   \mathbb{R}^4$.
These mappings are mutual inverses. Thus $\mathbb{C} \otimes_\mathbb{R} \mathbb{C}$ is free of rank 4 as a left $\mathbb{R}$ module.
Over a commutative ring, (finitely generated) free modules have a unique rank. Thus we have $\mathbb{C} \otimes_\mathbb{C} \mathbb{C} \not\cong_R \mathbb{C} \otimes_\mathbb{R} \mathbb{C}$.
A: Thinking of what your doing, it is more natural to show that they are not isomorphic as $\mathbb C$-vector spaces (even if the exercise does not say so). Indeed, you start with $\mathbb C$ and


*

*in one hand, you consider it as a $\mathbb C$-vector space and extend the scalar to $\mathbb C$, so basically do nothing : this is the isomorphism $\mathbb C \otimes_{\mathbb C} \mathbb C \simeq \mathbb C$ (as $\mathbb C$-vector spaces),

*in the other hand, you consider it as a $\mathbb R$-vector space and extend the scalar to the field $\mathbb C$ (you brutally decide that the scalar are now complex numbers, not just real numbers) : this is the isomorphism $\mathbb C \otimes_{\mathbb R} \mathbb C \simeq \mathbb C^2$ (as $\mathbb C$-vector spaces).
(Obviously $\mathbb C$-isomorphism leads $\mathbb R$-isomorphism and $\dim_{\mathbb R}(\mathbb C) \neq \dim_{\mathbb R}(\mathbb C^2)$, solving your exercise. But, in my humble opinion, this is not the important part of the exercise.)
