Dimension $\mathbb {R}(\mathbb {C})$ versus dimension $\mathbb {C}(\mathbb {R})$ I was reading notes about tensor products and the author said that $\operatorname{\dim}_{\mathbb{R}}(\mathbb{C}\otimes_{\mathbb{C}} \mathbb{C}) = 4$ while $\operatorname{\dim}_{\mathbb{C}}(\mathbb{C}\otimes_{\mathbb{R}} \mathbb{C})$ is not defined, but it is not clear to me why this is true. Could someone explain this to me please?
 A: As a prequel, if $V,W$ are two vector spaces over the same field $F$ then the tensor product $V \otimes_F W$ is also a field over $F$, and so the dimension of $V \otimes_F W$ over $F$ is defined. Let's call this operation "tensor product over $F$" (and its dimension over $F$ is the product of the dimensions of $V$ and $W$ over $F$).
So, for example, in the tensor product $\mathbb C \otimes_{\mathbb C} \mathbb C$, both $\mathbb C$ factors are vector spaces over $\mathbb C$, and you are tensoring over $\mathbb C$, the tensor product is also a vector space over $\mathbb C$, so this is tensor product over $\mathbb C$, and its dimension over $\mathbb C$ is defined (and is equal to $1 \times 1 = 1$).
To get to your question, in the tensor product $\mathbb C \otimes_{\mathbb R} \mathbb C$ both factors are vector spaces over $\mathbb R$, and you are tensoring over $\mathbb R$, so this is tensor product over $\mathbb R$. The dimension over $\mathbb R$ is defined (and is equal to $2 \cdot 2 = 4$).
But the "tensor product over $\mathbb C$" construction does not work on $\mathbb C \otimes_{\mathbb R} \mathbb C$, because $\mathbb R$ is not $\mathbb C$.
Nonetheless, there is a different construction (which I overlooked in my original version of this answer; thanks to @Thorgott for the correction). Namely if $V$ is a vector space over a field $F$, and if $E$ is an extension field of $F$, then $V \otimes_F E$ does indeed have a natural vector space structure over $E$. This is not the same operation as "tensor product over $F$", you can think of this instead as "extension of scalars from $F$ to $E$" (and its dimension over $E$ is equal to the dimension of $V$ over $F$).
So, $\mathbb C \otimes_{\mathbb R} \mathbb C$ may indeed be thought of as a vector space over $\mathbb C$ by using the operation of extension of scalars (and its dimension over $\mathbb C$ is equal to the dimension of $\mathbb C$ over $\mathbb R$ which is $2$).
