Here is the spaces I am trying to find their dimensions:
$1- \operatorname{dim}_{\mathbb C}(\mathbb C \otimes_{\mathbb R} \mathbb C)$
$2-\operatorname{dim}_{\mathbb R}(\mathbb C \otimes_{\mathbb R} \mathbb C)$
For the first one, I am convinced with what @LeeMosher told me here in his answer Dimension $\mathbb {R}(\mathbb {C})$ versus dimension $\mathbb {C}(\mathbb {R})$ , I just have a small question, which is:
Should not it be stated clearly that we are in the case of finite dimensional vector spaces? or it is understandable? as I am using the following theorem (which I do not know if it is also correct for infinite dimensional vector spaces or no):
"Let $V$ be a finite d.v.s over $F,$ and $F \subseteq K$ is a field, then $\operatorname{dim}_{F} V = \operatorname{dim}_{K} ( V \otimes_{ F} K).$"
For the second one, I am confused how to find its dimension and should I consider it as a tensor product or should it be an extension of scalars? Also, a friend of mine told me that the dimension of the second one is 1 but I could not understand why.
Could anyone help me in answering these questions please?
As stated for instance here, since $\mbox{dim}_{\mathbb{R}}(\mathbb{C}) = 2$, you have that $\mbox{dim}_{\mathbb{R}} \left( \mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \right) = \mbox{dim}_{\mathbb{R}}(\mathbb{C})^2 = 4$.
The first one is, in a sense, ill-defined, since $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}$ does not have a clear structure of $\mathbb{C}$-vector field (if you multiply an element by $i$, does it multiply the left-hand side ? or the right-hand side ?) However, once you fix this by defining it clearly, you must have $\mbox{dim}_{\mathbb{C}}\left( \mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \right) = \frac{1}{2} \mbox{dim}_{\mathbb{R}} \left( \mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \right) = 2$.