Are $\mathbb{C} \otimes _\mathbb{R} \mathbb{C}$ and $\mathbb{C} \otimes _\mathbb{C} \mathbb{C}$ isomorphic as $\mathbb{R}$-vector spaces? 
Are $\mathbb{C} \otimes _\mathbb{R} \mathbb{C}$ and $\mathbb{C} \otimes _\mathbb{C} \mathbb{C}$ isomorphic as $\mathbb{R}$-vector spaces?

I am having a very hard time at digesting tensor products and I do not know how to "compare" tensor products over different rings.
My hunch is that they are not isomorphic. It is easy to see that $\mathbb{C} \otimes _\mathbb{R} \mathbb{C}$ is an $\mathbb{R}$-vector space of dimension $4$. I suspect that $\mathbb{C} \otimes _\mathbb{C} \mathbb{C}$ is also an $\mathbb{R}$-vector space but of lower dimension, but I have no idea how to show this or if indeed this intuition is correct.
Thank you.
 A: Your intuition is indeed correct.
$\mathbb C\otimes_\mathbb R \mathbb C$: $\mathbb C$ as a real vector space is nothing by $\mathbb R^2$. $\mathbb R^2\otimes_{\mathbb R}\mathbb R^2\cong \mathbb R^4$ (you can write down a basis if you like).
On the other hand, very generally $A\otimes_A A\cong A$ always. Why? Well, the $\mathbb C$ in $\otimes_{\mathbb C}$ tells you that you can move scalars across. So you can make all of these tensors look like $v\otimes 1$. So again you can write down a basis. So this module is just $\mathbb C\cong \mathbb R^2$.
A: In this case the first one is saying that you can move any part of $\mathbb R$ from "side to side" - e.g. $2(1+i)\otimes 3i = (1+i)\otimes 6i$. However, it would not be true that $i\otimes 1 = 1 \otimes i$.
In the second case, $\mathbb C \otimes \mathbb C$, you can move elements of $\mathbb C$ (and hence also $\mathbb R$) from "side to side". e.g. $1\otimes i = i \otimes 1$.
I find this a relatively concrete way to think of it, even if not the best to generalise.
A: No, they are not. Your intuition is correct.
Given a ground field $k$, and two finite-dimensional $k$-vector spaces $V,W$, then $\dim_k(V \otimes_k W) = (\dim_k V)(\dim_k W)$. So:


*

*$\dim_{\mathbb C}(\mathbb C \otimes_{\mathbb C} \mathbb C) = (\dim_{\mathbb C} \mathbb  C)^2=  1$, therefore $\dim_{\mathbb R} (\mathbb C \otimes_{\mathbb C} \mathbb C)  = 2$.

*$\dim_{\mathbb R}(\mathbb C \otimes_{\mathbb R} \mathbb C) = (\dim_{\mathbb R} \mathbb  C)^2= 2^2 = 4$.


So the two are not isomorphic. In fact, $\mathbb C \otimes_{\mathbb C} \mathbb C \simeq \mathbb C$ canonically (and more generally $V \otimes_k k \simeq V$), while $\mathbb C \otimes_{\mathbb R} \mathbb C \simeq \mathbb C \oplus \mathbb C$.
