$\dim_{\mathbb k} A= \dim_{\mathbb k_1}(_{\mathbb k_1} A)\dim_{\mathbb k} \mathbb k_1 $. I want the proof of the following. $ \mathbb {k, k_1} $ are fields.
Consider $\mathbb k_1$, subalgebra of $ \mathbb k$-algebra $A$ . Then we can view $A$ as left $_{\mathbb k_1 }A$ or right $ A_{\mathbb k_1}$ vector space. Then we have $\dim_{\mathbb k} A= \dim_{\mathbb k_1}(_{\mathbb k_1} A)\dim_{\mathbb k} \mathbb k_1 $.
Since multiplication of dimensions is involved, I sense that tensor product is involved. But the two vector spaces are on different fields.
Reference for this will also be appreciated. 

I think the best way to do this is to take a $k_1$-basis of $A$, $e_1,...,e_n$; a $k$-basis of $k_1$, $f_1,...,f_d$, and show that the family $(f_ie_j)$ is a $k$-basis of $A$. From this the equality follows. 
