Currently I'm trying to get through "An Introduction to Algebraic K-theory" by C.A.Weibel.
In the third chapter there is the following example.
If $D$ is $d$-dimensional division algebra over its center $F$, then $d = n^2$. The field extension $E \supset F$ is called a splitting field over $D$ if $D \otimes_F E \cong M_n(E)$. For example any maximal subfield $E \subset D$ has $[E : n] = n$ and is a splitting field.
I am wondering why this is true (i.e., why $D \otimes_F E \cong M_n(E)$). I want to make a simple dimension check. The dimension of the left side of this equation is: $$\dim_F(D \otimes_F E) = \dim_F D \cdot \dim_F E = d \cdot n = n^3,$$ and the dimension of the right side is: $$\dim_F M_n(E) = \dim_F \mathcal L(E^n \to E^n) = \dim_F E^n \cdot \dim_F E^n = (n \cdot n)^2 = n^4,$$ which is not equal to $n^3$ unless $n = 1$.
Can you please point me to an incorrect deduction in the reasoning?
Your calculation of $\dim_F M_n(E)$ is incorrect. It should be $\dim_F M_n(E) = (\dim_E M_n(E))(\dim_F E) = n^2\cdot n = n^3$.
The issue lies in the meaning of $\mathcal{L}(E^n\to E^n)$. Is this the space of $E$-linear maps, or of $F$-linear maps? If the former, then its dimension is $n^3$, if the latter then the dimension is $n^4$.