Let $F_1$, $F_2$ be two fields, and $n_1$, $n_2>0$ be two integers, such that $M_{n_1}(F_1)\simeq M_{n_2}(F_2)$ as rings. Show that $n_1=n_2$ and $F_1\simeq F_2$.
I want to reduce the problem to $\text{char}(F)=0$ and $\text{char}(F)>0$.
For example, if $\text{char}(F_1)=\text{char}(F_2)=0$, we can assume that $$F_1=\mathbb{Q}(\alpha_1, \alpha_2, \dotsc, \alpha_{k}) \text{ and } F_2=\mathbb{Q}(\beta_1, \beta_2, \dotsc, \beta_{l}).$$
Then we can conclude that the number of bases of matrix ring $M_{n_1}(F_1)$ and $M_{n_2}(F_2)$ are $kn_1^2$ and $ln_2^2$, respectively. Since the two matrix rings are isomorphic, we have $kn_1^2=ln_2^2$.
Actually I don't think my idea works. Could you please help me? Thanks.
The centre of $M_n(F)$ is isomorphic to $F$, so the isomorphism type $M_n(F)$ determines $F$ up to isomorphism. As you say, you can then determine $n$ by a dimension count.