Problem. Suppose that $F$ is an uncountable algebraically closed field and $D$ is a division $F$-algebra. If $\dim_F D$ is countable, then $D=F$.
For each $x\in D$, $F(x)$ is a field extension of $F$ and therefore $[F(x)\colon F]\leqslant\aleph_0$. Since $F$ is algebraically closed, if $F(x)$ is a proper extension, it must be transcendental. Thus I think if the uncountableness of $F$ implies that the degree of any transcendental simple extension must be uncountable, then the problem is done, and this is where I am stuck. So I would like to ask whether I am in a right way, and what to do for the rest of the proof? Thanks in advance..