I know that there is a division algebra over $\mathbb{Q}$ such that it is algebraic and infinite dimensional over it's center i.e. $\mathbb{Q}$. But for construct this division algebra. we can use prime numbers.

I want to know a general method of construction for every arbitrary field. Actually, for an arbitrary field $F$, I want to build a division ring like $D$, such that $Z(D)=F$ and $[D:F]=\infty$ and $D$ be algebraic over $F$ (i.e. every element of $D$ be algebraic over $F$). For example by Hilbert's method for constructing division rings (from fields that have a non-trivial automorphism), we can make some division rings that have $Z(D)=F$ and $[D:F]=\infty$ but $D$ isn't algebraic over $F$. So I want to build division ring like Hilbert but be algebraic.

  • 3
    $\begingroup$ This is obviously impossible if $F$ is algebraically closed. $\endgroup$ – Eric Wofsey Sep 26 '15 at 4:31
  • $\begingroup$ By $D$, I mean division ring, not field. For example you can imagine the Quaternion over the complex number $\mathbb{C}$ which is algebraically closed. Even for fields we have $\mathbb{C}(x)$ over $\mathbb{C}$ is an infinite field extension. So it is not impossible. $\endgroup$ – MH.Fakharan Oct 6 '15 at 3:26
  • $\begingroup$ But $\mathbf{C}(x)$ is not algebraic over $\mathbf{C}$. $\endgroup$ – YCor Feb 8 at 13:23