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Feb 8, 2020 at 13:23 comment added YCor But $\mathbf{C}(x)$ is not algebraic over $\mathbf{C}$.
Oct 6, 2015 at 3:26 comment added MH.Fakharan 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.
Sep 26, 2015 at 4:31 comment added Eric Wofsey This is obviously impossible if $F$ is algebraically closed.
Sep 26, 2015 at 3:01 review Reopen votes
Sep 26, 2015 at 14:22
Sep 26, 2015 at 2:46 history edited MH.Fakharan CC BY-SA 3.0
I explained my aim of this question.
Sep 23, 2015 at 0:20 history closed Karl Schwede
Stefan Kohl
Neil Strickland
Jeremy Rickard
Jeremy Rouse
Needs details or clarity
Sep 22, 2015 at 18:08 review Close votes
Sep 23, 2015 at 0:31
Sep 22, 2015 at 16:43 review First posts
Sep 22, 2015 at 17:49
Sep 22, 2015 at 16:41 history asked MH.Fakharan CC BY-SA 3.0