Find a field extension $K=k(x,y)$ of transcendence degree 1 where $x\notin k(y)$, $y\notin k(x)$ and $K|k$ is purely transcendental

Asked Jun 14 '22 at 15:19

Active Jun 14 '22 at 15:19

Viewed 65 times

I saw this question online and it's actually a true or false question, the question is:

True of False: Let $K=k(x,y)$ be a field extension with transcendence degree 1. If $x\notin k(y)$ and $y\notin k(x)$, then $K|k$ isn't purely transcendental

I thought about it, it looks true, but i wasn't able to prove it and there's a part of me who thinks it's false and that there's a counterexample somewhere (i spent more time thinking about a counterexample to be honest). I was thinking about a separable field extension with non-zero characteristic, but i couldn't come up with anything.

asked Jun 14 '22 at 15:19

user8785084

1

how about $K=k(t)$ where $x=t^2$ and $y=t^3$? – Atticus Stonestrom Jun 14 '22 at 15:23
@AtticusStonestrom I don't think it works, because $k(t^2,t^3)\neq k(t)$. – user8785084 Jun 14 '22 at 15:40
1

why not? clearly $k(t^2,t^3)\subseteq k(t)$, since each of $t^2,t^3$ lie in $k(t)$. on the other hand, $t=t^3/t^2\in k(t^2,t^3)$, so also $k(t)\subseteq k(t^2,t^3)$ – Atticus Stonestrom Jun 14 '22 at 15:41
@AtticusStonestrom oh yeah, sorry, for a moment i thought about $k[t]$, not $k(t)$. You're right, thanks a lot! – user8785084 Jun 14 '22 at 15:43
1

no problem, easy mistake to make – happy it helped! :) – Atticus Stonestrom Jun 14 '22 at 15:43

Find a field extension $K=k(x,y)$ of transcendence degree 1 where $x\notin k(y)$, $y\notin k(x)$ and $K|k$ is purely transcendental

0 Answers0