Let $t$ and $x$ be indeterminates, and let $P\in\mathbb Q(t)[x]$ be irreducible. Adjoining a root of $P$ to $\mathbb Q(t)$ we obtain a function field $F/\mathbb Q$. Now suppose that $K$ is an algebraic extension of $\mathbb Q$. Regarding $P$ as an element of $K(t)[x]$, we can adjoin a root of $P$ to $K(t)$ to obtain a function field $E/K$. Under which conditions is it true that $E$ and $F$ have the same genus? In particular, will this hold if $P$ remains irreducible in $K(t)[x]$?



Genus of a function field after an extension
Let $X$ be the smooth projective curve with function field $F$. Then the base change $X\_K$ is the curve with function field $E$. The genera of $X$ and $X\_K$ are equal because $X\_K$ is regular since $\mathbb Q$ is a perfect field. Hence the genera of $E$ and $F$ are equal. 


I would be interested to see a proof relying purely on function field theory rather than algebraic geometry.


