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 $K(t)[x]$?
$\begingroup$
$\endgroup$
4
-
$\begingroup$ Can you say more about the extension $K/\mathbb{Q}$? For example, can I take $K$ to be equal to $F$? $\endgroup$– David TweedleCommented Mar 5, 2018 at 19:31
-
$\begingroup$ I'm mostly interested in the case where $K$ is algebraic over $\mathbb Q$. I've updated the question accordingly. $\endgroup$– 352506Commented Mar 5, 2018 at 19:40
-
$\begingroup$ The answer is basically always yes here. What definition of the genus of the function field are you using? If the genus of the underlying algebraic curve, the algebraic curves are the same. If not, it should be a straightforward computation. $\endgroup$– Will SawinCommented Mar 5, 2018 at 19:56
-
$\begingroup$ Yes, the genus can be interpreted as the genus of the corresponding smooth projective curve. However, the curve $X$ corresponding to $F$ is not the same as the curve corresponding to $E$; rather, the latter is the base change $X_K$. $\endgroup$– 352506Commented Mar 7, 2018 at 18:12
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
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.
-
$\begingroup$ See chapter 8 of Number Theory in Function Fields by Michael Rosen for a function field perspective. $\endgroup$ Commented Mar 7, 2018 at 19:22