Suppose we have an algebraic group $G$ defined over a field $k$. Suppose we consider the fraction field of $k[G]$. Is it possible to get a situation where this field is not a separable extension of $k$? An example would be very helpful.
$\begingroup$
$\endgroup$
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
The answer depends on what you mean by "algebraic group". If you mean algebraic group in the sense of Borel/Humphreys/Springer, the field of fractions of $k[G]$ is always separably generated, more-or-less by definition. If you mean algebraic group scheme, which must be integral for the question to make sense, then it doesn't have to be separably generated, as shown by Tom's example.
$\begingroup$
$\endgroup$
4
Yes, this is possible. (Notice that your question implicitly requires $k[G]$ to be an integral domain, otherwise you cannot take its fraction field.)
Let $k$ be a non-perfect field of characteristic $p$, and let $a \in k \setminus k^p$. Consider the algebraic group defined as a functor from $k$-algebras to groups via $$ G(R) = \{(r,r') \in R^2 \mid r^p = a{r'}^p \} $$ for all $R \in k\mathrm{-alg}$.
Then $k[G] \cong k[t,t']/(t^p-a{t'}^p)$, which is a domain whose fraction field $k(a^{1/p})(t)$ is visibly not separable over $k$.
answered May 7, 2014 at 8:02
-
$\begingroup$ I am curious: how is this a group? It looks like a $\mu_p$-torsor that is fppf-but-not-étale trivializable. $\endgroup$– S. Carnahan ♦Commented May 7, 2014 at 8:17
-
$\begingroup$ @S.Carnahan: The mistakes in the original version of this answer have been fixed. $\endgroup$ Commented May 7, 2014 at 8:50
-
$\begingroup$ @user76758: Thanks for the correction! $\endgroup$ Commented May 7, 2014 at 12:02
-
$\begingroup$ A thickened line of slope $a^{-1/p}$, very nice! $\endgroup$– S. Carnahan ♦Commented May 7, 2014 at 12:31