Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $G$ be an affine algebraic group. A character of $G$ is a morphism $G\to \mathbb G_m$. Let $X$ be the abelian group of all characters of $G$. Suppose this group is finitely generated, say by ${\chi_1,\ldots, \chi_n}$, with no $p$-torsion, where $p$ is the characteristic of the field. We have the following theorem. Suppose $Y$ is a subgroup of $X$ such that $X/Y$ has no $p$-torsion. If $\chi$ is a character having the property that

$$ \bigcap_{\eta\in Y}\ker\eta\subset \ker\chi,$$

then $\chi\in Y$.

I think I have a counterexample to this theorem, and I am not able to see why it does not work. Suppose $Y = \langle \{\chi_1,\chi_2\}\rangle$, where $\ker\chi_1\cap\ker\chi_2 = \{e\}$. Then, $X/Y\cong \langle\{\chi_3,\ldots,\chi_n\}\rangle$ has no $p$-torsion. Moreover, any character will contain $\ker\chi_1\cap\ker\chi_2$, but they are obviously not all in $Y$.

Can you please help me with the proof of this theorem?

share|cite|improve this question
Have you an example of such $Y$ ? – user18119 Feb 18 '13 at 9:26
For example, take $Y = \langle\{\chi_1, \chi_2\}\rangle$ as in my example. – user62727 Feb 18 '13 at 18:30
I meant have you an example with intersection of the kernels of $\chi_1$ and of $\chi_2$ reduced to $e$ and with $n>2$ ? – user18119 Feb 18 '13 at 20:27
Not really, I have trouble building some intuitions with these things. I am not sure anymore the statement is true in general. I think we need more assumptions on $G$. – user62727 Feb 18 '13 at 20:58
Each kernel is a subgroup of codimension $1$, hence the intersection has codimension $2$. If it is reduced to $e$, this implies that $G$ has dimension $2$. So $G/[G,G]$ has at most dimension $2$. Then I can't see how its group of characters can have rank $>2$. – user18119 Feb 18 '13 at 21:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.