1

Actually I have difficulties with the Section 7 Exercise 10 of the book J. E. Humphrey "Linear algebraic groups".

This exercises is about what is mentionned in the title. We have to make a demonstration by example and the book alrady gave one: Consider in $GL_2(\mathbb{C})$ the subgroup $H$ generated by the elements $\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right)$ and $\left( \begin{array}{cc} 1 & 1 \\ 0 & -1 \end{array} \right)$

In addition, I would like to find its Zariski closure $\overline{H}$ and I don't know how to.

Thank you in advance for any hint to solve this. K. Y.

Kal_Aki
  • 227
  • In a metric space $X$, a subset $S$ of $X$ is closed if and only if $\lim x_n \in S$ for every convergent sequence ${x_n}\subseteq S$. You've taken a divergent sequence so you can't conclude anything. – Rodrigo Dias Apr 03 '18 at 00:11
  • @rldias Yes, you're right. I had some doubt about it. So it is not the best argument for showing that it is not closed. Unfortunatly, I didn't manage to find another sequence. Could you help me for this? – Kal_Aki Apr 03 '18 at 00:36
  • In general, to find the Zariski closure of a set $A$, one needs to first find the algebraic equations $f$ for which $f(A)=0$. Then take all the points that satisfy all the equations. This will be the Zariski closure of $A$ – Exit path Apr 03 '18 at 01:01
  • Looks like this is more or less an exact duplicate of this oldie. As I answered that one I should not cast the first vote to close as a dupe though. Also, I endorse Clément's more general argument/hint (+1). – Jyrki Lahtonen Apr 03 '18 at 08:14
  • @JyrkiLahtonen Thank you for your answer. I did a research before asking my question but I kind of missed this topic. In addition to what you said, Since $\overline{H}=V(I(H))$. Indeed, we know that $H \subset \overline{H}$ then $I(\overline{H}) \subset I(H)$ then $V(I(H)) \subset V(I(\overline{H})) = \overline{H}$. And since $V(I(H))$ is a closed set that contains $H$, we have equality. A I right? – Kal_Aki Apr 03 '18 at 22:06

1 Answers1

2

First hint: Let $G$ be a subgroup of $GL_n(\mathbb{C})$, then if $G$ is both countable and Zariski closed, it is finite. Use this to deal with the exercise.

Second hint: Show that the kernel of the determinant map restricted to $H$ is of index $2$ in $H$ and explicitly compute its kernel. Deduce from this the connected component of the Zariski closure of $H$ and then $\overline{H}$.

Clément Guérin
  • 11,533