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Here is the problem:

Show by example that the subgroup of an algebraic group generated by two non-irreducible closed subsets need not be closed.

and a hint is given:

Use the cyclic subgroups of $GL(2, \mathbb{C})$ generated by $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ and $\begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix}$.

Now, let $G_1 = \{ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \} $ and $G_2 = \{ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix} \}$ (the cyclic subgroups generated by $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ and $\begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix}$ ). Both are closed and non-irreducible subsets (subgroups). The subgroup generated by $G_1$ and $G_2$ is $G = \{ \begin{pmatrix} 1 & n_1 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & n_2 \\ 0 & -1 \end{pmatrix} : n_1, n_2 \in \mathbb{Z} \} $. As $G$ is a set of discrete points in $GL(2, \mathbb{C} )$, can it be not closed?

In another way, set $G_1 = \{ \begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix}, \begin{pmatrix} b & 0 \\ 0 & -b \end{pmatrix} : a,b \in \mathbb{C}^* \} $ and $G_2 = \{ \begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix}, \begin{pmatrix} b & b \\ 0 & -b \end{pmatrix} : a,b \in \mathbb{C}^* \}$. They are closed and non-irreducible. And the subgroup of $GL(2, \mathbb{C})$ generated by $G_1$ and $G_2$ is $G= \{ \begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix}, \begin{pmatrix} b & 0 \\ 0 & -b \end{pmatrix}, \begin{pmatrix} c & c \\ 0 & -c \end{pmatrix}, \begin{pmatrix} d & d \\ 0 & d \end{pmatrix} : a,b,c,d \in \mathbb{C}^* \} $. Isn't it closed? And why?

I don't know if I am wrong somewhere.

To prove a variety is closed, we often make the target variety into the inverse image of a closed variety.

But how to prove a variety is not closed (in a certain larger one)?

Many thanks.

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    Yes, the groups are given the Zariski topology. Thanks.2011-07-13

1 Answers 1

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Assuming that you are supposed to use Zariski topology, then let's consider the ideal $I(G)$ of polynomials in $R=\mathbf{C}[x_{11},x_{12},x_{21},x_{22}]$ that vanish at all points of $G$. Clearly $x_{11}-1,x_{21},x_{22}^2-1\in I(G)$. Let $J$ be the ideal generated by these three polynomials. We shall prove that actually $J=I(G)$. As the generators of $J$ belong to $I(G)$, clearly $J\subseteq I(G)$. The reverse inclusion requires a little bit of work.

Let $q$ be an arbitrary polynomial in $I(G)$. The argument is based on the observation that we can write $q$ in the form $q=p+j$, where $j\in J$, and the other polynomial is of the form $p=f(x_{12})+x_{22}g(x_{12})$, with $f,g\in\mathbf{C}[x_{12}]$. This follows from the facts that

  1. For all positive integers $k$ the power $x_{11}^k\equiv 1\pmod J$, because the polynomial $x_{11}-1\in J$,
  2. Any multiple of $x_{21}$ is in $J$, because $x_{21}\in J$,
  3. For all positive integers $k$ the power $x_{22}^{k+2}\equiv x_{22}^k\pmod J$, because the polynomial $x_{22}^2-1\in J$. Applying this recursively we can replace a term with a high power of $x_{22}$ with another one, where the exponent is either zero or one. All this by moving within a single coset of $J$.

We have $ p\left(\begin{array}{cc}1&n\\0&1\end{array}\right)=f(n)+g(n), $ and $ p\left(\begin{array}{cr}1&n\\0&-1\end{array}\right)=f(n)-g(n). $ In order for both of these to vanish simultaneously we must have $f(n)=g(n)=0$. This must happen for all $n\in\mathbf{Z}$. This is possible if and only if $f=g=0$, because a non-zero polynomial has only finitely many zeros in $\mathbf{C}$. Therefore we must actually have $p=0$, so $q\in J$, and $I(G)=J$.

Thus the Zariski closure of $G$ consists of the zero set of polynomials in $J$, but this set $ V(I(G))=V(J)=\left\{\left(\begin{array}{cr}1&z\\0&\pm1\end{array}\right)\mid z\in\mathbf{C}\right\} $ is larger than $G$.

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    Thank you for all the instructions.$I$am sorry I don't know much about how to use this forum. But I am really very very glad to give my bounty, although very little, to someone who has been so helpful and given me so enlightenling and detailed answer. So, NO RETRACTION please. Thanks again.2011-10-17