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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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Since you tagged this question as being related to algebraic groups I might guess that you are supposed to use the Zariski topology as opposed to the usual topology. –  Jyrki Lahtonen Jul 13 '11 at 17:40
    
Yes, the groups are given the Zariski topology. Thanks. –  ShinyaSakai Jul 13 '11 at 18:13
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1 Answer 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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$G \subsetneqq \mathscr{V}(\mathscr{I}(G))$, so $G$ is not closed. Thank you very much. –  ShinyaSakai Jul 17 '11 at 13:48
    
Sincere thanks to you. I am sorry I wasn't careful enough when I read your answer. In fact there are still some points unclear to me. I think what has to be proved is $G\subsetneq \mathscr{V}(I)$, where $I=\mathscr{I}(G)$. Now what I have seen are $J\subseteq I$, and $G \subsetneq \mathscr{V}(J)$. But if I am not mistaken, when $J \subseteq I$, we have $\mathscr{V}(J) \supseteq \mathscr{V}(I)$. So, I cannot see $G \subsetneq \mathscr{V}(I)$ from $G\subsetneq \mathscr{V}(J)$ which contains $\mathscr{V}(I)$. I hope you will see this comment although this question is so old... –  ShinyaSakai Oct 16 '11 at 19:36
    
Is it true that $J$ in fact equals $I$? –  ShinyaSakai Oct 16 '11 at 19:49
    
@ShinyaSakai: The first point of my answer was that there cannot be a polynomial $p$ that is in $I(G)$, but is outside of $J$, so $J=I(G)$. The second point was that $V(J)=V(I(G))$ has $G$ as a strict subset. Therefore $G$ is not closed. BTW, the system will ping me automatically. If I hadn't been on-line, I would have seen your comment next time I log on. It's quite efficient :-) –  Jyrki Lahtonen Oct 16 '11 at 19:53
    
Thanks to the great INBOX! I've always been careless. This time, let me check your answer CAREFULLY to see if I have understood every point before I thank you one more time. –  ShinyaSakai Oct 16 '11 at 20:01
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