Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

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.

share|cite|improve this question
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
up vote 1 down vote accepted

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 $G$.

share|cite|improve this answer
$G \subsetneqq \mathscr{V}(\mathscr{I}(G))$, 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

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.