A closed $Y\subseteq\operatorname{GL}_3(\overline{\mathbb F_5(u)})$ with $P\in\overline{Y^m}\setminus Y^m$ for every $m\ge2$

Problem

Let $X$ be a Zariski-closed subset of a connected reductive linear algebraic group in positive characteristic. Must some power $X^c$, $c>1$, be closed?

$$X^c=\{x_1\cdots x_c:x_i\in X\},\qquad \exists c>1:\ \overline{X^c}=X^c\;?$$

The power consists of products of exactly $c$ factors.

The statement and the counterexample

Problem 16.28(a) of the Kourovka Notebook, proposed by E. P. Vdovin, asks whether every closed subset $X$ of a connected linear reductive algebraic group in positive characteristic has a closed power $X^c$ for some integer $c>1$ (1). Here

$$X^c=\{x_1\cdots x_c:x_1,\ldots,x_c\in X\}$$

denotes the set of products of exactly $c$ factors.

Linear algebraic groups are Zariski-closed matrix subgroups over an algebraically closed field. Reductivity means triviality of the connected normal unipotent radical (2). Closedness is taken in the matrix Zariski topology: closed subsets of $\mathrm{GL}_3(k)$ are intersections with polynomial zero sets in the nine entries.

Let $u$ be an indeterminate and put $k=\overline{\mathbb F_5(u)}$. We also write $u$ for its image in $k$. For $x\in k^\times$, define

$$B(x)=\begin{pmatrix}1&x&x^{-1}\\0&1&0\\0&0&u\end{pmatrix}, \qquad P=\operatorname{diag}(1,1,u^2),$$

and set

$$\tag{3} Z=\{\lambda I_3:\lambda\in k^\times\},\qquad D=\{\lambda B(x):\lambda,x\in k^\times\},\qquad Y=Z\cup D.$$

Theorem 1. The group $\mathrm{GL}_3(k)$ is connected and reductive. The subset $Y$ in (3) is closed and contains $I_3$. For every integer $m\geq2$,

$$P\in\overline{Y^m}\setminus Y^m.$$

The field $k$ is algebraically closed of characteristic five. For $a>b\geq0$, the equality $u^a=u^b$ would imply the false rational-function identity $u^{a-b}=1$. The embedding $\mathbb F_5(u)\hookrightarrow k$ therefore gives an element of infinite multiplicative order.

The ambient group

Affine nine-space over $k$ is irreducible. Indeed, two proper closed subsets admit nonzero vanishing polynomials; if they covered the space, the product would be a nonzero polynomial vanishing everywhere over the infinite field $k$, a contradiction. The nonempty open subset $\det\ne0$ is consequently irreducible. Thus $\mathrm{GL}_3(k)$ is connected. The following lemma proves its reductivity.

Lemma 2. Let $V$ be a finite-dimensional vector space over a field. If $H\trianglelefteq\mathrm{GL}(V)$ and every element of $H$ is unipotent, then $H=\{1\}$.

Applying Lemma 2 to the unipotent radical of $\mathrm{GL}_3(k)$ proves that this radical is trivial. Thus $\mathrm{GL}_3(k)$ is reductive.

Closedness and the obstruction

The scalar subgroup $Z$ is closed, being defined by vanishing of the off-diagonal entries and equality of the diagonal entries. The set $D$ is the zero locus in $\mathrm{GL}_3(k)$ of

$$\tag{9} \begin{gathered} a_{21}=a_{31}=a_{32}=a_{23}=0,\qquad a_{11}=a_{22},\qquad a_{33}=u a_{22},\\ a_{12}a_{13}=a_{22}^2. \end{gathered}$$

Indeed, these equations give $\det(a_{ij})=u a_{22}^3$. Invertibility forces $\lambda=a_{22}\ne0$. Dividing the matrix by $\lambda$ gives $B(x)$ with $x=a_{12}/\lambda\ne0$, because the last equation says $(a_{12}/\lambda)(a_{13}/\lambda)=1$. Conversely, every $\lambda B(x)$ satisfies (9). Thus $D$, and consequently $Y=Z\cup D$, is closed. Also $I_3\in Z$.

Fix $m\geq2$. Suppose a product of $m$ elements of $Y$ equals $P$, and let $j$ count the factors chosen from $D$. Since scalar matrices are central, this product can be written

$$\Lambda B(x_1)\cdots B(x_j),\qquad \Lambda\in k^\times.$$

Its diagonal is $(\Lambda,\Lambda,\Lambda u^j)$. Comparison with $P$ gives $\Lambda=1$ and $u^j=u^2$, hence $j=2$. But direct multiplication yields

$$B(x)B(y)= \begin{pmatrix}1&x+y&u/x+1/y\\0&1&0\\0&0&u^2\end{pmatrix}.$$

For this to equal $P$, one would need $y=-x$ and $(u-1)/x=0$, contradicting $u\ne1$. Therefore $P\notin Y^m$.

To prove the opposite closure membership, consider the polynomial matrix curve

$$\gamma(t)= \begin{pmatrix}1&0&(u-1)t\\0&1&0\\0&0&u^2\end{pmatrix} \qquad(t\in k).$$

Its determinant is the nonzero constant $u^2$, so it is defined in $\mathrm{GL}_3(k)$ for every $t$, and $\gamma(0)=P$. For $t\ne0$,

$$\gamma(t)=B(t^{-1})B(-t^{-1})\in Y^2\subseteq Y^m,$$

where the final inclusion inserts $m-2$ identity factors. The inverse image $\gamma^{-1}(\overline{Y^m})$ is closed, since substituting the polynomial entries of $\gamma$ into any matrix polynomial gives a polynomial in $t$. It contains $k\setminus\{0\}$, which is dense in the affine line over the infinite field $k$. Hence it also contains zero. Thus $P\in\overline{Y^m}$, completing the proof of Theorem 1.

The counterexample, including the field and ambient-group hypotheses, is formalised in Lean 4 with Mathlib; the proof source accompanies this article.

References

Preprint · Lean (GitHub)

1. E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, 21st edition, revision of 1 September 2026, Problem 16.28, arXiv:1401.0300.

2. J. S. Milne, Algebraic Groups: The Theory of Group Schemes of Finite Type over a Field, Cambridge Studies in Advanced Mathematics 170, Cambridge University Press, 2017.