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Let $k$ be an algebraically closed field with $\mathrm{char}(k)=p>0$. Let $U$ be a connected unipotent algebraic group over $k$.

Question: When $p$ is big enough, is it true that $Z_U(u)$ is connected for any $u\in U$, or at least $u\in Z_U(u)^o$ for any $u\in U$?

Remark: This is true when $U$ is the unipotent radical of a Borel subgroup of a reductive group with $p$ being good (so that a Springer homeomorphism exists).

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    $\begingroup$ Did you mean to ask whether, for a unipotent group scheme $U$ over $\mathbb{Z}$, the fibre $U_{\mathbb{F}_p}$ has connected centralisers for all sufficiently large $p$? This seems to me to be the more interesting question because one usually expects the fibres $U_{\mathbb{F}_p}$, for large $p$, to behave like groups over characteristic $0$ fields. There is no reason to expect this if $U$ is allowed to vary with $p$. $\endgroup$ – A Stasinski Jan 31 at 9:21
  • $\begingroup$ @AStasinski Yes, $p$ is expected to be irrelevant to the defining equations of $U$. The question stated in the current form is a bit misleading. $\endgroup$ – user148212 Jan 31 at 10:45
  • $\begingroup$ In that case kneidell's answer is not an answer to the intended question. $\endgroup$ – A Stasinski Jan 31 at 10:54
  • $\begingroup$ If your element and your group are globally defined (say, over $\mathbb Z$) then both $Z_U(u)$ and $Z_U(u)^\circ$ would be $\mathbb Z$ defined, hence so would the quotient group. Taking $p$'s larger than the exponent of this finite group, shouldn't this finite algebraic group have no points over $k$? $\endgroup$ – kneidell Feb 1 at 15:06
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The following is a counterexample which can be defined for arbitrarily large $p$'s.

Consider $U=\left\{ \begin{pmatrix}1&a&b\\&1&a^p\\&&1\end{pmatrix}:a,b\in k\right\}\subseteq\mathrm{GL}_3(k)$ and take $u=u_\lambda=\begin{pmatrix}1&\lambda\\&1&\lambda\\&&1\end{pmatrix}$ with $0\ne \lambda\in\mathbb{F}_p$ (i.e. $\lambda=\lambda^p$). Then one easily computes that $$\begin{pmatrix}1&a&b\\&1&a^p\\&&1\end{pmatrix}\in Z_U(u)\quad\iff\quad a=a^p, $$ and thus $Z_U(u)\simeq \mu_p\ltimes \mathbb G_a$ and $Z_U(u)^\circ=\begin{pmatrix}1&&*\\&1&\\&&1\end{pmatrix}\simeq \mathbb G_a$ (here $\mu_p$ is the group of $p$-th roots of $1$). In particular $u\notin Z_U(u)^\circ$.

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    $\begingroup$ Note that this gives a family of groups $U$ depending on $p$. See also my comment on the question. $\endgroup$ – A Stasinski Jan 31 at 9:14

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