Centralizers in Jacobson-Witt Lie algebras Recall the (Jacobson-)Witt Lie algebras in positive characteristic: $W(n,1)$ is the Lie algebra of derivations of $\Bbbk[X_1,\dots,X_n]/(X_1^p,\dots,X_n^p)$. (For simplicity; more generally, I'm interested in the $W(n,m_1,\dots,m_n)$ which are derivations of a ring of divided power series truncated at $X_1^{p^{m_1}},\dots,X_n^{p^{m_n}}$).
Q1. It seems that every element in $W(n,1)$ has a centralizer of dimension at least $n$. Is this known? I couldn't find it in my usual references, "Modular Lie algebras" by Seligman and by Strade, and I don't know how to prove it, but checked it in all cases I could think of, and don't see any particular form to the centralizer of an element.
Q2. Assuming Q1 true, how large can a subspace of elements with minimal centralizer be? I would like there to exist an $(n+1)$-dimensional subspace of $W(n,1)$ such that all non-zero elements of that subspace have $n$-dimensional centralizer.
 A: For $W(n,1)={\rm Der}(\mathcal{O}_n)$, defined over an algebraically filed $k$ of characteristic $p>2$, the smallest dimension of centralizers equals $n$ (here $\mathcal{O}_n$ is the $k$-algebra $k[X_1,\ldots, X_n]/(X_1^p,\ldots, X_n^p)$ of truncated polynomials in $n$ variables). Let $\mathcal{L}$ be a finite dimensional Lie algebra and denote by $\mathcal{L}_{\rm reg}$ the Zariski open subset of $\mathcal{L}$ consisting of all elements whose centralizer has the smallest possible dimension. Then it is known that 
$W(n,1)_{\rm reg}=W(n,1)\cap\mathfrak{gl}(\mathcal{O}_n)_{\rm reg}.$ Since
$\mathfrak{gl}(\mathcal{O}_n)\setminus\mathfrak{gl}(\mathcal{O}_n)_{\rm reg}$ is known to have codimension $3$ in $\mathfrak{gl}(\mathcal{O}_n)$, the above description implies that any $4$-dimensional subspace of $W(n,1)$ contains a non-zero element whose centralizer has dimension $>n$. So Q2 has a negative answer for all $n\ge 3$.
Let $r_n$ denote the codimension of the closed set $W(n,1)\setminus W(n,1)_{\rm reg}$ in $W(n,1)$. It is known (and not hard to see) that $r_n\ge 2$. On the other hand, the previous paragraph shows that $r_n\le 3$. Hence $r_n\in\{2,3\}$ for all $n$. If $p>3$, then one can check directly that $r_1=2$ (the case $(n,p)=(1,3)$ is exceptional as in characteristic three $W(1,1)\cong\mathfrak{sl}_2$). So it is reasonable to conjecture that $r_n=2$ for all $n\ge 2$ provided that $p>2$ and $(n,p)\ne (1,3)$.
