Embedding of $\mathfrak{sl}(2,F)$, char$F=2$, such that $x$ do not remain nilpotent

Question

If $L$ is a semi-simple linear Lie algebra inside $\mathfrak{gl}(V)$ then for any element of $L$, the Jordan decomposition (semi-simple + nilpotent) inside $L$ and in $\mathfrak{gl}(V)$ coincide.

Let $F$ be an algebraically closed field, but of characteristic $2$. Then $\mathfrak{sl}(2,F)$ is not semi-simple.

Qeustion: Can we embed $\mathfrak{sl}(2,F)$ in some $\mathfrak{gl}(V)$ so that $x=\begin{bmatrix} 0 & 1\\0 & 0\end{bmatrix}$ does not remain nilpotent in $\mathfrak{gl}(V)$?

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Assumptions: All vector space, representation spaces are finite dimensional, $F$ is algebraically closed,...

Answer 1

The Lie algebra $\mathfrak{sl}(2,F)$ in characteristic $2$, with basis $(x,y,z)$, has Lie bracket $[x,y]=z$, so is just the usual Heisenberg Lie algebra. Certainly the Heisenberg Lie algebra has a representation $\rho$ such that $\rho(x)$ is not nilpotent; just take the standard faithful representation of dimension $3$ and add the identity on the diagonal for the representing operator of $x$, i.e., $$ \rho(x)=\begin{pmatrix} 1 & 1 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \end{pmatrix},\; \rho(y)=\begin{pmatrix} 0 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 0 & 0 \end{pmatrix},\; \rho(z)=\begin{pmatrix} 0 & 0 & 1 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \end{pmatrix}. $$