I'm reading Springer's Linear algebraic group and I stuck with one exercise.
Let $k$ be a field of characteristic 2 and let $G = \mathrm{SL}_{2}$. Then it says that conjugacy class of the matrix $$ \begin{pmatrix} 1&1\\0&1\end{pmatrix} $$ is not closed in $G$. By hand, I can show that the conjugacy class is $$ \left\{ \begin{pmatrix} 1-ac & a^{2} \\ - c^{2} & ac+1 \end{pmatrix} = \begin{pmatrix} 1+ ac & a^{2} \\ c^{2} & 1+ac \end{pmatrix}\,:\, a, c\in k\right\}. $$ How can we show that the set is not topologically closed in $G$?
Hint: there exists a particularly simple matrix which isn't in this orbit which should be. You've missed one condition on $a,c$ when you say that $a,c\in k$ (can they both be zero at the same time?).