I've been playing with a certain matrix subring, but there is one step I am having trouble with.
Let $u=\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{pmatrix}$ in $M_3(\mathbb{Q})$ and let $x=\begin{pmatrix} u & 0 \\ 0 & u^2\end{pmatrix}$ and $y=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$, where $0$ and $1$ are the zero and identity matrices in $M_3(\mathbb{Q})$, so view $x$ and $y$ in $M_6(\mathbb{Q})$.
Let $R$ be the subring generated by $\mathbb{Q}$, $x$ and $y$. Then why is that if $x'$ is nilpotent in $R$, and $y'$ is such that $y'^2=0$, then $y'x'^2=0$?
Background work I've done: I've found that $x$ and $y$ satisfy the relations $ x^3=0=y^2\qquad yx=x^2y $ and that $\{1,x,x^2,y,xy,x^2y\}$ is a basis for $R$ as a $\mathbb{Q}$-vector space. The above fact in question will prove that $R$ has no anti-automorphisms, since $x^2y\neq 0$, but if $\varphi$ is some anti-automorphism, then $ \varphi(x^2y)=\varphi(y)\varphi(x^2)=\varphi(y)\varphi(x)^2=0 $ since $\varphi(y)^2=0$ and $\varphi(x)$ is nilpotent as the image of the nilpotent element $x$. But then $\varphi$ is not injective.