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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.

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Suppose $y'=aI+bx+cx^2+dy+exy+fx^2y, x'=a_1I+b_1x+c_1x^2+d_1y+e_1xy+f_1x^2y$.

Then the fact that $x',y'$ are nilpotents implies $a=a_1=0$ (otherwise the trace is not zero). Then by the relations you found it is easy to see $b=0$ (by expanding $0=y'^2$).

Now you can compute that $y'x'=(cc_1+da_1)x^2y$, and therefore $y'x'^2=(y'x')x'=0$.

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Thanks, I have it now. –  Buble Jun 20 '12 at 6:59
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