Question

M3[ Z ]has no nilpotent elements.

Answer 1

$M_3[Z]=\{\begin{pmatrix} a_{11} & a_{12}& a_{13} \\ a_{21} & a_{22}& a_{23} \\ a_{31} & a_{32}& a_{33} \\ \end{pmatrix},\\ a_{ij}\in Z\}$

Сonsider the basis

$E_{11}=\begin{pmatrix} 1 &0&0 \\ 0&0&0\\ 0&0&0 \end{pmatrix}, E_{12}=\begin{pmatrix} 0 &1&0 \\ 0&0&0\\ 0&0&0 \end{pmatrix}, \\ E_{13}=\begin{pmatrix} 0 &0&1 \\ 0&0&0\\ 0&0&0 \end{pmatrix}, E_{21}=\begin{pmatrix} 0&0&0 \\ 1&0&0\\ 0&0&0 \end{pmatrix}, \\ E_{22}=\begin{pmatrix} 0 &0&0 \\ 0&1&0\\ 0&0&0 \end{pmatrix}, E_{23}=\begin{pmatrix} 0 &0&0 \\ 0&0&1\\ 0&0&0 \end{pmatrix}, \\ E_{31}=\begin{pmatrix} 0 &0&0 \\ 0&0&0\\ 1&0&0 \end{pmatrix}, E_{32}=\begin{pmatrix} 0 &0&0 \\ 0&0&0\\ 0&1&0 \end{pmatrix}, \\ E_{33}=\begin{pmatrix} 0 &0&0 \\ 0&0&0\\ 0&0&1 \end{pmatrix}$

$A=\begin{pmatrix} a_{11} & a_{12}& a_{13} \\ a_{21} & a_{22}& a_{23} \\ a_{31} & a_{32}& a_{33} \\ \end{pmatrix}=a_{11}E_{11}+a_{12}E_{12}+\\ +a_{13}E_{13}+a_{21}E_{21}+a_{22}E_{22}+\\ +a_{23}E_{23}+a_{31}E_{31}+a_{32}E_{32}+\\ +a_{33}E_{33}$

Let's put together a Kelly table and  we denote

$O=\begin{pmatrix} 0 & 0&0 \\ 0 & 0&0\\ 0&0&0 \end{pmatrix}$

$\begin{matrix} &E_{11} &E_{12}&E_{13}&E_{21}&E_{22}&E_{23}&E_{31}&E_{32}&E_{33}\\ E_{11}& E_{11}&E_{12}&E_{13}&O&O&O&O&O&O\\ E_{12}& O&O&O&E_{11}&E_{12}&E_{13}&O&O&O\\ E_{13}& O&O&O&O&O&O&E_{11}&E_{12}&E_{13}\\ E_{21}& E_{21}&E_{22}&E_{23}&O&O&O&O&O&O\\ E_{22}& O&O&O&E_{21}&E_{22}&E_{23}&O&O&O\\ E_{23}& O&O&O&O&O&O&E_{21}&E_{22}&E_{23}\\ E_{31}& E_{31}&E_{32}&E_{33}&O&O&O&O&O&O\\ E_{32}& O&O&O&E_{31}&E_{32}&E_{33}&O&O&O\\ E_{33}& O&O&O&O&O&O&E_{31}&E_{32}&E_{33} \end{matrix}$

Let's show that $A^2\neq0$

$A^2=(a_{11}E_{11}+a_{12}E_{12}+\\ +a_{13}E_{13}+a_{21}E_{21}+a_{22}E_{22}+\\ +a_{23}E_{23}+a_{31}E_{31}+a_{32}E_{32}+\\ +a_{33}E_{33})(a_{11}E_{11}+a_{12}E_{12}+\\ +a_{13}E_{13}+a_{21}E_{21}+a_{22}E_{22}+\\ +a_{23}E_{23}+a_{31}E_{31}+a_{32}E_{32}+\\ +a_{33}E_{33})=\\ =a_{11}^2E_{11}+a_{11}a_{12}E_{12}+\\ +a_{11}a_{13}E_{13}+a_{12}a_{21}E_{11}+a_{12}a_{22}E_{12}+\\ +a_{12}a_{23}E_{13}+a_{13}a_{31}E_{11}+a_{13}a_{32}E_{12}+\\ +a_{13}a_{33}E_{13}+$

$+a_{21}a_{11}E_{21}+a_{21}a_{12}E_{22}+\\ +a_{21}a_{13}E_{23}+a_{22}a_{21}E_{21}+a_{22}^2E_{22}+\\ +a_{22}a_{23}E_{23}+a_{23}a_{31}E_{21}+a_{23}a_{32}E_{22}+\\ +a_{23}a_{33}E_{23}+\\ +a_{31}a_{11}E_{31}+a_{31}a_{12}E_{32}+\\ +a_{31}a_{13}E_{33}+a_{32}a_{21}E_{31}+a_{32}a_{22}E_{32}+\\ +a_{32}a_{23}E_{33}+a_{33}a_{31}E_{31}+a_{33}a_{32}E_{32}+\\ +a_{33}^2E_{33}=\\$

$=(a_{11}^2+a_{12}a_{21}+a_{13}a_{31})E_{11}+\\ +(a_{11}a_{12}+a_{12}a_{22}+a_{13}a_{32})E_{12}+\\ +(a_{11}a_{13}+a_{12}a_{23}+a_{13}a_{33})E_{13}=$

$+(a_{21}a_{11}+a_{22}a_{21}+a_{23}a_{31})E_{21}+\\ +(a_{21}a_{12}+a_{22}^2+a_{23}a_{32})E_{22}+\\ +(a_{21}a_{13}+a_{22}a_{23}+a_{23}a_{33})E_{23}+\\ +(a_{31}a_{11}+a_{32}a_{21}+a_{33}a_{31})E_{31}+\\ +(a_{31}a_{12}+a_{32}a_{22}+a_{33}a_{32})E_{32}+\\ +(a_{31}a_{13}+a_{32}a_{23}+a_{33}^2)E_{33}$

If $A^2=0$ then

$a_{11}^2+a_{12}a_{21}+a_{13}a_{31}=0\\ a_{11}a_{12}+a_{12}a_{22}+a_{13}a_{32}=0\\ a_{11}a_{13}+a_{12}a_{23}+a_{13}a_{33}=0\\ a_{21}a_{11}+a_{22}a_{21}+a_{23}a_{31}=0\\ a_{21}a_{12}+a_{22}^2+a_{23}a_{32}=0\\ a_{21}a_{13}+a_{22}a_{23}+a_{23}a_{33}=0\\ a_{31}a_{11}+a_{32}a_{21}+a_{33}a_{31}=0\\ a_{31}a_{12}+a_{32}a_{22}+a_{33}a_{32}=0\\ a_{31}a_{13}+a_{32}a_{23}+a_{33}^2=0$

This system has zero solution only $a_{ij}\in Z$ .

So $A^2\neq0$ .

Then $M_3[Z]$ has no nilpotent elements.