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**TheAbstractionist** Clearly $\displaystyle S=\left\{\begin{pmatrix}x & 0 & 0\\

0 & x & 0\\

0 & 0 & x

\end{pmatrix}:x\in F\right\}$ is an ideal of $\displaystyle R$ and the quotient ring $\displaystyle R/S$ is a commutative ring with identity $\displaystyle I_3+S.$

Let’s let $\displaystyle A(x,y,z)$ denote the element $\displaystyle \begin{pmatrix}x & 0 & y\\

0 & x & z\\

0 & 0 & x

\end{pmatrix}+S$ in $\displaystyle R/S.$

Then if $\displaystyle x\ne0,$ $\displaystyle A(x,y,z)\cdot A(x^{-1},-yx^{-2},-zx^{-2})=I_3+S.$ Hence $\displaystyle R/S$ is a field and so $\displaystyle S$ is a maximal ideal.