# Ideal of matrix ring

I'm going over some exercises and I'm not quite sure if I completely understand this one.

Let $$R=M_3(\mathbb{Q})$$, i.e. $$R$$ is the ring of all $$3\times3$$ matrices over rational numbers. Describe the minimal right ideal of $$R$$ containing the matrix $$\begin{bmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \\ \end{bmatrix}$$

M̶y̶ ̶g̶u̶e̶s̶s̶ ̶w̶a̶s̶ ̶t̶h̶a̶t̶ ̶I̶'̶m̶ ̶j̶u̶s̶t̶ ̶s̶u̶p̶p̶o̶s̶e̶d̶ ̶t̶o̶ ̶f̶i̶n̶d̶ ̶a̶ ̶m̶a̶t̶r̶i̶x̶ ̶i̶ ̶t̶h̶a̶t̶ ̶f̶i̶t̶s̶ ̶t̶h̶e̶ ̶d̶e̶f̶i̶n̶i̶t̶i̶o̶n̶ ̶o̶f̶ ̶a̶ ̶r̶i̶g̶h̶t̶ ̶i̶d̶e̶a̶l̶,̶ ̶w̶h̶i̶c̶h̶ ̶i̶s̶ ̶∀̶ ̶r̶ ̶∈̶ ̶R̶:̶ ̶i̶ ̶·̶ ̶r̶ ̶∈̶ ̶I̶ ̶(̶r̶ ̶b̶e̶i̶n̶g̶ ̶t̶h̶e̶ ̶g̶i̶v̶e̶n̶ ̶m̶a̶t̶r̶i̶x̶)̶ ̶a̶n̶d̶ ̶I̶ ̶=̶ ̶{̶(̶i̶)̶}̶ ̶.̶

But the more I tried to find something similar to this problem, the more I think that this is absolutely wrong. Can somebody help me with this and maybe explain how to get to the solution?

Thank you!

EDIT

My previous idea was a mix up of definitions, from the comments I now understand that I need to find $$I$$ (which contains the given matrix), but I'm still confused as to how to do so.

• Ideals are not matrices! Ideals are sets of matrices. You need to find the smallest $I \subseteq R$ such that $\left(\array{1&0&1\\1&1&0\\0&1&1}\right)\in I$ and for any $r \in R$ and any $i \in I$, $ir \in I$. – user3482749 Jan 4 '19 at 21:26
• I second what @user3482749 said. It seems that you don't quite understand the definition of ideals, so you need to go back and read it more carefully. What you stated as a "definition" has bits and pieces of the correct one, but the way it's stated doesn't make much sense... – zipirovich Jan 4 '19 at 22:25

The minimal right ideal of $$R$$ containing $$r\in R$$ is $$rR=\{rs:s\in R\}$$. In your case it is the set of all products of the form $$\begin{bmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \\ \end{bmatrix} \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{bmatrix}$$ where the second matrix varies through all elements of $$R=M_{3}(\mathbb{Q})$$.
• @Alex Call $r$ the given matrix. Then $r^{-1}$ exists and belongs to $R$, so among the elements of $rR$ there is $rr^{-1}=1$. – egreg Jan 4 '19 at 23:50