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!


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.

  • 3
    $\begingroup$ 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 $i \in I$, $ir \in I$. $\endgroup$ – user3482749 Jan 4 '19 at 21:26
  • $\begingroup$ 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... $\endgroup$ – 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 $R=M_{3}(\mathbb{Q})$.

Can this be described more easily, at least in this case? Well, the given matrix is…

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  • $\begingroup$ that's it? just the product of those two matrices? I should've known it wasn't anything too complicated, thank you very much $\endgroup$ – Alex Jan 4 '19 at 23:09
  • $\begingroup$ @Alex No, you didn't read carefully or I didn't explain myself too well: it is the set of all matrices of that form, infinitely many of them. But in this particular case, a property of the fixed matrix can be used. $\endgroup$ – egreg Jan 4 '19 at 23:13
  • $\begingroup$ no, sorry I have no idea where you're going with this $\endgroup$ – Alex Jan 4 '19 at 23:38
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    $\begingroup$ @Alex Call $r$ the given matrix. Then $r^{-1}$ exists and belongs to $R$, so among the elements of $rR$ there is $rr^{-1}=1$. $\endgroup$ – egreg Jan 4 '19 at 23:50
  • $\begingroup$ okay so the ideal will contain identity?? $\endgroup$ – Alex Jan 5 '19 at 0:14

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