Ideal of matrix ring












3














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 $3times3$ 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.






abstract-algebra matrices ring-theory ideals







share|cite|improve this question









New contributor




Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 3




    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 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 at 22:25
















3














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 $3times3$ 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.






abstract-algebra matrices ring-theory ideals







share|cite|improve this question









New contributor




Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 3




    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 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 at 22:25














3












3








3







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 $3times3$ 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.






abstract-algebra matrices ring-theory ideals







share|cite|improve this question









New contributor




Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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 $3times3$ 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.






abstract-algebra matrices ring-theory ideals




abstract-algebra matrices ring-theory ideals






share|cite|improve this question









New contributor




Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 22:41







Alex













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asked Jan 4 at 21:20









AlexAlex

304




304




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Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 3




    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 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 at 22:25














  • 3




    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 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 at 22:25








3




3




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 at 21:26






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 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 at 22:25




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 at 22:25










1 Answer
1






active

oldest

votes


















1














The minimal right ideal of $R$ containing $rin R$ is $rR={rs:sin R}$. 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…






share|cite|improve this answer























  • that's it? just the product of those two matrices? I should've known it wasn't anything too complicated, thank you very much
    – Alex
    Jan 4 at 23:09










  • @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.
    – egreg
    Jan 4 at 23:13












  • no, sorry I have no idea where you're going with this
    – Alex
    Jan 4 at 23:38






  • 2




    @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 at 23:50










  • okay so the ideal will contain identity??
    – Alex
    Jan 5 at 0:14











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














The minimal right ideal of $R$ containing $rin R$ is $rR={rs:sin R}$. 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…






share|cite|improve this answer























  • that's it? just the product of those two matrices? I should've known it wasn't anything too complicated, thank you very much
    – Alex
    Jan 4 at 23:09










  • @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.
    – egreg
    Jan 4 at 23:13












  • no, sorry I have no idea where you're going with this
    – Alex
    Jan 4 at 23:38






  • 2




    @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 at 23:50










  • okay so the ideal will contain identity??
    – Alex
    Jan 5 at 0:14
















1














The minimal right ideal of $R$ containing $rin R$ is $rR={rs:sin R}$. 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…






share|cite|improve this answer























  • that's it? just the product of those two matrices? I should've known it wasn't anything too complicated, thank you very much
    – Alex
    Jan 4 at 23:09










  • @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.
    – egreg
    Jan 4 at 23:13












  • no, sorry I have no idea where you're going with this
    – Alex
    Jan 4 at 23:38






  • 2




    @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 at 23:50










  • okay so the ideal will contain identity??
    – Alex
    Jan 5 at 0:14














1












1








1






The minimal right ideal of $R$ containing $rin R$ is $rR={rs:sin R}$. 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…






share|cite|improve this answer














The minimal right ideal of $R$ containing $rin R$ is $rR={rs:sin R}$. 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…







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 4 at 23:15

























answered Jan 4 at 22:50









egregegreg

179k1485202




179k1485202












  • that's it? just the product of those two matrices? I should've known it wasn't anything too complicated, thank you very much
    – Alex
    Jan 4 at 23:09










  • @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.
    – egreg
    Jan 4 at 23:13












  • no, sorry I have no idea where you're going with this
    – Alex
    Jan 4 at 23:38






  • 2




    @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 at 23:50










  • okay so the ideal will contain identity??
    – Alex
    Jan 5 at 0:14


















  • that's it? just the product of those two matrices? I should've known it wasn't anything too complicated, thank you very much
    – Alex
    Jan 4 at 23:09










  • @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.
    – egreg
    Jan 4 at 23:13












  • no, sorry I have no idea where you're going with this
    – Alex
    Jan 4 at 23:38






  • 2




    @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 at 23:50










  • okay so the ideal will contain identity??
    – Alex
    Jan 5 at 0:14
















that's it? just the product of those two matrices? I should've known it wasn't anything too complicated, thank you very much
– Alex
Jan 4 at 23:09




that's it? just the product of those two matrices? I should've known it wasn't anything too complicated, thank you very much
– Alex
Jan 4 at 23:09












@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.
– egreg
Jan 4 at 23:13






@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.
– egreg
Jan 4 at 23:13














no, sorry I have no idea where you're going with this
– Alex
Jan 4 at 23:38




no, sorry I have no idea where you're going with this
– Alex
Jan 4 at 23:38




2




2




@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 at 23:50




@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 at 23:50












okay so the ideal will contain identity??
– Alex
Jan 5 at 0:14




okay so the ideal will contain identity??
– Alex
Jan 5 at 0:14










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