Number of maximal ideals of a commutative ring with unity which is not a field.
$begingroup$
Let $R$ be a commutative ring with unity 1 which is not a field. Let $Isubset R$ be a proper ideal such that every element of $R$ not in $I$ is invertible in $R$. Then find the number of maximal ideals of $R$?
Now since $I$ does not contain any invertible element so $1$ does not belong to $I$, so $Inot=R$. But how to count the number of maximal ideals? I want hint to solve this problem. Can anyone provide me a hint please? Thank you.
abstract-algebra ring-theory maximal-and-prime-ideals
asked May 24 '18 at 6:36
Kushal BhuyanKushal Bhuyan
5,08021246
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add a comment |
$begingroup$
Let $R$ be a commutative ring with unity 1 which is not a field. Let $Isubset R$ be a proper ideal such that every element of $R$ not in $I$ is invertible in $R$. Then find the number of maximal ideals of $R$?
Now since $I$ does not contain any invertible element so $1$ does not belong to $I$, so $Inot=R$. But how to count the number of maximal ideals? I want hint to solve this problem. Can anyone provide me a hint please? Thank you.
abstract-algebra ring-theory maximal-and-prime-ideals
asked May 24 '18 at 6:36
Kushal BhuyanKushal Bhuyan
5,08021246
$endgroup$
add a comment |
$begingroup$
Let $R$ be a commutative ring with unity 1 which is not a field. Let $Isubset R$ be a proper ideal such that every element of $R$ not in $I$ is invertible in $R$. Then find the number of maximal ideals of $R$?
Now since $I$ does not contain any invertible element so $1$ does not belong to $I$, so $Inot=R$. But how to count the number of maximal ideals? I want hint to solve this problem. Can anyone provide me a hint please? Thank you.
abstract-algebra ring-theory maximal-and-prime-ideals
asked May 24 '18 at 6:36
Kushal BhuyanKushal Bhuyan
5,08021246
$endgroup$
Let $R$ be a commutative ring with unity 1 which is not a field. Let $Isubset R$ be a proper ideal such that every element of $R$ not in $I$ is invertible in $R$. Then find the number of maximal ideals of $R$?
Now since $I$ does not contain any invertible element so $1$ does not belong to $I$, so $Inot=R$. But how to count the number of maximal ideals? I want hint to solve this problem. Can anyone provide me a hint please? Thank you.
abstract-algebra ring-theory maximal-and-prime-ideals
abstract-algebra ring-theory maximal-and-prime-ideals
asked May 24 '18 at 6:36
Kushal BhuyanKushal Bhuyan
5,08021246
asked May 24 '18 at 6:36
Kushal BhuyanKushal Bhuyan
5,08021246
asked May 24 '18 at 6:36
Kushal BhuyanKushal Bhuyan
5,08021246
asked May 24 '18 at 6:36
Kushal BhuyanKushal Bhuyan
5,08021246
asked May 24 '18 at 6:36
Kushal BhuyanKushal Bhuyan
5,08021246
5,08021246
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Firstly, note that a proper ideal $J$ cannot contain invertible elements: in this case it would contain also 1, thus everything; but it was a proper ideal!
Now let $m$ be a maximal ideal. Suppose $m$ is not contained in $I$: by the assumptions, $m$ contains an invertible element, absurd for the above observations (a maximal ideal is proper!).
But then $m$ is ... and by maximality...
Can you conclude the proof?
answered May 24 '18 at 6:58
frame95frame95
533211
$endgroup$
$begingroup$
thanks. got it now.
$endgroup$
– Kushal Bhuyan
May 24 '18 at 7:10
add a comment |
$begingroup$
Let $A$ be any ideal, if $xin A$ but not in $I$ then $x$ is invertible but this is absurd. As a result, any ideal is inside $I$. There is only one maximal ideal which is $I$.
answered May 24 '18 at 6:40
BenBen
2,1831123
$endgroup$
$begingroup$
why "x is invertible but this is absurd"? Can't an element of an ideal invertible?
$endgroup$
– Kushal Bhuyan
May 24 '18 at 6:49
$begingroup$
Your ideal is proper
$endgroup$
– Tutankhamun
Aug 23 '18 at 10:12
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Firstly, note that a proper ideal $J$ cannot contain invertible elements: in this case it would contain also 1, thus everything; but it was a proper ideal!
Now let $m$ be a maximal ideal. Suppose $m$ is not contained in $I$: by the assumptions, $m$ contains an invertible element, absurd for the above observations (a maximal ideal is proper!).
But then $m$ is ... and by maximality...
Can you conclude the proof?
answered May 24 '18 at 6:58
frame95frame95
533211
$endgroup$
$begingroup$
thanks. got it now.
$endgroup$
– Kushal Bhuyan
May 24 '18 at 7:10
add a comment |
$begingroup$
Firstly, note that a proper ideal $J$ cannot contain invertible elements: in this case it would contain also 1, thus everything; but it was a proper ideal!
Now let $m$ be a maximal ideal. Suppose $m$ is not contained in $I$: by the assumptions, $m$ contains an invertible element, absurd for the above observations (a maximal ideal is proper!).
But then $m$ is ... and by maximality...
Can you conclude the proof?
answered May 24 '18 at 6:58
frame95frame95
533211
$endgroup$
$begingroup$
thanks. got it now.
$endgroup$
– Kushal Bhuyan
May 24 '18 at 7:10
add a comment |
$begingroup$
Firstly, note that a proper ideal $J$ cannot contain invertible elements: in this case it would contain also 1, thus everything; but it was a proper ideal!
Now let $m$ be a maximal ideal. Suppose $m$ is not contained in $I$: by the assumptions, $m$ contains an invertible element, absurd for the above observations (a maximal ideal is proper!).
But then $m$ is ... and by maximality...
Can you conclude the proof?
answered May 24 '18 at 6:58
frame95frame95
533211
$endgroup$
Firstly, note that a proper ideal $J$ cannot contain invertible elements: in this case it would contain also 1, thus everything; but it was a proper ideal!
Now let $m$ be a maximal ideal. Suppose $m$ is not contained in $I$: by the assumptions, $m$ contains an invertible element, absurd for the above observations (a maximal ideal is proper!).
But then $m$ is ... and by maximality...
Can you conclude the proof?
answered May 24 '18 at 6:58
frame95frame95
533211
answered May 24 '18 at 6:58
frame95frame95
533211
answered May 24 '18 at 6:58
frame95frame95
533211
answered May 24 '18 at 6:58
frame95frame95
533211
533211
$begingroup$
thanks. got it now.
$endgroup$
– Kushal Bhuyan
May 24 '18 at 7:10
add a comment |
$begingroup$
thanks. got it now.
$endgroup$
– Kushal Bhuyan
May 24 '18 at 7:10
$begingroup$
thanks. got it now.
$endgroup$
– Kushal Bhuyan
May 24 '18 at 7:10
$begingroup$
thanks. got it now.
$endgroup$
– Kushal Bhuyan
May 24 '18 at 7:10
add a comment |
$begingroup$
Let $A$ be any ideal, if $xin A$ but not in $I$ then $x$ is invertible but this is absurd. As a result, any ideal is inside $I$. There is only one maximal ideal which is $I$.
answered May 24 '18 at 6:40
BenBen
2,1831123
$endgroup$
$begingroup$
why "x is invertible but this is absurd"? Can't an element of an ideal invertible?
$endgroup$
– Kushal Bhuyan
May 24 '18 at 6:49
$begingroup$
Your ideal is proper
$endgroup$
– Tutankhamun
Aug 23 '18 at 10:12
add a comment |
$begingroup$
Let $A$ be any ideal, if $xin A$ but not in $I$ then $x$ is invertible but this is absurd. As a result, any ideal is inside $I$. There is only one maximal ideal which is $I$.
answered May 24 '18 at 6:40
BenBen
2,1831123
$endgroup$
$begingroup$
why "x is invertible but this is absurd"? Can't an element of an ideal invertible?
$endgroup$
– Kushal Bhuyan
May 24 '18 at 6:49
$begingroup$
Your ideal is proper
$endgroup$
– Tutankhamun
Aug 23 '18 at 10:12
add a comment |
$begingroup$
Let $A$ be any ideal, if $xin A$ but not in $I$ then $x$ is invertible but this is absurd. As a result, any ideal is inside $I$. There is only one maximal ideal which is $I$.
answered May 24 '18 at 6:40
BenBen
2,1831123
$endgroup$
Let $A$ be any ideal, if $xin A$ but not in $I$ then $x$ is invertible but this is absurd. As a result, any ideal is inside $I$. There is only one maximal ideal which is $I$.
answered May 24 '18 at 6:40
BenBen
2,1831123
answered May 24 '18 at 6:40
BenBen
2,1831123
answered May 24 '18 at 6:40
BenBen
2,1831123
answered May 24 '18 at 6:40
BenBen
2,1831123
2,1831123
$begingroup$
why "x is invertible but this is absurd"? Can't an element of an ideal invertible?
$endgroup$
– Kushal Bhuyan
May 24 '18 at 6:49
$begingroup$
Your ideal is proper
$endgroup$
– Tutankhamun
Aug 23 '18 at 10:12
add a comment |
$begingroup$
why "x is invertible but this is absurd"? Can't an element of an ideal invertible?
$endgroup$
– Kushal Bhuyan
May 24 '18 at 6:49
$begingroup$
Your ideal is proper
$endgroup$
– Tutankhamun
Aug 23 '18 at 10:12
$begingroup$
why "x is invertible but this is absurd"? Can't an element of an ideal invertible?
$endgroup$
– Kushal Bhuyan
May 24 '18 at 6:49
$begingroup$
why "x is invertible but this is absurd"? Can't an element of an ideal invertible?
$endgroup$
– Kushal Bhuyan
May 24 '18 at 6:49
$begingroup$
Your ideal is proper
$endgroup$
– Tutankhamun
Aug 23 '18 at 10:12
$begingroup$
Your ideal is proper
$endgroup$
– Tutankhamun
Aug 23 '18 at 10:12
add a comment |
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