Does this property of comaximal ideals always holds?Question on Comaximal IdealsUnital commutative ring and distinct maximal ideals.Where does the proof for commutative rings break down in the non-commutative ring when showing only two ideals implies the ring is a field?Direct-Sum Decomposition of an Artinian moduleProve that $m_1m_2ldots m_r=n_1n_2ldots n_s$ implies $r=s$ for distinct maximal idealsQuestion about maximal ideals in a commutative Artinian ringA property of associated prime idealsThe meaning of idempotents corresponding the standard basis in direct product of fieldsAre non-coprime ideals always contained in some prime ideal?Product of ideals equals intersection but they are not comaximal
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Does this property of comaximal ideals always holds?
Question on Comaximal IdealsUnital commutative ring and distinct maximal ideals.Where does the proof for commutative rings break down in the non-commutative ring when showing only two ideals implies the ring is a field?Direct-Sum Decomposition of an Artinian moduleProve that $m_1m_2ldots m_r=n_1n_2ldots n_s$ implies $r=s$ for distinct maximal idealsQuestion about maximal ideals in a commutative Artinian ringA property of associated prime idealsThe meaning of idempotents corresponding the standard basis in direct product of fieldsAre non-coprime ideals always contained in some prime ideal?Product of ideals equals intersection but they are not comaximal
$begingroup$
I am reading a paper in which the following result is used but I can’t see the proof of this.
let $R$ be a commutative ring with only two maximal ideals say $M_1$ and $M_2$. Suppose $m_1 in M_1$ be such that $m_1 notin M_2$ then can be always find $m_2 in M_2$ such that $m_1+m_2=1$
Any ideas?
abstract-algebra ring-theory commutative-algebra maximal-and-prime-ideals
asked 2 hours ago
Math LoverMath Lover
1,029315
$endgroup$
add a comment |
$begingroup$
I am reading a paper in which the following result is used but I can’t see the proof of this.
let $R$ be a commutative ring with only two maximal ideals say $M_1$ and $M_2$. Suppose $m_1 in M_1$ be such that $m_1 notin M_2$ then can be always find $m_2 in M_2$ such that $m_1+m_2=1$
Any ideas?
abstract-algebra ring-theory commutative-algebra maximal-and-prime-ideals
asked 2 hours ago
Math LoverMath Lover
1,029315
$endgroup$
$begingroup$
Consider the ideal generated by $M_2$ and $m_1$, this ideal must be $R=(1)$ since $M_2$ is maximal
$endgroup$
– B.Swan
1 hour ago
$begingroup$
@B.Swan this approach doesn't work, to see why try writing out the details
$endgroup$
– Alex Mathers
1 hour ago
1
$begingroup$
Set $I=(M_2 cup m_1) $, the ideal generated by $M_2$ and $m_1$. Elements of $I$ have the form $x+rm_1$, where $x in M_2$ and $r in R$. Since $m_1 notin M_2$ and $M_2$ maximal, it follows $I=R$. Thus there exists $s in R$ with $1=x+sm_1$. And I guess one gets stuck here. Sorry for the wrong approach and thanks for pointing it out.
$endgroup$
– B.Swan
1 hour ago
add a comment |
$begingroup$
I am reading a paper in which the following result is used but I can’t see the proof of this.
let $R$ be a commutative ring with only two maximal ideals say $M_1$ and $M_2$. Suppose $m_1 in M_1$ be such that $m_1 notin M_2$ then can be always find $m_2 in M_2$ such that $m_1+m_2=1$
Any ideas?
abstract-algebra ring-theory commutative-algebra maximal-and-prime-ideals
asked 2 hours ago
Math LoverMath Lover
1,029315
$endgroup$
I am reading a paper in which the following result is used but I can’t see the proof of this.
let $R$ be a commutative ring with only two maximal ideals say $M_1$ and $M_2$. Suppose $m_1 in M_1$ be such that $m_1 notin M_2$ then can be always find $m_2 in M_2$ such that $m_1+m_2=1$
Any ideas?
abstract-algebra ring-theory commutative-algebra maximal-and-prime-ideals
abstract-algebra ring-theory commutative-algebra maximal-and-prime-ideals
asked 2 hours ago
Math LoverMath Lover
1,029315
asked 2 hours ago
Math LoverMath Lover
1,029315
asked 2 hours ago
Math LoverMath Lover
1,029315
asked 2 hours ago
Math LoverMath Lover
1,029315
asked 2 hours ago
Math LoverMath Lover
1,029315
1,029315
$begingroup$
Consider the ideal generated by $M_2$ and $m_1$, this ideal must be $R=(1)$ since $M_2$ is maximal
$endgroup$
– B.Swan
1 hour ago
$begingroup$
@B.Swan this approach doesn't work, to see why try writing out the details
$endgroup$
– Alex Mathers
1 hour ago
1
$begingroup$
Set $I=(M_2 cup m_1) $, the ideal generated by $M_2$ and $m_1$. Elements of $I$ have the form $x+rm_1$, where $x in M_2$ and $r in R$. Since $m_1 notin M_2$ and $M_2$ maximal, it follows $I=R$. Thus there exists $s in R$ with $1=x+sm_1$. And I guess one gets stuck here. Sorry for the wrong approach and thanks for pointing it out.
$endgroup$
– B.Swan
1 hour ago
add a comment |
$begingroup$
Consider the ideal generated by $M_2$ and $m_1$, this ideal must be $R=(1)$ since $M_2$ is maximal
$endgroup$
– B.Swan
1 hour ago
$begingroup$
@B.Swan this approach doesn't work, to see why try writing out the details
$endgroup$
– Alex Mathers
1 hour ago
1
$begingroup$
Set $I=(M_2 cup m_1) $, the ideal generated by $M_2$ and $m_1$. Elements of $I$ have the form $x+rm_1$, where $x in M_2$ and $r in R$. Since $m_1 notin M_2$ and $M_2$ maximal, it follows $I=R$. Thus there exists $s in R$ with $1=x+sm_1$. And I guess one gets stuck here. Sorry for the wrong approach and thanks for pointing it out.
$endgroup$
– B.Swan
1 hour ago
$begingroup$
Consider the ideal generated by $M_2$ and $m_1$, this ideal must be $R=(1)$ since $M_2$ is maximal
$endgroup$
– B.Swan
1 hour ago
$begingroup$
Consider the ideal generated by $M_2$ and $m_1$, this ideal must be $R=(1)$ since $M_2$ is maximal
$endgroup$
– B.Swan
1 hour ago
$begingroup$
@B.Swan this approach doesn't work, to see why try writing out the details
$endgroup$
– Alex Mathers
1 hour ago
$begingroup$
@B.Swan this approach doesn't work, to see why try writing out the details
$endgroup$
– Alex Mathers
1 hour ago
1
1
$begingroup$
Set $I=(M_2 cup m_1) $, the ideal generated by $M_2$ and $m_1$. Elements of $I$ have the form $x+rm_1$, where $x in M_2$ and $r in R$. Since $m_1 notin M_2$ and $M_2$ maximal, it follows $I=R$. Thus there exists $s in R$ with $1=x+sm_1$. And I guess one gets stuck here. Sorry for the wrong approach and thanks for pointing it out.
$endgroup$
– B.Swan
1 hour ago
$begingroup$
Set $I=(M_2 cup m_1) $, the ideal generated by $M_2$ and $m_1$. Elements of $I$ have the form $x+rm_1$, where $x in M_2$ and $r in R$. Since $m_1 notin M_2$ and $M_2$ maximal, it follows $I=R$. Thus there exists $s in R$ with $1=x+sm_1$. And I guess one gets stuck here. Sorry for the wrong approach and thanks for pointing it out.
$endgroup$
– B.Swan
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
First notice that $1-m_1$ cannot be a unit, because this would imply $m_1$ is in the Jacobson radical of $R$, and in particular we would have $m_1in M_2$.
Now it follows that the ideal of $R$ generated by $1-m_1$ must be contained in a maximal ideal, but it cannot be contained in $M_1$ because then it would follow that $1in M_1$. Thus this ideal is contained in $M_2$ (the only other maximal ideal), i.e. you get $1-m_1in M_2$.
Edit: I think my reasoning for $1-m_1$ not being a unit is wrong (it seems we would need that $1-m_1x$ is a unit for every $xin R$ to conclude $m_1$ is in the Jacobson radical). The rest of the argument goes through, so I'm going to leave my answer up for a while in hopes that somebody can help figure that part out.
answered 1 hour ago
Alex MathersAlex Mathers
11.1k21344
$endgroup$
add a comment |
$begingroup$
Take $R=mathbbQtimesmathbbQ$, $M_1=mathbbQtimes0$, $M_2=0timesmathbbQ$, and $m_1=(2,0)in M_1setminus M_2$. Then $(1,1)inmathbbQtimesmathbbQ$ satisfies that $$(1,1)-(2,0)=(-1,1)notin M_2$$
Therefore, that property is not satisfied in general.
Maybe the property that they are really using is that there exist $ain M_1$ and $bin M_2$ such that $a+b=1$. Not arbitrary $a,b$. This other property is immediate by using the maximality of $M_1$ and $M_2$, which implies that $M_1+M_2=R$.
answered 1 hour ago
user647486user647486
413
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
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active
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active
oldest
votes
$begingroup$
First notice that $1-m_1$ cannot be a unit, because this would imply $m_1$ is in the Jacobson radical of $R$, and in particular we would have $m_1in M_2$.
Now it follows that the ideal of $R$ generated by $1-m_1$ must be contained in a maximal ideal, but it cannot be contained in $M_1$ because then it would follow that $1in M_1$. Thus this ideal is contained in $M_2$ (the only other maximal ideal), i.e. you get $1-m_1in M_2$.
Edit: I think my reasoning for $1-m_1$ not being a unit is wrong (it seems we would need that $1-m_1x$ is a unit for every $xin R$ to conclude $m_1$ is in the Jacobson radical). The rest of the argument goes through, so I'm going to leave my answer up for a while in hopes that somebody can help figure that part out.
answered 1 hour ago
Alex MathersAlex Mathers
11.1k21344
$endgroup$
add a comment |
$begingroup$
First notice that $1-m_1$ cannot be a unit, because this would imply $m_1$ is in the Jacobson radical of $R$, and in particular we would have $m_1in M_2$.
Now it follows that the ideal of $R$ generated by $1-m_1$ must be contained in a maximal ideal, but it cannot be contained in $M_1$ because then it would follow that $1in M_1$. Thus this ideal is contained in $M_2$ (the only other maximal ideal), i.e. you get $1-m_1in M_2$.
Edit: I think my reasoning for $1-m_1$ not being a unit is wrong (it seems we would need that $1-m_1x$ is a unit for every $xin R$ to conclude $m_1$ is in the Jacobson radical). The rest of the argument goes through, so I'm going to leave my answer up for a while in hopes that somebody can help figure that part out.
answered 1 hour ago
Alex MathersAlex Mathers
11.1k21344
$endgroup$
add a comment |
$begingroup$
First notice that $1-m_1$ cannot be a unit, because this would imply $m_1$ is in the Jacobson radical of $R$, and in particular we would have $m_1in M_2$.
Now it follows that the ideal of $R$ generated by $1-m_1$ must be contained in a maximal ideal, but it cannot be contained in $M_1$ because then it would follow that $1in M_1$. Thus this ideal is contained in $M_2$ (the only other maximal ideal), i.e. you get $1-m_1in M_2$.
Edit: I think my reasoning for $1-m_1$ not being a unit is wrong (it seems we would need that $1-m_1x$ is a unit for every $xin R$ to conclude $m_1$ is in the Jacobson radical). The rest of the argument goes through, so I'm going to leave my answer up for a while in hopes that somebody can help figure that part out.
answered 1 hour ago
Alex MathersAlex Mathers
11.1k21344
$endgroup$
First notice that $1-m_1$ cannot be a unit, because this would imply $m_1$ is in the Jacobson radical of $R$, and in particular we would have $m_1in M_2$.
Now it follows that the ideal of $R$ generated by $1-m_1$ must be contained in a maximal ideal, but it cannot be contained in $M_1$ because then it would follow that $1in M_1$. Thus this ideal is contained in $M_2$ (the only other maximal ideal), i.e. you get $1-m_1in M_2$.
Edit: I think my reasoning for $1-m_1$ not being a unit is wrong (it seems we would need that $1-m_1x$ is a unit for every $xin R$ to conclude $m_1$ is in the Jacobson radical). The rest of the argument goes through, so I'm going to leave my answer up for a while in hopes that somebody can help figure that part out.
answered 1 hour ago
Alex MathersAlex Mathers
11.1k21344
edited 1 hour ago
answered 1 hour ago
Alex MathersAlex Mathers
11.1k21344
answered 1 hour ago
Alex MathersAlex Mathers
11.1k21344
answered 1 hour ago
Alex MathersAlex Mathers
11.1k21344
11.1k21344
add a comment |
add a comment |
$begingroup$
Take $R=mathbbQtimesmathbbQ$, $M_1=mathbbQtimes0$, $M_2=0timesmathbbQ$, and $m_1=(2,0)in M_1setminus M_2$. Then $(1,1)inmathbbQtimesmathbbQ$ satisfies that $$(1,1)-(2,0)=(-1,1)notin M_2$$
Therefore, that property is not satisfied in general.
Maybe the property that they are really using is that there exist $ain M_1$ and $bin M_2$ such that $a+b=1$. Not arbitrary $a,b$. This other property is immediate by using the maximality of $M_1$ and $M_2$, which implies that $M_1+M_2=R$.
answered 1 hour ago
user647486user647486
413
$endgroup$
add a comment |
$begingroup$
Take $R=mathbbQtimesmathbbQ$, $M_1=mathbbQtimes0$, $M_2=0timesmathbbQ$, and $m_1=(2,0)in M_1setminus M_2$. Then $(1,1)inmathbbQtimesmathbbQ$ satisfies that $$(1,1)-(2,0)=(-1,1)notin M_2$$
Therefore, that property is not satisfied in general.
Maybe the property that they are really using is that there exist $ain M_1$ and $bin M_2$ such that $a+b=1$. Not arbitrary $a,b$. This other property is immediate by using the maximality of $M_1$ and $M_2$, which implies that $M_1+M_2=R$.
answered 1 hour ago
user647486user647486
413
$endgroup$
add a comment |
$begingroup$
Take $R=mathbbQtimesmathbbQ$, $M_1=mathbbQtimes0$, $M_2=0timesmathbbQ$, and $m_1=(2,0)in M_1setminus M_2$. Then $(1,1)inmathbbQtimesmathbbQ$ satisfies that $$(1,1)-(2,0)=(-1,1)notin M_2$$
Therefore, that property is not satisfied in general.
Maybe the property that they are really using is that there exist $ain M_1$ and $bin M_2$ such that $a+b=1$. Not arbitrary $a,b$. This other property is immediate by using the maximality of $M_1$ and $M_2$, which implies that $M_1+M_2=R$.
answered 1 hour ago
user647486user647486
413
$endgroup$
Take $R=mathbbQtimesmathbbQ$, $M_1=mathbbQtimes0$, $M_2=0timesmathbbQ$, and $m_1=(2,0)in M_1setminus M_2$. Then $(1,1)inmathbbQtimesmathbbQ$ satisfies that $$(1,1)-(2,0)=(-1,1)notin M_2$$
Therefore, that property is not satisfied in general.
Maybe the property that they are really using is that there exist $ain M_1$ and $bin M_2$ such that $a+b=1$. Not arbitrary $a,b$. This other property is immediate by using the maximality of $M_1$ and $M_2$, which implies that $M_1+M_2=R$.
answered 1 hour ago
user647486user647486
413
edited 15 mins ago
answered 1 hour ago
user647486user647486
413
answered 1 hour ago
user647486user647486
413
answered 1 hour ago
user647486user647486
413
413
add a comment |
add a comment |
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$begingroup$
Consider the ideal generated by $M_2$ and $m_1$, this ideal must be $R=(1)$ since $M_2$ is maximal
$endgroup$
– B.Swan
1 hour ago
$begingroup$
@B.Swan this approach doesn't work, to see why try writing out the details
$endgroup$
– Alex Mathers
1 hour ago
1
$begingroup$
Set $I=(M_2 cup m_1) $, the ideal generated by $M_2$ and $m_1$. Elements of $I$ have the form $x+rm_1$, where $x in M_2$ and $r in R$. Since $m_1 notin M_2$ and $M_2$ maximal, it follows $I=R$. Thus there exists $s in R$ with $1=x+sm_1$. And I guess one gets stuck here. Sorry for the wrong approach and thanks for pointing it out.
$endgroup$
– B.Swan
1 hour ago