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5

1

$begingroup$

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

share|cite|improve this question

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
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @B.Swan this approach doesn't work, to see why try writing out the details
  $endgroup$
  – Alex Mathers
  2 hours 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
  2 hours ago
  
  
  

  
  
  
  

add a comment | 

5

1

$begingroup$

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

share|cite|improve this question

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
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @B.Swan this approach doesn't work, to see why try writing out the details
  $endgroup$
  – Alex Mathers
  2 hours 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
  2 hours 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

share|cite|improve this question

asked 2 hours ago

Math LoverMath Lover

1,029315

$endgroup$

share|cite|improve this question

asked 2 hours ago

Math LoverMath Lover

1,029315

share|cite|improve this question

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

2 Answers
2

active

oldest

votes

5

$begingroup$

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.

share|cite|improve this answer

edited 2 hours ago

answered 2 hours ago

Alex MathersAlex Mathers

11.1k21344

$endgroup$

add a comment | 

4

$begingroup$

Take $R=mathbb{Q}timesmathbb{Q}$, $M_1=mathbb{Q}times{0}$, $M_2={0}timesmathbb{Q}$, and $m_1=(2,0)in M_1setminus M_2$. Then $(1,1)inmathbb{Q}timesmathbb{Q}$ 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 $M_1+M_2=R$.

share|cite|improve this answer

edited 38 mins ago

answered 1 hour ago

user647486user647486

613

$endgroup$

add a comment | 

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5

$begingroup$

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.

share|cite|improve this answer

edited 2 hours ago

answered 2 hours ago

Alex MathersAlex Mathers

11.1k21344

$endgroup$

add a comment | 

5

$begingroup$

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.

share|cite|improve this answer

edited 2 hours ago

answered 2 hours 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.

share|cite|improve this answer

edited 2 hours ago

answered 2 hours ago

Alex MathersAlex Mathers

11.1k21344

$endgroup$

share|cite|improve this answer

edited 2 hours ago

answered 2 hours ago

Alex MathersAlex Mathers

11.1k21344

answered 2 hours ago

Alex MathersAlex Mathers

11.1k21344

answered 2 hours ago

Alex MathersAlex Mathers

11.1k21344

4

$begingroup$

Take $R=mathbb{Q}timesmathbb{Q}$, $M_1=mathbb{Q}times{0}$, $M_2={0}timesmathbb{Q}$, and $m_1=(2,0)in M_1setminus M_2$. Then $(1,1)inmathbb{Q}timesmathbb{Q}$ 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 $M_1+M_2=R$.

share|cite|improve this answer

edited 38 mins ago

answered 1 hour ago

user647486user647486

613

$endgroup$

add a comment | 

4

$begingroup$

Take $R=mathbb{Q}timesmathbb{Q}$, $M_1=mathbb{Q}times{0}$, $M_2={0}timesmathbb{Q}$, and $m_1=(2,0)in M_1setminus M_2$. Then $(1,1)inmathbb{Q}timesmathbb{Q}$ 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 $M_1+M_2=R$.

share|cite|improve this answer

edited 38 mins ago

answered 1 hour ago

user647486user647486

613

$endgroup$

add a comment | 

$begingroup$

Take $R=mathbb{Q}timesmathbb{Q}$, $M_1=mathbb{Q}times{0}$, $M_2={0}timesmathbb{Q}$, and $m_1=(2,0)in M_1setminus M_2$. Then $(1,1)inmathbb{Q}timesmathbb{Q}$ 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 $M_1+M_2=R$.

share|cite|improve this answer

edited 38 mins ago

answered 1 hour ago

user647486user647486

613

$endgroup$

share|cite|improve this answer

edited 38 mins ago

answered 1 hour ago

user647486user647486

613

answered 1 hour ago

user647486user647486

613

answered 1 hour ago

user647486user647486

613