Ttyjukl

Know when to turn notes upside-down(eighth notes, sixteen notes, etc.)

Why do Australian milk farmers need to protest supermarkets' milk price?

Where is the 1/8 CR apprentice in Volo's Guide to Monsters?

Ban on all campaign finance?

My adviser wants to be the first author

Dot in front of file

Welcoming 2019 Pi day: How to draw the letter π?

Informing my boss about remarks from a nasty colleague

How to generate globally unique ids for different tables of the same database?

What is IP squat space

Does splitting a potentially monolithic application into several smaller ones help prevent bugs?

Why must traveling waves have the same amplitude to form a standing wave?

Can hydraulic brake levers get hot when brakes overheat?

Russian cases: A few examples, I'm really confused

An Accountant Seeks the Help of a Mathematician

Professor being mistaken for a grad student

Will a pinhole camera work with instant film?

Possible Leak In Concrete

Is it possible / allowed to upcast ritual spells?

Using "wallow" verb with object

How could a female member of a species produce eggs unto death?

What are some nice/clever ways to introduce the tonic's dominant seventh chord?

Why would a flight no longer considered airworthy be redirected like this?

How is the Swiss post e-voting system supposed to work, and how was it wrong?

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

4

$begingroup$

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 1 hour ago

Math LoverMath Lover

1,024315

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 | 

4

$begingroup$

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 1 hour ago

Math LoverMath Lover

1,024315

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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

share|cite|improve this question

asked 1 hour ago

Math LoverMath Lover

1,024315

$endgroup$

share|cite|improve this question

asked 1 hour ago

Math LoverMath Lover

1,024315

share|cite|improve this question

asked 1 hour ago

Math LoverMath Lover

1,024315

asked 1 hour ago

Math LoverMath Lover

1,024315

asked 1 hour ago

Math LoverMath Lover

1,024315

2 Answers
2

active

oldest

votes

4

$begingroup$

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 1 hour ago

answered 1 hour ago

Alex MathersAlex Mathers

11k21344

$endgroup$

add a comment | 

0

$begingroup$

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 $$(1,1)-(2,0)=(-1,1)notin M_2$$

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 $a,b$.

share|cite|improve this answer

answered 43 mins ago

user647486user647486

111

$endgroup$

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3148803%2fdoes-this-property-of-comaximal-ideals-always-holds%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

4

$begingroup$

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 1 hour ago

answered 1 hour ago

Alex MathersAlex Mathers

11k21344

$endgroup$

add a comment | 

4

$begingroup$

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 1 hour ago

answered 1 hour ago

Alex MathersAlex Mathers

11k21344

$endgroup$

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 1 hour ago

answered 1 hour ago

Alex MathersAlex Mathers

11k21344

$endgroup$

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

Alex MathersAlex Mathers

11k21344

answered 1 hour ago

Alex MathersAlex Mathers

11k21344

answered 1 hour ago

Alex MathersAlex Mathers

11k21344

0

$begingroup$

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 $$(1,1)-(2,0)=(-1,1)notin M_2$$

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 $a,b$.

share|cite|improve this answer

answered 43 mins ago

user647486user647486

111

$endgroup$

add a comment | 

0

$begingroup$

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 $$(1,1)-(2,0)=(-1,1)notin M_2$$

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 $a,b$.

share|cite|improve this answer

answered 43 mins ago

user647486user647486

111

$endgroup$

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 $$(1,1)-(2,0)=(-1,1)notin M_2$$

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 $a,b$.

share|cite|improve this answer

answered 43 mins ago

user647486user647486

111

$endgroup$

share|cite|improve this answer

answered 43 mins ago

user647486user647486

111

answered 43 mins ago

user647486user647486

111

answered 43 mins ago

user647486user647486

111