Can $Rx$ be contained in $I$ for some nonzero $xin S$, if $I:=langle Rsetminus Srangle$?

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Let $R$ be any ring and let any set of elements $Ssubset R$ (proper) be such that $0_R,1_Rin S$ with some other elements of $R$. Now, define $$I:=langle Rsetminus Srangle=langleain R: anotin Srangle.$$ That is, $I$ is generated by the set of elements not in $S$.

Question: Is there a possibility some left ideal $Rx$ to be contained in $I$ for some nonzero $xin S$?

ring-theory modules

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asked Jul 27 at 9:33

Sulayman

18717

add a comment | 

up vote
1
down vote

favorite

Let $R$ be any ring and let any set of elements $Ssubset R$ (proper) be such that $0_R,1_Rin S$ with some other elements of $R$. Now, define $$I:=langle Rsetminus Srangle=langleain R: anotin Srangle.$$ That is, $I$ is generated by the set of elements not in $S$.

Question: Is there a possibility some left ideal $Rx$ to be contained in $I$ for some nonzero $xin S$?

ring-theory modules

share|cite|improve this question

asked Jul 27 at 9:33

Sulayman

18717

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Let $R$ be any ring and let any set of elements $Ssubset R$ (proper) be such that $0_R,1_Rin S$ with some other elements of $R$. Now, define $$I:=langle Rsetminus Srangle=langleain R: anotin Srangle.$$ That is, $I$ is generated by the set of elements not in $S$.

Question: Is there a possibility some left ideal $Rx$ to be contained in $I$ for some nonzero $xin S$?

ring-theory modules

share|cite|improve this question

asked Jul 27 at 9:33

Sulayman

18717

Let $R$ be any ring and let any set of elements $Ssubset R$ (proper) be such that $0_R,1_Rin S$ with some other elements of $R$. Now, define $$I:=langle Rsetminus Srangle=langleain R: anotin Srangle.$$ That is, $I$ is generated by the set of elements not in $S$.

Question: Is there a possibility some left ideal $Rx$ to be contained in $I$ for some nonzero $xin S$?

ring-theory modules

share|cite|improve this question

asked Jul 27 at 9:33

Sulayman

18717

share|cite|improve this question

share|cite|improve this question

asked Jul 27 at 9:33

Sulayman

18717

asked Jul 27 at 9:33

Sulayman

18717

asked Jul 27 at 9:33

Sulayman

18717

18717

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
0
down vote



accepted

Let $R=mathbbZ$ and $S=mathbbZsetminus2$. Then $I=2mathbbZ$ and $4in Icap S$.

share|cite|improve this answer

edited Jul 27 at 14:40

answered Jul 27 at 10:21

egreg

164k1180187

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Suppose that there is an extra condition on $Rsetminus S$ that the unity can not be in $Rsetminus S$. i.e., $1notinRsetminus S$.
  – Sulayman
  Jul 27 at 12:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Sulayman You mean that $I$ is assumed to be proper?
  – egreg
  Jul 27 at 12:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, at all times, it is assumed that $Isubset R$ as a left ideal
  – Sulayman
  Jul 27 at 14:34
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Sulayman Then it's even easier.
  – egreg
  Jul 27 at 14:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, I am waiting for a concrete conclusion on whether there can be any possibility of some left ideal $Rx$ to be contained in $I$ for some nonzero $xin S$?
  – Sulayman
  Jul 27 at 14:49
  

  
  
  
  

 | 
show 4 more comments

up vote
0
down vote

Let $R=mathbbR^3times 3$ and let $S=leftE_i,jmid 1leq i,jleq 3rightcup left0_R,1_Rright$ where $E_i,j$ is the matrix with zeroes everywhere and a $1$ at position $(i,j)$. Then $I=leftlangle Rsetminus Srightrangle=R$. Clearly $RE_1,1$ is contained in $I$.

share|cite|improve this answer

answered Jul 27 at 10:04

Mathematician 42

8,17111437

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Suppose that there is an extra condition on $Rsetminus S$ that the unity can not be in $Rsetminus S$. i.e., $1notinRsetminus S$.
  – Sulayman
  Jul 27 at 12:10
  
  

  
  
  
  

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

up vote
0
down vote



accepted

Let $R=mathbbZ$ and $S=mathbbZsetminus2$. Then $I=2mathbbZ$ and $4in Icap S$.

share|cite|improve this answer

edited Jul 27 at 14:40

answered Jul 27 at 10:21

egreg

164k1180187

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Suppose that there is an extra condition on $Rsetminus S$ that the unity can not be in $Rsetminus S$. i.e., $1notinRsetminus S$.
  – Sulayman
  Jul 27 at 12:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Sulayman You mean that $I$ is assumed to be proper?
  – egreg
  Jul 27 at 12:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, at all times, it is assumed that $Isubset R$ as a left ideal
  – Sulayman
  Jul 27 at 14:34
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Sulayman Then it's even easier.
  – egreg
  Jul 27 at 14:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, I am waiting for a concrete conclusion on whether there can be any possibility of some left ideal $Rx$ to be contained in $I$ for some nonzero $xin S$?
  – Sulayman
  Jul 27 at 14:49
  

  
  
  
  

 | 
show 4 more comments

up vote
0
down vote



accepted

Let $R=mathbbZ$ and $S=mathbbZsetminus2$. Then $I=2mathbbZ$ and $4in Icap S$.

share|cite|improve this answer

edited Jul 27 at 14:40

answered Jul 27 at 10:21

egreg

164k1180187

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Suppose that there is an extra condition on $Rsetminus S$ that the unity can not be in $Rsetminus S$. i.e., $1notinRsetminus S$.
  – Sulayman
  Jul 27 at 12:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Sulayman You mean that $I$ is assumed to be proper?
  – egreg
  Jul 27 at 12:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, at all times, it is assumed that $Isubset R$ as a left ideal
  – Sulayman
  Jul 27 at 14:34
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Sulayman Then it's even easier.
  – egreg
  Jul 27 at 14:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, I am waiting for a concrete conclusion on whether there can be any possibility of some left ideal $Rx$ to be contained in $I$ for some nonzero $xin S$?
  – Sulayman
  Jul 27 at 14:49
  

  
  
  
  

 | 
show 4 more comments

up vote
0
down vote



accepted

up vote
0
down vote



accepted

Let $R=mathbbZ$ and $S=mathbbZsetminus2$. Then $I=2mathbbZ$ and $4in Icap S$.

share|cite|improve this answer

edited Jul 27 at 14:40

answered Jul 27 at 10:21

egreg

164k1180187

Let $R=mathbbZ$ and $S=mathbbZsetminus2$. Then $I=2mathbbZ$ and $4in Icap S$.

share|cite|improve this answer

edited Jul 27 at 14:40

answered Jul 27 at 10:21

egreg

164k1180187

share|cite|improve this answer

share|cite|improve this answer

edited Jul 27 at 14:40

edited Jul 27 at 14:40

edited Jul 27 at 14:40

answered Jul 27 at 10:21

egreg

164k1180187

answered Jul 27 at 10:21

egreg

164k1180187

answered Jul 27 at 10:21

egreg

164k1180187

164k1180187

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Suppose that there is an extra condition on $Rsetminus S$ that the unity can not be in $Rsetminus S$. i.e., $1notinRsetminus S$.
  – Sulayman
  Jul 27 at 12:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Sulayman You mean that $I$ is assumed to be proper?
  – egreg
  Jul 27 at 12:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, at all times, it is assumed that $Isubset R$ as a left ideal
  – Sulayman
  Jul 27 at 14:34
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Sulayman Then it's even easier.
  – egreg
  Jul 27 at 14:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, I am waiting for a concrete conclusion on whether there can be any possibility of some left ideal $Rx$ to be contained in $I$ for some nonzero $xin S$?
  – Sulayman
  Jul 27 at 14:49
  

  
  
  
  

 | 
show 4 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Suppose that there is an extra condition on $Rsetminus S$ that the unity can not be in $Rsetminus S$. i.e., $1notinRsetminus S$.
  – Sulayman
  Jul 27 at 12:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Sulayman You mean that $I$ is assumed to be proper?
  – egreg
  Jul 27 at 12:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, at all times, it is assumed that $Isubset R$ as a left ideal
  – Sulayman
  Jul 27 at 14:34
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Sulayman Then it's even easier.
  – egreg
  Jul 27 at 14:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, I am waiting for a concrete conclusion on whether there can be any possibility of some left ideal $Rx$ to be contained in $I$ for some nonzero $xin S$?
  – Sulayman
  Jul 27 at 14:49
  

  
  
  
  

Suppose that there is an extra condition on $Rsetminus S$ that the unity can not be in $Rsetminus S$. i.e., $1notinRsetminus S$.
– Sulayman
Jul 27 at 12:12

Suppose that there is an extra condition on $Rsetminus S$ that the unity can not be in $Rsetminus S$. i.e., $1notinRsetminus S$.
– Sulayman
Jul 27 at 12:12

@Sulayman You mean that $I$ is assumed to be proper?
– egreg
Jul 27 at 12:16

@Sulayman You mean that $I$ is assumed to be proper?
– egreg
Jul 27 at 12:16

Yes, at all times, it is assumed that $Isubset R$ as a left ideal
– Sulayman
Jul 27 at 14:34

Yes, at all times, it is assumed that $Isubset R$ as a left ideal
– Sulayman
Jul 27 at 14:34

@Sulayman Then it's even easier.
– egreg
Jul 27 at 14:40

@Sulayman Then it's even easier.
– egreg
Jul 27 at 14:40

Thanks, I am waiting for a concrete conclusion on whether there can be any possibility of some left ideal $Rx$ to be contained in $I$ for some nonzero $xin S$?
– Sulayman
Jul 27 at 14:49

Thanks, I am waiting for a concrete conclusion on whether there can be any possibility of some left ideal $Rx$ to be contained in $I$ for some nonzero $xin S$?
– Sulayman
Jul 27 at 14:49

 | 
show 4 more comments

up vote
0
down vote

Let $R=mathbbR^3times 3$ and let $S=leftE_i,jmid 1leq i,jleq 3rightcup left0_R,1_Rright$ where $E_i,j$ is the matrix with zeroes everywhere and a $1$ at position $(i,j)$. Then $I=leftlangle Rsetminus Srightrangle=R$. Clearly $RE_1,1$ is contained in $I$.

share|cite|improve this answer

answered Jul 27 at 10:04

Mathematician 42

8,17111437

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Suppose that there is an extra condition on $Rsetminus S$ that the unity can not be in $Rsetminus S$. i.e., $1notinRsetminus S$.
  – Sulayman
  Jul 27 at 12:10
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

Let $R=mathbbR^3times 3$ and let $S=leftE_i,jmid 1leq i,jleq 3rightcup left0_R,1_Rright$ where $E_i,j$ is the matrix with zeroes everywhere and a $1$ at position $(i,j)$. Then $I=leftlangle Rsetminus Srightrangle=R$. Clearly $RE_1,1$ is contained in $I$.

share|cite|improve this answer

answered Jul 27 at 10:04

Mathematician 42

8,17111437

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Suppose that there is an extra condition on $Rsetminus S$ that the unity can not be in $Rsetminus S$. i.e., $1notinRsetminus S$.
  – Sulayman
  Jul 27 at 12:10
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

Let $R=mathbbR^3times 3$ and let $S=leftE_i,jmid 1leq i,jleq 3rightcup left0_R,1_Rright$ where $E_i,j$ is the matrix with zeroes everywhere and a $1$ at position $(i,j)$. Then $I=leftlangle Rsetminus Srightrangle=R$. Clearly $RE_1,1$ is contained in $I$.

share|cite|improve this answer

answered Jul 27 at 10:04

Mathematician 42

8,17111437

Let $R=mathbbR^3times 3$ and let $S=leftE_i,jmid 1leq i,jleq 3rightcup left0_R,1_Rright$ where $E_i,j$ is the matrix with zeroes everywhere and a $1$ at position $(i,j)$. Then $I=leftlangle Rsetminus Srightrangle=R$. Clearly $RE_1,1$ is contained in $I$.

share|cite|improve this answer

answered Jul 27 at 10:04

Mathematician 42

8,17111437

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 10:04

Mathematician 42

8,17111437

answered Jul 27 at 10:04

Mathematician 42

8,17111437

answered Jul 27 at 10:04

Mathematician 42

8,17111437

8,17111437

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Suppose that there is an extra condition on $Rsetminus S$ that the unity can not be in $Rsetminus S$. i.e., $1notinRsetminus S$.
  – Sulayman
  Jul 27 at 12:10
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Suppose that there is an extra condition on $Rsetminus S$ that the unity can not be in $Rsetminus S$. i.e., $1notinRsetminus S$.
  – Sulayman
  Jul 27 at 12:10
  
  

  
  
  
  

Suppose that there is an extra condition on $Rsetminus S$ that the unity can not be in $Rsetminus S$. i.e., $1notinRsetminus S$.
– Sulayman
Jul 27 at 12:10

Suppose that there is an extra condition on $Rsetminus S$ that the unity can not be in $Rsetminus S$. i.e., $1notinRsetminus S$.
– Sulayman
Jul 27 at 12:10

add a comment | 

 

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