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Is Kaplansky's radical same as Jacobson radical?

Jacobson radical of $R[X]$, where $R$ is domainJacobson Radical and Finite Dimensional AlgebraJacobson radical subset of maximal idealJacobson Radical $J(R)$ is a proper idealJacobson radical of a commutative ringThe Jacobson radical of an IdealA question about Semisimple ring and their Jacobson radicalJacobson Radical of External Direct Sum of RingsIn homomorphic image of Jacobson ring Nilradical is equal to the Jacobson radical.Proving that the Jacobson radical of a (not necessarily unital) ring is contained in the intersection of all left modular ideals.

6

$begingroup$

For a ring $R$, Kaplansky defines $xcirc y= x+y+xy$ and an element $x$ is s.t.b. right quasi-regular (r.q.r) if $xcirc y=0$ for some element $y$. He defines radical to be set theoretic join (sum) of all right quasi regular (r.q.r) ideals where an ideal is r.q.r if all its elements are r.q.r.

Jacobson radical $J(R)$ is the intersection of maximal left (or right) ideals of a ring.

Does either of these radical implies other or are they completely unrelated? I can't see any connection here.

abstract-algebra ring-theory

share|cite|improve this question

edited Mar 15 at 1:59

blabla

asked Mar 15 at 1:55

blablablabla

481211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you explain if you want all these rings to have identity? You would have to for the intersection definition to be precise, and it simplifies the discussion using quasiregular elements a bit
  $endgroup$
  – rschwieb
  Mar 15 at 13:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @rschwieb yes rings have identity
  $endgroup$
  – blabla
  Mar 16 at 6:55
  

  
  
  
  

add a comment | 

6

$begingroup$

For a ring $R$, Kaplansky defines $xcirc y= x+y+xy$ and an element $x$ is s.t.b. right quasi-regular (r.q.r) if $xcirc y=0$ for some element $y$. He defines radical to be set theoretic join (sum) of all right quasi regular (r.q.r) ideals where an ideal is r.q.r if all its elements are r.q.r.

Jacobson radical $J(R)$ is the intersection of maximal left (or right) ideals of a ring.

Does either of these radical implies other or are they completely unrelated? I can't see any connection here.

abstract-algebra ring-theory

share|cite|improve this question

edited Mar 15 at 1:59

blabla

asked Mar 15 at 1:55

blablablabla

481211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you explain if you want all these rings to have identity? You would have to for the intersection definition to be precise, and it simplifies the discussion using quasiregular elements a bit
  $endgroup$
  – rschwieb
  Mar 15 at 13:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @rschwieb yes rings have identity
  $endgroup$
  – blabla
  Mar 16 at 6:55
  

  
  
  
  

add a comment | 

$begingroup$

For a ring $R$, Kaplansky defines $xcirc y= x+y+xy$ and an element $x$ is s.t.b. right quasi-regular (r.q.r) if $xcirc y=0$ for some element $y$. He defines radical to be set theoretic join (sum) of all right quasi regular (r.q.r) ideals where an ideal is r.q.r if all its elements are r.q.r.

Jacobson radical $J(R)$ is the intersection of maximal left (or right) ideals of a ring.

Does either of these radical implies other or are they completely unrelated? I can't see any connection here.

abstract-algebra ring-theory

share|cite|improve this question

edited Mar 15 at 1:59

blabla

asked Mar 15 at 1:55

blablablabla

481211

$endgroup$

share|cite|improve this question

edited Mar 15 at 1:59

blabla

asked Mar 15 at 1:55

blablablabla

481211

share|cite|improve this question

edited Mar 15 at 1:59

blabla

asked Mar 15 at 1:55

blablablabla

481211

asked Mar 15 at 1:55

blablablabla

481211

asked Mar 15 at 1:55

blablablabla

481211

1 Answer
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votes

3

$begingroup$

I'm assuming for the time being that you intended for all these rings to have identity, because the maximal left ideal definition needs to be refined when identity is gone. I'm pretty sure everything holds without identity, it just takes more care.

Here's some observations to work through:

1. 
   $x$ is r.q.r as you defined iff $x+1$ is right invertible in $R$.


2. The statement that "For $xin R$, there exists $y$ such that $x+y-xy=0$" is equivalent to $x-1$ being right invertible.


3. 
   

   These statements are equivalent:

   
   
   
   1. 
      $xr-1$ is right invertible for every $rin R$
   
   
   2. 
      $xr+1$ is right invertible for every $rin R$.
   
   
   3. 
      $1+xr$ is right invertible for every $rin R$.
   
   
   4. 
      $1-xr$ is right invertible for every $rin R$.
   
   



4. From the wiki article you can see that an element $xin R$ is in $J(R)$ iff $1-xr$ is right invertible for all $rin R$.

So from the last point, you can see that $J(R)$ itself has the property that all elements are r.q.r., and furthermore any element $x$ which lies in a r.q.r. ideal must have the property "$1-xr$ is right invertible for every $rin R$" and therefore such an ideal is contained in $J(R)$.

Then the sum of all r.q.r. ideals would have to be equal to $J(R)$.

share|cite|improve this answer

answered Mar 15 at 14:10

rschwiebrschwieb

107k12103251

$endgroup$

add a comment | 

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3

$begingroup$

I'm assuming for the time being that you intended for all these rings to have identity, because the maximal left ideal definition needs to be refined when identity is gone. I'm pretty sure everything holds without identity, it just takes more care.

Here's some observations to work through:

1. 
   $x$ is r.q.r as you defined iff $x+1$ is right invertible in $R$.


2. The statement that "For $xin R$, there exists $y$ such that $x+y-xy=0$" is equivalent to $x-1$ being right invertible.


3. 
   

   These statements are equivalent:

   
   
   
   1. 
      $xr-1$ is right invertible for every $rin R$
   
   
   2. 
      $xr+1$ is right invertible for every $rin R$.
   
   
   3. 
      $1+xr$ is right invertible for every $rin R$.
   
   
   4. 
      $1-xr$ is right invertible for every $rin R$.
   
   



4. From the wiki article you can see that an element $xin R$ is in $J(R)$ iff $1-xr$ is right invertible for all $rin R$.

So from the last point, you can see that $J(R)$ itself has the property that all elements are r.q.r., and furthermore any element $x$ which lies in a r.q.r. ideal must have the property "$1-xr$ is right invertible for every $rin R$" and therefore such an ideal is contained in $J(R)$.

Then the sum of all r.q.r. ideals would have to be equal to $J(R)$.

share|cite|improve this answer

answered Mar 15 at 14:10

rschwiebrschwieb

107k12103251

$endgroup$

add a comment | 

3

$begingroup$

I'm assuming for the time being that you intended for all these rings to have identity, because the maximal left ideal definition needs to be refined when identity is gone. I'm pretty sure everything holds without identity, it just takes more care.

Here's some observations to work through:

1. 
   $x$ is r.q.r as you defined iff $x+1$ is right invertible in $R$.


2. The statement that "For $xin R$, there exists $y$ such that $x+y-xy=0$" is equivalent to $x-1$ being right invertible.


3. 
   

   These statements are equivalent:

   
   
   
   1. 
      $xr-1$ is right invertible for every $rin R$
   
   
   2. 
      $xr+1$ is right invertible for every $rin R$.
   
   
   3. 
      $1+xr$ is right invertible for every $rin R$.
   
   
   4. 
      $1-xr$ is right invertible for every $rin R$.
   
   



4. From the wiki article you can see that an element $xin R$ is in $J(R)$ iff $1-xr$ is right invertible for all $rin R$.

So from the last point, you can see that $J(R)$ itself has the property that all elements are r.q.r., and furthermore any element $x$ which lies in a r.q.r. ideal must have the property "$1-xr$ is right invertible for every $rin R$" and therefore such an ideal is contained in $J(R)$.

Then the sum of all r.q.r. ideals would have to be equal to $J(R)$.

share|cite|improve this answer

answered Mar 15 at 14:10

rschwiebrschwieb

107k12103251

$endgroup$

add a comment | 

$begingroup$

I'm assuming for the time being that you intended for all these rings to have identity, because the maximal left ideal definition needs to be refined when identity is gone. I'm pretty sure everything holds without identity, it just takes more care.

Here's some observations to work through:

1. 
   $x$ is r.q.r as you defined iff $x+1$ is right invertible in $R$.


2. The statement that "For $xin R$, there exists $y$ such that $x+y-xy=0$" is equivalent to $x-1$ being right invertible.


3. 
   

   These statements are equivalent:

   
   
   
   1. 
      $xr-1$ is right invertible for every $rin R$
   
   
   2. 
      $xr+1$ is right invertible for every $rin R$.
   
   
   3. 
      $1+xr$ is right invertible for every $rin R$.
   
   
   4. 
      $1-xr$ is right invertible for every $rin R$.
   
   



4. From the wiki article you can see that an element $xin R$ is in $J(R)$ iff $1-xr$ is right invertible for all $rin R$.

So from the last point, you can see that $J(R)$ itself has the property that all elements are r.q.r., and furthermore any element $x$ which lies in a r.q.r. ideal must have the property "$1-xr$ is right invertible for every $rin R$" and therefore such an ideal is contained in $J(R)$.

Then the sum of all r.q.r. ideals would have to be equal to $J(R)$.

share|cite|improve this answer

answered Mar 15 at 14:10

rschwiebrschwieb

107k12103251

$endgroup$

share|cite|improve this answer

answered Mar 15 at 14:10

rschwiebrschwieb

107k12103251

answered Mar 15 at 14:10

rschwiebrschwieb

107k12103251

answered Mar 15 at 14:10

rschwiebrschwieb

107k12103251