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Idempotents and cyclic codes

1

$begingroup$

Let $C_1 = langle e_1(x) rangle$, $C_2 = langle e_2(x) rangle$ cyclic codes, where $e_1(x)$ and $e_2(x)$ are idempotents.

I know what cyclic codes and idempotents are, but why can one deduce the following: $C_1 subset C_2 Leftrightarrow e_1(x) e_2(x) = e_1(x)$?

abstract-algebra coding-theory

share|cite|improve this question

edited Mar 28 at 11:38

Siddharth Bhat

3,1821918

asked Mar 28 at 11:22

JohnDJohnD

338112

$endgroup$

add a comment | 

1

$begingroup$

Let $C_1 = langle e_1(x) rangle$, $C_2 = langle e_2(x) rangle$ cyclic codes, where $e_1(x)$ and $e_2(x)$ are idempotents.

I know what cyclic codes and idempotents are, but why can one deduce the following: $C_1 subset C_2 Leftrightarrow e_1(x) e_2(x) = e_1(x)$?

abstract-algebra coding-theory

share|cite|improve this question

edited Mar 28 at 11:38

Siddharth Bhat

3,1821918

asked Mar 28 at 11:22

JohnDJohnD

338112

$endgroup$

add a comment | 

$begingroup$

Let $C_1 = langle e_1(x) rangle$, $C_2 = langle e_2(x) rangle$ cyclic codes, where $e_1(x)$ and $e_2(x)$ are idempotents.

I know what cyclic codes and idempotents are, but why can one deduce the following: $C_1 subset C_2 Leftrightarrow e_1(x) e_2(x) = e_1(x)$?

abstract-algebra coding-theory

share|cite|improve this question

edited Mar 28 at 11:38

Siddharth Bhat

3,1821918

asked Mar 28 at 11:22

JohnDJohnD

338112

$endgroup$

share|cite|improve this question

edited Mar 28 at 11:38

Siddharth Bhat

3,1821918

asked Mar 28 at 11:22

JohnDJohnD

338112

share|cite|improve this question

edited Mar 28 at 11:38

Siddharth Bhat

3,1821918

asked Mar 28 at 11:22

JohnDJohnD

338112

edited Mar 28 at 11:38

Siddharth Bhat

3,1821918

edited Mar 28 at 11:38

Siddharth Bhat

3,1821918

asked Mar 28 at 11:22

JohnDJohnD

338112

asked Mar 28 at 11:22

JohnDJohnD

338112

2 Answers
2

active

oldest

votes

1

$begingroup$

If $C_1 subset C_2$, then note that $e_1 in C_1 in C_2$. Hence $e_1 e_2 = e_2 e_1 = e_1$ because $e_1$ is an idempotent for $C_1$, and we can interpret $e_2$ as an element of $C_1$

For the other direction, if $e_1 e_2 = e_1$, then let $x in C_1$. Now, $x = e_1 x = e_1 e_2 x = e_2 (e_1 x) = e_2 x$. Hence, $x = e_2 x$, and therefore $x in C_2$.

share|cite|improve this answer

answered Mar 28 at 11:33

Siddharth BhatSiddharth Bhat

3,1821918

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for this really fast answer! What about $C_1 cap C_2 = <e_1(x) e_2(x)> $ ? Why does this hold true?
  $endgroup$
  – JohnD
  Mar 28 at 11:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is from the properties of ideals. If $I = langle x rangle$, $J = langle y rangle$, then $I cap J = langle lcm(x, y) rangle$ in a principal ideal domain. You need to do some work to show that $lcm(e_1, e_2) = e_1e_2$ but that's the idea
  $endgroup$
  – Siddharth Bhat
  Mar 28 at 11:55
  
  

  
  
  
  

add a comment | 

1

$begingroup$

I believe you are talking about an idempotent $e=e(x)in F[x]/(x^n-1)$. The fact you are speaking of is actually true for idempotents in any commutative ring with identity.

Suppose $e,f$ are two idempotents in a commutative ring $R$.

Then it is elementary to show that $(e)cap (1-e)=0$ and $(f)cap (1-f)=0$.

Now if $(e)subseteq (f)$, then $e-ef=e(1-f)in (f)cap (1-f)=0$. Therefore $e=ef$.

In the other direction, suppose $e=ef$: then clearly $ein (f)$ and $(e)subseteq (f)$.

share|cite|improve this answer

answered Mar 28 at 15:03

rschwiebrschwieb

107k12103252

$endgroup$

add a comment | 

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1

$begingroup$

If $C_1 subset C_2$, then note that $e_1 in C_1 in C_2$. Hence $e_1 e_2 = e_2 e_1 = e_1$ because $e_1$ is an idempotent for $C_1$, and we can interpret $e_2$ as an element of $C_1$

For the other direction, if $e_1 e_2 = e_1$, then let $x in C_1$. Now, $x = e_1 x = e_1 e_2 x = e_2 (e_1 x) = e_2 x$. Hence, $x = e_2 x$, and therefore $x in C_2$.

share|cite|improve this answer

answered Mar 28 at 11:33

Siddharth BhatSiddharth Bhat

3,1821918

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for this really fast answer! What about $C_1 cap C_2 = <e_1(x) e_2(x)> $ ? Why does this hold true?
  $endgroup$
  – JohnD
  Mar 28 at 11:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is from the properties of ideals. If $I = langle x rangle$, $J = langle y rangle$, then $I cap J = langle lcm(x, y) rangle$ in a principal ideal domain. You need to do some work to show that $lcm(e_1, e_2) = e_1e_2$ but that's the idea
  $endgroup$
  – Siddharth Bhat
  Mar 28 at 11:55
  
  

  
  
  
  

add a comment | 

1

$begingroup$

If $C_1 subset C_2$, then note that $e_1 in C_1 in C_2$. Hence $e_1 e_2 = e_2 e_1 = e_1$ because $e_1$ is an idempotent for $C_1$, and we can interpret $e_2$ as an element of $C_1$

For the other direction, if $e_1 e_2 = e_1$, then let $x in C_1$. Now, $x = e_1 x = e_1 e_2 x = e_2 (e_1 x) = e_2 x$. Hence, $x = e_2 x$, and therefore $x in C_2$.

share|cite|improve this answer

answered Mar 28 at 11:33

Siddharth BhatSiddharth Bhat

3,1821918

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for this really fast answer! What about $C_1 cap C_2 = <e_1(x) e_2(x)> $ ? Why does this hold true?
  $endgroup$
  – JohnD
  Mar 28 at 11:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is from the properties of ideals. If $I = langle x rangle$, $J = langle y rangle$, then $I cap J = langle lcm(x, y) rangle$ in a principal ideal domain. You need to do some work to show that $lcm(e_1, e_2) = e_1e_2$ but that's the idea
  $endgroup$
  – Siddharth Bhat
  Mar 28 at 11:55
  
  

  
  
  
  

add a comment | 

$begingroup$

If $C_1 subset C_2$, then note that $e_1 in C_1 in C_2$. Hence $e_1 e_2 = e_2 e_1 = e_1$ because $e_1$ is an idempotent for $C_1$, and we can interpret $e_2$ as an element of $C_1$

For the other direction, if $e_1 e_2 = e_1$, then let $x in C_1$. Now, $x = e_1 x = e_1 e_2 x = e_2 (e_1 x) = e_2 x$. Hence, $x = e_2 x$, and therefore $x in C_2$.

share|cite|improve this answer

answered Mar 28 at 11:33

Siddharth BhatSiddharth Bhat

3,1821918

$endgroup$

share|cite|improve this answer

answered Mar 28 at 11:33

Siddharth BhatSiddharth Bhat

3,1821918

answered Mar 28 at 11:33

Siddharth BhatSiddharth Bhat

3,1821918

answered Mar 28 at 11:33

Siddharth BhatSiddharth Bhat

3,1821918

1

$begingroup$

I believe you are talking about an idempotent $e=e(x)in F[x]/(x^n-1)$. The fact you are speaking of is actually true for idempotents in any commutative ring with identity.

Suppose $e,f$ are two idempotents in a commutative ring $R$.

Then it is elementary to show that $(e)cap (1-e)=0$ and $(f)cap (1-f)=0$.

Now if $(e)subseteq (f)$, then $e-ef=e(1-f)in (f)cap (1-f)=0$. Therefore $e=ef$.

In the other direction, suppose $e=ef$: then clearly $ein (f)$ and $(e)subseteq (f)$.

share|cite|improve this answer

answered Mar 28 at 15:03

rschwiebrschwieb

107k12103252

$endgroup$

add a comment | 

1

$begingroup$

I believe you are talking about an idempotent $e=e(x)in F[x]/(x^n-1)$. The fact you are speaking of is actually true for idempotents in any commutative ring with identity.

Suppose $e,f$ are two idempotents in a commutative ring $R$.

Then it is elementary to show that $(e)cap (1-e)=0$ and $(f)cap (1-f)=0$.

Now if $(e)subseteq (f)$, then $e-ef=e(1-f)in (f)cap (1-f)=0$. Therefore $e=ef$.

In the other direction, suppose $e=ef$: then clearly $ein (f)$ and $(e)subseteq (f)$.

share|cite|improve this answer

answered Mar 28 at 15:03

rschwiebrschwieb

107k12103252

$endgroup$

add a comment | 

$begingroup$

I believe you are talking about an idempotent $e=e(x)in F[x]/(x^n-1)$. The fact you are speaking of is actually true for idempotents in any commutative ring with identity.

Suppose $e,f$ are two idempotents in a commutative ring $R$.

Then it is elementary to show that $(e)cap (1-e)=0$ and $(f)cap (1-f)=0$.

Now if $(e)subseteq (f)$, then $e-ef=e(1-f)in (f)cap (1-f)=0$. Therefore $e=ef$.

In the other direction, suppose $e=ef$: then clearly $ein (f)$ and $(e)subseteq (f)$.

share|cite|improve this answer

answered Mar 28 at 15:03

rschwiebrschwieb

107k12103252

$endgroup$

share|cite|improve this answer

answered Mar 28 at 15:03

rschwiebrschwieb

107k12103252

answered Mar 28 at 15:03

rschwiebrschwieb

107k12103252

answered Mar 28 at 15:03

rschwiebrschwieb

107k12103252