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About associate in abstract algebra [closed]

-2

$begingroup$

In $Bbb Z[sqrt-7]$, show that $$N(6+2 sqrt-7)=N(1+3 sqrt-7)$$
but $6+2 sqrt-7$ and $1+3 sqrt-7$ are not associates

I don't know how to solve this. Help me to solve this!

abstract-algebra ring-theory

share|cite|improve this question

edited Mar 24 at 6:04

Chinnapparaj R

6,34521029

asked Mar 24 at 5:37

Shivani ShahShivani Shah

6

$endgroup$

closed as off-topic by Arturo Magidin, Derek Holt, verret, YiFan, Shailesh Mar 25 at 0:24

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Arturo Magidin, Derek Holt, verret, YiFan, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  means they differ by a unit, I believe
  $endgroup$
  – Sean Nemetz
  Mar 24 at 5:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please use MathJax instead of posting an image.
  $endgroup$
  – Toby Mak
  Mar 24 at 5:53
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Shivani Shah $N(a+bsqrtd)=a^2-db^2.$ Show us your trying.
  $endgroup$
  – Michael Rozenberg
  Mar 24 at 6:21
  
  

  
  
  
  

add a comment | 

-2

$begingroup$

In $Bbb Z[sqrt-7]$, show that $$N(6+2 sqrt-7)=N(1+3 sqrt-7)$$
but $6+2 sqrt-7$ and $1+3 sqrt-7$ are not associates

I don't know how to solve this. Help me to solve this!

abstract-algebra ring-theory

share|cite|improve this question

edited Mar 24 at 6:04

Chinnapparaj R

6,34521029

asked Mar 24 at 5:37

Shivani ShahShivani Shah

6

$endgroup$

closed as off-topic by Arturo Magidin, Derek Holt, verret, YiFan, Shailesh Mar 25 at 0:24

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Arturo Magidin, Derek Holt, verret, YiFan, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  means they differ by a unit, I believe
  $endgroup$
  – Sean Nemetz
  Mar 24 at 5:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please use MathJax instead of posting an image.
  $endgroup$
  – Toby Mak
  Mar 24 at 5:53
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Shivani Shah $N(a+bsqrtd)=a^2-db^2.$ Show us your trying.
  $endgroup$
  – Michael Rozenberg
  Mar 24 at 6:21
  
  

  
  
  
  

add a comment | 

$begingroup$

In $Bbb Z[sqrt-7]$, show that $$N(6+2 sqrt-7)=N(1+3 sqrt-7)$$
but $6+2 sqrt-7$ and $1+3 sqrt-7$ are not associates

I don't know how to solve this. Help me to solve this!

abstract-algebra ring-theory

share|cite|improve this question

edited Mar 24 at 6:04

Chinnapparaj R

6,34521029

asked Mar 24 at 5:37

Shivani ShahShivani Shah

6

$endgroup$

share|cite|improve this question

edited Mar 24 at 6:04

Chinnapparaj R

6,34521029

asked Mar 24 at 5:37

Shivani ShahShivani Shah

6

share|cite|improve this question

edited Mar 24 at 6:04

Chinnapparaj R

6,34521029

asked Mar 24 at 5:37

Shivani ShahShivani Shah

6

edited Mar 24 at 6:04

Chinnapparaj R

6,34521029

edited Mar 24 at 6:04

Chinnapparaj R

6,34521029

asked Mar 24 at 5:37

Shivani ShahShivani Shah

6

asked Mar 24 at 5:37

Shivani ShahShivani Shah

6

1 Answer
1

active

oldest

votes

1

$begingroup$

Hints:

• $N(a+b sqrt-7)=vert a^2+7b^2 vert$


• If both are associates, then $$6+2 sqrt-7=u cdot[ 1+3 sqrt-7]$$ where $u$ is a unit in $Bbb Z[sqrt-7]$. But the only possible units in $Bbb Z[sqrt-7]$ are........? and so......?

share|cite|improve this answer

answered Mar 24 at 6:23

Chinnapparaj RChinnapparaj R

6,34521029

$endgroup$

add a comment | 

1

$begingroup$

Hints:

• $N(a+b sqrt-7)=vert a^2+7b^2 vert$


• If both are associates, then $$6+2 sqrt-7=u cdot[ 1+3 sqrt-7]$$ where $u$ is a unit in $Bbb Z[sqrt-7]$. But the only possible units in $Bbb Z[sqrt-7]$ are........? and so......?

share|cite|improve this answer

answered Mar 24 at 6:23

Chinnapparaj RChinnapparaj R

6,34521029

$endgroup$

add a comment | 

1

$begingroup$

Hints:

• $N(a+b sqrt-7)=vert a^2+7b^2 vert$


• If both are associates, then $$6+2 sqrt-7=u cdot[ 1+3 sqrt-7]$$ where $u$ is a unit in $Bbb Z[sqrt-7]$. But the only possible units in $Bbb Z[sqrt-7]$ are........? and so......?

share|cite|improve this answer

answered Mar 24 at 6:23

Chinnapparaj RChinnapparaj R

6,34521029

$endgroup$

add a comment | 

$begingroup$

Hints:

• $N(a+b sqrt-7)=vert a^2+7b^2 vert$


• If both are associates, then $$6+2 sqrt-7=u cdot[ 1+3 sqrt-7]$$ where $u$ is a unit in $Bbb Z[sqrt-7]$. But the only possible units in $Bbb Z[sqrt-7]$ are........? and so......?

share|cite|improve this answer

answered Mar 24 at 6:23

Chinnapparaj RChinnapparaj R

6,34521029

$endgroup$

share|cite|improve this answer

answered Mar 24 at 6:23

Chinnapparaj RChinnapparaj R

6,34521029

answered Mar 24 at 6:23

Chinnapparaj RChinnapparaj R

6,34521029

answered Mar 24 at 6:23

Chinnapparaj RChinnapparaj R

6,34521029