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Prove that $3$ is prime in $mathbb Q(i)$ but not in $mathbb Q(√6)$ [closed]

3

$begingroup$

I m confused a little bit helps me to get out of this
As I know if the norm of a given number is prime then that number is prime

algebraic-number-theory

share|cite|improve this question

edited Mar 26 at 10:12

YuiTo Cheng

2,43841037

asked Mar 26 at 9:02

lone aazam lone aazam

162

$endgroup$

closed as off-topic by Saad, Dietrich Burde, Riccardo.Alestra, YiFan, José Carlos Santos Mar 26 at 13:45

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." – Saad, Dietrich Burde, Riccardo.Alestra, YiFan, José Carlos Santos

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

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  This has been explained nicely on this site. First $3$ is prime in $Bbb Q(i)$, see here. In $Bbb Q(sqrt6)$ however, $3=(3+sqrt6)(3-sqrt6)$. For an almost duplicate of this, see here.
  $endgroup$
  – Dietrich Burde
  Mar 26 at 9:05
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  You probably mean $mathbbZ[i]$ and $mathbbZ[sqrt6]$ instead?
  $endgroup$
  – Henno Brandsma
  Mar 26 at 9:08
  

  
  
  
  

add a comment | 

3

$begingroup$

I m confused a little bit helps me to get out of this
As I know if the norm of a given number is prime then that number is prime

algebraic-number-theory

share|cite|improve this question

edited Mar 26 at 10:12

YuiTo Cheng

2,43841037

asked Mar 26 at 9:02

lone aazam lone aazam

162

$endgroup$

closed as off-topic by Saad, Dietrich Burde, Riccardo.Alestra, YiFan, José Carlos Santos Mar 26 at 13:45

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." – Saad, Dietrich Burde, Riccardo.Alestra, YiFan, José Carlos Santos

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

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  This has been explained nicely on this site. First $3$ is prime in $Bbb Q(i)$, see here. In $Bbb Q(sqrt6)$ however, $3=(3+sqrt6)(3-sqrt6)$. For an almost duplicate of this, see here.
  $endgroup$
  – Dietrich Burde
  Mar 26 at 9:05
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  You probably mean $mathbbZ[i]$ and $mathbbZ[sqrt6]$ instead?
  $endgroup$
  – Henno Brandsma
  Mar 26 at 9:08
  

  
  
  
  

add a comment | 

$begingroup$

I m confused a little bit helps me to get out of this
As I know if the norm of a given number is prime then that number is prime

algebraic-number-theory

share|cite|improve this question

edited Mar 26 at 10:12

YuiTo Cheng

2,43841037

asked Mar 26 at 9:02

lone aazam lone aazam

162

$endgroup$

share|cite|improve this question

edited Mar 26 at 10:12

YuiTo Cheng

2,43841037

asked Mar 26 at 9:02

lone aazam lone aazam

162

share|cite|improve this question

edited Mar 26 at 10:12

YuiTo Cheng

2,43841037

asked Mar 26 at 9:02

lone aazam lone aazam

162

edited Mar 26 at 10:12

YuiTo Cheng

2,43841037

edited Mar 26 at 10:12

YuiTo Cheng

2,43841037

asked Mar 26 at 9:02

lone aazam lone aazam

162

asked Mar 26 at 9:02

lone aazam lone aazam

162

1 Answer
1

active

oldest

votes

2

$begingroup$

We're working in general rings, not just in $mathbbZ$, which is where you first encountered the notion of a prime element.

An element $p$ of a (commutative unitary) ring $R$ is prime iff $forall x,y in R: p | xy implies p|x text or p|y$. Because we quantify over $R$ this very much depends on the ring we're working in. And (confusingly?) $3=1+1+1$ is a member of $mathbbZ$, $mathbbZ[i]$ and $mathbbZ[sqrt6]$ and many more rings and for all of these we can ask "is $3$ prime in that ring"? and we get different answers for different rings.

See the comments for links to previous answers, which I won't copy here.

share|cite|improve this answer

edited Mar 26 at 10:19

J. W. Tanner

4,8471420

answered Mar 26 at 9:13

Henno BrandsmaHenno Brandsma

116k349127

$endgroup$

add a comment | 

2

$begingroup$

We're working in general rings, not just in $mathbbZ$, which is where you first encountered the notion of a prime element.

An element $p$ of a (commutative unitary) ring $R$ is prime iff $forall x,y in R: p | xy implies p|x text or p|y$. Because we quantify over $R$ this very much depends on the ring we're working in. And (confusingly?) $3=1+1+1$ is a member of $mathbbZ$, $mathbbZ[i]$ and $mathbbZ[sqrt6]$ and many more rings and for all of these we can ask "is $3$ prime in that ring"? and we get different answers for different rings.

See the comments for links to previous answers, which I won't copy here.

share|cite|improve this answer

edited Mar 26 at 10:19

J. W. Tanner

4,8471420

answered Mar 26 at 9:13

Henno BrandsmaHenno Brandsma

116k349127

$endgroup$

add a comment | 

2

$begingroup$

We're working in general rings, not just in $mathbbZ$, which is where you first encountered the notion of a prime element.

An element $p$ of a (commutative unitary) ring $R$ is prime iff $forall x,y in R: p | xy implies p|x text or p|y$. Because we quantify over $R$ this very much depends on the ring we're working in. And (confusingly?) $3=1+1+1$ is a member of $mathbbZ$, $mathbbZ[i]$ and $mathbbZ[sqrt6]$ and many more rings and for all of these we can ask "is $3$ prime in that ring"? and we get different answers for different rings.

See the comments for links to previous answers, which I won't copy here.

share|cite|improve this answer

edited Mar 26 at 10:19

J. W. Tanner

4,8471420

answered Mar 26 at 9:13

Henno BrandsmaHenno Brandsma

116k349127

$endgroup$

add a comment | 

$begingroup$

We're working in general rings, not just in $mathbbZ$, which is where you first encountered the notion of a prime element.

An element $p$ of a (commutative unitary) ring $R$ is prime iff $forall x,y in R: p | xy implies p|x text or p|y$. Because we quantify over $R$ this very much depends on the ring we're working in. And (confusingly?) $3=1+1+1$ is a member of $mathbbZ$, $mathbbZ[i]$ and $mathbbZ[sqrt6]$ and many more rings and for all of these we can ask "is $3$ prime in that ring"? and we get different answers for different rings.

See the comments for links to previous answers, which I won't copy here.

share|cite|improve this answer

edited Mar 26 at 10:19

J. W. Tanner

4,8471420

answered Mar 26 at 9:13

Henno BrandsmaHenno Brandsma

116k349127

$endgroup$

share|cite|improve this answer

edited Mar 26 at 10:19

J. W. Tanner

4,8471420

answered Mar 26 at 9:13

Henno BrandsmaHenno Brandsma

116k349127

edited Mar 26 at 10:19

J. W. Tanner

4,8471420

edited Mar 26 at 10:19

J. W. Tanner

4,8471420

answered Mar 26 at 9:13

Henno BrandsmaHenno Brandsma

116k349127

answered Mar 26 at 9:13

Henno BrandsmaHenno Brandsma

116k349127