Prove that $3$ is prime in $mathbb Q(i)$ but not in $mathbb Q(√6)$ [closed] Announcing the...
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Prove that $3$ is prime in $mathbb Q(i)$ but not in $mathbb Q(√6)$ [closed]
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Prove 3 is prime in the ring $Bbb{Z[i]}$.Show that $3$ is not a prime in $mathbb Q[sqrt{7}]$Correspondence between valuations and prime idealsNorm of Prime IdealFalse proof: no element has a prime norm in the ring $mathbb Z[zeta_p]$possible norms of prime ideals in the class group of $K=mathbb{Q}(sqrt{-21})$confusion with calculating the ideal class group of a quadratic fieldCalculating the class numberFinding all the ideals of a given prime normDoes the prime ideal with its norm a power of prime exist? where $p$ is a prime number.If $mathcal{O}_{Bbb{Q}(sqrt{d})}$ has class number $2$ or higher, does that mean $sqrt{d}$ is irreducible but not prime?The prime ideal in a number ring
$begingroup$
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
edited Mar 26 at 10:12
YuiTo Cheng
2,65641037
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.
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
edited Mar 26 at 10:12
YuiTo Cheng
2,65641037
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 $Bbb Q(sqrt{6})$ however, $3=(3+sqrt{6})(3-sqrt{6})$. For an almost duplicate of this, see here.
$endgroup$
– Dietrich Burde
Mar 26 at 9:05
3
$begingroup$
You probably mean $mathbb{Z}[i]$ and $mathbb{Z}[sqrt{6}]$ 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
edited Mar 26 at 10:12
YuiTo Cheng
2,65641037
asked Mar 26 at 9:02
lone aazam lone aazam
162
$endgroup$
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
algebraic-number-theory
edited Mar 26 at 10:12
YuiTo Cheng
2,65641037
asked Mar 26 at 9:02
lone aazam lone aazam
162
edited Mar 26 at 10:12
YuiTo Cheng
2,65641037
asked Mar 26 at 9:02
lone aazam lone aazam
162
edited Mar 26 at 10:12
YuiTo Cheng
2,65641037
edited Mar 26 at 10:12
YuiTo Cheng
2,65641037
edited Mar 26 at 10:12
YuiTo Cheng
2,65641037
2,65641037
asked Mar 26 at 9:02
lone aazam lone aazam
162
asked Mar 26 at 9:02
lone aazam lone aazam
162
asked Mar 26 at 9:02
lone aazam lone aazam
162
162
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.
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 $Bbb Q(sqrt{6})$ however, $3=(3+sqrt{6})(3-sqrt{6})$. For an almost duplicate of this, see here.
$endgroup$
– Dietrich Burde
Mar 26 at 9:05
3
$begingroup$
You probably mean $mathbb{Z}[i]$ and $mathbb{Z}[sqrt{6}]$ instead?
$endgroup$
– Henno Brandsma
Mar 26 at 9:08
add a comment |
5
$begingroup$
This has been explained nicely on this site. First $3$ is prime in $Bbb Q(i)$, see here. In $Bbb Q(sqrt{6})$ however, $3=(3+sqrt{6})(3-sqrt{6})$. For an almost duplicate of this, see here.
$endgroup$
– Dietrich Burde
Mar 26 at 9:05
3
$begingroup$
You probably mean $mathbb{Z}[i]$ and $mathbb{Z}[sqrt{6}]$ instead?
$endgroup$
– Henno Brandsma
Mar 26 at 9:08
5
5
$begingroup$
This has been explained nicely on this site. First $3$ is prime in $Bbb Q(i)$, see here. In $Bbb Q(sqrt{6})$ however, $3=(3+sqrt{6})(3-sqrt{6})$. For an almost duplicate of this, see here.
$endgroup$
– Dietrich Burde
Mar 26 at 9:05
$begingroup$
This has been explained nicely on this site. First $3$ is prime in $Bbb Q(i)$, see here. In $Bbb Q(sqrt{6})$ however, $3=(3+sqrt{6})(3-sqrt{6})$. For an almost duplicate of this, see here.
$endgroup$
– Dietrich Burde
Mar 26 at 9:05
3
3
$begingroup$
You probably mean $mathbb{Z}[i]$ and $mathbb{Z}[sqrt{6}]$ instead?
$endgroup$
– Henno Brandsma
Mar 26 at 9:08
$begingroup$
You probably mean $mathbb{Z}[i]$ and $mathbb{Z}[sqrt{6}]$ instead?
$endgroup$
– Henno Brandsma
Mar 26 at 9:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We're working in general rings, not just in $mathbb{Z}$, 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 $mathbb{Z}$, $mathbb{Z}[i]$ and $mathbb{Z}[sqrt{6}]$ 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.
edited Mar 26 at 10:19
J. W. Tanner
5,0851520
answered Mar 26 at 9:13
Henno BrandsmaHenno Brandsma
117k350128
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We're working in general rings, not just in $mathbb{Z}$, 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 $mathbb{Z}$, $mathbb{Z}[i]$ and $mathbb{Z}[sqrt{6}]$ 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.
edited Mar 26 at 10:19
J. W. Tanner
5,0851520
answered Mar 26 at 9:13
Henno BrandsmaHenno Brandsma
117k350128
$endgroup$
add a comment |
$begingroup$
We're working in general rings, not just in $mathbb{Z}$, 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 $mathbb{Z}$, $mathbb{Z}[i]$ and $mathbb{Z}[sqrt{6}]$ 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.
edited Mar 26 at 10:19
J. W. Tanner
5,0851520
answered Mar 26 at 9:13
Henno BrandsmaHenno Brandsma
117k350128
$endgroup$
add a comment |
$begingroup$
We're working in general rings, not just in $mathbb{Z}$, 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 $mathbb{Z}$, $mathbb{Z}[i]$ and $mathbb{Z}[sqrt{6}]$ 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.
edited Mar 26 at 10:19
J. W. Tanner
5,0851520
answered Mar 26 at 9:13
Henno BrandsmaHenno Brandsma
117k350128
$endgroup$
We're working in general rings, not just in $mathbb{Z}$, 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 $mathbb{Z}$, $mathbb{Z}[i]$ and $mathbb{Z}[sqrt{6}]$ 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.
edited Mar 26 at 10:19
J. W. Tanner
5,0851520
answered Mar 26 at 9:13
Henno BrandsmaHenno Brandsma
117k350128
edited Mar 26 at 10:19
J. W. Tanner
5,0851520
edited Mar 26 at 10:19
J. W. Tanner
5,0851520
edited Mar 26 at 10:19
J. W. Tanner
5,0851520
5,0851520
answered Mar 26 at 9:13
Henno BrandsmaHenno Brandsma
117k350128
answered Mar 26 at 9:13
Henno BrandsmaHenno Brandsma
117k350128
answered Mar 26 at 9:13
Henno BrandsmaHenno Brandsma
117k350128
117k350128
add a comment |
add a comment |
5
$begingroup$
This has been explained nicely on this site. First $3$ is prime in $Bbb Q(i)$, see here. In $Bbb Q(sqrt{6})$ however, $3=(3+sqrt{6})(3-sqrt{6})$. For an almost duplicate of this, see here.
$endgroup$
– Dietrich Burde
Mar 26 at 9:05
3
$begingroup$
You probably mean $mathbb{Z}[i]$ and $mathbb{Z}[sqrt{6}]$ instead?
$endgroup$
– Henno Brandsma
Mar 26 at 9:08