Is this correct? $N(langle a, b rangle) = gcd(N(a), N(b))$
Given a ring $R$, not necessarily a principal ideal domain, and numbers $a, b in R$, is the norm of the ideal $langle a, b rangle$ the greatest common divisor of the norms of the numbers $a$ and $b$?
For example, given that $N(3) = 9$ and $N(1 + sqrt{-5}) = 6$, is $N(langle 3, 1 + sqrt{-5} rangle) = 3$? Or how about $N(3) = 9$ and $N(1 + sqrt{10}) = -9$, so $N(langle 3, 1 + sqrt{10} rangle) = 3$ also?
Related question: Norm of an ideal
algebraic-number-theory ideals
asked Dec 4 '18 at 22:00
Bob HappBob Happ
2531222
add a comment |
Given a ring $R$, not necessarily a principal ideal domain, and numbers $a, b in R$, is the norm of the ideal $langle a, b rangle$ the greatest common divisor of the norms of the numbers $a$ and $b$?
For example, given that $N(3) = 9$ and $N(1 + sqrt{-5}) = 6$, is $N(langle 3, 1 + sqrt{-5} rangle) = 3$? Or how about $N(3) = 9$ and $N(1 + sqrt{10}) = -9$, so $N(langle 3, 1 + sqrt{10} rangle) = 3$ also?
Related question: Norm of an ideal
algebraic-number-theory ideals
asked Dec 4 '18 at 22:00
Bob HappBob Happ
2531222
2
What is "the norm" of an element in an arbitrary ring? (Also, you do realize that the greatest common divisor of $9$ and $-9$ is $9$, not $3$, right?)
– Arturo Magidin
Dec 4 '18 at 22:20
@Arturo Thank you for pointing that out, I was mistaken. At first I was going to use $langle 2, sqrt{10} rangle$, so I would have had $gcd(4, -10) = 2$ instead.
– Bob Happ
Dec 4 '18 at 22:26
add a comment |
Given a ring $R$, not necessarily a principal ideal domain, and numbers $a, b in R$, is the norm of the ideal $langle a, b rangle$ the greatest common divisor of the norms of the numbers $a$ and $b$?
For example, given that $N(3) = 9$ and $N(1 + sqrt{-5}) = 6$, is $N(langle 3, 1 + sqrt{-5} rangle) = 3$? Or how about $N(3) = 9$ and $N(1 + sqrt{10}) = -9$, so $N(langle 3, 1 + sqrt{10} rangle) = 3$ also?
Related question: Norm of an ideal
algebraic-number-theory ideals
asked Dec 4 '18 at 22:00
Bob HappBob Happ
2531222
Given a ring $R$, not necessarily a principal ideal domain, and numbers $a, b in R$, is the norm of the ideal $langle a, b rangle$ the greatest common divisor of the norms of the numbers $a$ and $b$?
For example, given that $N(3) = 9$ and $N(1 + sqrt{-5}) = 6$, is $N(langle 3, 1 + sqrt{-5} rangle) = 3$? Or how about $N(3) = 9$ and $N(1 + sqrt{10}) = -9$, so $N(langle 3, 1 + sqrt{10} rangle) = 3$ also?
Related question: Norm of an ideal
algebraic-number-theory ideals
algebraic-number-theory ideals
asked Dec 4 '18 at 22:00
Bob HappBob Happ
2531222
asked Dec 4 '18 at 22:00
Bob HappBob Happ
2531222
asked Dec 4 '18 at 22:00
Bob HappBob Happ
2531222
asked Dec 4 '18 at 22:00
Bob HappBob Happ
2531222
asked Dec 4 '18 at 22:00
Bob HappBob Happ
2531222
2531222
2
What is "the norm" of an element in an arbitrary ring? (Also, you do realize that the greatest common divisor of $9$ and $-9$ is $9$, not $3$, right?)
– Arturo Magidin
Dec 4 '18 at 22:20
@Arturo Thank you for pointing that out, I was mistaken. At first I was going to use $langle 2, sqrt{10} rangle$, so I would have had $gcd(4, -10) = 2$ instead.
– Bob Happ
Dec 4 '18 at 22:26
add a comment |
2
What is "the norm" of an element in an arbitrary ring? (Also, you do realize that the greatest common divisor of $9$ and $-9$ is $9$, not $3$, right?)
– Arturo Magidin
Dec 4 '18 at 22:20
@Arturo Thank you for pointing that out, I was mistaken. At first I was going to use $langle 2, sqrt{10} rangle$, so I would have had $gcd(4, -10) = 2$ instead.
– Bob Happ
Dec 4 '18 at 22:26
2
2
What is "the norm" of an element in an arbitrary ring? (Also, you do realize that the greatest common divisor of $9$ and $-9$ is $9$, not $3$, right?)
– Arturo Magidin
Dec 4 '18 at 22:20
What is "the norm" of an element in an arbitrary ring? (Also, you do realize that the greatest common divisor of $9$ and $-9$ is $9$, not $3$, right?)
– Arturo Magidin
Dec 4 '18 at 22:20
@Arturo Thank you for pointing that out, I was mistaken. At first I was going to use $langle 2, sqrt{10} rangle$, so I would have had $gcd(4, -10) = 2$ instead.
– Bob Happ
Dec 4 '18 at 22:26
@Arturo Thank you for pointing that out, I was mistaken. At first I was going to use $langle 2, sqrt{10} rangle$, so I would have had $gcd(4, -10) = 2$ instead.
– Bob Happ
Dec 4 '18 at 22:26
add a comment |
1 Answer
1
active
oldest
votes
In general this is false. A counterezample in $R= Bbb Z[i]$ follows:
$$a= 2+i qquad b= 2-i$$
are prime elements since $N(a)=N(b)=5$. Note that $gcd (N(a), N(b))=5$.
However they are not associate each other, since $$ab^{-1} = frac{3}{5}- frac{4}{5}i notin R$$
In particular they are coprime, i.e. $$langle a,b rangle = R$$
Which implies
$$N( langle a, b rangle)=1 neq 5$$
The only thing you can say is that
$N( langle a, b rangle)$ divides both $N(a)$ and $N(b)$
i.e. it divides their $gcd$.
answered Dec 4 '18 at 22:38
CrostulCrostul
27.7k22352
add a comment |
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1 Answer
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1 Answer
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In general this is false. A counterezample in $R= Bbb Z[i]$ follows:
$$a= 2+i qquad b= 2-i$$
are prime elements since $N(a)=N(b)=5$. Note that $gcd (N(a), N(b))=5$.
However they are not associate each other, since $$ab^{-1} = frac{3}{5}- frac{4}{5}i notin R$$
In particular they are coprime, i.e. $$langle a,b rangle = R$$
Which implies
$$N( langle a, b rangle)=1 neq 5$$
The only thing you can say is that
$N( langle a, b rangle)$ divides both $N(a)$ and $N(b)$
i.e. it divides their $gcd$.
answered Dec 4 '18 at 22:38
CrostulCrostul
27.7k22352
add a comment |
In general this is false. A counterezample in $R= Bbb Z[i]$ follows:
$$a= 2+i qquad b= 2-i$$
are prime elements since $N(a)=N(b)=5$. Note that $gcd (N(a), N(b))=5$.
However they are not associate each other, since $$ab^{-1} = frac{3}{5}- frac{4}{5}i notin R$$
In particular they are coprime, i.e. $$langle a,b rangle = R$$
Which implies
$$N( langle a, b rangle)=1 neq 5$$
The only thing you can say is that
$N( langle a, b rangle)$ divides both $N(a)$ and $N(b)$
i.e. it divides their $gcd$.
answered Dec 4 '18 at 22:38
CrostulCrostul
27.7k22352
add a comment |
In general this is false. A counterezample in $R= Bbb Z[i]$ follows:
$$a= 2+i qquad b= 2-i$$
are prime elements since $N(a)=N(b)=5$. Note that $gcd (N(a), N(b))=5$.
However they are not associate each other, since $$ab^{-1} = frac{3}{5}- frac{4}{5}i notin R$$
In particular they are coprime, i.e. $$langle a,b rangle = R$$
Which implies
$$N( langle a, b rangle)=1 neq 5$$
The only thing you can say is that
$N( langle a, b rangle)$ divides both $N(a)$ and $N(b)$
i.e. it divides their $gcd$.
answered Dec 4 '18 at 22:38
CrostulCrostul
27.7k22352
In general this is false. A counterezample in $R= Bbb Z[i]$ follows:
$$a= 2+i qquad b= 2-i$$
are prime elements since $N(a)=N(b)=5$. Note that $gcd (N(a), N(b))=5$.
However they are not associate each other, since $$ab^{-1} = frac{3}{5}- frac{4}{5}i notin R$$
In particular they are coprime, i.e. $$langle a,b rangle = R$$
Which implies
$$N( langle a, b rangle)=1 neq 5$$
The only thing you can say is that
$N( langle a, b rangle)$ divides both $N(a)$ and $N(b)$
i.e. it divides their $gcd$.
answered Dec 4 '18 at 22:38
CrostulCrostul
27.7k22352
answered Dec 4 '18 at 22:38
CrostulCrostul
27.7k22352
answered Dec 4 '18 at 22:38
CrostulCrostul
27.7k22352
answered Dec 4 '18 at 22:38
CrostulCrostul
27.7k22352
27.7k22352
add a comment |
add a comment |
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2
What is "the norm" of an element in an arbitrary ring? (Also, you do realize that the greatest common divisor of $9$ and $-9$ is $9$, not $3$, right?)
– Arturo Magidin
Dec 4 '18 at 22:20
@Arturo Thank you for pointing that out, I was mistaken. At first I was going to use $langle 2, sqrt{10} rangle$, so I would have had $gcd(4, -10) = 2$ instead.
– Bob Happ
Dec 4 '18 at 22:26