Determine that an ideal is not a principal ideal
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I have the ring $mathbb{Z}[sqrt{-5}]$ and the function $phi: mathbb{Z}[sqrt{-5}] to mathbb{Z}/3mathbb{Z}$. I found the kernel of the function, namely $(3, 1-sqrt{-5})$. Now I am trying to prove that the kernel is not a principal ideal. How can I do this?
I know I could use norms to do it. But what do I need to compute to conclude that this ideal is not a principal ideal?
abstract-algebra ring-theory
edited Nov 17 at 16:51
Bernard
116k637108
asked Nov 17 at 16:48
Hans
587
add a comment |
up vote
0
down vote
favorite
I have the ring $mathbb{Z}[sqrt{-5}]$ and the function $phi: mathbb{Z}[sqrt{-5}] to mathbb{Z}/3mathbb{Z}$. I found the kernel of the function, namely $(3, 1-sqrt{-5})$. Now I am trying to prove that the kernel is not a principal ideal. How can I do this?
I know I could use norms to do it. But what do I need to compute to conclude that this ideal is not a principal ideal?
abstract-algebra ring-theory
edited Nov 17 at 16:51
Bernard
116k637108
asked Nov 17 at 16:48
Hans
587
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have the ring $mathbb{Z}[sqrt{-5}]$ and the function $phi: mathbb{Z}[sqrt{-5}] to mathbb{Z}/3mathbb{Z}$. I found the kernel of the function, namely $(3, 1-sqrt{-5})$. Now I am trying to prove that the kernel is not a principal ideal. How can I do this?
I know I could use norms to do it. But what do I need to compute to conclude that this ideal is not a principal ideal?
abstract-algebra ring-theory
edited Nov 17 at 16:51
Bernard
116k637108
asked Nov 17 at 16:48
Hans
587
I have the ring $mathbb{Z}[sqrt{-5}]$ and the function $phi: mathbb{Z}[sqrt{-5}] to mathbb{Z}/3mathbb{Z}$. I found the kernel of the function, namely $(3, 1-sqrt{-5})$. Now I am trying to prove that the kernel is not a principal ideal. How can I do this?
I know I could use norms to do it. But what do I need to compute to conclude that this ideal is not a principal ideal?
abstract-algebra ring-theory
abstract-algebra ring-theory
edited Nov 17 at 16:51
Bernard
116k637108
asked Nov 17 at 16:48
Hans
587
edited Nov 17 at 16:51
Bernard
116k637108
asked Nov 17 at 16:48
Hans
587
edited Nov 17 at 16:51
Bernard
116k637108
edited Nov 17 at 16:51
Bernard
116k637108
edited Nov 17 at 16:51
Bernard
116k637108
116k637108
asked Nov 17 at 16:48
Hans
587
asked Nov 17 at 16:48
Hans
587
asked Nov 17 at 16:48
Hans
587
587
add a comment |
add a comment |
1 Answer
1
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2
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The norm of a putative generator $alpha$ must divide that of $3$, which is $9$
and that of $1-sqrt{-5}$, namely $6$. Therefore $Nalpha=1$ or $3$.
If $Nalpha=1$ then $alpha=pm1$. That's impossible as $phi(pm1)ne0$.
But $Nalpha=3$ is impossible: there are no integers with $a^2+5b^2=3$.
answered Nov 17 at 16:51
Lord Shark the Unknown
98k958131
And if there exists two integers $a, b$ such that $a^2 + 5b^2 = x$, then it is a principal ideal?
– Hans
Nov 17 at 16:54
If there are two integers with $a^2+5b^2=x$ then there is a principal ideal of norm $x$ (but there might also be a non-principal ideal of norm $x$).
– Lord Shark the Unknown
Nov 17 at 16:56
And why is it needed that $phi(±1) neq 0$? Can't you say that there aren't two integers such that $a^2+5b^2=0$? (Which is true of course)
– Hans
Nov 17 at 18:35
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
The norm of a putative generator $alpha$ must divide that of $3$, which is $9$
and that of $1-sqrt{-5}$, namely $6$. Therefore $Nalpha=1$ or $3$.
If $Nalpha=1$ then $alpha=pm1$. That's impossible as $phi(pm1)ne0$.
But $Nalpha=3$ is impossible: there are no integers with $a^2+5b^2=3$.
answered Nov 17 at 16:51
Lord Shark the Unknown
98k958131
And if there exists two integers $a, b$ such that $a^2 + 5b^2 = x$, then it is a principal ideal?
– Hans
Nov 17 at 16:54
If there are two integers with $a^2+5b^2=x$ then there is a principal ideal of norm $x$ (but there might also be a non-principal ideal of norm $x$).
– Lord Shark the Unknown
Nov 17 at 16:56
And why is it needed that $phi(±1) neq 0$? Can't you say that there aren't two integers such that $a^2+5b^2=0$? (Which is true of course)
– Hans
Nov 17 at 18:35
add a comment |
up vote
2
down vote
The norm of a putative generator $alpha$ must divide that of $3$, which is $9$
and that of $1-sqrt{-5}$, namely $6$. Therefore $Nalpha=1$ or $3$.
If $Nalpha=1$ then $alpha=pm1$. That's impossible as $phi(pm1)ne0$.
But $Nalpha=3$ is impossible: there are no integers with $a^2+5b^2=3$.
answered Nov 17 at 16:51
Lord Shark the Unknown
98k958131
And if there exists two integers $a, b$ such that $a^2 + 5b^2 = x$, then it is a principal ideal?
– Hans
Nov 17 at 16:54
If there are two integers with $a^2+5b^2=x$ then there is a principal ideal of norm $x$ (but there might also be a non-principal ideal of norm $x$).
– Lord Shark the Unknown
Nov 17 at 16:56
And why is it needed that $phi(±1) neq 0$? Can't you say that there aren't two integers such that $a^2+5b^2=0$? (Which is true of course)
– Hans
Nov 17 at 18:35
add a comment |
up vote
2
down vote
up vote
2
down vote
The norm of a putative generator $alpha$ must divide that of $3$, which is $9$
and that of $1-sqrt{-5}$, namely $6$. Therefore $Nalpha=1$ or $3$.
If $Nalpha=1$ then $alpha=pm1$. That's impossible as $phi(pm1)ne0$.
But $Nalpha=3$ is impossible: there are no integers with $a^2+5b^2=3$.
answered Nov 17 at 16:51
Lord Shark the Unknown
98k958131
The norm of a putative generator $alpha$ must divide that of $3$, which is $9$
and that of $1-sqrt{-5}$, namely $6$. Therefore $Nalpha=1$ or $3$.
If $Nalpha=1$ then $alpha=pm1$. That's impossible as $phi(pm1)ne0$.
But $Nalpha=3$ is impossible: there are no integers with $a^2+5b^2=3$.
answered Nov 17 at 16:51
Lord Shark the Unknown
98k958131
answered Nov 17 at 16:51
Lord Shark the Unknown
98k958131
answered Nov 17 at 16:51
Lord Shark the Unknown
98k958131
answered Nov 17 at 16:51
Lord Shark the Unknown
98k958131
98k958131
And if there exists two integers $a, b$ such that $a^2 + 5b^2 = x$, then it is a principal ideal?
– Hans
Nov 17 at 16:54
If there are two integers with $a^2+5b^2=x$ then there is a principal ideal of norm $x$ (but there might also be a non-principal ideal of norm $x$).
– Lord Shark the Unknown
Nov 17 at 16:56
And why is it needed that $phi(±1) neq 0$? Can't you say that there aren't two integers such that $a^2+5b^2=0$? (Which is true of course)
– Hans
Nov 17 at 18:35
add a comment |
And if there exists two integers $a, b$ such that $a^2 + 5b^2 = x$, then it is a principal ideal?
– Hans
Nov 17 at 16:54
If there are two integers with $a^2+5b^2=x$ then there is a principal ideal of norm $x$ (but there might also be a non-principal ideal of norm $x$).
– Lord Shark the Unknown
Nov 17 at 16:56
And why is it needed that $phi(±1) neq 0$? Can't you say that there aren't two integers such that $a^2+5b^2=0$? (Which is true of course)
– Hans
Nov 17 at 18:35
And if there exists two integers $a, b$ such that $a^2 + 5b^2 = x$, then it is a principal ideal?
– Hans
Nov 17 at 16:54
And if there exists two integers $a, b$ such that $a^2 + 5b^2 = x$, then it is a principal ideal?
– Hans
Nov 17 at 16:54
If there are two integers with $a^2+5b^2=x$ then there is a principal ideal of norm $x$ (but there might also be a non-principal ideal of norm $x$).
– Lord Shark the Unknown
Nov 17 at 16:56
If there are two integers with $a^2+5b^2=x$ then there is a principal ideal of norm $x$ (but there might also be a non-principal ideal of norm $x$).
– Lord Shark the Unknown
Nov 17 at 16:56
And why is it needed that $phi(±1) neq 0$? Can't you say that there aren't two integers such that $a^2+5b^2=0$? (Which is true of course)
– Hans
Nov 17 at 18:35
And why is it needed that $phi(±1) neq 0$? Can't you say that there aren't two integers such that $a^2+5b^2=0$? (Which is true of course)
– Hans
Nov 17 at 18:35
add a comment |
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