Showing that an ideal is not principal in $mathbb{Z}[sqrt{-21}]$
$begingroup$
I am trying to show that the ideal
$$(2,sqrt{-21}-1)(3, sqrt{-21}) $$
is not principal in $mathbb{Z}[sqrt{-21}]$. Can anyone help with this?
number-theory algebraic-number-theory ideals principal-ideal-domains
edited Jan 8 at 23:54
J. W. Tanner
2,3751117
asked Jan 8 at 20:04
pullofthemoonpullofthemoon
815
$endgroup$
|
show 2 more comments
$begingroup$
I am trying to show that the ideal
$$(2,sqrt{-21}-1)(3, sqrt{-21}) $$
is not principal in $mathbb{Z}[sqrt{-21}]$. Can anyone help with this?
number-theory algebraic-number-theory ideals principal-ideal-domains
edited Jan 8 at 23:54
J. W. Tanner
2,3751117
asked Jan 8 at 20:04
pullofthemoonpullofthemoon
815
$endgroup$
$begingroup$
Hint: what is the norm of this ideal? Is there an element of this norm in the ring?
$endgroup$
– Wojowu
Jan 8 at 20:14
$begingroup$
The norm of this ideal would be 5, which implies there has to be an element $x in mathbb{Z}[sqrt{-21}]$ that has norm five. This is not possible. Is this enough?
$endgroup$
– pullofthemoon
Jan 8 at 20:20
1
$begingroup$
The norm is $6$, but yes, that argument works.
$endgroup$
– Wojowu
Jan 8 at 20:38
$begingroup$
How have you gotten that the norm is 6? I have that $(2,sqrt{-21}-1)(3, sqrt{-21})=mathfrak{p}_{5}$ where $mathfrak{p}_{5}=(5, sqrt{-21} -3)$. This has conjugate $(5, sqrt{-21} -2)$ since $mathfrak{p}_{5}overline{mathfrak{p}_{5}}$ is the prime factorisation of $(5)$, wouldn't this imply that the norm is 5?
$endgroup$
– pullofthemoon
Jan 8 at 20:49
$begingroup$
How is this product $mathfrak p_5$? Specifically, how is $5$ in the product of those ideals?
$endgroup$
– Wojowu
Jan 8 at 20:51
|
show 2 more comments
$begingroup$
I am trying to show that the ideal
$$(2,sqrt{-21}-1)(3, sqrt{-21}) $$
is not principal in $mathbb{Z}[sqrt{-21}]$. Can anyone help with this?
number-theory algebraic-number-theory ideals principal-ideal-domains
edited Jan 8 at 23:54
J. W. Tanner
2,3751117
asked Jan 8 at 20:04
pullofthemoonpullofthemoon
815
$endgroup$
I am trying to show that the ideal
$$(2,sqrt{-21}-1)(3, sqrt{-21}) $$
is not principal in $mathbb{Z}[sqrt{-21}]$. Can anyone help with this?
number-theory algebraic-number-theory ideals principal-ideal-domains
number-theory algebraic-number-theory ideals principal-ideal-domains
edited Jan 8 at 23:54
J. W. Tanner
2,3751117
asked Jan 8 at 20:04
pullofthemoonpullofthemoon
815
edited Jan 8 at 23:54
J. W. Tanner
2,3751117
asked Jan 8 at 20:04
pullofthemoonpullofthemoon
815
edited Jan 8 at 23:54
J. W. Tanner
2,3751117
edited Jan 8 at 23:54
J. W. Tanner
2,3751117
edited Jan 8 at 23:54
J. W. Tanner
2,3751117
2,3751117
asked Jan 8 at 20:04
pullofthemoonpullofthemoon
815
asked Jan 8 at 20:04
pullofthemoonpullofthemoon
815
asked Jan 8 at 20:04
pullofthemoonpullofthemoon
815
815
$begingroup$
Hint: what is the norm of this ideal? Is there an element of this norm in the ring?
$endgroup$
– Wojowu
Jan 8 at 20:14
$begingroup$
The norm of this ideal would be 5, which implies there has to be an element $x in mathbb{Z}[sqrt{-21}]$ that has norm five. This is not possible. Is this enough?
$endgroup$
– pullofthemoon
Jan 8 at 20:20
1
$begingroup$
The norm is $6$, but yes, that argument works.
$endgroup$
– Wojowu
Jan 8 at 20:38
$begingroup$
How have you gotten that the norm is 6? I have that $(2,sqrt{-21}-1)(3, sqrt{-21})=mathfrak{p}_{5}$ where $mathfrak{p}_{5}=(5, sqrt{-21} -3)$. This has conjugate $(5, sqrt{-21} -2)$ since $mathfrak{p}_{5}overline{mathfrak{p}_{5}}$ is the prime factorisation of $(5)$, wouldn't this imply that the norm is 5?
$endgroup$
– pullofthemoon
Jan 8 at 20:49
$begingroup$
How is this product $mathfrak p_5$? Specifically, how is $5$ in the product of those ideals?
$endgroup$
– Wojowu
Jan 8 at 20:51
|
show 2 more comments
$begingroup$
Hint: what is the norm of this ideal? Is there an element of this norm in the ring?
$endgroup$
– Wojowu
Jan 8 at 20:14
$begingroup$
The norm of this ideal would be 5, which implies there has to be an element $x in mathbb{Z}[sqrt{-21}]$ that has norm five. This is not possible. Is this enough?
$endgroup$
– pullofthemoon
Jan 8 at 20:20
1
$begingroup$
The norm is $6$, but yes, that argument works.
$endgroup$
– Wojowu
Jan 8 at 20:38
$begingroup$
How have you gotten that the norm is 6? I have that $(2,sqrt{-21}-1)(3, sqrt{-21})=mathfrak{p}_{5}$ where $mathfrak{p}_{5}=(5, sqrt{-21} -3)$. This has conjugate $(5, sqrt{-21} -2)$ since $mathfrak{p}_{5}overline{mathfrak{p}_{5}}$ is the prime factorisation of $(5)$, wouldn't this imply that the norm is 5?
$endgroup$
– pullofthemoon
Jan 8 at 20:49
$begingroup$
How is this product $mathfrak p_5$? Specifically, how is $5$ in the product of those ideals?
$endgroup$
– Wojowu
Jan 8 at 20:51
$begingroup$
Hint: what is the norm of this ideal? Is there an element of this norm in the ring?
$endgroup$
– Wojowu
Jan 8 at 20:14
$begingroup$
Hint: what is the norm of this ideal? Is there an element of this norm in the ring?
$endgroup$
– Wojowu
Jan 8 at 20:14
$begingroup$
The norm of this ideal would be 5, which implies there has to be an element $x in mathbb{Z}[sqrt{-21}]$ that has norm five. This is not possible. Is this enough?
$endgroup$
– pullofthemoon
Jan 8 at 20:20
$begingroup$
The norm of this ideal would be 5, which implies there has to be an element $x in mathbb{Z}[sqrt{-21}]$ that has norm five. This is not possible. Is this enough?
$endgroup$
– pullofthemoon
Jan 8 at 20:20
1
1
$begingroup$
The norm is $6$, but yes, that argument works.
$endgroup$
– Wojowu
Jan 8 at 20:38
$begingroup$
The norm is $6$, but yes, that argument works.
$endgroup$
– Wojowu
Jan 8 at 20:38
$begingroup$
How have you gotten that the norm is 6? I have that $(2,sqrt{-21}-1)(3, sqrt{-21})=mathfrak{p}_{5}$ where $mathfrak{p}_{5}=(5, sqrt{-21} -3)$. This has conjugate $(5, sqrt{-21} -2)$ since $mathfrak{p}_{5}overline{mathfrak{p}_{5}}$ is the prime factorisation of $(5)$, wouldn't this imply that the norm is 5?
$endgroup$
– pullofthemoon
Jan 8 at 20:49
$begingroup$
How have you gotten that the norm is 6? I have that $(2,sqrt{-21}-1)(3, sqrt{-21})=mathfrak{p}_{5}$ where $mathfrak{p}_{5}=(5, sqrt{-21} -3)$. This has conjugate $(5, sqrt{-21} -2)$ since $mathfrak{p}_{5}overline{mathfrak{p}_{5}}$ is the prime factorisation of $(5)$, wouldn't this imply that the norm is 5?
$endgroup$
– pullofthemoon
Jan 8 at 20:49
$begingroup$
How is this product $mathfrak p_5$? Specifically, how is $5$ in the product of those ideals?
$endgroup$
– Wojowu
Jan 8 at 20:51
$begingroup$
How is this product $mathfrak p_5$? Specifically, how is $5$ in the product of those ideals?
$endgroup$
– Wojowu
Jan 8 at 20:51
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The Very Useful Theorem that you want to use is that if an ideal $mathfrak a$ is principal, equal to $langle brangle$, then the norm of $mathfrak a$ is $mathbf N(b)$, where $mathbf N$ is the field-theoretic Norm map from the big field down to $Bbb Q$.
In your case, since an integral basis for the integers of $Bbb Q(sqrt{-21},)$ can be taken to be ${1,sqrt{-21},}$ (because $-21notequiv1pmod4$ ), so the Norm of the general integer $m+nsqrt{-21}$ is $m^2+21n^2$. I’m sure you can finish it all off from here.
answered Jan 9 at 2:19
LubinLubin
44.6k44586
$endgroup$
add a comment |
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$begingroup$
The Very Useful Theorem that you want to use is that if an ideal $mathfrak a$ is principal, equal to $langle brangle$, then the norm of $mathfrak a$ is $mathbf N(b)$, where $mathbf N$ is the field-theoretic Norm map from the big field down to $Bbb Q$.
In your case, since an integral basis for the integers of $Bbb Q(sqrt{-21},)$ can be taken to be ${1,sqrt{-21},}$ (because $-21notequiv1pmod4$ ), so the Norm of the general integer $m+nsqrt{-21}$ is $m^2+21n^2$. I’m sure you can finish it all off from here.
answered Jan 9 at 2:19
LubinLubin
44.6k44586
$endgroup$
add a comment |
$begingroup$
The Very Useful Theorem that you want to use is that if an ideal $mathfrak a$ is principal, equal to $langle brangle$, then the norm of $mathfrak a$ is $mathbf N(b)$, where $mathbf N$ is the field-theoretic Norm map from the big field down to $Bbb Q$.
In your case, since an integral basis for the integers of $Bbb Q(sqrt{-21},)$ can be taken to be ${1,sqrt{-21},}$ (because $-21notequiv1pmod4$ ), so the Norm of the general integer $m+nsqrt{-21}$ is $m^2+21n^2$. I’m sure you can finish it all off from here.
answered Jan 9 at 2:19
LubinLubin
44.6k44586
$endgroup$
add a comment |
$begingroup$
The Very Useful Theorem that you want to use is that if an ideal $mathfrak a$ is principal, equal to $langle brangle$, then the norm of $mathfrak a$ is $mathbf N(b)$, where $mathbf N$ is the field-theoretic Norm map from the big field down to $Bbb Q$.
In your case, since an integral basis for the integers of $Bbb Q(sqrt{-21},)$ can be taken to be ${1,sqrt{-21},}$ (because $-21notequiv1pmod4$ ), so the Norm of the general integer $m+nsqrt{-21}$ is $m^2+21n^2$. I’m sure you can finish it all off from here.
answered Jan 9 at 2:19
LubinLubin
44.6k44586
$endgroup$
The Very Useful Theorem that you want to use is that if an ideal $mathfrak a$ is principal, equal to $langle brangle$, then the norm of $mathfrak a$ is $mathbf N(b)$, where $mathbf N$ is the field-theoretic Norm map from the big field down to $Bbb Q$.
In your case, since an integral basis for the integers of $Bbb Q(sqrt{-21},)$ can be taken to be ${1,sqrt{-21},}$ (because $-21notequiv1pmod4$ ), so the Norm of the general integer $m+nsqrt{-21}$ is $m^2+21n^2$. I’m sure you can finish it all off from here.
answered Jan 9 at 2:19
LubinLubin
44.6k44586
answered Jan 9 at 2:19
LubinLubin
44.6k44586
answered Jan 9 at 2:19
LubinLubin
44.6k44586
answered Jan 9 at 2:19
LubinLubin
44.6k44586
44.6k44586
add a comment |
add a comment |
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$begingroup$
Hint: what is the norm of this ideal? Is there an element of this norm in the ring?
$endgroup$
– Wojowu
Jan 8 at 20:14
$begingroup$
The norm of this ideal would be 5, which implies there has to be an element $x in mathbb{Z}[sqrt{-21}]$ that has norm five. This is not possible. Is this enough?
$endgroup$
– pullofthemoon
Jan 8 at 20:20
1
$begingroup$
The norm is $6$, but yes, that argument works.
$endgroup$
– Wojowu
Jan 8 at 20:38
$begingroup$
How have you gotten that the norm is 6? I have that $(2,sqrt{-21}-1)(3, sqrt{-21})=mathfrak{p}_{5}$ where $mathfrak{p}_{5}=(5, sqrt{-21} -3)$. This has conjugate $(5, sqrt{-21} -2)$ since $mathfrak{p}_{5}overline{mathfrak{p}_{5}}$ is the prime factorisation of $(5)$, wouldn't this imply that the norm is 5?
$endgroup$
– pullofthemoon
Jan 8 at 20:49
$begingroup$
How is this product $mathfrak p_5$? Specifically, how is $5$ in the product of those ideals?
$endgroup$
– Wojowu
Jan 8 at 20:51