Showing that an ideal is not principal in $mathbb{Z}[sqrt{-21}]$












1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 23:54









J. W. Tanner

1117




1117










asked Jan 8 at 20:04









pullofthemoonpullofthemoon

665




665












  • $begingroup$
    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










  • $begingroup$
    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










1 Answer
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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.






share|cite|improve this answer









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    1 Answer
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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.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 9 at 2:19









        LubinLubin

        44k44585




        44k44585






























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