Showing that an ideal in $mathbb{Z}[sqrt{-21}]$ is principal












2












$begingroup$


I have $mathfrak{a}=(5,sqrt{-21}-2).$
Can anyone tell me why $mathfrak{a}^2$ is principal?
I have multiplied the ideals out to obtain
$$(5, 5sqrt{-21}-10,-17-4sqrt{-21}) $$
How does this reduce to a principal ideal?






number-theory algebraic-number-theory ideals principal-ideal-domains







share|cite|improve this question











$endgroup$












  • $begingroup$
    The generators you have found are not correct; see my answer.
    $endgroup$
    – Servaes
    Jan 9 at 16:29










  • $begingroup$
    Yeah, $5notin mathfrak{a}^2.$
    $endgroup$
    – Thomas Andrews
    Jan 9 at 16:35
















2












$begingroup$


I have $mathfrak{a}=(5,sqrt{-21}-2).$
Can anyone tell me why $mathfrak{a}^2$ is principal?
I have multiplied the ideals out to obtain
$$(5, 5sqrt{-21}-10,-17-4sqrt{-21}) $$
How does this reduce to a principal ideal?






number-theory algebraic-number-theory ideals principal-ideal-domains







share|cite|improve this question











$endgroup$












  • $begingroup$
    The generators you have found are not correct; see my answer.
    $endgroup$
    – Servaes
    Jan 9 at 16:29










  • $begingroup$
    Yeah, $5notin mathfrak{a}^2.$
    $endgroup$
    – Thomas Andrews
    Jan 9 at 16:35














2












2








2





$begingroup$


I have $mathfrak{a}=(5,sqrt{-21}-2).$
Can anyone tell me why $mathfrak{a}^2$ is principal?
I have multiplied the ideals out to obtain
$$(5, 5sqrt{-21}-10,-17-4sqrt{-21}) $$
How does this reduce to a principal ideal?






number-theory algebraic-number-theory ideals principal-ideal-domains







share|cite|improve this question











$endgroup$




I have $mathfrak{a}=(5,sqrt{-21}-2).$
Can anyone tell me why $mathfrak{a}^2$ is principal?
I have multiplied the ideals out to obtain
$$(5, 5sqrt{-21}-10,-17-4sqrt{-21}) $$
How does this reduce to a principal ideal?






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 9 at 16:30









Servaes

23.4k33893




23.4k33893










asked Jan 9 at 16:18









pullofthemoonpullofthemoon

815




815












  • $begingroup$
    The generators you have found are not correct; see my answer.
    $endgroup$
    – Servaes
    Jan 9 at 16:29










  • $begingroup$
    Yeah, $5notin mathfrak{a}^2.$
    $endgroup$
    – Thomas Andrews
    Jan 9 at 16:35


















  • $begingroup$
    The generators you have found are not correct; see my answer.
    $endgroup$
    – Servaes
    Jan 9 at 16:29










  • $begingroup$
    Yeah, $5notin mathfrak{a}^2.$
    $endgroup$
    – Thomas Andrews
    Jan 9 at 16:35
















$begingroup$
The generators you have found are not correct; see my answer.
$endgroup$
– Servaes
Jan 9 at 16:29




$begingroup$
The generators you have found are not correct; see my answer.
$endgroup$
– Servaes
Jan 9 at 16:29












$begingroup$
Yeah, $5notin mathfrak{a}^2.$
$endgroup$
– Thomas Andrews
Jan 9 at 16:35




$begingroup$
Yeah, $5notin mathfrak{a}^2.$
$endgroup$
– Thomas Andrews
Jan 9 at 16:35










1 Answer
1






active

oldest

votes


















2












$begingroup$

The ideal $mathfrak{a}^2$ is the ideal generated by the products of the generators of $mathfrak{a}$, hence
$$mathfrak{a}^2=(5^2,5(sqrt{-21}-2),(sqrt{-21}-2)^2)=(25,-10+5sqrt{-21},-17-4sqrt{-21}).$$
The latter two generators are clearly both multiples of $sqrt{-21}-2$, and the identity
$$-(sqrt{-21}+2)(sqrt{-21}-2)=25,$$
shows that the first generator is also a multiple of $sqrt{-21}-2$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    You also have to show that $sqrt{-21}-2$ is in the ideal, of course.
    $endgroup$
    – Thomas Andrews
    Jan 9 at 16:33










  • $begingroup$
    Just add the last-named three generators, @ThomasAndrews.
    $endgroup$
    – Lubin
    Jan 9 at 22:00










  • $begingroup$
    @ThomasAndrews Thanks, that is certainly worth making explicit.
    $endgroup$
    – Servaes
    Jan 10 at 10:46











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1 Answer
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active

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active

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2












$begingroup$

The ideal $mathfrak{a}^2$ is the ideal generated by the products of the generators of $mathfrak{a}$, hence
$$mathfrak{a}^2=(5^2,5(sqrt{-21}-2),(sqrt{-21}-2)^2)=(25,-10+5sqrt{-21},-17-4sqrt{-21}).$$
The latter two generators are clearly both multiples of $sqrt{-21}-2$, and the identity
$$-(sqrt{-21}+2)(sqrt{-21}-2)=25,$$
shows that the first generator is also a multiple of $sqrt{-21}-2$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    You also have to show that $sqrt{-21}-2$ is in the ideal, of course.
    $endgroup$
    – Thomas Andrews
    Jan 9 at 16:33










  • $begingroup$
    Just add the last-named three generators, @ThomasAndrews.
    $endgroup$
    – Lubin
    Jan 9 at 22:00










  • $begingroup$
    @ThomasAndrews Thanks, that is certainly worth making explicit.
    $endgroup$
    – Servaes
    Jan 10 at 10:46
















2












$begingroup$

The ideal $mathfrak{a}^2$ is the ideal generated by the products of the generators of $mathfrak{a}$, hence
$$mathfrak{a}^2=(5^2,5(sqrt{-21}-2),(sqrt{-21}-2)^2)=(25,-10+5sqrt{-21},-17-4sqrt{-21}).$$
The latter two generators are clearly both multiples of $sqrt{-21}-2$, and the identity
$$-(sqrt{-21}+2)(sqrt{-21}-2)=25,$$
shows that the first generator is also a multiple of $sqrt{-21}-2$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    You also have to show that $sqrt{-21}-2$ is in the ideal, of course.
    $endgroup$
    – Thomas Andrews
    Jan 9 at 16:33










  • $begingroup$
    Just add the last-named three generators, @ThomasAndrews.
    $endgroup$
    – Lubin
    Jan 9 at 22:00










  • $begingroup$
    @ThomasAndrews Thanks, that is certainly worth making explicit.
    $endgroup$
    – Servaes
    Jan 10 at 10:46














2












2








2





$begingroup$

The ideal $mathfrak{a}^2$ is the ideal generated by the products of the generators of $mathfrak{a}$, hence
$$mathfrak{a}^2=(5^2,5(sqrt{-21}-2),(sqrt{-21}-2)^2)=(25,-10+5sqrt{-21},-17-4sqrt{-21}).$$
The latter two generators are clearly both multiples of $sqrt{-21}-2$, and the identity
$$-(sqrt{-21}+2)(sqrt{-21}-2)=25,$$
shows that the first generator is also a multiple of $sqrt{-21}-2$.






share|cite|improve this answer











$endgroup$



The ideal $mathfrak{a}^2$ is the ideal generated by the products of the generators of $mathfrak{a}$, hence
$$mathfrak{a}^2=(5^2,5(sqrt{-21}-2),(sqrt{-21}-2)^2)=(25,-10+5sqrt{-21},-17-4sqrt{-21}).$$
The latter two generators are clearly both multiples of $sqrt{-21}-2$, and the identity
$$-(sqrt{-21}+2)(sqrt{-21}-2)=25,$$
shows that the first generator is also a multiple of $sqrt{-21}-2$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 9 at 16:29

























answered Jan 9 at 16:22









ServaesServaes

23.4k33893




23.4k33893








  • 1




    $begingroup$
    You also have to show that $sqrt{-21}-2$ is in the ideal, of course.
    $endgroup$
    – Thomas Andrews
    Jan 9 at 16:33










  • $begingroup$
    Just add the last-named three generators, @ThomasAndrews.
    $endgroup$
    – Lubin
    Jan 9 at 22:00










  • $begingroup$
    @ThomasAndrews Thanks, that is certainly worth making explicit.
    $endgroup$
    – Servaes
    Jan 10 at 10:46














  • 1




    $begingroup$
    You also have to show that $sqrt{-21}-2$ is in the ideal, of course.
    $endgroup$
    – Thomas Andrews
    Jan 9 at 16:33










  • $begingroup$
    Just add the last-named three generators, @ThomasAndrews.
    $endgroup$
    – Lubin
    Jan 9 at 22:00










  • $begingroup$
    @ThomasAndrews Thanks, that is certainly worth making explicit.
    $endgroup$
    – Servaes
    Jan 10 at 10:46








1




1




$begingroup$
You also have to show that $sqrt{-21}-2$ is in the ideal, of course.
$endgroup$
– Thomas Andrews
Jan 9 at 16:33




$begingroup$
You also have to show that $sqrt{-21}-2$ is in the ideal, of course.
$endgroup$
– Thomas Andrews
Jan 9 at 16:33












$begingroup$
Just add the last-named three generators, @ThomasAndrews.
$endgroup$
– Lubin
Jan 9 at 22:00




$begingroup$
Just add the last-named three generators, @ThomasAndrews.
$endgroup$
– Lubin
Jan 9 at 22:00












$begingroup$
@ThomasAndrews Thanks, that is certainly worth making explicit.
$endgroup$
– Servaes
Jan 10 at 10:46




$begingroup$
@ThomasAndrews Thanks, that is certainly worth making explicit.
$endgroup$
– Servaes
Jan 10 at 10:46


















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