Mixing
Showing that an ideal in $mathbb{Z}[sqrt{-21}]$ is principal
$begingroup$
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
edited Jan 9 at 16:30
Servaes
23.4k33893
asked Jan 9 at 16:18
pullofthemoonpullofthemoon
815
$endgroup$
add a comment |
$begingroup$
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
edited Jan 9 at 16:30
Servaes
23.4k33893
asked Jan 9 at 16:18
pullofthemoonpullofthemoon
815
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Jan 9 at 16:30
Servaes
23.4k33893
asked Jan 9 at 16:18
pullofthemoonpullofthemoon
815
$endgroup$
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
edited Jan 9 at 16:30
Servaes
23.4k33893
asked Jan 9 at 16:18
pullofthemoonpullofthemoon
815
edited Jan 9 at 16:30
Servaes
23.4k33893
asked Jan 9 at 16:18
pullofthemoonpullofthemoon
815
edited Jan 9 at 16:30
Servaes
23.4k33893
edited Jan 9 at 16:30
Servaes
23.4k33893
edited Jan 9 at 16:30
Servaes
23.4k33893
23.4k33893
asked Jan 9 at 16:18
pullofthemoonpullofthemoon
815
asked Jan 9 at 16:18
pullofthemoonpullofthemoon
815
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $sqrt{-21}-2$.
answered Jan 9 at 16:22
ServaesServaes
23.4k33893
$endgroup$
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
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $sqrt{-21}-2$.
answered Jan 9 at 16:22
ServaesServaes
23.4k33893
$endgroup$
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
add a comment |
$begingroup$
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 $sqrt{-21}-2$.
answered Jan 9 at 16:22
ServaesServaes
23.4k33893
$endgroup$
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
add a comment |
$begingroup$
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 $sqrt{-21}-2$.
answered Jan 9 at 16:22
ServaesServaes
23.4k33893
$endgroup$
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 $sqrt{-21}-2$.
answered Jan 9 at 16:22
ServaesServaes
23.4k33893
edited Jan 9 at 16:29
answered Jan 9 at 16:22
ServaesServaes
23.4k33893
answered Jan 9 at 16:22
ServaesServaes
23.4k33893
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
add a comment |
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
add a comment |
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$begingroup$
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