Associated elements in $mathbb{Z}[frac{1+sqrt{-3}}{2}]$
$begingroup$
I got four elements and want to check whether these elements are associated or not. For
$alpha = frac{1+sqrt{-3}}{2}$
my elements are:
$a_1 = 2 - alpha = frac{3 - sqrt{-3}}{2}$
$a_2 = 1 - 2alpha = -sqrt{-3}$
$ a_3 = 3 + 2alpha = 4 + sqrt{-3}$
$ a_4 = 3 - 2alpha = 2 - sqrt{-3}$
I just calculated $frac{a_1}{a_2} = frac{1-sqrt{-3}}{2}$ and $frac{a_2}{a_1} = frac{-1+sqrt{-3}}{2}$. All other pairs $frac{a_1}{a_3}, frac{a_1}{a_4},frac{a_2}{a_3} frac{a_2}{a_4}, frac{a_3}{a_4}$ give an odd denominator which is not possble in $mathbb{Z}[frac{1+sqrt{-3}}{2}]$ (except the denomitor 1), therefore only $a_1$ and $a_2$ are associated... is this right? Or do I miss something?
Thx for any help on this
abstract-algebra ring-theory associativity euclidean-domain eisenstein-integers
edited Jan 8 at 22:09
Batominovski
33.1k33293
asked Jan 8 at 19:53
franzfffranzff
111
$endgroup$
add a comment |
$begingroup$
I got four elements and want to check whether these elements are associated or not. For
$alpha = frac{1+sqrt{-3}}{2}$
my elements are:
$a_1 = 2 - alpha = frac{3 - sqrt{-3}}{2}$
$a_2 = 1 - 2alpha = -sqrt{-3}$
$ a_3 = 3 + 2alpha = 4 + sqrt{-3}$
$ a_4 = 3 - 2alpha = 2 - sqrt{-3}$
I just calculated $frac{a_1}{a_2} = frac{1-sqrt{-3}}{2}$ and $frac{a_2}{a_1} = frac{-1+sqrt{-3}}{2}$. All other pairs $frac{a_1}{a_3}, frac{a_1}{a_4},frac{a_2}{a_3} frac{a_2}{a_4}, frac{a_3}{a_4}$ give an odd denominator which is not possble in $mathbb{Z}[frac{1+sqrt{-3}}{2}]$ (except the denomitor 1), therefore only $a_1$ and $a_2$ are associated... is this right? Or do I miss something?
Thx for any help on this
abstract-algebra ring-theory associativity euclidean-domain eisenstein-integers
edited Jan 8 at 22:09
Batominovski
33.1k33293
asked Jan 8 at 19:53
franzfffranzff
111
$endgroup$
add a comment |
$begingroup$
I got four elements and want to check whether these elements are associated or not. For
$alpha = frac{1+sqrt{-3}}{2}$
my elements are:
$a_1 = 2 - alpha = frac{3 - sqrt{-3}}{2}$
$a_2 = 1 - 2alpha = -sqrt{-3}$
$ a_3 = 3 + 2alpha = 4 + sqrt{-3}$
$ a_4 = 3 - 2alpha = 2 - sqrt{-3}$
I just calculated $frac{a_1}{a_2} = frac{1-sqrt{-3}}{2}$ and $frac{a_2}{a_1} = frac{-1+sqrt{-3}}{2}$. All other pairs $frac{a_1}{a_3}, frac{a_1}{a_4},frac{a_2}{a_3} frac{a_2}{a_4}, frac{a_3}{a_4}$ give an odd denominator which is not possble in $mathbb{Z}[frac{1+sqrt{-3}}{2}]$ (except the denomitor 1), therefore only $a_1$ and $a_2$ are associated... is this right? Or do I miss something?
Thx for any help on this
abstract-algebra ring-theory associativity euclidean-domain eisenstein-integers
edited Jan 8 at 22:09
Batominovski
33.1k33293
asked Jan 8 at 19:53
franzfffranzff
111
$endgroup$
I got four elements and want to check whether these elements are associated or not. For
$alpha = frac{1+sqrt{-3}}{2}$
my elements are:
$a_1 = 2 - alpha = frac{3 - sqrt{-3}}{2}$
$a_2 = 1 - 2alpha = -sqrt{-3}$
$ a_3 = 3 + 2alpha = 4 + sqrt{-3}$
$ a_4 = 3 - 2alpha = 2 - sqrt{-3}$
I just calculated $frac{a_1}{a_2} = frac{1-sqrt{-3}}{2}$ and $frac{a_2}{a_1} = frac{-1+sqrt{-3}}{2}$. All other pairs $frac{a_1}{a_3}, frac{a_1}{a_4},frac{a_2}{a_3} frac{a_2}{a_4}, frac{a_3}{a_4}$ give an odd denominator which is not possble in $mathbb{Z}[frac{1+sqrt{-3}}{2}]$ (except the denomitor 1), therefore only $a_1$ and $a_2$ are associated... is this right? Or do I miss something?
Thx for any help on this
abstract-algebra ring-theory associativity euclidean-domain eisenstein-integers
abstract-algebra ring-theory associativity euclidean-domain eisenstein-integers
edited Jan 8 at 22:09
Batominovski
33.1k33293
asked Jan 8 at 19:53
franzfffranzff
111
edited Jan 8 at 22:09
Batominovski
33.1k33293
asked Jan 8 at 19:53
franzfffranzff
111
edited Jan 8 at 22:09
Batominovski
33.1k33293
edited Jan 8 at 22:09
Batominovski
33.1k33293
edited Jan 8 at 22:09
Batominovski
33.1k33293
33.1k33293
asked Jan 8 at 19:53
franzfffranzff
111
asked Jan 8 at 19:53
franzfffranzff
111
asked Jan 8 at 19:53
franzfffranzff
111
111
add a comment |
add a comment |
1 Answer
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$begingroup$
I would just compute the norms first. We have
$$N(a_1)=frac{3^2+3}{4}=3,$$
$$N(a_2)=3,$$
$$N(a_3)=4^2+3=19,$$
and
$$N(a_4)=2^2+3=7.$$
So the only candidates to be conjugates are $a_1$ and $a_2$, which you have verified to be the case.
$endgroup$
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1 Answer
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active
oldest
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1 Answer
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active
oldest
votes
active
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votes
$begingroup$
I would just compute the norms first. We have
$$N(a_1)=frac{3^2+3}{4}=3,$$
$$N(a_2)=3,$$
$$N(a_3)=4^2+3=19,$$
and
$$N(a_4)=2^2+3=7.$$
So the only candidates to be conjugates are $a_1$ and $a_2$, which you have verified to be the case.
$endgroup$
add a comment |
$begingroup$
I would just compute the norms first. We have
$$N(a_1)=frac{3^2+3}{4}=3,$$
$$N(a_2)=3,$$
$$N(a_3)=4^2+3=19,$$
and
$$N(a_4)=2^2+3=7.$$
So the only candidates to be conjugates are $a_1$ and $a_2$, which you have verified to be the case.
$endgroup$
add a comment |
$begingroup$
I would just compute the norms first. We have
$$N(a_1)=frac{3^2+3}{4}=3,$$
$$N(a_2)=3,$$
$$N(a_3)=4^2+3=19,$$
and
$$N(a_4)=2^2+3=7.$$
So the only candidates to be conjugates are $a_1$ and $a_2$, which you have verified to be the case.
$endgroup$
I would just compute the norms first. We have
$$N(a_1)=frac{3^2+3}{4}=3,$$
$$N(a_2)=3,$$
$$N(a_3)=4^2+3=19,$$
and
$$N(a_4)=2^2+3=7.$$
So the only candidates to be conjugates are $a_1$ and $a_2$, which you have verified to be the case.
answered Jan 8 at 21:02
user614671
add a comment |
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