Mixing
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The set of the form $ m+nsqrt{-3} $, $ m, ninmathbb Z $ or $ m, n $ are both halves of odd integers is a...
$begingroup$
Let $ D $ be the set of complex numbers of the form $ m+nsqrt{-3} $ where $ m $ and $ n $ are either both in $ mathbb Z $ or are both halves of odd integers. Show that $ D $ is a Euclidean domain relative to $ delta(m+nsqrt{-3})=m^2+3n^2 $.
This is an exercise on Page 149, Basic Algebra I, Jacobson. It is easy to show that $ D $ is a domain. And we have already known that $ mathbb Z[sqrt{-3}] $ is not a Euclidean domain (See: Is $mathbb Z[sqrt{-3}]$ Euclidean under some other norm? ). However, $ mathbb Z [sqrt 3] $ is a Euclidean domain (See: https://math.stackexchange.com/a/1154089/549397 ).
I have checked that $ delta $ is a homomorphism from $ D $ to $ mathbb Z $. My question is:
Why the method of proving $ mathbb Z[sqrt 3] $ is a Euclidean domain is not applicable to $ D $, even, not applicable to $ mathbb Z[sqrt -3] $ ?
abstract-algebra number-theory ring-theory euclidean-domain
edited Jan 20 at 9:45
user26857
39.3k124183
asked Jan 16 at 6:20
user549397user549397
1,5101418
$endgroup$
add a comment |
$begingroup$
Let $ D $ be the set of complex numbers of the form $ m+nsqrt{-3} $ where $ m $ and $ n $ are either both in $ mathbb Z $ or are both halves of odd integers. Show that $ D $ is a Euclidean domain relative to $ delta(m+nsqrt{-3})=m^2+3n^2 $.
This is an exercise on Page 149, Basic Algebra I, Jacobson. It is easy to show that $ D $ is a domain. And we have already known that $ mathbb Z[sqrt{-3}] $ is not a Euclidean domain (See: Is $mathbb Z[sqrt{-3}]$ Euclidean under some other norm? ). However, $ mathbb Z [sqrt 3] $ is a Euclidean domain (See: https://math.stackexchange.com/a/1154089/549397 ).
I have checked that $ delta $ is a homomorphism from $ D $ to $ mathbb Z $. My question is:
Why the method of proving $ mathbb Z[sqrt 3] $ is a Euclidean domain is not applicable to $ D $, even, not applicable to $ mathbb Z[sqrt -3] $ ?
abstract-algebra number-theory ring-theory euclidean-domain
edited Jan 20 at 9:45
user26857
39.3k124183
asked Jan 16 at 6:20
user549397user549397
1,5101418
$endgroup$
$begingroup$
Well, why do you assume it isn't?
$endgroup$
– Eric Wofsey
Jan 16 at 6:29
$begingroup$
@EricWofsey You mean the method proving $ mathbb Z [sqrt 3]$ ? Since $ mathbb Z [sqrt -3] $ is not a Euclidean domain
$endgroup$
– user549397
Jan 16 at 6:33
$begingroup$
Right, that makes sense for $mathbb{Z}[sqrt{-3}]$, but what about $D$?
$endgroup$
– Eric Wofsey
Jan 16 at 6:34
add a comment |
$begingroup$
Let $ D $ be the set of complex numbers of the form $ m+nsqrt{-3} $ where $ m $ and $ n $ are either both in $ mathbb Z $ or are both halves of odd integers. Show that $ D $ is a Euclidean domain relative to $ delta(m+nsqrt{-3})=m^2+3n^2 $.
This is an exercise on Page 149, Basic Algebra I, Jacobson. It is easy to show that $ D $ is a domain. And we have already known that $ mathbb Z[sqrt{-3}] $ is not a Euclidean domain (See: Is $mathbb Z[sqrt{-3}]$ Euclidean under some other norm? ). However, $ mathbb Z [sqrt 3] $ is a Euclidean domain (See: https://math.stackexchange.com/a/1154089/549397 ).
I have checked that $ delta $ is a homomorphism from $ D $ to $ mathbb Z $. My question is:
Why the method of proving $ mathbb Z[sqrt 3] $ is a Euclidean domain is not applicable to $ D $, even, not applicable to $ mathbb Z[sqrt -3] $ ?
abstract-algebra number-theory ring-theory euclidean-domain
edited Jan 20 at 9:45
user26857
39.3k124183
asked Jan 16 at 6:20
user549397user549397
1,5101418
$endgroup$
Let $ D $ be the set of complex numbers of the form $ m+nsqrt{-3} $ where $ m $ and $ n $ are either both in $ mathbb Z $ or are both halves of odd integers. Show that $ D $ is a Euclidean domain relative to $ delta(m+nsqrt{-3})=m^2+3n^2 $.
This is an exercise on Page 149, Basic Algebra I, Jacobson. It is easy to show that $ D $ is a domain. And we have already known that $ mathbb Z[sqrt{-3}] $ is not a Euclidean domain (See: Is $mathbb Z[sqrt{-3}]$ Euclidean under some other norm? ). However, $ mathbb Z [sqrt 3] $ is a Euclidean domain (See: https://math.stackexchange.com/a/1154089/549397 ).
I have checked that $ delta $ is a homomorphism from $ D $ to $ mathbb Z $. My question is:
Why the method of proving $ mathbb Z[sqrt 3] $ is a Euclidean domain is not applicable to $ D $, even, not applicable to $ mathbb Z[sqrt -3] $ ?
abstract-algebra number-theory ring-theory euclidean-domain
abstract-algebra number-theory ring-theory euclidean-domain
edited Jan 20 at 9:45
user26857
39.3k124183
asked Jan 16 at 6:20
user549397user549397
1,5101418
edited Jan 20 at 9:45
user26857
39.3k124183
asked Jan 16 at 6:20
user549397user549397
1,5101418
edited Jan 20 at 9:45
user26857
39.3k124183
edited Jan 20 at 9:45
user26857
39.3k124183
edited Jan 20 at 9:45
user26857
39.3k124183
39.3k124183
asked Jan 16 at 6:20
user549397user549397
1,5101418
asked Jan 16 at 6:20
user549397user549397
1,5101418
asked Jan 16 at 6:20
user549397user549397
1,5101418
1,5101418
$begingroup$
Well, why do you assume it isn't?
$endgroup$
– Eric Wofsey
Jan 16 at 6:29
$begingroup$
@EricWofsey You mean the method proving $ mathbb Z [sqrt 3]$ ? Since $ mathbb Z [sqrt -3] $ is not a Euclidean domain
$endgroup$
– user549397
Jan 16 at 6:33
$begingroup$
Right, that makes sense for $mathbb{Z}[sqrt{-3}]$, but what about $D$?
$endgroup$
– Eric Wofsey
Jan 16 at 6:34
add a comment |
$begingroup$
Well, why do you assume it isn't?
$endgroup$
– Eric Wofsey
Jan 16 at 6:29
$begingroup$
@EricWofsey You mean the method proving $ mathbb Z [sqrt 3]$ ? Since $ mathbb Z [sqrt -3] $ is not a Euclidean domain
$endgroup$
– user549397
Jan 16 at 6:33
$begingroup$
Right, that makes sense for $mathbb{Z}[sqrt{-3}]$, but what about $D$?
$endgroup$
– Eric Wofsey
Jan 16 at 6:34
$begingroup$
Well, why do you assume it isn't?
$endgroup$
– Eric Wofsey
Jan 16 at 6:29
$begingroup$
Well, why do you assume it isn't?
$endgroup$
– Eric Wofsey
Jan 16 at 6:29
$begingroup$
@EricWofsey You mean the method proving $ mathbb Z [sqrt 3]$ ? Since $ mathbb Z [sqrt -3] $ is not a Euclidean domain
$endgroup$
– user549397
Jan 16 at 6:33
$begingroup$
@EricWofsey You mean the method proving $ mathbb Z [sqrt 3]$ ? Since $ mathbb Z [sqrt -3] $ is not a Euclidean domain
$endgroup$
– user549397
Jan 16 at 6:33
$begingroup$
Right, that makes sense for $mathbb{Z}[sqrt{-3}]$, but what about $D$?
$endgroup$
– Eric Wofsey
Jan 16 at 6:34
$begingroup$
Right, that makes sense for $mathbb{Z}[sqrt{-3}]$, but what about $D$?
$endgroup$
– Eric Wofsey
Jan 16 at 6:34
add a comment |
1 Answer
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oldest
votes
$begingroup$
Well, the way to answer this question is just to go through the proof for $mathbb{Z}[sqrt{3}]$ step by step and see where something special about its norm is used. The only place that happens is in the red inequality below from the end of the proof you linked:
begin{align*}
N(gamma) &= N(beta cdot theta) \
&= N(beta) cdot N(theta) \
&= N(beta) cdot vert (p - r)^2 - 3 (q - s)^2 vert \
&color{red}leq N(beta) cdot max{ (p - r)^2, 3(p - s)^2} \
&leqfrac34 N(beta)\
&< N(beta).
end{align*}
Going through these steps with your function $delta$ on $D$ or $mathbb{Z}[sqrt{-3}]$, we would instead have
begin{align*}
delta(gamma) &= delta(beta cdot theta) \
&= delta(beta) cdot delta(theta) \
&= delta(beta) cdot vert (p - r)^2 color{red}+ 3 (q - s)^2 vert. \
end{align*}
That plus sign instead of a minus sign makes all the difference, since we can no longer say that $vert (p - r)^2+ 3 (q - s)^2 vert$ is at most $max{ (p - r)^2, 3(p - s)^2}$. Instead the best bound we can get on $vert (p - r)^2+ 3 (q - s)^2 vert$ using $|p-r|leq 1/2$, $|q-s|leq 1/2$ is $1$ which lets us conclude $delta(gamma)leq delta(beta)$ instead of $delta(gamma)<delta(beta)$ as we need.
Note, though, that this doesn't mean there isn't any way to adapt the proof for $D$; you'll just have to make more changes. In particular, you'll need to use the fact that the coefficients of elements of $D$ can sometimes be half-integers to make a better choice of $p$ and $q$ in some cases.
answered Jan 16 at 6:36
Eric WofseyEric Wofsey
187k14215344
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Well, the way to answer this question is just to go through the proof for $mathbb{Z}[sqrt{3}]$ step by step and see where something special about its norm is used. The only place that happens is in the red inequality below from the end of the proof you linked:
begin{align*}
N(gamma) &= N(beta cdot theta) \
&= N(beta) cdot N(theta) \
&= N(beta) cdot vert (p - r)^2 - 3 (q - s)^2 vert \
&color{red}leq N(beta) cdot max{ (p - r)^2, 3(p - s)^2} \
&leqfrac34 N(beta)\
&< N(beta).
end{align*}
Going through these steps with your function $delta$ on $D$ or $mathbb{Z}[sqrt{-3}]$, we would instead have
begin{align*}
delta(gamma) &= delta(beta cdot theta) \
&= delta(beta) cdot delta(theta) \
&= delta(beta) cdot vert (p - r)^2 color{red}+ 3 (q - s)^2 vert. \
end{align*}
That plus sign instead of a minus sign makes all the difference, since we can no longer say that $vert (p - r)^2+ 3 (q - s)^2 vert$ is at most $max{ (p - r)^2, 3(p - s)^2}$. Instead the best bound we can get on $vert (p - r)^2+ 3 (q - s)^2 vert$ using $|p-r|leq 1/2$, $|q-s|leq 1/2$ is $1$ which lets us conclude $delta(gamma)leq delta(beta)$ instead of $delta(gamma)<delta(beta)$ as we need.
Note, though, that this doesn't mean there isn't any way to adapt the proof for $D$; you'll just have to make more changes. In particular, you'll need to use the fact that the coefficients of elements of $D$ can sometimes be half-integers to make a better choice of $p$ and $q$ in some cases.
answered Jan 16 at 6:36
Eric WofseyEric Wofsey
187k14215344
$endgroup$
add a comment |
$begingroup$
Well, the way to answer this question is just to go through the proof for $mathbb{Z}[sqrt{3}]$ step by step and see where something special about its norm is used. The only place that happens is in the red inequality below from the end of the proof you linked:
begin{align*}
N(gamma) &= N(beta cdot theta) \
&= N(beta) cdot N(theta) \
&= N(beta) cdot vert (p - r)^2 - 3 (q - s)^2 vert \
&color{red}leq N(beta) cdot max{ (p - r)^2, 3(p - s)^2} \
&leqfrac34 N(beta)\
&< N(beta).
end{align*}
Going through these steps with your function $delta$ on $D$ or $mathbb{Z}[sqrt{-3}]$, we would instead have
begin{align*}
delta(gamma) &= delta(beta cdot theta) \
&= delta(beta) cdot delta(theta) \
&= delta(beta) cdot vert (p - r)^2 color{red}+ 3 (q - s)^2 vert. \
end{align*}
That plus sign instead of a minus sign makes all the difference, since we can no longer say that $vert (p - r)^2+ 3 (q - s)^2 vert$ is at most $max{ (p - r)^2, 3(p - s)^2}$. Instead the best bound we can get on $vert (p - r)^2+ 3 (q - s)^2 vert$ using $|p-r|leq 1/2$, $|q-s|leq 1/2$ is $1$ which lets us conclude $delta(gamma)leq delta(beta)$ instead of $delta(gamma)<delta(beta)$ as we need.
Note, though, that this doesn't mean there isn't any way to adapt the proof for $D$; you'll just have to make more changes. In particular, you'll need to use the fact that the coefficients of elements of $D$ can sometimes be half-integers to make a better choice of $p$ and $q$ in some cases.
answered Jan 16 at 6:36
Eric WofseyEric Wofsey
187k14215344
$endgroup$
add a comment |
$begingroup$
Well, the way to answer this question is just to go through the proof for $mathbb{Z}[sqrt{3}]$ step by step and see where something special about its norm is used. The only place that happens is in the red inequality below from the end of the proof you linked:
begin{align*}
N(gamma) &= N(beta cdot theta) \
&= N(beta) cdot N(theta) \
&= N(beta) cdot vert (p - r)^2 - 3 (q - s)^2 vert \
&color{red}leq N(beta) cdot max{ (p - r)^2, 3(p - s)^2} \
&leqfrac34 N(beta)\
&< N(beta).
end{align*}
Going through these steps with your function $delta$ on $D$ or $mathbb{Z}[sqrt{-3}]$, we would instead have
begin{align*}
delta(gamma) &= delta(beta cdot theta) \
&= delta(beta) cdot delta(theta) \
&= delta(beta) cdot vert (p - r)^2 color{red}+ 3 (q - s)^2 vert. \
end{align*}
That plus sign instead of a minus sign makes all the difference, since we can no longer say that $vert (p - r)^2+ 3 (q - s)^2 vert$ is at most $max{ (p - r)^2, 3(p - s)^2}$. Instead the best bound we can get on $vert (p - r)^2+ 3 (q - s)^2 vert$ using $|p-r|leq 1/2$, $|q-s|leq 1/2$ is $1$ which lets us conclude $delta(gamma)leq delta(beta)$ instead of $delta(gamma)<delta(beta)$ as we need.
Note, though, that this doesn't mean there isn't any way to adapt the proof for $D$; you'll just have to make more changes. In particular, you'll need to use the fact that the coefficients of elements of $D$ can sometimes be half-integers to make a better choice of $p$ and $q$ in some cases.
answered Jan 16 at 6:36
Eric WofseyEric Wofsey
187k14215344
$endgroup$
Well, the way to answer this question is just to go through the proof for $mathbb{Z}[sqrt{3}]$ step by step and see where something special about its norm is used. The only place that happens is in the red inequality below from the end of the proof you linked:
begin{align*}
N(gamma) &= N(beta cdot theta) \
&= N(beta) cdot N(theta) \
&= N(beta) cdot vert (p - r)^2 - 3 (q - s)^2 vert \
&color{red}leq N(beta) cdot max{ (p - r)^2, 3(p - s)^2} \
&leqfrac34 N(beta)\
&< N(beta).
end{align*}
Going through these steps with your function $delta$ on $D$ or $mathbb{Z}[sqrt{-3}]$, we would instead have
begin{align*}
delta(gamma) &= delta(beta cdot theta) \
&= delta(beta) cdot delta(theta) \
&= delta(beta) cdot vert (p - r)^2 color{red}+ 3 (q - s)^2 vert. \
end{align*}
That plus sign instead of a minus sign makes all the difference, since we can no longer say that $vert (p - r)^2+ 3 (q - s)^2 vert$ is at most $max{ (p - r)^2, 3(p - s)^2}$. Instead the best bound we can get on $vert (p - r)^2+ 3 (q - s)^2 vert$ using $|p-r|leq 1/2$, $|q-s|leq 1/2$ is $1$ which lets us conclude $delta(gamma)leq delta(beta)$ instead of $delta(gamma)<delta(beta)$ as we need.
Note, though, that this doesn't mean there isn't any way to adapt the proof for $D$; you'll just have to make more changes. In particular, you'll need to use the fact that the coefficients of elements of $D$ can sometimes be half-integers to make a better choice of $p$ and $q$ in some cases.
answered Jan 16 at 6:36
Eric WofseyEric Wofsey
187k14215344
answered Jan 16 at 6:36
Eric WofseyEric Wofsey
187k14215344
answered Jan 16 at 6:36
Eric WofseyEric Wofsey
187k14215344
answered Jan 16 at 6:36
Eric WofseyEric Wofsey
187k14215344
187k14215344
add a comment |
add a comment |
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$begingroup$
Well, why do you assume it isn't?
$endgroup$
– Eric Wofsey
Jan 16 at 6:29
$begingroup$
@EricWofsey You mean the method proving $ mathbb Z [sqrt 3]$ ? Since $ mathbb Z [sqrt -3] $ is not a Euclidean domain
$endgroup$
– user549397
Jan 16 at 6:33
$begingroup$
Right, that makes sense for $mathbb{Z}[sqrt{-3}]$, but what about $D$?
$endgroup$
– Eric Wofsey
Jan 16 at 6:34