Mixing
Show that $ | mathbb Z[sqrt{-1}]/(a+bsqrt{-1}) |=a^2+b^2 $ [duplicate]
$begingroup$
This question already has an answer here:
Quotient rings of Gaussian integers [duplicate]
3 answers
Equivalence of two characterizations of the norm of an algebraic integer.
3 answers
Let $ a+bsqrt{-1} $ be a non-zero element of the ring of Gaussian integers $ mathbb Z[sqrt{-1}]. $ Show that $ | mathbb Z[sqrt{-1}]/(a+bsqrt{-1}) |=a^2+b^2 $. [Jacobson, Basic Algebra, P202.3]
My attempt:
Since $ mathbb{Z}[sqrt{-1}] $ is a p.i.d. and we can factor $ a+bsqrt{-1} $ into primes: $ a+bsqrt{-1}=p_1^{j_1}p_2^{j_2}...p_n^{j_n} $, $ p_i $ are primes in $ mathbb Z[sqrt {-1}] $, then we only have to prove the case for the primary quotient submodule in the following sense:
$$ | mathbb Z[sqrt{-1}]/(p_i^{j_i}) |=a^2+b^2, ,text{where $ a+bi=p_i^{j_i} $}. $$
From here I am stuck......
abstract-algebra number-theory modules algebraic-number-theory gaussian-integers
asked Jan 27 at 14:51
user549397user549397
1,5511418
$endgroup$
marked as duplicate by Bill Dubuque abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 27 at 15:18
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Quotient rings of Gaussian integers [duplicate]
3 answers
Equivalence of two characterizations of the norm of an algebraic integer.
3 answers
Let $ a+bsqrt{-1} $ be a non-zero element of the ring of Gaussian integers $ mathbb Z[sqrt{-1}]. $ Show that $ | mathbb Z[sqrt{-1}]/(a+bsqrt{-1}) |=a^2+b^2 $. [Jacobson, Basic Algebra, P202.3]
My attempt:
Since $ mathbb{Z}[sqrt{-1}] $ is a p.i.d. and we can factor $ a+bsqrt{-1} $ into primes: $ a+bsqrt{-1}=p_1^{j_1}p_2^{j_2}...p_n^{j_n} $, $ p_i $ are primes in $ mathbb Z[sqrt {-1}] $, then we only have to prove the case for the primary quotient submodule in the following sense:
$$ | mathbb Z[sqrt{-1}]/(p_i^{j_i}) |=a^2+b^2, ,text{where $ a+bi=p_i^{j_i} $}. $$
From here I am stuck......
abstract-algebra number-theory modules algebraic-number-theory gaussian-integers
asked Jan 27 at 14:51
user549397user549397
1,5511418
$endgroup$
marked as duplicate by Bill Dubuque abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 27 at 15:18
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Quotient rings of Gaussian integers [duplicate]
3 answers
Equivalence of two characterizations of the norm of an algebraic integer.
3 answers
Let $ a+bsqrt{-1} $ be a non-zero element of the ring of Gaussian integers $ mathbb Z[sqrt{-1}]. $ Show that $ | mathbb Z[sqrt{-1}]/(a+bsqrt{-1}) |=a^2+b^2 $. [Jacobson, Basic Algebra, P202.3]
My attempt:
Since $ mathbb{Z}[sqrt{-1}] $ is a p.i.d. and we can factor $ a+bsqrt{-1} $ into primes: $ a+bsqrt{-1}=p_1^{j_1}p_2^{j_2}...p_n^{j_n} $, $ p_i $ are primes in $ mathbb Z[sqrt {-1}] $, then we only have to prove the case for the primary quotient submodule in the following sense:
$$ | mathbb Z[sqrt{-1}]/(p_i^{j_i}) |=a^2+b^2, ,text{where $ a+bi=p_i^{j_i} $}. $$
From here I am stuck......
abstract-algebra number-theory modules algebraic-number-theory gaussian-integers
asked Jan 27 at 14:51
user549397user549397
1,5511418
$endgroup$
This question already has an answer here:
Quotient rings of Gaussian integers [duplicate]
3 answers
Equivalence of two characterizations of the norm of an algebraic integer.
3 answers
Let $ a+bsqrt{-1} $ be a non-zero element of the ring of Gaussian integers $ mathbb Z[sqrt{-1}]. $ Show that $ | mathbb Z[sqrt{-1}]/(a+bsqrt{-1}) |=a^2+b^2 $. [Jacobson, Basic Algebra, P202.3]
My attempt:
Since $ mathbb{Z}[sqrt{-1}] $ is a p.i.d. and we can factor $ a+bsqrt{-1} $ into primes: $ a+bsqrt{-1}=p_1^{j_1}p_2^{j_2}...p_n^{j_n} $, $ p_i $ are primes in $ mathbb Z[sqrt {-1}] $, then we only have to prove the case for the primary quotient submodule in the following sense:
$$ | mathbb Z[sqrt{-1}]/(p_i^{j_i}) |=a^2+b^2, ,text{where $ a+bi=p_i^{j_i} $}. $$
From here I am stuck......
This question already has an answer here:
Quotient rings of Gaussian integers [duplicate]
3 answers
Equivalence of two characterizations of the norm of an algebraic integer.
3 answers
abstract-algebra number-theory modules algebraic-number-theory gaussian-integers
abstract-algebra number-theory modules algebraic-number-theory gaussian-integers
asked Jan 27 at 14:51
user549397user549397
1,5511418
asked Jan 27 at 14:51
user549397user549397
1,5511418
asked Jan 27 at 14:51
user549397user549397
1,5511418
asked Jan 27 at 14:51
user549397user549397
1,5511418
asked Jan 27 at 14:51
user549397user549397
1,5511418
1,5511418
marked as duplicate by Bill Dubuque abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 27 at 15:18
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Bill Dubuque abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 27 at 15:18
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The ideal $(a+bsqrt{-1})$ forms a sublattice of $Bbb Z[sqrt{-1}]$ in the plane. The side length of a mesh square is clearly $sqrt{a^2+b^2}$, so the area of a mesh is $a^2+b^2$ and hence contains essentially $a^2+b^2$ mesh squares of the $Bbb Z[sqrt{-1}]$ lattice.
edited Jan 27 at 17:30
Thomas Shelby
4,4092726
answered Jan 27 at 15:01
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The ideal $(a+bsqrt{-1})$ forms a sublattice of $Bbb Z[sqrt{-1}]$ in the plane. The side length of a mesh square is clearly $sqrt{a^2+b^2}$, so the area of a mesh is $a^2+b^2$ and hence contains essentially $a^2+b^2$ mesh squares of the $Bbb Z[sqrt{-1}]$ lattice.
edited Jan 27 at 17:30
Thomas Shelby
4,4092726
answered Jan 27 at 15:01
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
add a comment |
$begingroup$
The ideal $(a+bsqrt{-1})$ forms a sublattice of $Bbb Z[sqrt{-1}]$ in the plane. The side length of a mesh square is clearly $sqrt{a^2+b^2}$, so the area of a mesh is $a^2+b^2$ and hence contains essentially $a^2+b^2$ mesh squares of the $Bbb Z[sqrt{-1}]$ lattice.
edited Jan 27 at 17:30
Thomas Shelby
4,4092726
answered Jan 27 at 15:01
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
add a comment |
$begingroup$
The ideal $(a+bsqrt{-1})$ forms a sublattice of $Bbb Z[sqrt{-1}]$ in the plane. The side length of a mesh square is clearly $sqrt{a^2+b^2}$, so the area of a mesh is $a^2+b^2$ and hence contains essentially $a^2+b^2$ mesh squares of the $Bbb Z[sqrt{-1}]$ lattice.
edited Jan 27 at 17:30
Thomas Shelby
4,4092726
answered Jan 27 at 15:01
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
The ideal $(a+bsqrt{-1})$ forms a sublattice of $Bbb Z[sqrt{-1}]$ in the plane. The side length of a mesh square is clearly $sqrt{a^2+b^2}$, so the area of a mesh is $a^2+b^2$ and hence contains essentially $a^2+b^2$ mesh squares of the $Bbb Z[sqrt{-1}]$ lattice.
edited Jan 27 at 17:30
Thomas Shelby
4,4092726
answered Jan 27 at 15:01
Hagen von EitzenHagen von Eitzen
283k23272507
edited Jan 27 at 17:30
Thomas Shelby
4,4092726
edited Jan 27 at 17:30
Thomas Shelby
4,4092726
edited Jan 27 at 17:30
Thomas Shelby
4,4092726
4,4092726
answered Jan 27 at 15:01
Hagen von EitzenHagen von Eitzen
283k23272507
answered Jan 27 at 15:01
Hagen von EitzenHagen von Eitzen
283k23272507
answered Jan 27 at 15:01
Hagen von EitzenHagen von Eitzen
283k23272507
283k23272507
add a comment |
add a comment |
