Why $R/(a+bi)$ is a finite field with characteristic $p$ where $a^2+b^2=p$ [duplicate]
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This question already has an answer here:
Quotient ring of Gaussian integers
6 answers
Suppose $R=mathbb{Z}[i]$ is Gaussian domain , and $a^2+b^2=p$ is prime. Denote $alpha=a+bi$ and $I=(alpha)$. Prove $R/I$ is a finite field with characteristic p.
My attempts: First prove $alpha $ is irreducible, hence $I$ is a maximal ideal and it follows $R/I$ is a field. And $R=mathbb{Z}[i]$ is an Euclidean domain. Hence for $xin mathbb{Z}[i],x=qalpha+r$ where $N(r)lt N(alpha)$ , so there are finite elements in $R/I$. Thus $R/I$ is a finite filed. But I can't figure out why the characteristic is $p$.
Thanks for your hints.
abstract-algebra finite-fields
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marked as duplicate by André 3000, Namaste
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Jan 3 at 18:47
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 ring of Gaussian integers
6 answers
Suppose $R=mathbb{Z}[i]$ is Gaussian domain , and $a^2+b^2=p$ is prime. Denote $alpha=a+bi$ and $I=(alpha)$. Prove $R/I$ is a finite field with characteristic p.
My attempts: First prove $alpha $ is irreducible, hence $I$ is a maximal ideal and it follows $R/I$ is a field. And $R=mathbb{Z}[i]$ is an Euclidean domain. Hence for $xin mathbb{Z}[i],x=qalpha+r$ where $N(r)lt N(alpha)$ , so there are finite elements in $R/I$. Thus $R/I$ is a finite filed. But I can't figure out why the characteristic is $p$.
Thanks for your hints.
abstract-algebra finite-fields
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marked as duplicate by André 3000, Namaste
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Jan 3 at 18:47
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.
1
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$R/I$ has $p$ elements.
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– Lord Shark the Unknown
Jan 3 at 17:13
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always $mathbb Z[i]/(a+bi)cong mathbb Z/N(a+bi)mathbb Z$ where $(a,b)=1$
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– Mustafa
Jan 3 at 17:14
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How to determine the number of elements in $R/I$?
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– Jaqen Chou
Jan 3 at 17:18
add a comment |
$begingroup$
This question already has an answer here:
Quotient ring of Gaussian integers
6 answers
Suppose $R=mathbb{Z}[i]$ is Gaussian domain , and $a^2+b^2=p$ is prime. Denote $alpha=a+bi$ and $I=(alpha)$. Prove $R/I$ is a finite field with characteristic p.
My attempts: First prove $alpha $ is irreducible, hence $I$ is a maximal ideal and it follows $R/I$ is a field. And $R=mathbb{Z}[i]$ is an Euclidean domain. Hence for $xin mathbb{Z}[i],x=qalpha+r$ where $N(r)lt N(alpha)$ , so there are finite elements in $R/I$. Thus $R/I$ is a finite filed. But I can't figure out why the characteristic is $p$.
Thanks for your hints.
abstract-algebra finite-fields
$endgroup$
This question already has an answer here:
Quotient ring of Gaussian integers
6 answers
Suppose $R=mathbb{Z}[i]$ is Gaussian domain , and $a^2+b^2=p$ is prime. Denote $alpha=a+bi$ and $I=(alpha)$. Prove $R/I$ is a finite field with characteristic p.
My attempts: First prove $alpha $ is irreducible, hence $I$ is a maximal ideal and it follows $R/I$ is a field. And $R=mathbb{Z}[i]$ is an Euclidean domain. Hence for $xin mathbb{Z}[i],x=qalpha+r$ where $N(r)lt N(alpha)$ , so there are finite elements in $R/I$. Thus $R/I$ is a finite filed. But I can't figure out why the characteristic is $p$.
Thanks for your hints.
This question already has an answer here:
Quotient ring of Gaussian integers
6 answers
abstract-algebra finite-fields
abstract-algebra finite-fields
asked Jan 3 at 17:13
Jaqen ChouJaqen Chou
460110
460110
marked as duplicate by André 3000, Namaste
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Jan 3 at 18:47
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marked as duplicate by André 3000, Namaste
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Jan 3 at 18:47
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.
1
$begingroup$
$R/I$ has $p$ elements.
$endgroup$
– Lord Shark the Unknown
Jan 3 at 17:13
$begingroup$
always $mathbb Z[i]/(a+bi)cong mathbb Z/N(a+bi)mathbb Z$ where $(a,b)=1$
$endgroup$
– Mustafa
Jan 3 at 17:14
$begingroup$
How to determine the number of elements in $R/I$?
$endgroup$
– Jaqen Chou
Jan 3 at 17:18
add a comment |
1
$begingroup$
$R/I$ has $p$ elements.
$endgroup$
– Lord Shark the Unknown
Jan 3 at 17:13
$begingroup$
always $mathbb Z[i]/(a+bi)cong mathbb Z/N(a+bi)mathbb Z$ where $(a,b)=1$
$endgroup$
– Mustafa
Jan 3 at 17:14
$begingroup$
How to determine the number of elements in $R/I$?
$endgroup$
– Jaqen Chou
Jan 3 at 17:18
1
1
$begingroup$
$R/I$ has $p$ elements.
$endgroup$
– Lord Shark the Unknown
Jan 3 at 17:13
$begingroup$
$R/I$ has $p$ elements.
$endgroup$
– Lord Shark the Unknown
Jan 3 at 17:13
$begingroup$
always $mathbb Z[i]/(a+bi)cong mathbb Z/N(a+bi)mathbb Z$ where $(a,b)=1$
$endgroup$
– Mustafa
Jan 3 at 17:14
$begingroup$
always $mathbb Z[i]/(a+bi)cong mathbb Z/N(a+bi)mathbb Z$ where $(a,b)=1$
$endgroup$
– Mustafa
Jan 3 at 17:14
$begingroup$
How to determine the number of elements in $R/I$?
$endgroup$
– Jaqen Chou
Jan 3 at 17:18
$begingroup$
How to determine the number of elements in $R/I$?
$endgroup$
– Jaqen Chou
Jan 3 at 17:18
add a comment |
1 Answer
1
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$p = a^2+b^2 = bar alpha alpha in I$ implies $p=0$ in $R/I$.
If you know that $R/I$ is a field, then it must have characteristic $p$.
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$p = a^2+b^2 = bar alpha alpha in I$ implies $p=0$ in $R/I$.
If you know that $R/I$ is a field, then it must have characteristic $p$.
$endgroup$
add a comment |
$begingroup$
$p = a^2+b^2 = bar alpha alpha in I$ implies $p=0$ in $R/I$.
If you know that $R/I$ is a field, then it must have characteristic $p$.
$endgroup$
add a comment |
$begingroup$
$p = a^2+b^2 = bar alpha alpha in I$ implies $p=0$ in $R/I$.
If you know that $R/I$ is a field, then it must have characteristic $p$.
$endgroup$
$p = a^2+b^2 = bar alpha alpha in I$ implies $p=0$ in $R/I$.
If you know that $R/I$ is a field, then it must have characteristic $p$.
answered Jan 3 at 17:19
lhflhf
166k10171398
166k10171398
add a comment |
add a comment |
1
$begingroup$
$R/I$ has $p$ elements.
$endgroup$
– Lord Shark the Unknown
Jan 3 at 17:13
$begingroup$
always $mathbb Z[i]/(a+bi)cong mathbb Z/N(a+bi)mathbb Z$ where $(a,b)=1$
$endgroup$
– Mustafa
Jan 3 at 17:14
$begingroup$
How to determine the number of elements in $R/I$?
$endgroup$
– Jaqen Chou
Jan 3 at 17:18