Ideals in Gaussian integers
Let $R:=mathbb{Z}[i]$. Prove that every nonzero prime ideal
$mathfrak{P}$ of $R$ belongs to one of the following families:
$mathfrak{P}=(1+i)R$
$mathfrak{P}=(a+bi)R$ where $a,binmathbb{Z}$ and $a^2+b^2$ is an odd prime $p$ which is congruent to $1$ modulo $4$
$mathfrak{P}=pR$ where $p$ is an odd prime which is congruent to $3$ modulo $4$.
Hint: in case 3), let $alphain R$ be written as $c+id$ with $c,dinmathbb{Z}$ and suppose $alphanotinmathfrak{P}$. Consider
$alphaoverline{alpha}=c^2+d^2$; prove that $p$ does not divide
$c^2+d^2$, so that there exists an integer $e$ such that
$(c^2+d^2)e=1bmod p$. Conclude that
$alphacdotoverline{alpha}e=1bmodmathfrak{P}$.
I can't understand the hints that i'm given. I've proved that $p$ doesn't divide $c^2+d^2$, in fact $p=3bmod 4$ implies that $p$ is also a Gaussian prime, so if it divides $c^2+d^2=(c+di)(c-di)$ then it should divide one of the two factors, which is impossible.
Hence, being $p$ a rational prime, not dividing $c^2+d^2$, it must be coprime to $c^2+d^2$ so that there exists $e$ such that etc. etc.
And now? I have proven that $alpha$ is invertible modulo $mathfrak{P}$. How can i use this?
ring-theory prime-numbers ideals
edited Nov 29 at 9:22
Klangen
1,52111332
asked May 30 '13 at 10:12
bateman
1,8401020
add a comment |
Let $R:=mathbb{Z}[i]$. Prove that every nonzero prime ideal
$mathfrak{P}$ of $R$ belongs to one of the following families:
$mathfrak{P}=(1+i)R$
$mathfrak{P}=(a+bi)R$ where $a,binmathbb{Z}$ and $a^2+b^2$ is an odd prime $p$ which is congruent to $1$ modulo $4$
$mathfrak{P}=pR$ where $p$ is an odd prime which is congruent to $3$ modulo $4$.
Hint: in case 3), let $alphain R$ be written as $c+id$ with $c,dinmathbb{Z}$ and suppose $alphanotinmathfrak{P}$. Consider
$alphaoverline{alpha}=c^2+d^2$; prove that $p$ does not divide
$c^2+d^2$, so that there exists an integer $e$ such that
$(c^2+d^2)e=1bmod p$. Conclude that
$alphacdotoverline{alpha}e=1bmodmathfrak{P}$.
I can't understand the hints that i'm given. I've proved that $p$ doesn't divide $c^2+d^2$, in fact $p=3bmod 4$ implies that $p$ is also a Gaussian prime, so if it divides $c^2+d^2=(c+di)(c-di)$ then it should divide one of the two factors, which is impossible.
Hence, being $p$ a rational prime, not dividing $c^2+d^2$, it must be coprime to $c^2+d^2$ so that there exists $e$ such that etc. etc.
And now? I have proven that $alpha$ is invertible modulo $mathfrak{P}$. How can i use this?
ring-theory prime-numbers ideals
edited Nov 29 at 9:22
Klangen
1,52111332
asked May 30 '13 at 10:12
bateman
1,8401020
This shows that $R/pR$ is a field, so $pR$ is a maximal (hence prime) ideal.
– user26857
Oct 23 '16 at 16:21
Have you proven this theorem?
– Ninja
Oct 20 '17 at 10:49
add a comment |
Let $R:=mathbb{Z}[i]$. Prove that every nonzero prime ideal
$mathfrak{P}$ of $R$ belongs to one of the following families:
$mathfrak{P}=(1+i)R$
$mathfrak{P}=(a+bi)R$ where $a,binmathbb{Z}$ and $a^2+b^2$ is an odd prime $p$ which is congruent to $1$ modulo $4$
$mathfrak{P}=pR$ where $p$ is an odd prime which is congruent to $3$ modulo $4$.
Hint: in case 3), let $alphain R$ be written as $c+id$ with $c,dinmathbb{Z}$ and suppose $alphanotinmathfrak{P}$. Consider
$alphaoverline{alpha}=c^2+d^2$; prove that $p$ does not divide
$c^2+d^2$, so that there exists an integer $e$ such that
$(c^2+d^2)e=1bmod p$. Conclude that
$alphacdotoverline{alpha}e=1bmodmathfrak{P}$.
I can't understand the hints that i'm given. I've proved that $p$ doesn't divide $c^2+d^2$, in fact $p=3bmod 4$ implies that $p$ is also a Gaussian prime, so if it divides $c^2+d^2=(c+di)(c-di)$ then it should divide one of the two factors, which is impossible.
Hence, being $p$ a rational prime, not dividing $c^2+d^2$, it must be coprime to $c^2+d^2$ so that there exists $e$ such that etc. etc.
And now? I have proven that $alpha$ is invertible modulo $mathfrak{P}$. How can i use this?
ring-theory prime-numbers ideals
edited Nov 29 at 9:22
Klangen
1,52111332
asked May 30 '13 at 10:12
bateman
1,8401020
Let $R:=mathbb{Z}[i]$. Prove that every nonzero prime ideal
$mathfrak{P}$ of $R$ belongs to one of the following families:
$mathfrak{P}=(1+i)R$
$mathfrak{P}=(a+bi)R$ where $a,binmathbb{Z}$ and $a^2+b^2$ is an odd prime $p$ which is congruent to $1$ modulo $4$
$mathfrak{P}=pR$ where $p$ is an odd prime which is congruent to $3$ modulo $4$.
Hint: in case 3), let $alphain R$ be written as $c+id$ with $c,dinmathbb{Z}$ and suppose $alphanotinmathfrak{P}$. Consider
$alphaoverline{alpha}=c^2+d^2$; prove that $p$ does not divide
$c^2+d^2$, so that there exists an integer $e$ such that
$(c^2+d^2)e=1bmod p$. Conclude that
$alphacdotoverline{alpha}e=1bmodmathfrak{P}$.
I can't understand the hints that i'm given. I've proved that $p$ doesn't divide $c^2+d^2$, in fact $p=3bmod 4$ implies that $p$ is also a Gaussian prime, so if it divides $c^2+d^2=(c+di)(c-di)$ then it should divide one of the two factors, which is impossible.
Hence, being $p$ a rational prime, not dividing $c^2+d^2$, it must be coprime to $c^2+d^2$ so that there exists $e$ such that etc. etc.
And now? I have proven that $alpha$ is invertible modulo $mathfrak{P}$. How can i use this?
ring-theory prime-numbers ideals
ring-theory prime-numbers ideals
edited Nov 29 at 9:22
Klangen
1,52111332
asked May 30 '13 at 10:12
bateman
1,8401020
edited Nov 29 at 9:22
Klangen
1,52111332
asked May 30 '13 at 10:12
bateman
1,8401020
edited Nov 29 at 9:22
Klangen
1,52111332
edited Nov 29 at 9:22
Klangen
1,52111332
edited Nov 29 at 9:22
Klangen
1,52111332
1,52111332
asked May 30 '13 at 10:12
bateman
1,8401020
asked May 30 '13 at 10:12
bateman
1,8401020
asked May 30 '13 at 10:12
bateman
1,8401020
1,8401020
This shows that $R/pR$ is a field, so $pR$ is a maximal (hence prime) ideal.
– user26857
Oct 23 '16 at 16:21
Have you proven this theorem?
– Ninja
Oct 20 '17 at 10:49
add a comment |
This shows that $R/pR$ is a field, so $pR$ is a maximal (hence prime) ideal.
– user26857
Oct 23 '16 at 16:21
Have you proven this theorem?
– Ninja
Oct 20 '17 at 10:49
This shows that $R/pR$ is a field, so $pR$ is a maximal (hence prime) ideal.
– user26857
Oct 23 '16 at 16:21
This shows that $R/pR$ is a field, so $pR$ is a maximal (hence prime) ideal.
– user26857
Oct 23 '16 at 16:21
Have you proven this theorem?
– Ninja
Oct 20 '17 at 10:49
Have you proven this theorem?
– Ninja
Oct 20 '17 at 10:49
add a comment |
1 Answer
1
active
oldest
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The definition of a prime ideal $P subset R$ is that for any $x,y in R$ such that $xy in P$ either $x in P$ or $y in P$ (or both).
In your case, let $x,y in R$ be such that $xy in P$. Assume for a contradiction that neither $xin P$ nor $y in P $. Then both $x$ and $y$ are invertible mod $P$(as you've already proved) and therefore so is their product! So their product cannot possible lie in $P$(it would imply $1 in P$) unless $P=R$ which is clearly impossible.
answered Nov 29 at 9:48
Sorin Tirc
94710
add a comment |
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1 Answer
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The definition of a prime ideal $P subset R$ is that for any $x,y in R$ such that $xy in P$ either $x in P$ or $y in P$ (or both).
In your case, let $x,y in R$ be such that $xy in P$. Assume for a contradiction that neither $xin P$ nor $y in P $. Then both $x$ and $y$ are invertible mod $P$(as you've already proved) and therefore so is their product! So their product cannot possible lie in $P$(it would imply $1 in P$) unless $P=R$ which is clearly impossible.
answered Nov 29 at 9:48
Sorin Tirc
94710
add a comment |
The definition of a prime ideal $P subset R$ is that for any $x,y in R$ such that $xy in P$ either $x in P$ or $y in P$ (or both).
In your case, let $x,y in R$ be such that $xy in P$. Assume for a contradiction that neither $xin P$ nor $y in P $. Then both $x$ and $y$ are invertible mod $P$(as you've already proved) and therefore so is their product! So their product cannot possible lie in $P$(it would imply $1 in P$) unless $P=R$ which is clearly impossible.
answered Nov 29 at 9:48
Sorin Tirc
94710
add a comment |
The definition of a prime ideal $P subset R$ is that for any $x,y in R$ such that $xy in P$ either $x in P$ or $y in P$ (or both).
In your case, let $x,y in R$ be such that $xy in P$. Assume for a contradiction that neither $xin P$ nor $y in P $. Then both $x$ and $y$ are invertible mod $P$(as you've already proved) and therefore so is their product! So their product cannot possible lie in $P$(it would imply $1 in P$) unless $P=R$ which is clearly impossible.
answered Nov 29 at 9:48
Sorin Tirc
94710
The definition of a prime ideal $P subset R$ is that for any $x,y in R$ such that $xy in P$ either $x in P$ or $y in P$ (or both).
In your case, let $x,y in R$ be such that $xy in P$. Assume for a contradiction that neither $xin P$ nor $y in P $. Then both $x$ and $y$ are invertible mod $P$(as you've already proved) and therefore so is their product! So their product cannot possible lie in $P$(it would imply $1 in P$) unless $P=R$ which is clearly impossible.
answered Nov 29 at 9:48
Sorin Tirc
94710
answered Nov 29 at 9:48
Sorin Tirc
94710
answered Nov 29 at 9:48
Sorin Tirc
94710
answered Nov 29 at 9:48
Sorin Tirc
94710
94710
add a comment |
add a comment |
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This shows that $R/pR$ is a field, so $pR$ is a maximal (hence prime) ideal.
– user26857
Oct 23 '16 at 16:21
Have you proven this theorem?
– Ninja
Oct 20 '17 at 10:49