Mixing
![Approximating probability of Cumulative Sum of random variables samples from Uniform[-1,1]](https://lh5.googleusercontent.com/-PBP73op69lo/AAAAAAAAAAI/AAAAAAAAUcQ/o9MZrYfCMak/s72-c/photo.jpg?sz=32)
Prove that every nonzero prime ideal is maximal in $mathbb{Z}[sqrt{d}]$
$begingroup$
$d in mathbb{Z}$ is a square-free integer ($d ne 1$, and $d$ has no factors of the form $c^2$ except $c = pm 1$), and let $R=mathbb{Z}[sqrt{d}]= { a+bsqrt{d} mid a,b in mathbb{Z} }$. Prove that every nonzero prime ideal $P subset R$ is a maximal ideal.
I have a possible outline which I think is good enough to follow.
I think that we need to first prove that every ideal $I subset R$ is finitely generated.
So if $I$ is non-zero, then $I cap mathbb{Z}$ is a non-zero ideal in $mathbb{Z}$.
Then I need to find $I cap mathbb{Z} = { xa mid a in mathbb{Z} }$ for some $x in mathbb{Z}$. That way if I let $J$ be the set of all integers $b$ such that $a+bsqrt{d} in I$ for some $ain mathbb{Z}$, then if there exists a integer $y$ such that $J={ yt mid tin mathbb{Z} }$, then there must exist $s in mathbb{Z}$ such that $s+ysqrt{d} in I$.
Then all I need to show is that $I = ( x,s+ysqrt{d} )$.
Now I need to derive that the factor ring $R / P$ is a finite ring without zero divisors, also finite, then since every finite integral domain is a field, every prime ideal $P subset R$ is a maximal ideal, then I'll be done.
abstract-algebra ring-theory field-theory ideals prime-factorization
edited Jan 29 at 14:40
enedil
1,479720
asked Dec 15 '13 at 2:31
PandaManPandaMan
1,19911333
$endgroup$
add a comment |
$begingroup$
$d in mathbb{Z}$ is a square-free integer ($d ne 1$, and $d$ has no factors of the form $c^2$ except $c = pm 1$), and let $R=mathbb{Z}[sqrt{d}]= { a+bsqrt{d} mid a,b in mathbb{Z} }$. Prove that every nonzero prime ideal $P subset R$ is a maximal ideal.
I have a possible outline which I think is good enough to follow.
I think that we need to first prove that every ideal $I subset R$ is finitely generated.
So if $I$ is non-zero, then $I cap mathbb{Z}$ is a non-zero ideal in $mathbb{Z}$.
Then I need to find $I cap mathbb{Z} = { xa mid a in mathbb{Z} }$ for some $x in mathbb{Z}$. That way if I let $J$ be the set of all integers $b$ such that $a+bsqrt{d} in I$ for some $ain mathbb{Z}$, then if there exists a integer $y$ such that $J={ yt mid tin mathbb{Z} }$, then there must exist $s in mathbb{Z}$ such that $s+ysqrt{d} in I$.
Then all I need to show is that $I = ( x,s+ysqrt{d} )$.
Now I need to derive that the factor ring $R / P$ is a finite ring without zero divisors, also finite, then since every finite integral domain is a field, every prime ideal $P subset R$ is a maximal ideal, then I'll be done.
abstract-algebra ring-theory field-theory ideals prime-factorization
edited Jan 29 at 14:40
enedil
1,479720
asked Dec 15 '13 at 2:31
PandaManPandaMan
1,19911333
$endgroup$
$begingroup$
It's not important for $d$ to be squarefree. The result is true for any $d$ that is not a square (including ${mathbf Z}[sqrt{12}]$, for instance).
$endgroup$
– KCd
Dec 15 '13 at 3:59
add a comment |
$begingroup$
$d in mathbb{Z}$ is a square-free integer ($d ne 1$, and $d$ has no factors of the form $c^2$ except $c = pm 1$), and let $R=mathbb{Z}[sqrt{d}]= { a+bsqrt{d} mid a,b in mathbb{Z} }$. Prove that every nonzero prime ideal $P subset R$ is a maximal ideal.
I have a possible outline which I think is good enough to follow.
I think that we need to first prove that every ideal $I subset R$ is finitely generated.
So if $I$ is non-zero, then $I cap mathbb{Z}$ is a non-zero ideal in $mathbb{Z}$.
Then I need to find $I cap mathbb{Z} = { xa mid a in mathbb{Z} }$ for some $x in mathbb{Z}$. That way if I let $J$ be the set of all integers $b$ such that $a+bsqrt{d} in I$ for some $ain mathbb{Z}$, then if there exists a integer $y$ such that $J={ yt mid tin mathbb{Z} }$, then there must exist $s in mathbb{Z}$ such that $s+ysqrt{d} in I$.
Then all I need to show is that $I = ( x,s+ysqrt{d} )$.
Now I need to derive that the factor ring $R / P$ is a finite ring without zero divisors, also finite, then since every finite integral domain is a field, every prime ideal $P subset R$ is a maximal ideal, then I'll be done.
abstract-algebra ring-theory field-theory ideals prime-factorization
edited Jan 29 at 14:40
enedil
1,479720
asked Dec 15 '13 at 2:31
PandaManPandaMan
1,19911333
$endgroup$
$d in mathbb{Z}$ is a square-free integer ($d ne 1$, and $d$ has no factors of the form $c^2$ except $c = pm 1$), and let $R=mathbb{Z}[sqrt{d}]= { a+bsqrt{d} mid a,b in mathbb{Z} }$. Prove that every nonzero prime ideal $P subset R$ is a maximal ideal.
I have a possible outline which I think is good enough to follow.
I think that we need to first prove that every ideal $I subset R$ is finitely generated.
So if $I$ is non-zero, then $I cap mathbb{Z}$ is a non-zero ideal in $mathbb{Z}$.
Then I need to find $I cap mathbb{Z} = { xa mid a in mathbb{Z} }$ for some $x in mathbb{Z}$. That way if I let $J$ be the set of all integers $b$ such that $a+bsqrt{d} in I$ for some $ain mathbb{Z}$, then if there exists a integer $y$ such that $J={ yt mid tin mathbb{Z} }$, then there must exist $s in mathbb{Z}$ such that $s+ysqrt{d} in I$.
Then all I need to show is that $I = ( x,s+ysqrt{d} )$.
Now I need to derive that the factor ring $R / P$ is a finite ring without zero divisors, also finite, then since every finite integral domain is a field, every prime ideal $P subset R$ is a maximal ideal, then I'll be done.
abstract-algebra ring-theory field-theory ideals prime-factorization
abstract-algebra ring-theory field-theory ideals prime-factorization
edited Jan 29 at 14:40
enedil
1,479720
asked Dec 15 '13 at 2:31
PandaManPandaMan
1,19911333
edited Jan 29 at 14:40
enedil
1,479720
asked Dec 15 '13 at 2:31
PandaManPandaMan
1,19911333
edited Jan 29 at 14:40
enedil
1,479720
edited Jan 29 at 14:40
enedil
1,479720
edited Jan 29 at 14:40
enedil
1,479720
1,479720
asked Dec 15 '13 at 2:31
PandaManPandaMan
1,19911333
asked Dec 15 '13 at 2:31
PandaManPandaMan
1,19911333
asked Dec 15 '13 at 2:31
PandaManPandaMan
1,19911333
1,19911333
$begingroup$
It's not important for $d$ to be squarefree. The result is true for any $d$ that is not a square (including ${mathbf Z}[sqrt{12}]$, for instance).
$endgroup$
– KCd
Dec 15 '13 at 3:59
add a comment |
$begingroup$
It's not important for $d$ to be squarefree. The result is true for any $d$ that is not a square (including ${mathbf Z}[sqrt{12}]$, for instance).
$endgroup$
– KCd
Dec 15 '13 at 3:59
$begingroup$
It's not important for $d$ to be squarefree. The result is true for any $d$ that is not a square (including ${mathbf Z}[sqrt{12}]$, for instance).
$endgroup$
– KCd
Dec 15 '13 at 3:59
$begingroup$
It's not important for $d$ to be squarefree. The result is true for any $d$ that is not a square (including ${mathbf Z}[sqrt{12}]$, for instance).
$endgroup$
– KCd
Dec 15 '13 at 3:59
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Sounds good. In order to show that $I cap Bbb Z$ is a non-zero ideal, it is enough to notice that for $a+bsqrt{d}in I$ you have $(a+bsqrt{d})(a-bsqrt{d}) = a^2 -db^2 =:nin Bbb Z cap I$. Now by writing $R = Bbb Z[X]/(X^2-d)$ you can show that $R/P$ is a quotient of the finite ring $(Bbb Z / nBbb Z)[X]/(X^2-d)$ where $X^2-d$ denotes the reduced polynomial in $(Bbb Z/n Bbb Z)[X]$. So indeed, $R/P$ is finite without zero divisors.
answered Dec 15 '13 at 3:05
benhbenh
6,1061424
$endgroup$
1
$begingroup$
You omitted a crucial point: $nne 0$ because $ldots$
$endgroup$
– Bill Dubuque
Dec 18 '13 at 20:12
$begingroup$
Well, yes... but even if $d$ is a square we can choose a,b appropriately (e.g. by $a mapsto a+1$) to get $n neq 0$.
$endgroup$
– benh
Dec 18 '13 at 23:13
add a comment |
$begingroup$
If the ideal is prime, almost by definition the quotient has no zero divisors.
On the other hand, since $R$ is a finite generated abelian group, the quotient $R/P$ is also a finitely generated abelian group, and to show it is finite it is enough to show that $R/P$ has finite exponent.
If $P$ is non-zero, there is a non-zero element $x=a+bsqrt d$ in $P$, and then $e=a^2-db^2=(a-bsqrt d)xin P$; you can check easily that $eneq0$. It follows that the product of every element of $R/P$ by $e$ is zero, and therefore the exponent of the abelian group $R/P$ divides $e$.
answered Dec 18 '13 at 4:53
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7158290
$endgroup$
$begingroup$
I'm still unsure as to why the factor ring has no zero divisors. I know it may seem obvious but I can't seem to make the connection.
$endgroup$
– BlakeM
Dec 18 '13 at 19:15
1
$begingroup$
Suppose $a$ and $b$ are elements of $R$ such that the product in $R/P$ of their classes is zero. This means that $ab$ is an element of $P$ and, since the ideeal is prime, that one of $a$ or $b$ is in $P$.
$endgroup$
– Mariano Suárez-Álvarez
Dec 18 '13 at 19:17
$begingroup$
Great, this makes sense to me.
$endgroup$
– BlakeM
Dec 18 '13 at 19:25
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Sounds good. In order to show that $I cap Bbb Z$ is a non-zero ideal, it is enough to notice that for $a+bsqrt{d}in I$ you have $(a+bsqrt{d})(a-bsqrt{d}) = a^2 -db^2 =:nin Bbb Z cap I$. Now by writing $R = Bbb Z[X]/(X^2-d)$ you can show that $R/P$ is a quotient of the finite ring $(Bbb Z / nBbb Z)[X]/(X^2-d)$ where $X^2-d$ denotes the reduced polynomial in $(Bbb Z/n Bbb Z)[X]$. So indeed, $R/P$ is finite without zero divisors.
answered Dec 15 '13 at 3:05
benhbenh
6,1061424
$endgroup$
1
$begingroup$
You omitted a crucial point: $nne 0$ because $ldots$
$endgroup$
– Bill Dubuque
Dec 18 '13 at 20:12
$begingroup$
Well, yes... but even if $d$ is a square we can choose a,b appropriately (e.g. by $a mapsto a+1$) to get $n neq 0$.
$endgroup$
– benh
Dec 18 '13 at 23:13
add a comment |
$begingroup$
Sounds good. In order to show that $I cap Bbb Z$ is a non-zero ideal, it is enough to notice that for $a+bsqrt{d}in I$ you have $(a+bsqrt{d})(a-bsqrt{d}) = a^2 -db^2 =:nin Bbb Z cap I$. Now by writing $R = Bbb Z[X]/(X^2-d)$ you can show that $R/P$ is a quotient of the finite ring $(Bbb Z / nBbb Z)[X]/(X^2-d)$ where $X^2-d$ denotes the reduced polynomial in $(Bbb Z/n Bbb Z)[X]$. So indeed, $R/P$ is finite without zero divisors.
answered Dec 15 '13 at 3:05
benhbenh
6,1061424
$endgroup$
1
$begingroup$
You omitted a crucial point: $nne 0$ because $ldots$
$endgroup$
– Bill Dubuque
Dec 18 '13 at 20:12
$begingroup$
Well, yes... but even if $d$ is a square we can choose a,b appropriately (e.g. by $a mapsto a+1$) to get $n neq 0$.
$endgroup$
– benh
Dec 18 '13 at 23:13
add a comment |
$begingroup$
Sounds good. In order to show that $I cap Bbb Z$ is a non-zero ideal, it is enough to notice that for $a+bsqrt{d}in I$ you have $(a+bsqrt{d})(a-bsqrt{d}) = a^2 -db^2 =:nin Bbb Z cap I$. Now by writing $R = Bbb Z[X]/(X^2-d)$ you can show that $R/P$ is a quotient of the finite ring $(Bbb Z / nBbb Z)[X]/(X^2-d)$ where $X^2-d$ denotes the reduced polynomial in $(Bbb Z/n Bbb Z)[X]$. So indeed, $R/P$ is finite without zero divisors.
answered Dec 15 '13 at 3:05
benhbenh
6,1061424
$endgroup$
Sounds good. In order to show that $I cap Bbb Z$ is a non-zero ideal, it is enough to notice that for $a+bsqrt{d}in I$ you have $(a+bsqrt{d})(a-bsqrt{d}) = a^2 -db^2 =:nin Bbb Z cap I$. Now by writing $R = Bbb Z[X]/(X^2-d)$ you can show that $R/P$ is a quotient of the finite ring $(Bbb Z / nBbb Z)[X]/(X^2-d)$ where $X^2-d$ denotes the reduced polynomial in $(Bbb Z/n Bbb Z)[X]$. So indeed, $R/P$ is finite without zero divisors.
answered Dec 15 '13 at 3:05
benhbenh
6,1061424
edited Jan 5 '14 at 2:44
answered Dec 15 '13 at 3:05
benhbenh
6,1061424
answered Dec 15 '13 at 3:05
benhbenh
6,1061424
answered Dec 15 '13 at 3:05
benhbenh
6,1061424
6,1061424
1
$begingroup$
You omitted a crucial point: $nne 0$ because $ldots$
$endgroup$
– Bill Dubuque
Dec 18 '13 at 20:12
$begingroup$
Well, yes... but even if $d$ is a square we can choose a,b appropriately (e.g. by $a mapsto a+1$) to get $n neq 0$.
$endgroup$
– benh
Dec 18 '13 at 23:13
add a comment |
1
$begingroup$
You omitted a crucial point: $nne 0$ because $ldots$
$endgroup$
– Bill Dubuque
Dec 18 '13 at 20:12
$begingroup$
Well, yes... but even if $d$ is a square we can choose a,b appropriately (e.g. by $a mapsto a+1$) to get $n neq 0$.
$endgroup$
– benh
Dec 18 '13 at 23:13
1
1
$begingroup$
You omitted a crucial point: $nne 0$ because $ldots$
$endgroup$
– Bill Dubuque
Dec 18 '13 at 20:12
$begingroup$
You omitted a crucial point: $nne 0$ because $ldots$
$endgroup$
– Bill Dubuque
Dec 18 '13 at 20:12
$begingroup$
Well, yes... but even if $d$ is a square we can choose a,b appropriately (e.g. by $a mapsto a+1$) to get $n neq 0$.
$endgroup$
– benh
Dec 18 '13 at 23:13
$begingroup$
Well, yes... but even if $d$ is a square we can choose a,b appropriately (e.g. by $a mapsto a+1$) to get $n neq 0$.
$endgroup$
– benh
Dec 18 '13 at 23:13
add a comment |
$begingroup$
If the ideal is prime, almost by definition the quotient has no zero divisors.
On the other hand, since $R$ is a finite generated abelian group, the quotient $R/P$ is also a finitely generated abelian group, and to show it is finite it is enough to show that $R/P$ has finite exponent.
If $P$ is non-zero, there is a non-zero element $x=a+bsqrt d$ in $P$, and then $e=a^2-db^2=(a-bsqrt d)xin P$; you can check easily that $eneq0$. It follows that the product of every element of $R/P$ by $e$ is zero, and therefore the exponent of the abelian group $R/P$ divides $e$.
answered Dec 18 '13 at 4:53
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7158290
$endgroup$
$begingroup$
I'm still unsure as to why the factor ring has no zero divisors. I know it may seem obvious but I can't seem to make the connection.
$endgroup$
– BlakeM
Dec 18 '13 at 19:15
1
$begingroup$
Suppose $a$ and $b$ are elements of $R$ such that the product in $R/P$ of their classes is zero. This means that $ab$ is an element of $P$ and, since the ideeal is prime, that one of $a$ or $b$ is in $P$.
$endgroup$
– Mariano Suárez-Álvarez
Dec 18 '13 at 19:17
$begingroup$
Great, this makes sense to me.
$endgroup$
– BlakeM
Dec 18 '13 at 19:25
add a comment |
$begingroup$
If the ideal is prime, almost by definition the quotient has no zero divisors.
On the other hand, since $R$ is a finite generated abelian group, the quotient $R/P$ is also a finitely generated abelian group, and to show it is finite it is enough to show that $R/P$ has finite exponent.
If $P$ is non-zero, there is a non-zero element $x=a+bsqrt d$ in $P$, and then $e=a^2-db^2=(a-bsqrt d)xin P$; you can check easily that $eneq0$. It follows that the product of every element of $R/P$ by $e$ is zero, and therefore the exponent of the abelian group $R/P$ divides $e$.
answered Dec 18 '13 at 4:53
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7158290
$endgroup$
$begingroup$
I'm still unsure as to why the factor ring has no zero divisors. I know it may seem obvious but I can't seem to make the connection.
$endgroup$
– BlakeM
Dec 18 '13 at 19:15
1
$begingroup$
Suppose $a$ and $b$ are elements of $R$ such that the product in $R/P$ of their classes is zero. This means that $ab$ is an element of $P$ and, since the ideeal is prime, that one of $a$ or $b$ is in $P$.
$endgroup$
– Mariano Suárez-Álvarez
Dec 18 '13 at 19:17
$begingroup$
Great, this makes sense to me.
$endgroup$
– BlakeM
Dec 18 '13 at 19:25
add a comment |
$begingroup$
If the ideal is prime, almost by definition the quotient has no zero divisors.
On the other hand, since $R$ is a finite generated abelian group, the quotient $R/P$ is also a finitely generated abelian group, and to show it is finite it is enough to show that $R/P$ has finite exponent.
If $P$ is non-zero, there is a non-zero element $x=a+bsqrt d$ in $P$, and then $e=a^2-db^2=(a-bsqrt d)xin P$; you can check easily that $eneq0$. It follows that the product of every element of $R/P$ by $e$ is zero, and therefore the exponent of the abelian group $R/P$ divides $e$.
answered Dec 18 '13 at 4:53
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7158290
$endgroup$
If the ideal is prime, almost by definition the quotient has no zero divisors.
On the other hand, since $R$ is a finite generated abelian group, the quotient $R/P$ is also a finitely generated abelian group, and to show it is finite it is enough to show that $R/P$ has finite exponent.
If $P$ is non-zero, there is a non-zero element $x=a+bsqrt d$ in $P$, and then $e=a^2-db^2=(a-bsqrt d)xin P$; you can check easily that $eneq0$. It follows that the product of every element of $R/P$ by $e$ is zero, and therefore the exponent of the abelian group $R/P$ divides $e$.
answered Dec 18 '13 at 4:53
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7158290
answered Dec 18 '13 at 4:53
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7158290
answered Dec 18 '13 at 4:53
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7158290
answered Dec 18 '13 at 4:53
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7158290
112k7158290
$begingroup$
I'm still unsure as to why the factor ring has no zero divisors. I know it may seem obvious but I can't seem to make the connection.
$endgroup$
– BlakeM
Dec 18 '13 at 19:15
1
$begingroup$
Suppose $a$ and $b$ are elements of $R$ such that the product in $R/P$ of their classes is zero. This means that $ab$ is an element of $P$ and, since the ideeal is prime, that one of $a$ or $b$ is in $P$.
$endgroup$
– Mariano Suárez-Álvarez
Dec 18 '13 at 19:17
$begingroup$
Great, this makes sense to me.
$endgroup$
– BlakeM
Dec 18 '13 at 19:25
add a comment |
$begingroup$
I'm still unsure as to why the factor ring has no zero divisors. I know it may seem obvious but I can't seem to make the connection.
$endgroup$
– BlakeM
Dec 18 '13 at 19:15
1
$begingroup$
Suppose $a$ and $b$ are elements of $R$ such that the product in $R/P$ of their classes is zero. This means that $ab$ is an element of $P$ and, since the ideeal is prime, that one of $a$ or $b$ is in $P$.
$endgroup$
– Mariano Suárez-Álvarez
Dec 18 '13 at 19:17
$begingroup$
Great, this makes sense to me.
$endgroup$
– BlakeM
Dec 18 '13 at 19:25
$begingroup$
I'm still unsure as to why the factor ring has no zero divisors. I know it may seem obvious but I can't seem to make the connection.
$endgroup$
– BlakeM
Dec 18 '13 at 19:15
$begingroup$
I'm still unsure as to why the factor ring has no zero divisors. I know it may seem obvious but I can't seem to make the connection.
$endgroup$
– BlakeM
Dec 18 '13 at 19:15
1
1
$begingroup$
Suppose $a$ and $b$ are elements of $R$ such that the product in $R/P$ of their classes is zero. This means that $ab$ is an element of $P$ and, since the ideeal is prime, that one of $a$ or $b$ is in $P$.
$endgroup$
– Mariano Suárez-Álvarez
Dec 18 '13 at 19:17
$begingroup$
Suppose $a$ and $b$ are elements of $R$ such that the product in $R/P$ of their classes is zero. This means that $ab$ is an element of $P$ and, since the ideeal is prime, that one of $a$ or $b$ is in $P$.
$endgroup$
– Mariano Suárez-Álvarez
Dec 18 '13 at 19:17
$begingroup$
Great, this makes sense to me.
$endgroup$
– BlakeM
Dec 18 '13 at 19:25
$begingroup$
Great, this makes sense to me.
$endgroup$
– BlakeM
Dec 18 '13 at 19:25
add a comment |
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$begingroup$
It's not important for $d$ to be squarefree. The result is true for any $d$ that is not a square (including ${mathbf Z}[sqrt{12}]$, for instance).
$endgroup$
– KCd
Dec 15 '13 at 3:59