Mixing
Noetherian prime localizations does not imply that whole ring is Noetherian
$begingroup$
We know the following fact: If $A$ is Noetherian ring and $S$ is multiplicatively closed subset of $A$ then $S^{-1}A$ is also Noetherian ring.
Corollary: If $A$ is Noetherian ring then for any prime ideal $P$ the ring $A_P$ is also Noetherian.
However, the converse of this corollary is not true.
Let's demonstrate it in the following example: Take $R=prod limits_{i=1}^{infty}mathbb{Z}_2$ and we see that $R$ is not Noetherian because it is not finitely generated.
We see that for any $xin R$ we have $x^2=x$, i.e. $R$ is a Boolean ring. Let $P$ be a prime ideal in $R$ then we can show that $R/Pcong mathbb{Z}_2$ which means that $R/P$ is a field. Hence $P$ is maximal ideal.
Thus maximal and prime ideals in $R$ are the same
Now we want to consider the localizations $R_P$ for any prime ideal $P$.
I have the following two questions (I have spent couple of hours but I was not able to answer these questions).
1) How to show that $R$ is local ring? i.e. there is only one maximal ideal (i.e. one prime ideal).
2) If we can show this then $P=0$ because intersection of all prime ideals is the set of nilpotent elements in $R$ but $R$ being boolean has no nonzero nilpotent elements. Hence $P=0$.
How it follows that $R_P$ is field?
I know that this question is already exists in forum. But they are not detailed. Could anyone help me to asnwer these questions?
Would be very grateful for help! Also please do not duplicate this question!
abstract-algebra ring-theory commutative-algebra
asked Jan 29 at 14:38
ZFRZFR
5,29331540
$endgroup$
add a comment |
$begingroup$
We know the following fact: If $A$ is Noetherian ring and $S$ is multiplicatively closed subset of $A$ then $S^{-1}A$ is also Noetherian ring.
Corollary: If $A$ is Noetherian ring then for any prime ideal $P$ the ring $A_P$ is also Noetherian.
However, the converse of this corollary is not true.
Let's demonstrate it in the following example: Take $R=prod limits_{i=1}^{infty}mathbb{Z}_2$ and we see that $R$ is not Noetherian because it is not finitely generated.
We see that for any $xin R$ we have $x^2=x$, i.e. $R$ is a Boolean ring. Let $P$ be a prime ideal in $R$ then we can show that $R/Pcong mathbb{Z}_2$ which means that $R/P$ is a field. Hence $P$ is maximal ideal.
Thus maximal and prime ideals in $R$ are the same
Now we want to consider the localizations $R_P$ for any prime ideal $P$.
I have the following two questions (I have spent couple of hours but I was not able to answer these questions).
1) How to show that $R$ is local ring? i.e. there is only one maximal ideal (i.e. one prime ideal).
2) If we can show this then $P=0$ because intersection of all prime ideals is the set of nilpotent elements in $R$ but $R$ being boolean has no nonzero nilpotent elements. Hence $P=0$.
How it follows that $R_P$ is field?
I know that this question is already exists in forum. But they are not detailed. Could anyone help me to asnwer these questions?
Would be very grateful for help! Also please do not duplicate this question!
abstract-algebra ring-theory commutative-algebra
asked Jan 29 at 14:38
ZFRZFR
5,29331540
$endgroup$
add a comment |
$begingroup$
We know the following fact: If $A$ is Noetherian ring and $S$ is multiplicatively closed subset of $A$ then $S^{-1}A$ is also Noetherian ring.
Corollary: If $A$ is Noetherian ring then for any prime ideal $P$ the ring $A_P$ is also Noetherian.
However, the converse of this corollary is not true.
Let's demonstrate it in the following example: Take $R=prod limits_{i=1}^{infty}mathbb{Z}_2$ and we see that $R$ is not Noetherian because it is not finitely generated.
We see that for any $xin R$ we have $x^2=x$, i.e. $R$ is a Boolean ring. Let $P$ be a prime ideal in $R$ then we can show that $R/Pcong mathbb{Z}_2$ which means that $R/P$ is a field. Hence $P$ is maximal ideal.
Thus maximal and prime ideals in $R$ are the same
Now we want to consider the localizations $R_P$ for any prime ideal $P$.
I have the following two questions (I have spent couple of hours but I was not able to answer these questions).
1) How to show that $R$ is local ring? i.e. there is only one maximal ideal (i.e. one prime ideal).
2) If we can show this then $P=0$ because intersection of all prime ideals is the set of nilpotent elements in $R$ but $R$ being boolean has no nonzero nilpotent elements. Hence $P=0$.
How it follows that $R_P$ is field?
I know that this question is already exists in forum. But they are not detailed. Could anyone help me to asnwer these questions?
Would be very grateful for help! Also please do not duplicate this question!
abstract-algebra ring-theory commutative-algebra
asked Jan 29 at 14:38
ZFRZFR
5,29331540
$endgroup$
We know the following fact: If $A$ is Noetherian ring and $S$ is multiplicatively closed subset of $A$ then $S^{-1}A$ is also Noetherian ring.
Corollary: If $A$ is Noetherian ring then for any prime ideal $P$ the ring $A_P$ is also Noetherian.
However, the converse of this corollary is not true.
Let's demonstrate it in the following example: Take $R=prod limits_{i=1}^{infty}mathbb{Z}_2$ and we see that $R$ is not Noetherian because it is not finitely generated.
We see that for any $xin R$ we have $x^2=x$, i.e. $R$ is a Boolean ring. Let $P$ be a prime ideal in $R$ then we can show that $R/Pcong mathbb{Z}_2$ which means that $R/P$ is a field. Hence $P$ is maximal ideal.
Thus maximal and prime ideals in $R$ are the same
Now we want to consider the localizations $R_P$ for any prime ideal $P$.
I have the following two questions (I have spent couple of hours but I was not able to answer these questions).
1) How to show that $R$ is local ring? i.e. there is only one maximal ideal (i.e. one prime ideal).
2) If we can show this then $P=0$ because intersection of all prime ideals is the set of nilpotent elements in $R$ but $R$ being boolean has no nonzero nilpotent elements. Hence $P=0$.
How it follows that $R_P$ is field?
I know that this question is already exists in forum. But they are not detailed. Could anyone help me to asnwer these questions?
Would be very grateful for help! Also please do not duplicate this question!
abstract-algebra ring-theory commutative-algebra
abstract-algebra ring-theory commutative-algebra
asked Jan 29 at 14:38
ZFRZFR
5,29331540
asked Jan 29 at 14:38
ZFRZFR
5,29331540
asked Jan 29 at 14:38
ZFRZFR
5,29331540
asked Jan 29 at 14:38
ZFRZFR
5,29331540
asked Jan 29 at 14:38
ZFRZFR
5,29331540
5,29331540
add a comment |
add a comment |
1 Answer
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$begingroup$
How to show that $R$ is local ring? i.e. there is only one maximal ideal (i.e. one prime ideal).
I guess you mean how to show $R_P$ is a local ring? This is the first thing you learn about localizations: localizing at the complement of a prime ideal results in a local ring. Just suppose that $M$ is a maximal ideal of $R_P$ different from $PR_P={frac{p}{s}mid pin P, snotin P}$. Then there is some $frac{m}{s}in Msetminus PR_P$, and that entails that $mnotin P$. But $frac{m}{1}$ is a unit in $R_P$, so $frac{m}{s}$ is a unit and $M=R_P$, a contradiction.
If we can show this then $P={0}$ because intersection of all prime ideals is the set of nilpotent elements in $R$ but $R$ being boolean has no nonzero nilpotent elements. Hence $P={0}$.
What? There are infinitely many nonzero prime ideals in $R$. I don't know how you expect this to work.
How it follows that $R_P$ is field?
On one hand, $R_P$ is a local ring. On the other hand, everything in it is idempotent: for $frac{a}{b}frac{a}{b}=frac{aa}{bb}=frac{a}{b}$ for any $frac{a}{b}in R_P$. But in a local ring, idempotents must be $0$ or $1$. Therefore everything in $R_P$ is the additive or multiplicative identity: therefore $R_P=F_2$.
answered Jan 29 at 15:28
rschwiebrschwieb
107k12103252
$endgroup$
$begingroup$
Your answer indeed very nice! After a long battle I guess that I've finally got it. +1
$endgroup$
– ZFR
Jan 29 at 17:31
add a comment |
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$begingroup$
How to show that $R$ is local ring? i.e. there is only one maximal ideal (i.e. one prime ideal).
I guess you mean how to show $R_P$ is a local ring? This is the first thing you learn about localizations: localizing at the complement of a prime ideal results in a local ring. Just suppose that $M$ is a maximal ideal of $R_P$ different from $PR_P={frac{p}{s}mid pin P, snotin P}$. Then there is some $frac{m}{s}in Msetminus PR_P$, and that entails that $mnotin P$. But $frac{m}{1}$ is a unit in $R_P$, so $frac{m}{s}$ is a unit and $M=R_P$, a contradiction.
If we can show this then $P={0}$ because intersection of all prime ideals is the set of nilpotent elements in $R$ but $R$ being boolean has no nonzero nilpotent elements. Hence $P={0}$.
What? There are infinitely many nonzero prime ideals in $R$. I don't know how you expect this to work.
How it follows that $R_P$ is field?
On one hand, $R_P$ is a local ring. On the other hand, everything in it is idempotent: for $frac{a}{b}frac{a}{b}=frac{aa}{bb}=frac{a}{b}$ for any $frac{a}{b}in R_P$. But in a local ring, idempotents must be $0$ or $1$. Therefore everything in $R_P$ is the additive or multiplicative identity: therefore $R_P=F_2$.
answered Jan 29 at 15:28
rschwiebrschwieb
107k12103252
$endgroup$
$begingroup$
Your answer indeed very nice! After a long battle I guess that I've finally got it. +1
$endgroup$
– ZFR
Jan 29 at 17:31
add a comment |
$begingroup$
How to show that $R$ is local ring? i.e. there is only one maximal ideal (i.e. one prime ideal).
I guess you mean how to show $R_P$ is a local ring? This is the first thing you learn about localizations: localizing at the complement of a prime ideal results in a local ring. Just suppose that $M$ is a maximal ideal of $R_P$ different from $PR_P={frac{p}{s}mid pin P, snotin P}$. Then there is some $frac{m}{s}in Msetminus PR_P$, and that entails that $mnotin P$. But $frac{m}{1}$ is a unit in $R_P$, so $frac{m}{s}$ is a unit and $M=R_P$, a contradiction.
If we can show this then $P={0}$ because intersection of all prime ideals is the set of nilpotent elements in $R$ but $R$ being boolean has no nonzero nilpotent elements. Hence $P={0}$.
What? There are infinitely many nonzero prime ideals in $R$. I don't know how you expect this to work.
How it follows that $R_P$ is field?
On one hand, $R_P$ is a local ring. On the other hand, everything in it is idempotent: for $frac{a}{b}frac{a}{b}=frac{aa}{bb}=frac{a}{b}$ for any $frac{a}{b}in R_P$. But in a local ring, idempotents must be $0$ or $1$. Therefore everything in $R_P$ is the additive or multiplicative identity: therefore $R_P=F_2$.
answered Jan 29 at 15:28
rschwiebrschwieb
107k12103252
$endgroup$
$begingroup$
Your answer indeed very nice! After a long battle I guess that I've finally got it. +1
$endgroup$
– ZFR
Jan 29 at 17:31
add a comment |
$begingroup$
How to show that $R$ is local ring? i.e. there is only one maximal ideal (i.e. one prime ideal).
I guess you mean how to show $R_P$ is a local ring? This is the first thing you learn about localizations: localizing at the complement of a prime ideal results in a local ring. Just suppose that $M$ is a maximal ideal of $R_P$ different from $PR_P={frac{p}{s}mid pin P, snotin P}$. Then there is some $frac{m}{s}in Msetminus PR_P$, and that entails that $mnotin P$. But $frac{m}{1}$ is a unit in $R_P$, so $frac{m}{s}$ is a unit and $M=R_P$, a contradiction.
If we can show this then $P={0}$ because intersection of all prime ideals is the set of nilpotent elements in $R$ but $R$ being boolean has no nonzero nilpotent elements. Hence $P={0}$.
What? There are infinitely many nonzero prime ideals in $R$. I don't know how you expect this to work.
How it follows that $R_P$ is field?
On one hand, $R_P$ is a local ring. On the other hand, everything in it is idempotent: for $frac{a}{b}frac{a}{b}=frac{aa}{bb}=frac{a}{b}$ for any $frac{a}{b}in R_P$. But in a local ring, idempotents must be $0$ or $1$. Therefore everything in $R_P$ is the additive or multiplicative identity: therefore $R_P=F_2$.
answered Jan 29 at 15:28
rschwiebrschwieb
107k12103252
$endgroup$
How to show that $R$ is local ring? i.e. there is only one maximal ideal (i.e. one prime ideal).
I guess you mean how to show $R_P$ is a local ring? This is the first thing you learn about localizations: localizing at the complement of a prime ideal results in a local ring. Just suppose that $M$ is a maximal ideal of $R_P$ different from $PR_P={frac{p}{s}mid pin P, snotin P}$. Then there is some $frac{m}{s}in Msetminus PR_P$, and that entails that $mnotin P$. But $frac{m}{1}$ is a unit in $R_P$, so $frac{m}{s}$ is a unit and $M=R_P$, a contradiction.
If we can show this then $P={0}$ because intersection of all prime ideals is the set of nilpotent elements in $R$ but $R$ being boolean has no nonzero nilpotent elements. Hence $P={0}$.
What? There are infinitely many nonzero prime ideals in $R$. I don't know how you expect this to work.
How it follows that $R_P$ is field?
On one hand, $R_P$ is a local ring. On the other hand, everything in it is idempotent: for $frac{a}{b}frac{a}{b}=frac{aa}{bb}=frac{a}{b}$ for any $frac{a}{b}in R_P$. But in a local ring, idempotents must be $0$ or $1$. Therefore everything in $R_P$ is the additive or multiplicative identity: therefore $R_P=F_2$.
answered Jan 29 at 15:28
rschwiebrschwieb
107k12103252
edited Jan 29 at 15:34
answered Jan 29 at 15:28
rschwiebrschwieb
107k12103252
answered Jan 29 at 15:28
rschwiebrschwieb
107k12103252
answered Jan 29 at 15:28
rschwiebrschwieb
107k12103252
107k12103252
$begingroup$
Your answer indeed very nice! After a long battle I guess that I've finally got it. +1
$endgroup$
– ZFR
Jan 29 at 17:31
add a comment |
$begingroup$
Your answer indeed very nice! After a long battle I guess that I've finally got it. +1
$endgroup$
– ZFR
Jan 29 at 17:31
$begingroup$
Your answer indeed very nice! After a long battle I guess that I've finally got it. +1
$endgroup$
– ZFR
Jan 29 at 17:31
$begingroup$
Your answer indeed very nice! After a long battle I guess that I've finally got it. +1
$endgroup$
– ZFR
Jan 29 at 17:31
add a comment |
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