Can a non-PID noetherian integral domain have all PID localizations?

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If a noetherian integral domain R has all its localizations by maximal ideals being PIDs, does it imply that R itself is a PID?

Thank you for your help.

ring-theory commutative-algebra principal-ideal-domains localization noetherian

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asked Jul 15 at 14:34

UnrealVillager

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If a noetherian integral domain R has all its localizations by maximal ideals being PIDs, does it imply that R itself is a PID?

Thank you for your help.

ring-theory commutative-algebra principal-ideal-domains localization noetherian

share|cite|improve this question

asked Jul 15 at 14:34

UnrealVillager

256

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

If a noetherian integral domain R has all its localizations by maximal ideals being PIDs, does it imply that R itself is a PID?

Thank you for your help.

ring-theory commutative-algebra principal-ideal-domains localization noetherian

share|cite|improve this question

asked Jul 15 at 14:34

UnrealVillager

256

If a noetherian integral domain R has all its localizations by maximal ideals being PIDs, does it imply that R itself is a PID?

Thank you for your help.

ring-theory commutative-algebra principal-ideal-domains localization noetherian

share|cite|improve this question

asked Jul 15 at 14:34

UnrealVillager

256

share|cite|improve this question

share|cite|improve this question

asked Jul 15 at 14:34

UnrealVillager

256

asked Jul 15 at 14:34

UnrealVillager

256

asked Jul 15 at 14:34

UnrealVillager

256

256

add a comment | 

add a comment | 

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Short Version: No, an example would be the ring $R=mathbbZ[sqrt5i]$. I'll explain in the long version, why this example was chosen and why it does not satisfy your criteria.

Motivation: The idea is to first give a name to a Noetherian domain $A$ of Krull dimension one (if localization at every maximal is a PID, then the dimension is necessarily one), such that localization at every prime (not just maximal) ideal is a PID. This, as we will see, is a Dedekind domain. It is also known that if a Dedekind domain is a UFD, this it is a PID. So I have translated (a stronger version of) your question into:

Q: Does there exists a Dedekind domain that is not a UFD? A: Yes!

Long Version:
If you consult Atiyah & MacDonald, chapter 9 (Valuation Rings), you will see the following statements:

Proposition 9.2: Let $A$ be a Noetherian local domain of [Krull] dimension one, $mathfrakm$ its maximal ideal, $k=A/mathfrakm$
its residue field. Then the following are equivalent:

1. $A$ is a discrete valuation ring;


2. $A$ is integrally closed;


3. $mathfrakm$ is a principal ideal;


4. $dim mathfrakm/mathfrakm^2=1$;


5. Every non-zero ideal is a power of $mathfrakm$.


6. There exists $xin A$ such that every non-zero ideal is of the form $(x^k)$, $kgeq 0$.

In particular, if a Noetherian local domain of dimension one $A$ is a PID then $A$ is a DVR (by $3to 1$) and conversely, if $A$ is a DVR then $A$ is a PID (by $1to 6$).

Now we define a Dedekind domain:

Proposition 9.3: Let $A$ be a Noetherian domain of [Krull] dimension one. Then the following are equivalent:

1. $A$ is integrally closed;


2. Every primary ideal in $A$ is a prime power;


3. Every local ring $A_mathfrakp$ ($mathfrakpneq 0$) is a DVR;

A ring satisfying any (so all) of these conditions is called a Dedekind domain.

A very famous example of Dedekind domains are the following group (for a proof see here, proposition 13.5)

Let $mathbbQsubset K$ be a finite field extension [This is called a number field], and let $R$ be the integral closure of $mathbbZ$ in $K$. Then $R$ is a Dedekind domain.

This is how the example in the short version is made. I'll leave it to you to show $mathbbZ[sqrt5i]$ is not a UFD.

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answered Jul 15 at 18:04

Hamed

4,421421

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  Hamed, thanks a lot for such an extensive reply.
  – UnrealVillager
  Jul 15 at 18:36
  

  
  
  
  

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1 Answer
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votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

Short Version: No, an example would be the ring $R=mathbbZ[sqrt5i]$. I'll explain in the long version, why this example was chosen and why it does not satisfy your criteria.

Motivation: The idea is to first give a name to a Noetherian domain $A$ of Krull dimension one (if localization at every maximal is a PID, then the dimension is necessarily one), such that localization at every prime (not just maximal) ideal is a PID. This, as we will see, is a Dedekind domain. It is also known that if a Dedekind domain is a UFD, this it is a PID. So I have translated (a stronger version of) your question into:

Q: Does there exists a Dedekind domain that is not a UFD? A: Yes!

Long Version:
If you consult Atiyah & MacDonald, chapter 9 (Valuation Rings), you will see the following statements:

Proposition 9.2: Let $A$ be a Noetherian local domain of [Krull] dimension one, $mathfrakm$ its maximal ideal, $k=A/mathfrakm$
its residue field. Then the following are equivalent:

1. $A$ is a discrete valuation ring;


2. $A$ is integrally closed;


3. $mathfrakm$ is a principal ideal;


4. $dim mathfrakm/mathfrakm^2=1$;


5. Every non-zero ideal is a power of $mathfrakm$.


6. There exists $xin A$ such that every non-zero ideal is of the form $(x^k)$, $kgeq 0$.

In particular, if a Noetherian local domain of dimension one $A$ is a PID then $A$ is a DVR (by $3to 1$) and conversely, if $A$ is a DVR then $A$ is a PID (by $1to 6$).

Now we define a Dedekind domain:

Proposition 9.3: Let $A$ be a Noetherian domain of [Krull] dimension one. Then the following are equivalent:

1. $A$ is integrally closed;


2. Every primary ideal in $A$ is a prime power;


3. Every local ring $A_mathfrakp$ ($mathfrakpneq 0$) is a DVR;

A ring satisfying any (so all) of these conditions is called a Dedekind domain.

A very famous example of Dedekind domains are the following group (for a proof see here, proposition 13.5)

Let $mathbbQsubset K$ be a finite field extension [This is called a number field], and let $R$ be the integral closure of $mathbbZ$ in $K$. Then $R$ is a Dedekind domain.

This is how the example in the short version is made. I'll leave it to you to show $mathbbZ[sqrt5i]$ is not a UFD.

share|cite|improve this answer

answered Jul 15 at 18:04

Hamed

4,421421

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hamed, thanks a lot for such an extensive reply.
  – UnrealVillager
  Jul 15 at 18:36
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

Short Version: No, an example would be the ring $R=mathbbZ[sqrt5i]$. I'll explain in the long version, why this example was chosen and why it does not satisfy your criteria.

Motivation: The idea is to first give a name to a Noetherian domain $A$ of Krull dimension one (if localization at every maximal is a PID, then the dimension is necessarily one), such that localization at every prime (not just maximal) ideal is a PID. This, as we will see, is a Dedekind domain. It is also known that if a Dedekind domain is a UFD, this it is a PID. So I have translated (a stronger version of) your question into:

Q: Does there exists a Dedekind domain that is not a UFD? A: Yes!

Long Version:
If you consult Atiyah & MacDonald, chapter 9 (Valuation Rings), you will see the following statements:

Proposition 9.2: Let $A$ be a Noetherian local domain of [Krull] dimension one, $mathfrakm$ its maximal ideal, $k=A/mathfrakm$
its residue field. Then the following are equivalent:

1. $A$ is a discrete valuation ring;


2. $A$ is integrally closed;


3. $mathfrakm$ is a principal ideal;


4. $dim mathfrakm/mathfrakm^2=1$;


5. Every non-zero ideal is a power of $mathfrakm$.


6. There exists $xin A$ such that every non-zero ideal is of the form $(x^k)$, $kgeq 0$.

In particular, if a Noetherian local domain of dimension one $A$ is a PID then $A$ is a DVR (by $3to 1$) and conversely, if $A$ is a DVR then $A$ is a PID (by $1to 6$).

Now we define a Dedekind domain:

Proposition 9.3: Let $A$ be a Noetherian domain of [Krull] dimension one. Then the following are equivalent:

1. $A$ is integrally closed;


2. Every primary ideal in $A$ is a prime power;


3. Every local ring $A_mathfrakp$ ($mathfrakpneq 0$) is a DVR;

A ring satisfying any (so all) of these conditions is called a Dedekind domain.

A very famous example of Dedekind domains are the following group (for a proof see here, proposition 13.5)

Let $mathbbQsubset K$ be a finite field extension [This is called a number field], and let $R$ be the integral closure of $mathbbZ$ in $K$. Then $R$ is a Dedekind domain.

This is how the example in the short version is made. I'll leave it to you to show $mathbbZ[sqrt5i]$ is not a UFD.

share|cite|improve this answer

answered Jul 15 at 18:04

Hamed

4,421421

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hamed, thanks a lot for such an extensive reply.
  – UnrealVillager
  Jul 15 at 18:36
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

Short Version: No, an example would be the ring $R=mathbbZ[sqrt5i]$. I'll explain in the long version, why this example was chosen and why it does not satisfy your criteria.

Motivation: The idea is to first give a name to a Noetherian domain $A$ of Krull dimension one (if localization at every maximal is a PID, then the dimension is necessarily one), such that localization at every prime (not just maximal) ideal is a PID. This, as we will see, is a Dedekind domain. It is also known that if a Dedekind domain is a UFD, this it is a PID. So I have translated (a stronger version of) your question into:

Q: Does there exists a Dedekind domain that is not a UFD? A: Yes!

Long Version:
If you consult Atiyah & MacDonald, chapter 9 (Valuation Rings), you will see the following statements:

Proposition 9.2: Let $A$ be a Noetherian local domain of [Krull] dimension one, $mathfrakm$ its maximal ideal, $k=A/mathfrakm$
its residue field. Then the following are equivalent:

1. $A$ is a discrete valuation ring;


2. $A$ is integrally closed;


3. $mathfrakm$ is a principal ideal;


4. $dim mathfrakm/mathfrakm^2=1$;


5. Every non-zero ideal is a power of $mathfrakm$.


6. There exists $xin A$ such that every non-zero ideal is of the form $(x^k)$, $kgeq 0$.

In particular, if a Noetherian local domain of dimension one $A$ is a PID then $A$ is a DVR (by $3to 1$) and conversely, if $A$ is a DVR then $A$ is a PID (by $1to 6$).

Now we define a Dedekind domain:

Proposition 9.3: Let $A$ be a Noetherian domain of [Krull] dimension one. Then the following are equivalent:

1. $A$ is integrally closed;


2. Every primary ideal in $A$ is a prime power;


3. Every local ring $A_mathfrakp$ ($mathfrakpneq 0$) is a DVR;

A ring satisfying any (so all) of these conditions is called a Dedekind domain.

A very famous example of Dedekind domains are the following group (for a proof see here, proposition 13.5)

Let $mathbbQsubset K$ be a finite field extension [This is called a number field], and let $R$ be the integral closure of $mathbbZ$ in $K$. Then $R$ is a Dedekind domain.

This is how the example in the short version is made. I'll leave it to you to show $mathbbZ[sqrt5i]$ is not a UFD.

share|cite|improve this answer

answered Jul 15 at 18:04

Hamed

4,421421

Short Version: No, an example would be the ring $R=mathbbZ[sqrt5i]$. I'll explain in the long version, why this example was chosen and why it does not satisfy your criteria.

Motivation: The idea is to first give a name to a Noetherian domain $A$ of Krull dimension one (if localization at every maximal is a PID, then the dimension is necessarily one), such that localization at every prime (not just maximal) ideal is a PID. This, as we will see, is a Dedekind domain. It is also known that if a Dedekind domain is a UFD, this it is a PID. So I have translated (a stronger version of) your question into:

Q: Does there exists a Dedekind domain that is not a UFD? A: Yes!

Long Version:
If you consult Atiyah & MacDonald, chapter 9 (Valuation Rings), you will see the following statements:

Proposition 9.2: Let $A$ be a Noetherian local domain of [Krull] dimension one, $mathfrakm$ its maximal ideal, $k=A/mathfrakm$
its residue field. Then the following are equivalent:

1. $A$ is a discrete valuation ring;


2. $A$ is integrally closed;


3. $mathfrakm$ is a principal ideal;


4. $dim mathfrakm/mathfrakm^2=1$;


5. Every non-zero ideal is a power of $mathfrakm$.


6. There exists $xin A$ such that every non-zero ideal is of the form $(x^k)$, $kgeq 0$.

In particular, if a Noetherian local domain of dimension one $A$ is a PID then $A$ is a DVR (by $3to 1$) and conversely, if $A$ is a DVR then $A$ is a PID (by $1to 6$).

Now we define a Dedekind domain:

Proposition 9.3: Let $A$ be a Noetherian domain of [Krull] dimension one. Then the following are equivalent:

1. $A$ is integrally closed;


2. Every primary ideal in $A$ is a prime power;


3. Every local ring $A_mathfrakp$ ($mathfrakpneq 0$) is a DVR;

A ring satisfying any (so all) of these conditions is called a Dedekind domain.

A very famous example of Dedekind domains are the following group (for a proof see here, proposition 13.5)

Let $mathbbQsubset K$ be a finite field extension [This is called a number field], and let $R$ be the integral closure of $mathbbZ$ in $K$. Then $R$ is a Dedekind domain.

This is how the example in the short version is made. I'll leave it to you to show $mathbbZ[sqrt5i]$ is not a UFD.

share|cite|improve this answer

answered Jul 15 at 18:04

Hamed

4,421421

share|cite|improve this answer

share|cite|improve this answer

answered Jul 15 at 18:04

Hamed

4,421421

answered Jul 15 at 18:04

Hamed

4,421421

answered Jul 15 at 18:04

Hamed

4,421421

4,421421

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hamed, thanks a lot for such an extensive reply.
  – UnrealVillager
  Jul 15 at 18:36
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hamed, thanks a lot for such an extensive reply.
  – UnrealVillager
  Jul 15 at 18:36
  

  
  
  
  

Hamed, thanks a lot for such an extensive reply.
– UnrealVillager
Jul 15 at 18:36

Hamed, thanks a lot for such an extensive reply.
– UnrealVillager
Jul 15 at 18:36

add a comment | 

 

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