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The localization of the ring $mathbb{Z} times mathbb{Z}$ at every prime ideal is an integral domain

If $A$ is a prime ideal on $Rtimes S$, then $A = P times S$, where P is a prime ideal on R or $A = Rtimes Q$, where $Q$ is a prime ideal on $S$.Localization at a prime ideal in $mathbb{Z}/6mathbb{Z}$If the localization of a ring $R$ at every prime ideal is an integral domain, must $R$ be an integral domain?If $P$ is a prime ideal of an integral domain $D$, then is $D$ equal to its localization at $P$?When is a local, reduced, (commutative) ring an integral domain?Maximal ideals and prime ideals of $mathbb{Z}/2 times mathbb{Z}/2$?Localization of a one-dimensional noetherian integral domain$R$ be a commutative ring with unity such that every prime ideal contains no non-zero zero divisor , then is $R$ an integral domain?Prime ideal and maximal ideal of an integral domainPower of maximal ideal in the localization of $k[x,y]$ at $(x,y)$why prime ideal and maximal ideal are same in finite integral domain?

2

1

$begingroup$

I want to show that the localization of the ring $mathbb{Z} times mathbb{Z}$ at every prime ideal is an integral domain.

$mathbb{Z} times mathbb{Z}$ is not an integral domain since $(0,1)cdot(1,0)=(0,0)$. I think $(mathbb{Z} times mathbb{Z})_P$ wouldn't be an integral domain for similar reason. Can anyone explain what difference the localization can make here? Thanks in advance for your help.

abstract-algebra commutative-algebra modules localization integral-domain

share|cite|improve this question

edited Mar 10 at 18:07

user26857

39.4k124183

asked Mar 10 at 2:13

AndrewAndrew

37318

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You say “at its prime ideal”. Are you saying there is only one prime idea? If so, what is it?
  $endgroup$
  – Arturo Magidin
  Mar 10 at 2:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Arturo Magidin sorry I meant every prime ideal
  $endgroup$
  – Andrew
  Mar 10 at 2:18
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Then edit the question...
  $endgroup$
  – Arturo Magidin
  Mar 10 at 2:20
  

  
  
  
  

add a comment | 

2

1

$begingroup$

I want to show that the localization of the ring $mathbb{Z} times mathbb{Z}$ at every prime ideal is an integral domain.

$mathbb{Z} times mathbb{Z}$ is not an integral domain since $(0,1)cdot(1,0)=(0,0)$. I think $(mathbb{Z} times mathbb{Z})_P$ wouldn't be an integral domain for similar reason. Can anyone explain what difference the localization can make here? Thanks in advance for your help.

abstract-algebra commutative-algebra modules localization integral-domain

share|cite|improve this question

edited Mar 10 at 18:07

user26857

39.4k124183

asked Mar 10 at 2:13

AndrewAndrew

37318

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You say “at its prime ideal”. Are you saying there is only one prime idea? If so, what is it?
  $endgroup$
  – Arturo Magidin
  Mar 10 at 2:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Arturo Magidin sorry I meant every prime ideal
  $endgroup$
  – Andrew
  Mar 10 at 2:18
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Then edit the question...
  $endgroup$
  – Arturo Magidin
  Mar 10 at 2:20
  

  
  
  
  

add a comment | 

$begingroup$

I want to show that the localization of the ring $mathbb{Z} times mathbb{Z}$ at every prime ideal is an integral domain.

$mathbb{Z} times mathbb{Z}$ is not an integral domain since $(0,1)cdot(1,0)=(0,0)$. I think $(mathbb{Z} times mathbb{Z})_P$ wouldn't be an integral domain for similar reason. Can anyone explain what difference the localization can make here? Thanks in advance for your help.

abstract-algebra commutative-algebra modules localization integral-domain

share|cite|improve this question

edited Mar 10 at 18:07

user26857

39.4k124183

asked Mar 10 at 2:13

AndrewAndrew

37318

$endgroup$

share|cite|improve this question

edited Mar 10 at 18:07

user26857

39.4k124183

asked Mar 10 at 2:13

AndrewAndrew

37318

share|cite|improve this question

edited Mar 10 at 18:07

user26857

39.4k124183

asked Mar 10 at 2:13

AndrewAndrew

37318

edited Mar 10 at 18:07

user26857

39.4k124183

edited Mar 10 at 18:07

user26857

39.4k124183

asked Mar 10 at 2:13

AndrewAndrew

37318

asked Mar 10 at 2:13

AndrewAndrew

37318

1 Answer
1

active

oldest

votes

3

$begingroup$

Every localization of $Bbb{Z}timesBbb{Z}$ is an integral domain because every localization of $Bbb{Z}$ is an integral domain, because $Bbb{Z}$ is an integral domain. This follows immediately from the following simple fact:

If $R$ and $S$ are commutative unital rings and $Psubset Rtimes S$ is a prime ideal, then, after switching $R$ and $S$ if necessary, we have $P=Qtimes S$, where $Qsubset R$ is a prime ideal, from which it follows that
$$(Rtimes S)_Pcong R_Q.$$

share|cite|improve this answer

answered Mar 10 at 2:20

ServaesServaes

28.3k34099

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your answer. This might be helpful if anyone is wondering why $P=Qtimes S$: math.stackexchange.com/questions/2235763/…
  $endgroup$
  – Andrew
  Mar 10 at 3:03
  
  
  

  
  
  
  

add a comment | 

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3

$begingroup$

Every localization of $Bbb{Z}timesBbb{Z}$ is an integral domain because every localization of $Bbb{Z}$ is an integral domain, because $Bbb{Z}$ is an integral domain. This follows immediately from the following simple fact:

If $R$ and $S$ are commutative unital rings and $Psubset Rtimes S$ is a prime ideal, then, after switching $R$ and $S$ if necessary, we have $P=Qtimes S$, where $Qsubset R$ is a prime ideal, from which it follows that
$$(Rtimes S)_Pcong R_Q.$$

share|cite|improve this answer

answered Mar 10 at 2:20

ServaesServaes

28.3k34099

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your answer. This might be helpful if anyone is wondering why $P=Qtimes S$: math.stackexchange.com/questions/2235763/…
  $endgroup$
  – Andrew
  Mar 10 at 3:03
  
  
  

  
  
  
  

add a comment | 

3

$begingroup$

Every localization of $Bbb{Z}timesBbb{Z}$ is an integral domain because every localization of $Bbb{Z}$ is an integral domain, because $Bbb{Z}$ is an integral domain. This follows immediately from the following simple fact:

If $R$ and $S$ are commutative unital rings and $Psubset Rtimes S$ is a prime ideal, then, after switching $R$ and $S$ if necessary, we have $P=Qtimes S$, where $Qsubset R$ is a prime ideal, from which it follows that
$$(Rtimes S)_Pcong R_Q.$$

share|cite|improve this answer

answered Mar 10 at 2:20

ServaesServaes

28.3k34099

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your answer. This might be helpful if anyone is wondering why $P=Qtimes S$: math.stackexchange.com/questions/2235763/…
  $endgroup$
  – Andrew
  Mar 10 at 3:03
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Every localization of $Bbb{Z}timesBbb{Z}$ is an integral domain because every localization of $Bbb{Z}$ is an integral domain, because $Bbb{Z}$ is an integral domain. This follows immediately from the following simple fact:

If $R$ and $S$ are commutative unital rings and $Psubset Rtimes S$ is a prime ideal, then, after switching $R$ and $S$ if necessary, we have $P=Qtimes S$, where $Qsubset R$ is a prime ideal, from which it follows that
$$(Rtimes S)_Pcong R_Q.$$

share|cite|improve this answer

answered Mar 10 at 2:20

ServaesServaes

28.3k34099

$endgroup$

share|cite|improve this answer

answered Mar 10 at 2:20

ServaesServaes

28.3k34099

answered Mar 10 at 2:20

ServaesServaes

28.3k34099

answered Mar 10 at 2:20

ServaesServaes

28.3k34099