Another version of the Chinese Remainder Theorem

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Let $R$ be a ring, $I,J$ ideals such that $I+J=R$. The Chinese Remainder Theorem claims that if $Icap J=(0)$, then $Rsimeq R/Itimes R/J$.

I remember seeing a version of the theorem for the case of principal ideals $I=(p), J=(q)$ which assumes instead of the comaximality of $I$ and $J$ (and probably of the condition on the intersection) that $I$ and $J$) that $p$ and $q$ are coprime elements in $R$. (I'm not sure that my statement is precise.) This version is easier to use sometimes, e.g. it looks easier for me to prove that $x-1$ and $x^2+x+1$ are coprime than to prove that $(x-1)+(x^2+x+1)=(1)$ in $mathbb R[x]$.

So my questions are: whether my statement is precise, and how does it follow from the general Chinese remainder theorem?

abstract-algebra number-theory ring-theory ideals factoring

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asked Jul 19 at 22:50

user437309

556212

add a comment | 

up vote
0
down vote

favorite

Let $R$ be a ring, $I,J$ ideals such that $I+J=R$. The Chinese Remainder Theorem claims that if $Icap J=(0)$, then $Rsimeq R/Itimes R/J$.

I remember seeing a version of the theorem for the case of principal ideals $I=(p), J=(q)$ which assumes instead of the comaximality of $I$ and $J$ (and probably of the condition on the intersection) that $I$ and $J$) that $p$ and $q$ are coprime elements in $R$. (I'm not sure that my statement is precise.) This version is easier to use sometimes, e.g. it looks easier for me to prove that $x-1$ and $x^2+x+1$ are coprime than to prove that $(x-1)+(x^2+x+1)=(1)$ in $mathbb R[x]$.

So my questions are: whether my statement is precise, and how does it follow from the general Chinese remainder theorem?

abstract-algebra number-theory ring-theory ideals factoring

share|cite|improve this question

asked Jul 19 at 22:50

user437309

556212

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Let $R$ be a ring, $I,J$ ideals such that $I+J=R$. The Chinese Remainder Theorem claims that if $Icap J=(0)$, then $Rsimeq R/Itimes R/J$.

I remember seeing a version of the theorem for the case of principal ideals $I=(p), J=(q)$ which assumes instead of the comaximality of $I$ and $J$ (and probably of the condition on the intersection) that $I$ and $J$) that $p$ and $q$ are coprime elements in $R$. (I'm not sure that my statement is precise.) This version is easier to use sometimes, e.g. it looks easier for me to prove that $x-1$ and $x^2+x+1$ are coprime than to prove that $(x-1)+(x^2+x+1)=(1)$ in $mathbb R[x]$.

So my questions are: whether my statement is precise, and how does it follow from the general Chinese remainder theorem?

abstract-algebra number-theory ring-theory ideals factoring

share|cite|improve this question

asked Jul 19 at 22:50

user437309

556212

Let $R$ be a ring, $I,J$ ideals such that $I+J=R$. The Chinese Remainder Theorem claims that if $Icap J=(0)$, then $Rsimeq R/Itimes R/J$.

I remember seeing a version of the theorem for the case of principal ideals $I=(p), J=(q)$ which assumes instead of the comaximality of $I$ and $J$ (and probably of the condition on the intersection) that $I$ and $J$) that $p$ and $q$ are coprime elements in $R$. (I'm not sure that my statement is precise.) This version is easier to use sometimes, e.g. it looks easier for me to prove that $x-1$ and $x^2+x+1$ are coprime than to prove that $(x-1)+(x^2+x+1)=(1)$ in $mathbb R[x]$.

So my questions are: whether my statement is precise, and how does it follow from the general Chinese remainder theorem?

abstract-algebra number-theory ring-theory ideals factoring

share|cite|improve this question

asked Jul 19 at 22:50

user437309

556212

share|cite|improve this question

share|cite|improve this question

asked Jul 19 at 22:50

user437309

556212

asked Jul 19 at 22:50

user437309

556212

asked Jul 19 at 22:50

user437309

556212

556212

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
1
down vote

The problem is what do you mean by the statement: "$x$ and $y$ are coprime elements in a ring $R$". Let's stick to integers for now. There are two different ways you can think about coprime integers. Two integers $x,yin mathbbZ$ are said to be coprime if one of the two equivalent conditions are satisfied.

1. The greatest common divisor of $x,y$ is equal to one.


2. There exists $a,bin mathbbZ$ such that $ax+by=1$ (Bezout's identity)

The problem with definition (1) is that there are many rings out there with no such thing as a greatest common divisor. Say $mathbbC[x]/(x^2)$ is not even a domain. Or $mathbbZ[-sqrt3]$ which is a domain but the two elements $x=4=2times 2=(1+sqrt-3)(1-sqrt-3)$ and $y=(1+sqrt-3)times 2$ have no gcd. A commutative ring admitting a gcd for any pair of elements is called a GCD domain.

Your example of $mathbbR[x]$ is a GCD domain (actually, more strongly, a principal ideal domain, PID). But even if a ring $R$ is a GCD domain, it does not mean $R$ admits a Bezout identity. An example would be $mathbbR[x,y]$ which is a GCD domain (actually a unique factorization domain, UFD) which does not admit a Bezout identity. A ring that admits a Bezout identity is called a Bezout domain. All Bezout domains are GCD domains. Alternatively, a Bezout domain can be defined as follows:

An integral domain $R$ is called a Bezout domain if given any two elements $x,y$, the sum of ideals $(x)+(y)$ is again principal.

So you can see that in a Bezout domain, $x,y$ are coprime if and only if the ideals $(x), (y)$ are coprime (Two ideals $I,Jsubset R$ are called coprime if $I+J=R$). However, one can define coprime ideals for all commutative rings, and coprime elements only for some rings. In a sense then, the way coprime ideals are defined is the natural generalization of how we expect "coprime" to be.

Now the Chinese Reminder theorem says

If the ideals $I,Jsubset R$ are comaximal, then $R/Icap Jsimeq R/Itimes R/J$.

If we assume $R$ is a Bezout domain, and $x,yin R$ coprime, then $(x)cap (y)=(xy)$ (which is not zero y the way). And Chinese remainder theorem, in this case, becomes
$$
R/(xy)simeq R/(x)times R/(y)
$$
So sure, if in your ring of interest it makes sense to talk about coprime elements, you can go with your special version of the theorem. But if there is no such thing, well... I hope, this sheds some light on the matter.

share|cite|improve this answer

edited Jul 19 at 23:35

answered Jul 19 at 23:28

Hamed

4,391421

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I thought the property $I+J=R$ is called comaximality (not coprimeness)?
  – user437309
  Jul 19 at 23:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Both terminologies are used interchangeably.
  – Hamed
  Jul 19 at 23:35
  

  
  
  
  

add a comment | 

up vote
0
down vote

The abstract Chinese remainder theorem that I know is this (stated for two ideals):

Let $R$ be a (commutative) ring, $I, J$ two coprime ideals in $R$, i.e. satisfying$I+J=R$. Then $IJ=Icap J$, and we have an isomorphism
beginalign
R/Icap J&xrightarrow;sim;R/Itimes R/J\
r+Icap J&longmapsto (r+ I, r+ J)
endalign

share|cite|improve this answer

edited Jul 19 at 23:29

Kaj Hansen

26.9k43679

answered Jul 19 at 23:26

Bernard

110k635103

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

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active

oldest

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votes

up vote
1
down vote

The problem is what do you mean by the statement: "$x$ and $y$ are coprime elements in a ring $R$". Let's stick to integers for now. There are two different ways you can think about coprime integers. Two integers $x,yin mathbbZ$ are said to be coprime if one of the two equivalent conditions are satisfied.

1. The greatest common divisor of $x,y$ is equal to one.


2. There exists $a,bin mathbbZ$ such that $ax+by=1$ (Bezout's identity)

The problem with definition (1) is that there are many rings out there with no such thing as a greatest common divisor. Say $mathbbC[x]/(x^2)$ is not even a domain. Or $mathbbZ[-sqrt3]$ which is a domain but the two elements $x=4=2times 2=(1+sqrt-3)(1-sqrt-3)$ and $y=(1+sqrt-3)times 2$ have no gcd. A commutative ring admitting a gcd for any pair of elements is called a GCD domain.

Your example of $mathbbR[x]$ is a GCD domain (actually, more strongly, a principal ideal domain, PID). But even if a ring $R$ is a GCD domain, it does not mean $R$ admits a Bezout identity. An example would be $mathbbR[x,y]$ which is a GCD domain (actually a unique factorization domain, UFD) which does not admit a Bezout identity. A ring that admits a Bezout identity is called a Bezout domain. All Bezout domains are GCD domains. Alternatively, a Bezout domain can be defined as follows:

An integral domain $R$ is called a Bezout domain if given any two elements $x,y$, the sum of ideals $(x)+(y)$ is again principal.

So you can see that in a Bezout domain, $x,y$ are coprime if and only if the ideals $(x), (y)$ are coprime (Two ideals $I,Jsubset R$ are called coprime if $I+J=R$). However, one can define coprime ideals for all commutative rings, and coprime elements only for some rings. In a sense then, the way coprime ideals are defined is the natural generalization of how we expect "coprime" to be.

Now the Chinese Reminder theorem says

If the ideals $I,Jsubset R$ are comaximal, then $R/Icap Jsimeq R/Itimes R/J$.

If we assume $R$ is a Bezout domain, and $x,yin R$ coprime, then $(x)cap (y)=(xy)$ (which is not zero y the way). And Chinese remainder theorem, in this case, becomes
$$
R/(xy)simeq R/(x)times R/(y)
$$
So sure, if in your ring of interest it makes sense to talk about coprime elements, you can go with your special version of the theorem. But if there is no such thing, well... I hope, this sheds some light on the matter.

share|cite|improve this answer

edited Jul 19 at 23:35

answered Jul 19 at 23:28

Hamed

4,391421

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I thought the property $I+J=R$ is called comaximality (not coprimeness)?
  – user437309
  Jul 19 at 23:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Both terminologies are used interchangeably.
  – Hamed
  Jul 19 at 23:35
  

  
  
  
  

add a comment | 

up vote
1
down vote

The problem is what do you mean by the statement: "$x$ and $y$ are coprime elements in a ring $R$". Let's stick to integers for now. There are two different ways you can think about coprime integers. Two integers $x,yin mathbbZ$ are said to be coprime if one of the two equivalent conditions are satisfied.

1. The greatest common divisor of $x,y$ is equal to one.


2. There exists $a,bin mathbbZ$ such that $ax+by=1$ (Bezout's identity)

The problem with definition (1) is that there are many rings out there with no such thing as a greatest common divisor. Say $mathbbC[x]/(x^2)$ is not even a domain. Or $mathbbZ[-sqrt3]$ which is a domain but the two elements $x=4=2times 2=(1+sqrt-3)(1-sqrt-3)$ and $y=(1+sqrt-3)times 2$ have no gcd. A commutative ring admitting a gcd for any pair of elements is called a GCD domain.

Your example of $mathbbR[x]$ is a GCD domain (actually, more strongly, a principal ideal domain, PID). But even if a ring $R$ is a GCD domain, it does not mean $R$ admits a Bezout identity. An example would be $mathbbR[x,y]$ which is a GCD domain (actually a unique factorization domain, UFD) which does not admit a Bezout identity. A ring that admits a Bezout identity is called a Bezout domain. All Bezout domains are GCD domains. Alternatively, a Bezout domain can be defined as follows:

An integral domain $R$ is called a Bezout domain if given any two elements $x,y$, the sum of ideals $(x)+(y)$ is again principal.

So you can see that in a Bezout domain, $x,y$ are coprime if and only if the ideals $(x), (y)$ are coprime (Two ideals $I,Jsubset R$ are called coprime if $I+J=R$). However, one can define coprime ideals for all commutative rings, and coprime elements only for some rings. In a sense then, the way coprime ideals are defined is the natural generalization of how we expect "coprime" to be.

Now the Chinese Reminder theorem says

If the ideals $I,Jsubset R$ are comaximal, then $R/Icap Jsimeq R/Itimes R/J$.

If we assume $R$ is a Bezout domain, and $x,yin R$ coprime, then $(x)cap (y)=(xy)$ (which is not zero y the way). And Chinese remainder theorem, in this case, becomes
$$
R/(xy)simeq R/(x)times R/(y)
$$
So sure, if in your ring of interest it makes sense to talk about coprime elements, you can go with your special version of the theorem. But if there is no such thing, well... I hope, this sheds some light on the matter.

share|cite|improve this answer

edited Jul 19 at 23:35

answered Jul 19 at 23:28

Hamed

4,391421

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I thought the property $I+J=R$ is called comaximality (not coprimeness)?
  – user437309
  Jul 19 at 23:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Both terminologies are used interchangeably.
  – Hamed
  Jul 19 at 23:35
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

The problem is what do you mean by the statement: "$x$ and $y$ are coprime elements in a ring $R$". Let's stick to integers for now. There are two different ways you can think about coprime integers. Two integers $x,yin mathbbZ$ are said to be coprime if one of the two equivalent conditions are satisfied.

1. The greatest common divisor of $x,y$ is equal to one.


2. There exists $a,bin mathbbZ$ such that $ax+by=1$ (Bezout's identity)

The problem with definition (1) is that there are many rings out there with no such thing as a greatest common divisor. Say $mathbbC[x]/(x^2)$ is not even a domain. Or $mathbbZ[-sqrt3]$ which is a domain but the two elements $x=4=2times 2=(1+sqrt-3)(1-sqrt-3)$ and $y=(1+sqrt-3)times 2$ have no gcd. A commutative ring admitting a gcd for any pair of elements is called a GCD domain.

Your example of $mathbbR[x]$ is a GCD domain (actually, more strongly, a principal ideal domain, PID). But even if a ring $R$ is a GCD domain, it does not mean $R$ admits a Bezout identity. An example would be $mathbbR[x,y]$ which is a GCD domain (actually a unique factorization domain, UFD) which does not admit a Bezout identity. A ring that admits a Bezout identity is called a Bezout domain. All Bezout domains are GCD domains. Alternatively, a Bezout domain can be defined as follows:

An integral domain $R$ is called a Bezout domain if given any two elements $x,y$, the sum of ideals $(x)+(y)$ is again principal.

So you can see that in a Bezout domain, $x,y$ are coprime if and only if the ideals $(x), (y)$ are coprime (Two ideals $I,Jsubset R$ are called coprime if $I+J=R$). However, one can define coprime ideals for all commutative rings, and coprime elements only for some rings. In a sense then, the way coprime ideals are defined is the natural generalization of how we expect "coprime" to be.

Now the Chinese Reminder theorem says

If the ideals $I,Jsubset R$ are comaximal, then $R/Icap Jsimeq R/Itimes R/J$.

If we assume $R$ is a Bezout domain, and $x,yin R$ coprime, then $(x)cap (y)=(xy)$ (which is not zero y the way). And Chinese remainder theorem, in this case, becomes
$$
R/(xy)simeq R/(x)times R/(y)
$$
So sure, if in your ring of interest it makes sense to talk about coprime elements, you can go with your special version of the theorem. But if there is no such thing, well... I hope, this sheds some light on the matter.

share|cite|improve this answer

edited Jul 19 at 23:35

answered Jul 19 at 23:28

Hamed

4,391421

The problem is what do you mean by the statement: "$x$ and $y$ are coprime elements in a ring $R$". Let's stick to integers for now. There are two different ways you can think about coprime integers. Two integers $x,yin mathbbZ$ are said to be coprime if one of the two equivalent conditions are satisfied.

1. The greatest common divisor of $x,y$ is equal to one.


2. There exists $a,bin mathbbZ$ such that $ax+by=1$ (Bezout's identity)

The problem with definition (1) is that there are many rings out there with no such thing as a greatest common divisor. Say $mathbbC[x]/(x^2)$ is not even a domain. Or $mathbbZ[-sqrt3]$ which is a domain but the two elements $x=4=2times 2=(1+sqrt-3)(1-sqrt-3)$ and $y=(1+sqrt-3)times 2$ have no gcd. A commutative ring admitting a gcd for any pair of elements is called a GCD domain.

Your example of $mathbbR[x]$ is a GCD domain (actually, more strongly, a principal ideal domain, PID). But even if a ring $R$ is a GCD domain, it does not mean $R$ admits a Bezout identity. An example would be $mathbbR[x,y]$ which is a GCD domain (actually a unique factorization domain, UFD) which does not admit a Bezout identity. A ring that admits a Bezout identity is called a Bezout domain. All Bezout domains are GCD domains. Alternatively, a Bezout domain can be defined as follows:

An integral domain $R$ is called a Bezout domain if given any two elements $x,y$, the sum of ideals $(x)+(y)$ is again principal.

So you can see that in a Bezout domain, $x,y$ are coprime if and only if the ideals $(x), (y)$ are coprime (Two ideals $I,Jsubset R$ are called coprime if $I+J=R$). However, one can define coprime ideals for all commutative rings, and coprime elements only for some rings. In a sense then, the way coprime ideals are defined is the natural generalization of how we expect "coprime" to be.

Now the Chinese Reminder theorem says

If the ideals $I,Jsubset R$ are comaximal, then $R/Icap Jsimeq R/Itimes R/J$.

If we assume $R$ is a Bezout domain, and $x,yin R$ coprime, then $(x)cap (y)=(xy)$ (which is not zero y the way). And Chinese remainder theorem, in this case, becomes
$$
R/(xy)simeq R/(x)times R/(y)
$$
So sure, if in your ring of interest it makes sense to talk about coprime elements, you can go with your special version of the theorem. But if there is no such thing, well... I hope, this sheds some light on the matter.

share|cite|improve this answer

edited Jul 19 at 23:35

answered Jul 19 at 23:28

Hamed

4,391421

share|cite|improve this answer

share|cite|improve this answer

edited Jul 19 at 23:35

edited Jul 19 at 23:35

edited Jul 19 at 23:35

answered Jul 19 at 23:28

Hamed

4,391421

answered Jul 19 at 23:28

Hamed

4,391421

answered Jul 19 at 23:28

Hamed

4,391421

4,391421

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I thought the property $I+J=R$ is called comaximality (not coprimeness)?
  – user437309
  Jul 19 at 23:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Both terminologies are used interchangeably.
  – Hamed
  Jul 19 at 23:35
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I thought the property $I+J=R$ is called comaximality (not coprimeness)?
  – user437309
  Jul 19 at 23:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Both terminologies are used interchangeably.
  – Hamed
  Jul 19 at 23:35
  

  
  
  
  

I thought the property $I+J=R$ is called comaximality (not coprimeness)?
– user437309
Jul 19 at 23:34

I thought the property $I+J=R$ is called comaximality (not coprimeness)?
– user437309
Jul 19 at 23:34

Both terminologies are used interchangeably.
– Hamed
Jul 19 at 23:35

Both terminologies are used interchangeably.
– Hamed
Jul 19 at 23:35

add a comment | 

up vote
0
down vote

The abstract Chinese remainder theorem that I know is this (stated for two ideals):

Let $R$ be a (commutative) ring, $I, J$ two coprime ideals in $R$, i.e. satisfying$I+J=R$. Then $IJ=Icap J$, and we have an isomorphism
beginalign
R/Icap J&xrightarrow;sim;R/Itimes R/J\
r+Icap J&longmapsto (r+ I, r+ J)
endalign

share|cite|improve this answer

edited Jul 19 at 23:29

Kaj Hansen

26.9k43679

answered Jul 19 at 23:26

Bernard

110k635103

add a comment | 

up vote
0
down vote

The abstract Chinese remainder theorem that I know is this (stated for two ideals):

Let $R$ be a (commutative) ring, $I, J$ two coprime ideals in $R$, i.e. satisfying$I+J=R$. Then $IJ=Icap J$, and we have an isomorphism
beginalign
R/Icap J&xrightarrow;sim;R/Itimes R/J\
r+Icap J&longmapsto (r+ I, r+ J)
endalign

share|cite|improve this answer

edited Jul 19 at 23:29

Kaj Hansen

26.9k43679

answered Jul 19 at 23:26

Bernard

110k635103

add a comment | 

up vote
0
down vote

up vote
0
down vote

The abstract Chinese remainder theorem that I know is this (stated for two ideals):

Let $R$ be a (commutative) ring, $I, J$ two coprime ideals in $R$, i.e. satisfying$I+J=R$. Then $IJ=Icap J$, and we have an isomorphism
beginalign
R/Icap J&xrightarrow;sim;R/Itimes R/J\
r+Icap J&longmapsto (r+ I, r+ J)
endalign

share|cite|improve this answer

edited Jul 19 at 23:29

Kaj Hansen

26.9k43679

answered Jul 19 at 23:26

Bernard

110k635103

The abstract Chinese remainder theorem that I know is this (stated for two ideals):

Let $R$ be a (commutative) ring, $I, J$ two coprime ideals in $R$, i.e. satisfying$I+J=R$. Then $IJ=Icap J$, and we have an isomorphism
beginalign
R/Icap J&xrightarrow;sim;R/Itimes R/J\
r+Icap J&longmapsto (r+ I, r+ J)
endalign

share|cite|improve this answer

edited Jul 19 at 23:29

Kaj Hansen

26.9k43679

answered Jul 19 at 23:26

Bernard

110k635103

share|cite|improve this answer

share|cite|improve this answer

edited Jul 19 at 23:29

Kaj Hansen

26.9k43679

edited Jul 19 at 23:29

Kaj Hansen

26.9k43679

edited Jul 19 at 23:29

Kaj Hansen

26.9k43679

26.9k43679

answered Jul 19 at 23:26

Bernard

110k635103

answered Jul 19 at 23:26

Bernard

110k635103

answered Jul 19 at 23:26

Bernard

110k635103

110k635103

add a comment | 

add a comment | 

 

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