Constructing an integral domain with a specific subring

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What tools are available for constructing various (noncommutative) domains with a given subdomain? That is, how can I begin to examine various domains which contain the domain $S$, other than obviously $S$ itself?

If we needed only a ring $R$ with a subring $S$, we could take direct sums of $S$ with other rings, but unfortunately the direct sum does not preserve the domain property, so this construction method is not useful.

abstract-algebra ring-theory noncommutative-algebra

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edited Jul 17 at 23:11

Mike Pierce

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asked Jul 17 at 20:52

MightyTyGuy

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up vote
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down vote

favorite

What tools are available for constructing various (noncommutative) domains with a given subdomain? That is, how can I begin to examine various domains which contain the domain $S$, other than obviously $S$ itself?

If we needed only a ring $R$ with a subring $S$, we could take direct sums of $S$ with other rings, but unfortunately the direct sum does not preserve the domain property, so this construction method is not useful.

abstract-algebra ring-theory noncommutative-algebra

share|cite|improve this question

edited Jul 17 at 23:11

Mike Pierce

11k93574

asked Jul 17 at 20:52

MightyTyGuy

1,020314

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

What tools are available for constructing various (noncommutative) domains with a given subdomain? That is, how can I begin to examine various domains which contain the domain $S$, other than obviously $S$ itself?

If we needed only a ring $R$ with a subring $S$, we could take direct sums of $S$ with other rings, but unfortunately the direct sum does not preserve the domain property, so this construction method is not useful.

abstract-algebra ring-theory noncommutative-algebra

share|cite|improve this question

edited Jul 17 at 23:11

Mike Pierce

11k93574

asked Jul 17 at 20:52

MightyTyGuy

1,020314

What tools are available for constructing various (noncommutative) domains with a given subdomain? That is, how can I begin to examine various domains which contain the domain $S$, other than obviously $S$ itself?

If we needed only a ring $R$ with a subring $S$, we could take direct sums of $S$ with other rings, but unfortunately the direct sum does not preserve the domain property, so this construction method is not useful.

abstract-algebra ring-theory noncommutative-algebra

share|cite|improve this question

edited Jul 17 at 23:11

Mike Pierce

11k93574

asked Jul 17 at 20:52

MightyTyGuy

1,020314

share|cite|improve this question

share|cite|improve this question

edited Jul 17 at 23:11

Mike Pierce

11k93574

edited Jul 17 at 23:11

Mike Pierce

11k93574

edited Jul 17 at 23:11

Mike Pierce

11k93574

11k93574

asked Jul 17 at 20:52

MightyTyGuy

1,020314

asked Jul 17 at 20:52

MightyTyGuy

1,020314

asked Jul 17 at 20:52

MightyTyGuy

1,020314

1,020314

add a comment | 

add a comment | 

1 Answer
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In the more familiar commutative case, you could look at certain quotients of polynomial rings over $S$. The quotient $S[x]/I$ will contain a copy of $S$ so long as $I$ trivially intersects the zeroth graded component of $S[x].$ Furthermore, if $S$ is commutative and unital, $S[x]/I$ will be a domain if and only if $I$ is a prime ideal. This will give you many examples of commutative domains that have $S$ as a subring.

This recipe only has to be fudged a little bit to allow for noncommutative quotient rings. Instead of starting with a polynomial ring over $S$, start with a free algebra over $S$ like $Slangle x,yrangle$ instead. Then the quotient $Slangle x, y rangle/I$ will be a domain if and only if $I$ is a completely prime ideal.

share|cite|improve this answer

edited Jul 17 at 23:35

answered Jul 17 at 22:24

Mike Pierce

11k93574

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How can one take the free algebra over $S$ if $S$ is not commutative?
  – MightyTyGuy
  Jul 18 at 21:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MightyTyGuy With great difficulty. If $S$ is not commutative then the usual "free algebra" construction only gets you the structure of an $S$-module. For multiplication in the free algebra, or generally multiplication in any $S$-algebra $A$, for $s in S$ and $a,b in A$ you need to have $scdot(ab) = (scdot a)b = a(scdot b)$. You see $s$ has to "commute with" $a$. You can read the comments in this question to learn more.
  – Mike Pierce
  Jul 19 at 6:39
  
  

  
  
  
  

add a comment | 

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1 Answer
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1 Answer
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up vote
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down vote

In the more familiar commutative case, you could look at certain quotients of polynomial rings over $S$. The quotient $S[x]/I$ will contain a copy of $S$ so long as $I$ trivially intersects the zeroth graded component of $S[x].$ Furthermore, if $S$ is commutative and unital, $S[x]/I$ will be a domain if and only if $I$ is a prime ideal. This will give you many examples of commutative domains that have $S$ as a subring.

This recipe only has to be fudged a little bit to allow for noncommutative quotient rings. Instead of starting with a polynomial ring over $S$, start with a free algebra over $S$ like $Slangle x,yrangle$ instead. Then the quotient $Slangle x, y rangle/I$ will be a domain if and only if $I$ is a completely prime ideal.

share|cite|improve this answer

edited Jul 17 at 23:35

answered Jul 17 at 22:24

Mike Pierce

11k93574

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How can one take the free algebra over $S$ if $S$ is not commutative?
  – MightyTyGuy
  Jul 18 at 21:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MightyTyGuy With great difficulty. If $S$ is not commutative then the usual "free algebra" construction only gets you the structure of an $S$-module. For multiplication in the free algebra, or generally multiplication in any $S$-algebra $A$, for $s in S$ and $a,b in A$ you need to have $scdot(ab) = (scdot a)b = a(scdot b)$. You see $s$ has to "commute with" $a$. You can read the comments in this question to learn more.
  – Mike Pierce
  Jul 19 at 6:39
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

In the more familiar commutative case, you could look at certain quotients of polynomial rings over $S$. The quotient $S[x]/I$ will contain a copy of $S$ so long as $I$ trivially intersects the zeroth graded component of $S[x].$ Furthermore, if $S$ is commutative and unital, $S[x]/I$ will be a domain if and only if $I$ is a prime ideal. This will give you many examples of commutative domains that have $S$ as a subring.

This recipe only has to be fudged a little bit to allow for noncommutative quotient rings. Instead of starting with a polynomial ring over $S$, start with a free algebra over $S$ like $Slangle x,yrangle$ instead. Then the quotient $Slangle x, y rangle/I$ will be a domain if and only if $I$ is a completely prime ideal.

share|cite|improve this answer

edited Jul 17 at 23:35

answered Jul 17 at 22:24

Mike Pierce

11k93574

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How can one take the free algebra over $S$ if $S$ is not commutative?
  – MightyTyGuy
  Jul 18 at 21:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MightyTyGuy With great difficulty. If $S$ is not commutative then the usual "free algebra" construction only gets you the structure of an $S$-module. For multiplication in the free algebra, or generally multiplication in any $S$-algebra $A$, for $s in S$ and $a,b in A$ you need to have $scdot(ab) = (scdot a)b = a(scdot b)$. You see $s$ has to "commute with" $a$. You can read the comments in this question to learn more.
  – Mike Pierce
  Jul 19 at 6:39
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

In the more familiar commutative case, you could look at certain quotients of polynomial rings over $S$. The quotient $S[x]/I$ will contain a copy of $S$ so long as $I$ trivially intersects the zeroth graded component of $S[x].$ Furthermore, if $S$ is commutative and unital, $S[x]/I$ will be a domain if and only if $I$ is a prime ideal. This will give you many examples of commutative domains that have $S$ as a subring.

This recipe only has to be fudged a little bit to allow for noncommutative quotient rings. Instead of starting with a polynomial ring over $S$, start with a free algebra over $S$ like $Slangle x,yrangle$ instead. Then the quotient $Slangle x, y rangle/I$ will be a domain if and only if $I$ is a completely prime ideal.

share|cite|improve this answer

edited Jul 17 at 23:35

answered Jul 17 at 22:24

Mike Pierce

11k93574

In the more familiar commutative case, you could look at certain quotients of polynomial rings over $S$. The quotient $S[x]/I$ will contain a copy of $S$ so long as $I$ trivially intersects the zeroth graded component of $S[x].$ Furthermore, if $S$ is commutative and unital, $S[x]/I$ will be a domain if and only if $I$ is a prime ideal. This will give you many examples of commutative domains that have $S$ as a subring.

This recipe only has to be fudged a little bit to allow for noncommutative quotient rings. Instead of starting with a polynomial ring over $S$, start with a free algebra over $S$ like $Slangle x,yrangle$ instead. Then the quotient $Slangle x, y rangle/I$ will be a domain if and only if $I$ is a completely prime ideal.

share|cite|improve this answer

edited Jul 17 at 23:35

answered Jul 17 at 22:24

Mike Pierce

11k93574

share|cite|improve this answer

share|cite|improve this answer

edited Jul 17 at 23:35

edited Jul 17 at 23:35

edited Jul 17 at 23:35

answered Jul 17 at 22:24

Mike Pierce

11k93574

answered Jul 17 at 22:24

Mike Pierce

11k93574

answered Jul 17 at 22:24

Mike Pierce

11k93574

11k93574

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How can one take the free algebra over $S$ if $S$ is not commutative?
  – MightyTyGuy
  Jul 18 at 21:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MightyTyGuy With great difficulty. If $S$ is not commutative then the usual "free algebra" construction only gets you the structure of an $S$-module. For multiplication in the free algebra, or generally multiplication in any $S$-algebra $A$, for $s in S$ and $a,b in A$ you need to have $scdot(ab) = (scdot a)b = a(scdot b)$. You see $s$ has to "commute with" $a$. You can read the comments in this question to learn more.
  – Mike Pierce
  Jul 19 at 6:39
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How can one take the free algebra over $S$ if $S$ is not commutative?
  – MightyTyGuy
  Jul 18 at 21:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MightyTyGuy With great difficulty. If $S$ is not commutative then the usual "free algebra" construction only gets you the structure of an $S$-module. For multiplication in the free algebra, or generally multiplication in any $S$-algebra $A$, for $s in S$ and $a,b in A$ you need to have $scdot(ab) = (scdot a)b = a(scdot b)$. You see $s$ has to "commute with" $a$. You can read the comments in this question to learn more.
  – Mike Pierce
  Jul 19 at 6:39
  
  

  
  
  
  

How can one take the free algebra over $S$ if $S$ is not commutative?
– MightyTyGuy
Jul 18 at 21:23

How can one take the free algebra over $S$ if $S$ is not commutative?
– MightyTyGuy
Jul 18 at 21:23

@MightyTyGuy With great difficulty. If $S$ is not commutative then the usual "free algebra" construction only gets you the structure of an $S$-module. For multiplication in the free algebra, or generally multiplication in any $S$-algebra $A$, for $s in S$ and $a,b in A$ you need to have $scdot(ab) = (scdot a)b = a(scdot b)$. You see $s$ has to "commute with" $a$. You can read the comments in this question to learn more.
– Mike Pierce
Jul 19 at 6:39

@MightyTyGuy With great difficulty. If $S$ is not commutative then the usual "free algebra" construction only gets you the structure of an $S$-module. For multiplication in the free algebra, or generally multiplication in any $S$-algebra $A$, for $s in S$ and $a,b in A$ you need to have $scdot(ab) = (scdot a)b = a(scdot b)$. You see $s$ has to "commute with" $a$. You can read the comments in this question to learn more.
– Mike Pierce
Jul 19 at 6:39

add a comment | 

 

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