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Subring of a UFD [duplicate]

Give an example of a UFD having a subring which is not a UFD.How do I prove whether something is a Euclidean domain?Subring of $mathbb Z[x]$ is UFD or not?GCD in a subring is GCD in a bigger ring?Give an example of a UFD having a subring which is not a UFD.Is a polynomial ring over a UFD in countably many variables a UFD?Is a subring a UFD?True/False regarding PIDs and UFDsIdeals generated by irreducible elements in a UFD.Which of the following statements about UFDs and PIDs are true?Determining whether a subring with a certain property is a UFDQuotient of UFD

1

1

$begingroup$

This question already has an answer here:

• 
  Give an example of a UFD having a subring which is not a UFD.
  
  5 answers
  
  

Let $R$ be a UFD with $1$, are the following two statements true?

• If $S$ is a subring of $R$, then $S$ is also a UFD;




• if $S$ is a subring of $R$, and $1in S$, then $S$ is UFD.

abstract-algebra ring-theory unique-factorization-domains

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asked Mar 24 at 14:29

qinrqinr

767

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marked as duplicate by Cameron Buie, Eric Wofsey abstract-algebra
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Mar 24 at 15:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Doesn’t subring automatically have $1$?
  $endgroup$
  – J. W. Tanner
  Mar 24 at 14:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @J.W.Tanner: It depends on the text. Some don't require rings to have a unit, so (for example) $2n:ninBbb Z$ is a ring with the usual integer operations.
  $endgroup$
  – Cameron Buie
  Mar 24 at 14:54
  

  
  
  
  

add a comment | 

1

1

$begingroup$

This question already has an answer here:

• 
  Give an example of a UFD having a subring which is not a UFD.
  
  5 answers
  
  

Let $R$ be a UFD with $1$, are the following two statements true?

• If $S$ is a subring of $R$, then $S$ is also a UFD;




• if $S$ is a subring of $R$, and $1in S$, then $S$ is UFD.

abstract-algebra ring-theory unique-factorization-domains

share|cite|improve this question

asked Mar 24 at 14:29

qinrqinr

767

$endgroup$

marked as duplicate by Cameron Buie, Eric Wofsey abstract-algebra
Users with the  abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.

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Mar 24 at 15:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Doesn’t subring automatically have $1$?
  $endgroup$
  – J. W. Tanner
  Mar 24 at 14:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @J.W.Tanner: It depends on the text. Some don't require rings to have a unit, so (for example) $2n:ninBbb Z$ is a ring with the usual integer operations.
  $endgroup$
  – Cameron Buie
  Mar 24 at 14:54
  

  
  
  
  

add a comment | 

$begingroup$

This question already has an answer here:

• 
  Give an example of a UFD having a subring which is not a UFD.
  
  5 answers
  
  

Let $R$ be a UFD with $1$, are the following two statements true?

• If $S$ is a subring of $R$, then $S$ is also a UFD;




• if $S$ is a subring of $R$, and $1in S$, then $S$ is UFD.

abstract-algebra ring-theory unique-factorization-domains

share|cite|improve this question

asked Mar 24 at 14:29

qinrqinr

767

$endgroup$

share|cite|improve this question

asked Mar 24 at 14:29

qinrqinr

767

share|cite|improve this question

asked Mar 24 at 14:29

qinrqinr

767

asked Mar 24 at 14:29

qinrqinr

767

asked Mar 24 at 14:29

qinrqinr

767

3 Answers
3

active

oldest

votes

3

$begingroup$

Not in general. For example, note that $R=Bbb C$ is a field, so trivially a UFD. However, consider $S=Bbb Z[2i]:=x+2iy:x,yinBbb Z.$ One can show fairly easily that $S$ is a subring of $R$ with unity. However, while $2$ is irreducible in $S,$ it is not prime in $S$. Hence, while $S$ is an integral domain (as a subring with unity of an integral domain), it is not a UFD. See here for more information on how we might go about proving this (and feel free to ask me questions if you're stuck).

share|cite|improve this answer

edited Mar 24 at 14:53

answered Mar 24 at 14:41

Cameron BuieCameron Buie

88.8k773163

$endgroup$

add a comment | 

1

$begingroup$

Generalizing the examples given by Bill Dubuque and Cameron Buie, we can use the fact that a UFD is normal (that is, it is integrally closed in its field of fractions) to give us a strategy for finding counterexamples.

Namely, take some UFD $R$, consider its field of fractions $K$, and then find a (unital) subring $Ssubseteq R$ such that the field of fractions of $S$ is still $K$, and $S$ is not integrally closed, i.e., there is $omegain Ksetminus S$ which satisfies a monic polynomial over $S$.

This gives us the counterexamples from both of the other examples. Cameron Buie's has $R=BbbZ[i]$, $S=BbbZ[2i]$, $omega=i$. Bill Dubuque's has $R=BbbZ[sqrtn]$, $S=BbbZ[2sqrtn]$, and $omega = sqrtn$.

Both of these examples are of rings of integers, however, we can also find a subring of a polynomial ring that is not a UFD via this strategy. Consider $R=k[t]$, $S=k[t^2,t^3]$, $omega = t$. We have $omega^2-t^2=0$, so $omega$ is integral over $S$. Note that $t$ is in the fraction field of $S$, since $t=fract^3t^2$.

share|cite|improve this answer

answered Mar 24 at 15:17

jgonjgon

18.3k32345

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In case it was not obvious I chose the Rational Root form of integral closure since many beginners are not familiar with the more general ring-theoretic notion of integral closure (which is not really any more general here).
  $endgroup$
  – Bill Dubuque
  Mar 24 at 15:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Bill Dubuque, I upvoted both your answer and Cameron Buie's, I wrote this answer because I felt that it wasn't necessarily clear how those examples were arrived at. It's possible that this answer will be less useful to the OP, but I did think that it might be useful to someone. :)
  $endgroup$
  – jgon
  Mar 24 at 15:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No doubt. I didn't elaborate since it's already been done in many prior answers (and the CRUDE folks will soon be deleting this thread due to the "quality" of the question)
  $endgroup$
  – Bill Dubuque
  Mar 24 at 15:41
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Pick any quadratic number ring UFD $,R = Bbb Z[sqrt n],$ and let $,S = Bbb Z[2sqrt n]subset R.,$ Notice that $,w = (2sqrtn)/2 = sqrtn,$ is a fraction over $S$ and $,wnotin S,$ but $w$ is a root of the monic polynomial $,x^2 - n, ,$ contra the Rational Root Test (which is valid in any UFD). So $S$ is not a UFD.

share|cite|improve this answer

edited Mar 24 at 15:26

answered Mar 24 at 15:05

Bill DubuqueBill Dubuque

217k29204668

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you switch $R$ and $S$?
  $endgroup$
  – J. W. Tanner
  Mar 24 at 15:23
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @J.W.Tanner Yes, typos now fixed, thanks.
  $endgroup$
  – Bill Dubuque
  Mar 24 at 15:26
  

  
  
  
  

add a comment | 

3

$begingroup$

Not in general. For example, note that $R=Bbb C$ is a field, so trivially a UFD. However, consider $S=Bbb Z[2i]:=x+2iy:x,yinBbb Z.$ One can show fairly easily that $S$ is a subring of $R$ with unity. However, while $2$ is irreducible in $S,$ it is not prime in $S$. Hence, while $S$ is an integral domain (as a subring with unity of an integral domain), it is not a UFD. See here for more information on how we might go about proving this (and feel free to ask me questions if you're stuck).

share|cite|improve this answer

edited Mar 24 at 14:53

answered Mar 24 at 14:41

Cameron BuieCameron Buie

88.8k773163

$endgroup$

add a comment | 

3

$begingroup$

Not in general. For example, note that $R=Bbb C$ is a field, so trivially a UFD. However, consider $S=Bbb Z[2i]:=x+2iy:x,yinBbb Z.$ One can show fairly easily that $S$ is a subring of $R$ with unity. However, while $2$ is irreducible in $S,$ it is not prime in $S$. Hence, while $S$ is an integral domain (as a subring with unity of an integral domain), it is not a UFD. See here for more information on how we might go about proving this (and feel free to ask me questions if you're stuck).

share|cite|improve this answer

edited Mar 24 at 14:53

answered Mar 24 at 14:41

Cameron BuieCameron Buie

88.8k773163

$endgroup$

add a comment | 

$begingroup$

Not in general. For example, note that $R=Bbb C$ is a field, so trivially a UFD. However, consider $S=Bbb Z[2i]:=x+2iy:x,yinBbb Z.$ One can show fairly easily that $S$ is a subring of $R$ with unity. However, while $2$ is irreducible in $S,$ it is not prime in $S$. Hence, while $S$ is an integral domain (as a subring with unity of an integral domain), it is not a UFD. See here for more information on how we might go about proving this (and feel free to ask me questions if you're stuck).

share|cite|improve this answer

edited Mar 24 at 14:53

answered Mar 24 at 14:41

Cameron BuieCameron Buie

88.8k773163

$endgroup$

share|cite|improve this answer

edited Mar 24 at 14:53

answered Mar 24 at 14:41

Cameron BuieCameron Buie

88.8k773163

answered Mar 24 at 14:41

Cameron BuieCameron Buie

88.8k773163

answered Mar 24 at 14:41

Cameron BuieCameron Buie

88.8k773163

1

$begingroup$

Generalizing the examples given by Bill Dubuque and Cameron Buie, we can use the fact that a UFD is normal (that is, it is integrally closed in its field of fractions) to give us a strategy for finding counterexamples.

Namely, take some UFD $R$, consider its field of fractions $K$, and then find a (unital) subring $Ssubseteq R$ such that the field of fractions of $S$ is still $K$, and $S$ is not integrally closed, i.e., there is $omegain Ksetminus S$ which satisfies a monic polynomial over $S$.

This gives us the counterexamples from both of the other examples. Cameron Buie's has $R=BbbZ[i]$, $S=BbbZ[2i]$, $omega=i$. Bill Dubuque's has $R=BbbZ[sqrtn]$, $S=BbbZ[2sqrtn]$, and $omega = sqrtn$.

Both of these examples are of rings of integers, however, we can also find a subring of a polynomial ring that is not a UFD via this strategy. Consider $R=k[t]$, $S=k[t^2,t^3]$, $omega = t$. We have $omega^2-t^2=0$, so $omega$ is integral over $S$. Note that $t$ is in the fraction field of $S$, since $t=fract^3t^2$.

share|cite|improve this answer

answered Mar 24 at 15:17

jgonjgon

18.3k32345

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In case it was not obvious I chose the Rational Root form of integral closure since many beginners are not familiar with the more general ring-theoretic notion of integral closure (which is not really any more general here).
  $endgroup$
  – Bill Dubuque
  Mar 24 at 15:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Bill Dubuque, I upvoted both your answer and Cameron Buie's, I wrote this answer because I felt that it wasn't necessarily clear how those examples were arrived at. It's possible that this answer will be less useful to the OP, but I did think that it might be useful to someone. :)
  $endgroup$
  – jgon
  Mar 24 at 15:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No doubt. I didn't elaborate since it's already been done in many prior answers (and the CRUDE folks will soon be deleting this thread due to the "quality" of the question)
  $endgroup$
  – Bill Dubuque
  Mar 24 at 15:41
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Generalizing the examples given by Bill Dubuque and Cameron Buie, we can use the fact that a UFD is normal (that is, it is integrally closed in its field of fractions) to give us a strategy for finding counterexamples.

Namely, take some UFD $R$, consider its field of fractions $K$, and then find a (unital) subring $Ssubseteq R$ such that the field of fractions of $S$ is still $K$, and $S$ is not integrally closed, i.e., there is $omegain Ksetminus S$ which satisfies a monic polynomial over $S$.

This gives us the counterexamples from both of the other examples. Cameron Buie's has $R=BbbZ[i]$, $S=BbbZ[2i]$, $omega=i$. Bill Dubuque's has $R=BbbZ[sqrtn]$, $S=BbbZ[2sqrtn]$, and $omega = sqrtn$.

Both of these examples are of rings of integers, however, we can also find a subring of a polynomial ring that is not a UFD via this strategy. Consider $R=k[t]$, $S=k[t^2,t^3]$, $omega = t$. We have $omega^2-t^2=0$, so $omega$ is integral over $S$. Note that $t$ is in the fraction field of $S$, since $t=fract^3t^2$.

share|cite|improve this answer

answered Mar 24 at 15:17

jgonjgon

18.3k32345

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In case it was not obvious I chose the Rational Root form of integral closure since many beginners are not familiar with the more general ring-theoretic notion of integral closure (which is not really any more general here).
  $endgroup$
  – Bill Dubuque
  Mar 24 at 15:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Bill Dubuque, I upvoted both your answer and Cameron Buie's, I wrote this answer because I felt that it wasn't necessarily clear how those examples were arrived at. It's possible that this answer will be less useful to the OP, but I did think that it might be useful to someone. :)
  $endgroup$
  – jgon
  Mar 24 at 15:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No doubt. I didn't elaborate since it's already been done in many prior answers (and the CRUDE folks will soon be deleting this thread due to the "quality" of the question)
  $endgroup$
  – Bill Dubuque
  Mar 24 at 15:41
  
  

  
  
  
  

add a comment | 

$begingroup$

Generalizing the examples given by Bill Dubuque and Cameron Buie, we can use the fact that a UFD is normal (that is, it is integrally closed in its field of fractions) to give us a strategy for finding counterexamples.

Namely, take some UFD $R$, consider its field of fractions $K$, and then find a (unital) subring $Ssubseteq R$ such that the field of fractions of $S$ is still $K$, and $S$ is not integrally closed, i.e., there is $omegain Ksetminus S$ which satisfies a monic polynomial over $S$.

This gives us the counterexamples from both of the other examples. Cameron Buie's has $R=BbbZ[i]$, $S=BbbZ[2i]$, $omega=i$. Bill Dubuque's has $R=BbbZ[sqrtn]$, $S=BbbZ[2sqrtn]$, and $omega = sqrtn$.

Both of these examples are of rings of integers, however, we can also find a subring of a polynomial ring that is not a UFD via this strategy. Consider $R=k[t]$, $S=k[t^2,t^3]$, $omega = t$. We have $omega^2-t^2=0$, so $omega$ is integral over $S$. Note that $t$ is in the fraction field of $S$, since $t=fract^3t^2$.

share|cite|improve this answer

answered Mar 24 at 15:17

jgonjgon

18.3k32345

$endgroup$

share|cite|improve this answer

answered Mar 24 at 15:17

jgonjgon

18.3k32345

answered Mar 24 at 15:17

jgonjgon

18.3k32345

answered Mar 24 at 15:17

jgonjgon

18.3k32345

1

$begingroup$

Pick any quadratic number ring UFD $,R = Bbb Z[sqrt n],$ and let $,S = Bbb Z[2sqrt n]subset R.,$ Notice that $,w = (2sqrtn)/2 = sqrtn,$ is a fraction over $S$ and $,wnotin S,$ but $w$ is a root of the monic polynomial $,x^2 - n, ,$ contra the Rational Root Test (which is valid in any UFD). So $S$ is not a UFD.

share|cite|improve this answer

edited Mar 24 at 15:26

answered Mar 24 at 15:05

Bill DubuqueBill Dubuque

217k29204668

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you switch $R$ and $S$?
  $endgroup$
  – J. W. Tanner
  Mar 24 at 15:23
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @J.W.Tanner Yes, typos now fixed, thanks.
  $endgroup$
  – Bill Dubuque
  Mar 24 at 15:26
  

  
  
  
  

add a comment | 

1

$begingroup$

Pick any quadratic number ring UFD $,R = Bbb Z[sqrt n],$ and let $,S = Bbb Z[2sqrt n]subset R.,$ Notice that $,w = (2sqrtn)/2 = sqrtn,$ is a fraction over $S$ and $,wnotin S,$ but $w$ is a root of the monic polynomial $,x^2 - n, ,$ contra the Rational Root Test (which is valid in any UFD). So $S$ is not a UFD.

share|cite|improve this answer

edited Mar 24 at 15:26

answered Mar 24 at 15:05

Bill DubuqueBill Dubuque

217k29204668

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you switch $R$ and $S$?
  $endgroup$
  – J. W. Tanner
  Mar 24 at 15:23
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @J.W.Tanner Yes, typos now fixed, thanks.
  $endgroup$
  – Bill Dubuque
  Mar 24 at 15:26
  

  
  
  
  

add a comment | 

$begingroup$

Pick any quadratic number ring UFD $,R = Bbb Z[sqrt n],$ and let $,S = Bbb Z[2sqrt n]subset R.,$ Notice that $,w = (2sqrtn)/2 = sqrtn,$ is a fraction over $S$ and $,wnotin S,$ but $w$ is a root of the monic polynomial $,x^2 - n, ,$ contra the Rational Root Test (which is valid in any UFD). So $S$ is not a UFD.

share|cite|improve this answer

edited Mar 24 at 15:26

answered Mar 24 at 15:05

Bill DubuqueBill Dubuque

217k29204668

$endgroup$

share|cite|improve this answer

edited Mar 24 at 15:26

answered Mar 24 at 15:05

Bill DubuqueBill Dubuque

217k29204668

answered Mar 24 at 15:05

Bill DubuqueBill Dubuque

217k29204668

answered Mar 24 at 15:05

Bill DubuqueBill Dubuque

217k29204668