about flat $R$-modules with different conditions [closed]
Need to know if the following statements are true or false:
Let $S$ be a multiplicative set in $R$. Then $(S^{-1}R)[x,y,z]$ is always a flat $R$-module.
Let $I,J subset R$ be proper ideals of $R$ such that $I+J=R$. Then $R/I$ is always flat $R$-module.
Let $R[[x]]$ be the formal power series ring over $R$. Then $R[[x]]/left<sum^{infty}_{i=2} x^iright>$ is always flat $R$-module.
$R/N(R)$ is always a flat $R$-module. ($N(R)$ the nilradical). False. Seen here. Sure?
abstract-algebra commutative-algebra modules flatness
edited Nov 22 at 19:43
user26857
39.2k123882
asked Nov 22 at 13:37
idriskameni
769
closed as off-topic by user26857, rschwieb, Saad, max_zorn, Brahadeesh Nov 23 at 6:28
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user26857, rschwieb, Saad, max_zorn, Brahadeesh
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Need to know if the following statements are true or false:
Let $S$ be a multiplicative set in $R$. Then $(S^{-1}R)[x,y,z]$ is always a flat $R$-module.
Let $I,J subset R$ be proper ideals of $R$ such that $I+J=R$. Then $R/I$ is always flat $R$-module.
Let $R[[x]]$ be the formal power series ring over $R$. Then $R[[x]]/left<sum^{infty}_{i=2} x^iright>$ is always flat $R$-module.
$R/N(R)$ is always a flat $R$-module. ($N(R)$ the nilradical). False. Seen here. Sure?
abstract-algebra commutative-algebra modules flatness
edited Nov 22 at 19:43
user26857
39.2k123882
asked Nov 22 at 13:37
idriskameni
769
closed as off-topic by user26857, rschwieb, Saad, max_zorn, Brahadeesh Nov 23 at 6:28
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user26857, rschwieb, Saad, max_zorn, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
Need to know if the following statements are true or false:
Let $S$ be a multiplicative set in $R$. Then $(S^{-1}R)[x,y,z]$ is always a flat $R$-module.
Let $I,J subset R$ be proper ideals of $R$ such that $I+J=R$. Then $R/I$ is always flat $R$-module.
Let $R[[x]]$ be the formal power series ring over $R$. Then $R[[x]]/left<sum^{infty}_{i=2} x^iright>$ is always flat $R$-module.
$R/N(R)$ is always a flat $R$-module. ($N(R)$ the nilradical). False. Seen here. Sure?
abstract-algebra commutative-algebra modules flatness
edited Nov 22 at 19:43
user26857
39.2k123882
asked Nov 22 at 13:37
idriskameni
769
Need to know if the following statements are true or false:
Let $S$ be a multiplicative set in $R$. Then $(S^{-1}R)[x,y,z]$ is always a flat $R$-module.
Let $I,J subset R$ be proper ideals of $R$ such that $I+J=R$. Then $R/I$ is always flat $R$-module.
Let $R[[x]]$ be the formal power series ring over $R$. Then $R[[x]]/left<sum^{infty}_{i=2} x^iright>$ is always flat $R$-module.
$R/N(R)$ is always a flat $R$-module. ($N(R)$ the nilradical). False. Seen here. Sure?
abstract-algebra commutative-algebra modules flatness
abstract-algebra commutative-algebra modules flatness
edited Nov 22 at 19:43
user26857
39.2k123882
asked Nov 22 at 13:37
idriskameni
769
edited Nov 22 at 19:43
user26857
39.2k123882
asked Nov 22 at 13:37
idriskameni
769
edited Nov 22 at 19:43
user26857
39.2k123882
edited Nov 22 at 19:43
user26857
39.2k123882
edited Nov 22 at 19:43
user26857
39.2k123882
39.2k123882
asked Nov 22 at 13:37
idriskameni
769
asked Nov 22 at 13:37
idriskameni
769
asked Nov 22 at 13:37
idriskameni
769
769
closed as off-topic by user26857, rschwieb, Saad, max_zorn, Brahadeesh Nov 23 at 6:28
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user26857, rschwieb, Saad, max_zorn, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by user26857, rschwieb, Saad, max_zorn, Brahadeesh Nov 23 at 6:28
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user26857, rschwieb, Saad, max_zorn, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
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1 Answer
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I assume you are working over commutative rings.
This is correct. $S^{-1}R$ is flat over $R$ (localizations are flat) and if $Ato Bto C$ are ring homomorphisms with $Ato B$ and $Bto C$ are flat, then so is $Ato C$. Since $S^{-1}Rto S^{-1}[x,y,z]$ is flat, you are done.
This is false. Take $R=mathbb{Z}$, $I=pR, J=qR$ where $p,q$ are distinct primes.
This is correct. $f=sum_{n=2}^{infty} x^n=x^2u$ where $u=sum_{n=0}^{infty} x^n$. Easy to check that $u$ is a unit. So, $R[[x]]/(f)=R[[x]]/(x^2)$ which is free module over $R$ with basis $1,x$.
This is false. For example, let $R=mathbb{Z}/(p^2)$, $p$ a prime. Then $R/N(R)=R/(p)$ is not flat over $R$.
edited Nov 22 at 21:37
user26857
39.2k123882
answered Nov 22 at 18:23
Mohan
11.5k1817
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
I assume you are working over commutative rings.
This is correct. $S^{-1}R$ is flat over $R$ (localizations are flat) and if $Ato Bto C$ are ring homomorphisms with $Ato B$ and $Bto C$ are flat, then so is $Ato C$. Since $S^{-1}Rto S^{-1}[x,y,z]$ is flat, you are done.
This is false. Take $R=mathbb{Z}$, $I=pR, J=qR$ where $p,q$ are distinct primes.
This is correct. $f=sum_{n=2}^{infty} x^n=x^2u$ where $u=sum_{n=0}^{infty} x^n$. Easy to check that $u$ is a unit. So, $R[[x]]/(f)=R[[x]]/(x^2)$ which is free module over $R$ with basis $1,x$.
This is false. For example, let $R=mathbb{Z}/(p^2)$, $p$ a prime. Then $R/N(R)=R/(p)$ is not flat over $R$.
edited Nov 22 at 21:37
user26857
39.2k123882
answered Nov 22 at 18:23
Mohan
11.5k1817
add a comment |
I assume you are working over commutative rings.
This is correct. $S^{-1}R$ is flat over $R$ (localizations are flat) and if $Ato Bto C$ are ring homomorphisms with $Ato B$ and $Bto C$ are flat, then so is $Ato C$. Since $S^{-1}Rto S^{-1}[x,y,z]$ is flat, you are done.
This is false. Take $R=mathbb{Z}$, $I=pR, J=qR$ where $p,q$ are distinct primes.
This is correct. $f=sum_{n=2}^{infty} x^n=x^2u$ where $u=sum_{n=0}^{infty} x^n$. Easy to check that $u$ is a unit. So, $R[[x]]/(f)=R[[x]]/(x^2)$ which is free module over $R$ with basis $1,x$.
This is false. For example, let $R=mathbb{Z}/(p^2)$, $p$ a prime. Then $R/N(R)=R/(p)$ is not flat over $R$.
edited Nov 22 at 21:37
user26857
39.2k123882
answered Nov 22 at 18:23
Mohan
11.5k1817
add a comment |
I assume you are working over commutative rings.
This is correct. $S^{-1}R$ is flat over $R$ (localizations are flat) and if $Ato Bto C$ are ring homomorphisms with $Ato B$ and $Bto C$ are flat, then so is $Ato C$. Since $S^{-1}Rto S^{-1}[x,y,z]$ is flat, you are done.
This is false. Take $R=mathbb{Z}$, $I=pR, J=qR$ where $p,q$ are distinct primes.
This is correct. $f=sum_{n=2}^{infty} x^n=x^2u$ where $u=sum_{n=0}^{infty} x^n$. Easy to check that $u$ is a unit. So, $R[[x]]/(f)=R[[x]]/(x^2)$ which is free module over $R$ with basis $1,x$.
This is false. For example, let $R=mathbb{Z}/(p^2)$, $p$ a prime. Then $R/N(R)=R/(p)$ is not flat over $R$.
edited Nov 22 at 21:37
user26857
39.2k123882
answered Nov 22 at 18:23
Mohan
11.5k1817
I assume you are working over commutative rings.
This is correct. $S^{-1}R$ is flat over $R$ (localizations are flat) and if $Ato Bto C$ are ring homomorphisms with $Ato B$ and $Bto C$ are flat, then so is $Ato C$. Since $S^{-1}Rto S^{-1}[x,y,z]$ is flat, you are done.
This is false. Take $R=mathbb{Z}$, $I=pR, J=qR$ where $p,q$ are distinct primes.
This is correct. $f=sum_{n=2}^{infty} x^n=x^2u$ where $u=sum_{n=0}^{infty} x^n$. Easy to check that $u$ is a unit. So, $R[[x]]/(f)=R[[x]]/(x^2)$ which is free module over $R$ with basis $1,x$.
This is false. For example, let $R=mathbb{Z}/(p^2)$, $p$ a prime. Then $R/N(R)=R/(p)$ is not flat over $R$.
edited Nov 22 at 21:37
user26857
39.2k123882
answered Nov 22 at 18:23
Mohan
11.5k1817
edited Nov 22 at 21:37
user26857
39.2k123882
edited Nov 22 at 21:37
user26857
39.2k123882
edited Nov 22 at 21:37
user26857
39.2k123882
39.2k123882
answered Nov 22 at 18:23
Mohan
11.5k1817
answered Nov 22 at 18:23
Mohan
11.5k1817
answered Nov 22 at 18:23
Mohan
11.5k1817
11.5k1817
add a comment |
add a comment |
