A Noetherian ring which is not a Noetherian $R$-module

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Let $R$ be a Noetherian integral domain and $omegain Rbackslash0$ be a non-unit. For the multiplicative set $S=omega^n:ninmathbbZ^+$ of $R$, let $R_omega=S^-1R$ be the ring of fractions of $R$ by $S$. Prove that $R_omega$ is a Noetherian ring, but not a Noetherian $R$-module.

My attempt: Proving that $R_omega$ is a Noetherian ring is easy, since $R_omega cong R[x]/(xomega-1)$ and $R[x]$ is a Noetherian ring. However, I'm not sure if my proof of the latter statement is correct. To obtain a contradiction, I'd like to prove that $R_omega$ is not finitely generated as an $R$-module. Below is my solution.

Suppose $R_omegacong R[x]/(xomega-1)$ is a finitely generated $R$-module. Then there exists $f_1,...,f_nin R[x]$ such that $langle f_1,...,f_nrangle=R[x]/(xomega-1)$.(Here I'm abusing the notation; each $f_i$ means the coset $f_i+(xomega-1)$.) Let $deg(f_i)= m_i$. Now let $M=maxm_1,...,m_n$. Then $x^M+1notin langle f_1,...,f_nrangle$. If it were, then there would be $r_1 ,..., r_n in R$ and $gin R[x]$ such that $x^M+1= sum_i=1^nr_i f_i + (xomega-1)gin R[x]$. Comparing the coefficients of the both sides, the leading coefficient of $g$ must be $1/omega$, which is absurd since $omega$ is a non-unit.

Is my solution correct?

commutative-algebra noetherian

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edited Jul 15 at 13:29

asked Jul 15 at 13:22

bellcircle

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Let $R$ be a Noetherian integral domain and $omegain Rbackslash0$ be a non-unit. For the multiplicative set $S=omega^n:ninmathbbZ^+$ of $R$, let $R_omega=S^-1R$ be the ring of fractions of $R$ by $S$. Prove that $R_omega$ is a Noetherian ring, but not a Noetherian $R$-module.

My attempt: Proving that $R_omega$ is a Noetherian ring is easy, since $R_omega cong R[x]/(xomega-1)$ and $R[x]$ is a Noetherian ring. However, I'm not sure if my proof of the latter statement is correct. To obtain a contradiction, I'd like to prove that $R_omega$ is not finitely generated as an $R$-module. Below is my solution.

Suppose $R_omegacong R[x]/(xomega-1)$ is a finitely generated $R$-module. Then there exists $f_1,...,f_nin R[x]$ such that $langle f_1,...,f_nrangle=R[x]/(xomega-1)$.(Here I'm abusing the notation; each $f_i$ means the coset $f_i+(xomega-1)$.) Let $deg(f_i)= m_i$. Now let $M=maxm_1,...,m_n$. Then $x^M+1notin langle f_1,...,f_nrangle$. If it were, then there would be $r_1 ,..., r_n in R$ and $gin R[x]$ such that $x^M+1= sum_i=1^nr_i f_i + (xomega-1)gin R[x]$. Comparing the coefficients of the both sides, the leading coefficient of $g$ must be $1/omega$, which is absurd since $omega$ is a non-unit.

Is my solution correct?

commutative-algebra noetherian

share|cite|improve this question

edited Jul 15 at 13:29

asked Jul 15 at 13:22

bellcircle

1,123311

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Let $R$ be a Noetherian integral domain and $omegain Rbackslash0$ be a non-unit. For the multiplicative set $S=omega^n:ninmathbbZ^+$ of $R$, let $R_omega=S^-1R$ be the ring of fractions of $R$ by $S$. Prove that $R_omega$ is a Noetherian ring, but not a Noetherian $R$-module.

My attempt: Proving that $R_omega$ is a Noetherian ring is easy, since $R_omega cong R[x]/(xomega-1)$ and $R[x]$ is a Noetherian ring. However, I'm not sure if my proof of the latter statement is correct. To obtain a contradiction, I'd like to prove that $R_omega$ is not finitely generated as an $R$-module. Below is my solution.

Suppose $R_omegacong R[x]/(xomega-1)$ is a finitely generated $R$-module. Then there exists $f_1,...,f_nin R[x]$ such that $langle f_1,...,f_nrangle=R[x]/(xomega-1)$.(Here I'm abusing the notation; each $f_i$ means the coset $f_i+(xomega-1)$.) Let $deg(f_i)= m_i$. Now let $M=maxm_1,...,m_n$. Then $x^M+1notin langle f_1,...,f_nrangle$. If it were, then there would be $r_1 ,..., r_n in R$ and $gin R[x]$ such that $x^M+1= sum_i=1^nr_i f_i + (xomega-1)gin R[x]$. Comparing the coefficients of the both sides, the leading coefficient of $g$ must be $1/omega$, which is absurd since $omega$ is a non-unit.

Is my solution correct?

commutative-algebra noetherian

share|cite|improve this question

edited Jul 15 at 13:29

asked Jul 15 at 13:22

bellcircle

1,123311

Let $R$ be a Noetherian integral domain and $omegain Rbackslash0$ be a non-unit. For the multiplicative set $S=omega^n:ninmathbbZ^+$ of $R$, let $R_omega=S^-1R$ be the ring of fractions of $R$ by $S$. Prove that $R_omega$ is a Noetherian ring, but not a Noetherian $R$-module.

My attempt: Proving that $R_omega$ is a Noetherian ring is easy, since $R_omega cong R[x]/(xomega-1)$ and $R[x]$ is a Noetherian ring. However, I'm not sure if my proof of the latter statement is correct. To obtain a contradiction, I'd like to prove that $R_omega$ is not finitely generated as an $R$-module. Below is my solution.

Suppose $R_omegacong R[x]/(xomega-1)$ is a finitely generated $R$-module. Then there exists $f_1,...,f_nin R[x]$ such that $langle f_1,...,f_nrangle=R[x]/(xomega-1)$.(Here I'm abusing the notation; each $f_i$ means the coset $f_i+(xomega-1)$.) Let $deg(f_i)= m_i$. Now let $M=maxm_1,...,m_n$. Then $x^M+1notin langle f_1,...,f_nrangle$. If it were, then there would be $r_1 ,..., r_n in R$ and $gin R[x]$ such that $x^M+1= sum_i=1^nr_i f_i + (xomega-1)gin R[x]$. Comparing the coefficients of the both sides, the leading coefficient of $g$ must be $1/omega$, which is absurd since $omega$ is a non-unit.

Is my solution correct?

commutative-algebra noetherian

share|cite|improve this question

edited Jul 15 at 13:29

asked Jul 15 at 13:22

bellcircle

1,123311

share|cite|improve this question

share|cite|improve this question

edited Jul 15 at 13:29

edited Jul 15 at 13:29

edited Jul 15 at 13:29

asked Jul 15 at 13:22

bellcircle

1,123311

asked Jul 15 at 13:22

bellcircle

1,123311

asked Jul 15 at 13:22

bellcircle

1,123311

1,123311

add a comment | 

add a comment | 

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The proof that $R_omega$ is Noetherian is good.

Let $x_1/omega^k,dots,x_n/omega^k$ be a finite set in $R_omega$. It's not restrictive to assume the denominators are the same $k>0$, with $x_iin R$. Let's show that
$$
frac1omega^k+1notinfracx_1omega^kR+dots+fracx_nomega^kR
$$
Assume the contrary; then we can write
$$
frac1omega^k+1=fracromega^k
$$
in $R_omega$, for some $rin R$, which in turn implies
$$
fracr1=frac1omega
$$
As $R$ is an an integral domain, this implies $romega=1$.

Note that the assumption $R$ is a domain is necessary. For instance, if $omega$ is nilpotent, then $R_omega=0$. But this is not required: for $R=mathbbZ/6mathbbZ$ and $omega=bar2$, the $R$-module $R_omega$ is generated by $1/bar4$.

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answered Jul 15 at 13:58

egreg

164k1180187

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The proof that $R_omega$ is Noetherian is good.

Let $x_1/omega^k,dots,x_n/omega^k$ be a finite set in $R_omega$. It's not restrictive to assume the denominators are the same $k>0$, with $x_iin R$. Let's show that
$$
frac1omega^k+1notinfracx_1omega^kR+dots+fracx_nomega^kR
$$
Assume the contrary; then we can write
$$
frac1omega^k+1=fracromega^k
$$
in $R_omega$, for some $rin R$, which in turn implies
$$
fracr1=frac1omega
$$
As $R$ is an an integral domain, this implies $romega=1$.

Note that the assumption $R$ is a domain is necessary. For instance, if $omega$ is nilpotent, then $R_omega=0$. But this is not required: for $R=mathbbZ/6mathbbZ$ and $omega=bar2$, the $R$-module $R_omega$ is generated by $1/bar4$.

share|cite|improve this answer

answered Jul 15 at 13:58

egreg

164k1180187

add a comment | 

up vote
0
down vote

The proof that $R_omega$ is Noetherian is good.

Let $x_1/omega^k,dots,x_n/omega^k$ be a finite set in $R_omega$. It's not restrictive to assume the denominators are the same $k>0$, with $x_iin R$. Let's show that
$$
frac1omega^k+1notinfracx_1omega^kR+dots+fracx_nomega^kR
$$
Assume the contrary; then we can write
$$
frac1omega^k+1=fracromega^k
$$
in $R_omega$, for some $rin R$, which in turn implies
$$
fracr1=frac1omega
$$
As $R$ is an an integral domain, this implies $romega=1$.

Note that the assumption $R$ is a domain is necessary. For instance, if $omega$ is nilpotent, then $R_omega=0$. But this is not required: for $R=mathbbZ/6mathbbZ$ and $omega=bar2$, the $R$-module $R_omega$ is generated by $1/bar4$.

share|cite|improve this answer

answered Jul 15 at 13:58

egreg

164k1180187

add a comment | 

up vote
0
down vote

up vote
0
down vote

The proof that $R_omega$ is Noetherian is good.

Let $x_1/omega^k,dots,x_n/omega^k$ be a finite set in $R_omega$. It's not restrictive to assume the denominators are the same $k>0$, with $x_iin R$. Let's show that
$$
frac1omega^k+1notinfracx_1omega^kR+dots+fracx_nomega^kR
$$
Assume the contrary; then we can write
$$
frac1omega^k+1=fracromega^k
$$
in $R_omega$, for some $rin R$, which in turn implies
$$
fracr1=frac1omega
$$
As $R$ is an an integral domain, this implies $romega=1$.

Note that the assumption $R$ is a domain is necessary. For instance, if $omega$ is nilpotent, then $R_omega=0$. But this is not required: for $R=mathbbZ/6mathbbZ$ and $omega=bar2$, the $R$-module $R_omega$ is generated by $1/bar4$.

share|cite|improve this answer

answered Jul 15 at 13:58

egreg

164k1180187

The proof that $R_omega$ is Noetherian is good.

Let $x_1/omega^k,dots,x_n/omega^k$ be a finite set in $R_omega$. It's not restrictive to assume the denominators are the same $k>0$, with $x_iin R$. Let's show that
$$
frac1omega^k+1notinfracx_1omega^kR+dots+fracx_nomega^kR
$$
Assume the contrary; then we can write
$$
frac1omega^k+1=fracromega^k
$$
in $R_omega$, for some $rin R$, which in turn implies
$$
fracr1=frac1omega
$$
As $R$ is an an integral domain, this implies $romega=1$.

Note that the assumption $R$ is a domain is necessary. For instance, if $omega$ is nilpotent, then $R_omega=0$. But this is not required: for $R=mathbbZ/6mathbbZ$ and $omega=bar2$, the $R$-module $R_omega$ is generated by $1/bar4$.

share|cite|improve this answer

answered Jul 15 at 13:58

egreg

164k1180187

share|cite|improve this answer

share|cite|improve this answer

answered Jul 15 at 13:58

egreg

164k1180187

answered Jul 15 at 13:58

egreg

164k1180187

answered Jul 15 at 13:58

egreg

164k1180187

164k1180187

add a comment | 

add a comment | 

 

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