Mixing
There exists an element $x$ of $M$ such that $W^{-1}Rdfrac{x}{1}=W^{-1}Rz$ and $operatorname{ann}_R...
$begingroup$
Let $R$ be a Noetherian ring, $W$ a multiplicative subset of $R$, $M$ an $R$-module, and $z$ an element of $W^{-1}M$. Show that there exists an element $x$ of $M$ so that $W^{-1}Rdfrac{x}{1}=W^{-1}Rz$ and $operatorname{ann}_R x=operatorname{ann}_{W^{-1}R} zcap R$ (i.e. $Rxhookrightarrow W^{-1}Rdfrac{x}{1}=W^{-1}Rz$).
For the first part: Let $zin W^{-1}M$ and $xin M$. Well then $dfrac{x}{1}in W^{-1}M$. Since $z$ is an arbitrary element of $W^{-1}M$ and $R$ is Noetherian then $W^{-1}Rdfrac{x}{1}=W^{-1}Rz$. Is this correct?
Also how do I begin the following part? I know that $operatorname{ann}_R(M)={ain R | aM=0}$. But how do I use that when I have intersections?
commutative-algebra modules ideals localization
edited Jan 25 at 21:58
user26857
39.4k124183
asked Jan 25 at 17:11
Username UnknownUsername Unknown
1,17742259
$endgroup$
|
show 1 more comment
$begingroup$
Let $R$ be a Noetherian ring, $W$ a multiplicative subset of $R$, $M$ an $R$-module, and $z$ an element of $W^{-1}M$. Show that there exists an element $x$ of $M$ so that $W^{-1}Rdfrac{x}{1}=W^{-1}Rz$ and $operatorname{ann}_R x=operatorname{ann}_{W^{-1}R} zcap R$ (i.e. $Rxhookrightarrow W^{-1}Rdfrac{x}{1}=W^{-1}Rz$).
For the first part: Let $zin W^{-1}M$ and $xin M$. Well then $dfrac{x}{1}in W^{-1}M$. Since $z$ is an arbitrary element of $W^{-1}M$ and $R$ is Noetherian then $W^{-1}Rdfrac{x}{1}=W^{-1}Rz$. Is this correct?
Also how do I begin the following part? I know that $operatorname{ann}_R(M)={ain R | aM=0}$. But how do I use that when I have intersections?
commutative-algebra modules ideals localization
edited Jan 25 at 21:58
user26857
39.4k124183
asked Jan 25 at 17:11
Username UnknownUsername Unknown
1,17742259
$endgroup$
$begingroup$
1) I found no meaning to your argument for the first part. 2) The element $x$ has to be found. In order to use that $R$ is Noetherian I suggest to consider the ideal $I=mathrm{ann}_{W^{-1}R} zcap R$ of $R$. Pick a finite system of generators for $I$ and use it to construct an appropriate $x$.
$endgroup$
– user26857
Jan 25 at 21:58
$begingroup$
So I show prove the second item first to then apply it to the first item.
$endgroup$
– Username Unknown
Jan 26 at 1:48
$begingroup$
For the first claim it's easy to find such $x$. If $z=m/w$ then $x=m$ works. The second condition is the problem.
$endgroup$
– user26857
Jan 26 at 8:23
$begingroup$
So how should I proceed?
$endgroup$
– Username Unknown
Jan 27 at 18:11
$begingroup$
Just do what I suggested you in the first comment.
$endgroup$
– user26857
Jan 27 at 22:29
|
show 1 more comment
$begingroup$
Let $R$ be a Noetherian ring, $W$ a multiplicative subset of $R$, $M$ an $R$-module, and $z$ an element of $W^{-1}M$. Show that there exists an element $x$ of $M$ so that $W^{-1}Rdfrac{x}{1}=W^{-1}Rz$ and $operatorname{ann}_R x=operatorname{ann}_{W^{-1}R} zcap R$ (i.e. $Rxhookrightarrow W^{-1}Rdfrac{x}{1}=W^{-1}Rz$).
For the first part: Let $zin W^{-1}M$ and $xin M$. Well then $dfrac{x}{1}in W^{-1}M$. Since $z$ is an arbitrary element of $W^{-1}M$ and $R$ is Noetherian then $W^{-1}Rdfrac{x}{1}=W^{-1}Rz$. Is this correct?
Also how do I begin the following part? I know that $operatorname{ann}_R(M)={ain R | aM=0}$. But how do I use that when I have intersections?
commutative-algebra modules ideals localization
edited Jan 25 at 21:58
user26857
39.4k124183
asked Jan 25 at 17:11
Username UnknownUsername Unknown
1,17742259
$endgroup$
Let $R$ be a Noetherian ring, $W$ a multiplicative subset of $R$, $M$ an $R$-module, and $z$ an element of $W^{-1}M$. Show that there exists an element $x$ of $M$ so that $W^{-1}Rdfrac{x}{1}=W^{-1}Rz$ and $operatorname{ann}_R x=operatorname{ann}_{W^{-1}R} zcap R$ (i.e. $Rxhookrightarrow W^{-1}Rdfrac{x}{1}=W^{-1}Rz$).
For the first part: Let $zin W^{-1}M$ and $xin M$. Well then $dfrac{x}{1}in W^{-1}M$. Since $z$ is an arbitrary element of $W^{-1}M$ and $R$ is Noetherian then $W^{-1}Rdfrac{x}{1}=W^{-1}Rz$. Is this correct?
Also how do I begin the following part? I know that $operatorname{ann}_R(M)={ain R | aM=0}$. But how do I use that when I have intersections?
commutative-algebra modules ideals localization
commutative-algebra modules ideals localization
edited Jan 25 at 21:58
user26857
39.4k124183
asked Jan 25 at 17:11
Username UnknownUsername Unknown
1,17742259
edited Jan 25 at 21:58
user26857
39.4k124183
asked Jan 25 at 17:11
Username UnknownUsername Unknown
1,17742259
edited Jan 25 at 21:58
user26857
39.4k124183
edited Jan 25 at 21:58
user26857
39.4k124183
edited Jan 25 at 21:58
user26857
39.4k124183
39.4k124183
asked Jan 25 at 17:11
Username UnknownUsername Unknown
1,17742259
asked Jan 25 at 17:11
Username UnknownUsername Unknown
1,17742259
asked Jan 25 at 17:11
Username UnknownUsername Unknown
1,17742259
1,17742259
$begingroup$
1) I found no meaning to your argument for the first part. 2) The element $x$ has to be found. In order to use that $R$ is Noetherian I suggest to consider the ideal $I=mathrm{ann}_{W^{-1}R} zcap R$ of $R$. Pick a finite system of generators for $I$ and use it to construct an appropriate $x$.
$endgroup$
– user26857
Jan 25 at 21:58
$begingroup$
So I show prove the second item first to then apply it to the first item.
$endgroup$
– Username Unknown
Jan 26 at 1:48
$begingroup$
For the first claim it's easy to find such $x$. If $z=m/w$ then $x=m$ works. The second condition is the problem.
$endgroup$
– user26857
Jan 26 at 8:23
$begingroup$
So how should I proceed?
$endgroup$
– Username Unknown
Jan 27 at 18:11
$begingroup$
Just do what I suggested you in the first comment.
$endgroup$
– user26857
Jan 27 at 22:29
|
show 1 more comment
$begingroup$
1) I found no meaning to your argument for the first part. 2) The element $x$ has to be found. In order to use that $R$ is Noetherian I suggest to consider the ideal $I=mathrm{ann}_{W^{-1}R} zcap R$ of $R$. Pick a finite system of generators for $I$ and use it to construct an appropriate $x$.
$endgroup$
– user26857
Jan 25 at 21:58
$begingroup$
So I show prove the second item first to then apply it to the first item.
$endgroup$
– Username Unknown
Jan 26 at 1:48
$begingroup$
For the first claim it's easy to find such $x$. If $z=m/w$ then $x=m$ works. The second condition is the problem.
$endgroup$
– user26857
Jan 26 at 8:23
$begingroup$
So how should I proceed?
$endgroup$
– Username Unknown
Jan 27 at 18:11
$begingroup$
Just do what I suggested you in the first comment.
$endgroup$
– user26857
Jan 27 at 22:29
$begingroup$
1) I found no meaning to your argument for the first part. 2) The element $x$ has to be found. In order to use that $R$ is Noetherian I suggest to consider the ideal $I=mathrm{ann}_{W^{-1}R} zcap R$ of $R$. Pick a finite system of generators for $I$ and use it to construct an appropriate $x$.
$endgroup$
– user26857
Jan 25 at 21:58
$begingroup$
1) I found no meaning to your argument for the first part. 2) The element $x$ has to be found. In order to use that $R$ is Noetherian I suggest to consider the ideal $I=mathrm{ann}_{W^{-1}R} zcap R$ of $R$. Pick a finite system of generators for $I$ and use it to construct an appropriate $x$.
$endgroup$
– user26857
Jan 25 at 21:58
$begingroup$
So I show prove the second item first to then apply it to the first item.
$endgroup$
– Username Unknown
Jan 26 at 1:48
$begingroup$
So I show prove the second item first to then apply it to the first item.
$endgroup$
– Username Unknown
Jan 26 at 1:48
$begingroup$
For the first claim it's easy to find such $x$. If $z=m/w$ then $x=m$ works. The second condition is the problem.
$endgroup$
– user26857
Jan 26 at 8:23
$begingroup$
For the first claim it's easy to find such $x$. If $z=m/w$ then $x=m$ works. The second condition is the problem.
$endgroup$
– user26857
Jan 26 at 8:23
$begingroup$
So how should I proceed?
$endgroup$
– Username Unknown
Jan 27 at 18:11
$begingroup$
So how should I proceed?
$endgroup$
– Username Unknown
Jan 27 at 18:11
$begingroup$
Just do what I suggested you in the first comment.
$endgroup$
– user26857
Jan 27 at 22:29
$begingroup$
Just do what I suggested you in the first comment.
$endgroup$
– user26857
Jan 27 at 22:29
|
show 1 more comment
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$begingroup$
1) I found no meaning to your argument for the first part. 2) The element $x$ has to be found. In order to use that $R$ is Noetherian I suggest to consider the ideal $I=mathrm{ann}_{W^{-1}R} zcap R$ of $R$. Pick a finite system of generators for $I$ and use it to construct an appropriate $x$.
$endgroup$
– user26857
Jan 25 at 21:58
$begingroup$
So I show prove the second item first to then apply it to the first item.
$endgroup$
– Username Unknown
Jan 26 at 1:48
$begingroup$
For the first claim it's easy to find such $x$. If $z=m/w$ then $x=m$ works. The second condition is the problem.
$endgroup$
– user26857
Jan 26 at 8:23
$begingroup$
So how should I proceed?
$endgroup$
– Username Unknown
Jan 27 at 18:11
$begingroup$
Just do what I suggested you in the first comment.
$endgroup$
– user26857
Jan 27 at 22:29