Reduced rings and simple modules
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I'm having trouble getting my intuition pumping and the details in my head on the Jacobson radical (intersection of maximal ideals or intersection of annihilators of simple modules). What I'm interested in showing is the following:
$$rad R/U = (rad R)/U$$
where $U$ is an ideal contained in $radR$. That is, an ideal contained in every maximal left ideal. It seems intuitive to me. For if, $U$ is contained in every maximal left ideal, then it annihilates every simple module of $R$. So, when we peel that stuff out of the radical it should be the same as modding it out of the ring in the first place but I'm having trouble with the specific details.
Further, from here we wish to conclude that $R$ and $R/radR$ have the same simple modules. It is clear that a simple module for $R$ is certainly a module for $R/radR$. But showing that it is simple just because I removed it's annihilator is unclear to me. Any help is appreciated. Thanks! A somewhat relevant post
Minimal ideals in a reduced ring
modules semi-simple-rings
asked Dec 14 '18 at 2:12
Aaron ZolotorAaron Zolotor
1,632525
$endgroup$
add a comment |
$begingroup$
I'm having trouble getting my intuition pumping and the details in my head on the Jacobson radical (intersection of maximal ideals or intersection of annihilators of simple modules). What I'm interested in showing is the following:
$$rad R/U = (rad R)/U$$
where $U$ is an ideal contained in $radR$. That is, an ideal contained in every maximal left ideal. It seems intuitive to me. For if, $U$ is contained in every maximal left ideal, then it annihilates every simple module of $R$. So, when we peel that stuff out of the radical it should be the same as modding it out of the ring in the first place but I'm having trouble with the specific details.
Further, from here we wish to conclude that $R$ and $R/radR$ have the same simple modules. It is clear that a simple module for $R$ is certainly a module for $R/radR$. But showing that it is simple just because I removed it's annihilator is unclear to me. Any help is appreciated. Thanks! A somewhat relevant post
Minimal ideals in a reduced ring
modules semi-simple-rings
asked Dec 14 '18 at 2:12
Aaron ZolotorAaron Zolotor
1,632525
$endgroup$
add a comment |
$begingroup$
I'm having trouble getting my intuition pumping and the details in my head on the Jacobson radical (intersection of maximal ideals or intersection of annihilators of simple modules). What I'm interested in showing is the following:
$$rad R/U = (rad R)/U$$
where $U$ is an ideal contained in $radR$. That is, an ideal contained in every maximal left ideal. It seems intuitive to me. For if, $U$ is contained in every maximal left ideal, then it annihilates every simple module of $R$. So, when we peel that stuff out of the radical it should be the same as modding it out of the ring in the first place but I'm having trouble with the specific details.
Further, from here we wish to conclude that $R$ and $R/radR$ have the same simple modules. It is clear that a simple module for $R$ is certainly a module for $R/radR$. But showing that it is simple just because I removed it's annihilator is unclear to me. Any help is appreciated. Thanks! A somewhat relevant post
Minimal ideals in a reduced ring
modules semi-simple-rings
asked Dec 14 '18 at 2:12
Aaron ZolotorAaron Zolotor
1,632525
$endgroup$
I'm having trouble getting my intuition pumping and the details in my head on the Jacobson radical (intersection of maximal ideals or intersection of annihilators of simple modules). What I'm interested in showing is the following:
$$rad R/U = (rad R)/U$$
where $U$ is an ideal contained in $radR$. That is, an ideal contained in every maximal left ideal. It seems intuitive to me. For if, $U$ is contained in every maximal left ideal, then it annihilates every simple module of $R$. So, when we peel that stuff out of the radical it should be the same as modding it out of the ring in the first place but I'm having trouble with the specific details.
Further, from here we wish to conclude that $R$ and $R/radR$ have the same simple modules. It is clear that a simple module for $R$ is certainly a module for $R/radR$. But showing that it is simple just because I removed it's annihilator is unclear to me. Any help is appreciated. Thanks! A somewhat relevant post
Minimal ideals in a reduced ring
modules semi-simple-rings
modules semi-simple-rings
asked Dec 14 '18 at 2:12
Aaron ZolotorAaron Zolotor
1,632525
asked Dec 14 '18 at 2:12
Aaron ZolotorAaron Zolotor
1,632525
asked Dec 14 '18 at 2:12
Aaron ZolotorAaron Zolotor
1,632525
asked Dec 14 '18 at 2:12
Aaron ZolotorAaron Zolotor
1,632525
asked Dec 14 '18 at 2:12
Aaron ZolotorAaron Zolotor
1,632525
1,632525
add a comment |
add a comment |
1 Answer
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Here is all you need:
An $R/I$ module is always an $R$ module. In fact, the $R/I$ modules are precisely the $R$ modules annihilated by $I$, and the structure is the obvious one: $(r+I)cdot m = rcdot m$, where the left side is the $R/I$ action and the right side is the $R$ action.
Judging from what you've written, you're aware of at least part of this, but it looks like you might not see the ramifications.
If $N$ is an $R$ submodule of $M$ and $I$ annihilates $M$, then $I$ also clearly annihilates $N$. Therefore $N$ is also an $R/I$ submodule of the $R/I$ module $M$. You see, the submodules of an $R$ module annihilated by $I$ correspond with the $R/I$ submodules exactly.
Therefore if $M$ is a simple $R$ module annihilated by $I$, it is a simple $R/I$ module as well. That's why we say the simple modules of $R$ and $R/I$ are the same whenever $Isubseteq J(R)$.
Another heuristic: by being contained in $J(R)$, $R$ and $R/I$ have "the same maximal ideals" (they correspond exactly) and so they should have the same set of quotients by maximal ideals (simple modules.)
answered Dec 16 '18 at 14:40
rschwiebrschwieb
107k12102251
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Here is all you need:
An $R/I$ module is always an $R$ module. In fact, the $R/I$ modules are precisely the $R$ modules annihilated by $I$, and the structure is the obvious one: $(r+I)cdot m = rcdot m$, where the left side is the $R/I$ action and the right side is the $R$ action.
Judging from what you've written, you're aware of at least part of this, but it looks like you might not see the ramifications.
If $N$ is an $R$ submodule of $M$ and $I$ annihilates $M$, then $I$ also clearly annihilates $N$. Therefore $N$ is also an $R/I$ submodule of the $R/I$ module $M$. You see, the submodules of an $R$ module annihilated by $I$ correspond with the $R/I$ submodules exactly.
Therefore if $M$ is a simple $R$ module annihilated by $I$, it is a simple $R/I$ module as well. That's why we say the simple modules of $R$ and $R/I$ are the same whenever $Isubseteq J(R)$.
Another heuristic: by being contained in $J(R)$, $R$ and $R/I$ have "the same maximal ideals" (they correspond exactly) and so they should have the same set of quotients by maximal ideals (simple modules.)
answered Dec 16 '18 at 14:40
rschwiebrschwieb
107k12102251
$endgroup$
add a comment |
$begingroup$
Here is all you need:
An $R/I$ module is always an $R$ module. In fact, the $R/I$ modules are precisely the $R$ modules annihilated by $I$, and the structure is the obvious one: $(r+I)cdot m = rcdot m$, where the left side is the $R/I$ action and the right side is the $R$ action.
Judging from what you've written, you're aware of at least part of this, but it looks like you might not see the ramifications.
If $N$ is an $R$ submodule of $M$ and $I$ annihilates $M$, then $I$ also clearly annihilates $N$. Therefore $N$ is also an $R/I$ submodule of the $R/I$ module $M$. You see, the submodules of an $R$ module annihilated by $I$ correspond with the $R/I$ submodules exactly.
Therefore if $M$ is a simple $R$ module annihilated by $I$, it is a simple $R/I$ module as well. That's why we say the simple modules of $R$ and $R/I$ are the same whenever $Isubseteq J(R)$.
Another heuristic: by being contained in $J(R)$, $R$ and $R/I$ have "the same maximal ideals" (they correspond exactly) and so they should have the same set of quotients by maximal ideals (simple modules.)
answered Dec 16 '18 at 14:40
rschwiebrschwieb
107k12102251
$endgroup$
add a comment |
$begingroup$
Here is all you need:
An $R/I$ module is always an $R$ module. In fact, the $R/I$ modules are precisely the $R$ modules annihilated by $I$, and the structure is the obvious one: $(r+I)cdot m = rcdot m$, where the left side is the $R/I$ action and the right side is the $R$ action.
Judging from what you've written, you're aware of at least part of this, but it looks like you might not see the ramifications.
If $N$ is an $R$ submodule of $M$ and $I$ annihilates $M$, then $I$ also clearly annihilates $N$. Therefore $N$ is also an $R/I$ submodule of the $R/I$ module $M$. You see, the submodules of an $R$ module annihilated by $I$ correspond with the $R/I$ submodules exactly.
Therefore if $M$ is a simple $R$ module annihilated by $I$, it is a simple $R/I$ module as well. That's why we say the simple modules of $R$ and $R/I$ are the same whenever $Isubseteq J(R)$.
Another heuristic: by being contained in $J(R)$, $R$ and $R/I$ have "the same maximal ideals" (they correspond exactly) and so they should have the same set of quotients by maximal ideals (simple modules.)
answered Dec 16 '18 at 14:40
rschwiebrschwieb
107k12102251
$endgroup$
Here is all you need:
An $R/I$ module is always an $R$ module. In fact, the $R/I$ modules are precisely the $R$ modules annihilated by $I$, and the structure is the obvious one: $(r+I)cdot m = rcdot m$, where the left side is the $R/I$ action and the right side is the $R$ action.
Judging from what you've written, you're aware of at least part of this, but it looks like you might not see the ramifications.
If $N$ is an $R$ submodule of $M$ and $I$ annihilates $M$, then $I$ also clearly annihilates $N$. Therefore $N$ is also an $R/I$ submodule of the $R/I$ module $M$. You see, the submodules of an $R$ module annihilated by $I$ correspond with the $R/I$ submodules exactly.
Therefore if $M$ is a simple $R$ module annihilated by $I$, it is a simple $R/I$ module as well. That's why we say the simple modules of $R$ and $R/I$ are the same whenever $Isubseteq J(R)$.
Another heuristic: by being contained in $J(R)$, $R$ and $R/I$ have "the same maximal ideals" (they correspond exactly) and so they should have the same set of quotients by maximal ideals (simple modules.)
answered Dec 16 '18 at 14:40
rschwiebrschwieb
107k12102251
answered Dec 16 '18 at 14:40
rschwiebrschwieb
107k12102251
answered Dec 16 '18 at 14:40
rschwiebrschwieb
107k12102251
answered Dec 16 '18 at 14:40
rschwiebrschwieb
107k12102251
107k12102251
add a comment |
add a comment |
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