Are Finitely generated modules over a ring also finitely generated over a subring containing the identity?Is every subring of a field over $mathbbQ$ a finitely generated $mathbbZ$-module?Integral Dependence & Finitely Generated Modulessubalgebras and finitely generated modulesAre polynomial rings finitely generated modules over the base ring?Finitely generated modulesFinitely generated modules over PID'sFinitely generated semisimple modulesAre free submodules of finitely generated modules finitely generated?Are finitely generated submodules of finitely generated free modules free?A submodule of finitely generated modules
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Are Finitely generated modules over a ring also finitely generated over a subring containing the identity?
Is every subring of a field over $mathbbQ$ a finitely generated $mathbbZ$-module?Integral Dependence & Finitely Generated Modulessubalgebras and finitely generated modulesAre polynomial rings finitely generated modules over the base ring?Finitely generated modulesFinitely generated modules over PID'sFinitely generated semisimple modulesAre free submodules of finitely generated modules finitely generated?Are finitely generated submodules of finitely generated free modules free?A submodule of finitely generated modules
$begingroup$
Let us consider a finitely generated R-module M and let S be a subring of R. Is it true that M is a finitely generated S-module?
abstract-algebra modules
asked May 2 at 20:06
TheWandererTheWanderer
1,88011029
$endgroup$
add a comment |
$begingroup$
Let us consider a finitely generated R-module M and let S be a subring of R. Is it true that M is a finitely generated S-module?
abstract-algebra modules
asked May 2 at 20:06
TheWandererTheWanderer
1,88011029
$endgroup$
3
$begingroup$
Take $M=R$ where $R$ is a $S$-algebra that it not finitely generated.
$endgroup$
– Captain Lama
May 2 at 20:09
1
$begingroup$
Take $R=M=mathbbZ[x_1,x_2,x_3,ldots]$ and $S=mathbbZ$.
$endgroup$
– Arturo Magidin
May 2 at 20:10
2
$begingroup$
Your proposed counterexample is wrong: every finitely generated $mathbbZ$-module is also finitely generated as a $2mathbbZ$-module (since it is already finitely generated simply as an abelian group). Note that if you're talking about non-unital modules, the submodule generated by a set is not simply the set of linear combinations with coefficients in the ring, since you can also take integral linear combinations of the generators.
$endgroup$
– Eric Wofsey
May 2 at 20:12
add a comment |
$begingroup$
Let us consider a finitely generated R-module M and let S be a subring of R. Is it true that M is a finitely generated S-module?
abstract-algebra modules
asked May 2 at 20:06
TheWandererTheWanderer
1,88011029
$endgroup$
Let us consider a finitely generated R-module M and let S be a subring of R. Is it true that M is a finitely generated S-module?
abstract-algebra modules
abstract-algebra modules
asked May 2 at 20:06
TheWandererTheWanderer
1,88011029
asked May 2 at 20:06
TheWandererTheWanderer
1,88011029
edited May 2 at 20:13
TheWanderer
asked May 2 at 20:06
TheWandererTheWanderer
1,88011029
asked May 2 at 20:06
TheWandererTheWanderer
1,88011029
asked May 2 at 20:06
TheWandererTheWanderer
1,88011029
1,88011029
3
$begingroup$
Take $M=R$ where $R$ is a $S$-algebra that it not finitely generated.
$endgroup$
– Captain Lama
May 2 at 20:09
1
$begingroup$
Take $R=M=mathbbZ[x_1,x_2,x_3,ldots]$ and $S=mathbbZ$.
$endgroup$
– Arturo Magidin
May 2 at 20:10
2
$begingroup$
Your proposed counterexample is wrong: every finitely generated $mathbbZ$-module is also finitely generated as a $2mathbbZ$-module (since it is already finitely generated simply as an abelian group). Note that if you're talking about non-unital modules, the submodule generated by a set is not simply the set of linear combinations with coefficients in the ring, since you can also take integral linear combinations of the generators.
$endgroup$
– Eric Wofsey
May 2 at 20:12
add a comment |
3
$begingroup$
Take $M=R$ where $R$ is a $S$-algebra that it not finitely generated.
$endgroup$
– Captain Lama
May 2 at 20:09
1
$begingroup$
Take $R=M=mathbbZ[x_1,x_2,x_3,ldots]$ and $S=mathbbZ$.
$endgroup$
– Arturo Magidin
May 2 at 20:10
2
$begingroup$
Your proposed counterexample is wrong: every finitely generated $mathbbZ$-module is also finitely generated as a $2mathbbZ$-module (since it is already finitely generated simply as an abelian group). Note that if you're talking about non-unital modules, the submodule generated by a set is not simply the set of linear combinations with coefficients in the ring, since you can also take integral linear combinations of the generators.
$endgroup$
– Eric Wofsey
May 2 at 20:12
3
3
$begingroup$
Take $M=R$ where $R$ is a $S$-algebra that it not finitely generated.
$endgroup$
– Captain Lama
May 2 at 20:09
$begingroup$
Take $M=R$ where $R$ is a $S$-algebra that it not finitely generated.
$endgroup$
– Captain Lama
May 2 at 20:09
1
1
$begingroup$
Take $R=M=mathbbZ[x_1,x_2,x_3,ldots]$ and $S=mathbbZ$.
$endgroup$
– Arturo Magidin
May 2 at 20:10
$begingroup$
Take $R=M=mathbbZ[x_1,x_2,x_3,ldots]$ and $S=mathbbZ$.
$endgroup$
– Arturo Magidin
May 2 at 20:10
2
2
$begingroup$
Your proposed counterexample is wrong: every finitely generated $mathbbZ$-module is also finitely generated as a $2mathbbZ$-module (since it is already finitely generated simply as an abelian group). Note that if you're talking about non-unital modules, the submodule generated by a set is not simply the set of linear combinations with coefficients in the ring, since you can also take integral linear combinations of the generators.
$endgroup$
– Eric Wofsey
May 2 at 20:12
$begingroup$
Your proposed counterexample is wrong: every finitely generated $mathbbZ$-module is also finitely generated as a $2mathbbZ$-module (since it is already finitely generated simply as an abelian group). Note that if you're talking about non-unital modules, the submodule generated by a set is not simply the set of linear combinations with coefficients in the ring, since you can also take integral linear combinations of the generators.
$endgroup$
– Eric Wofsey
May 2 at 20:12
add a comment |
2 Answers
2
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$begingroup$
The answer is no in general: this would imply that any ring, being finitely generated over itselfn would be finitely generated over any of its subrings.
Easy counter-example: $mathbf Q$ is not finitely generated over $mathbf Z$.
answered May 2 at 20:11
BernardBernard
125k743119
$endgroup$
add a comment |
$begingroup$
Consider the counterexample of $R=M=mathbbQ$, with $S=mathbbZ$.
answered May 2 at 20:10
Zev ChonolesZev Chonoles
111k16235432
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
The answer is no in general: this would imply that any ring, being finitely generated over itselfn would be finitely generated over any of its subrings.
Easy counter-example: $mathbf Q$ is not finitely generated over $mathbf Z$.
answered May 2 at 20:11
BernardBernard
125k743119
$endgroup$
add a comment |
$begingroup$
The answer is no in general: this would imply that any ring, being finitely generated over itselfn would be finitely generated over any of its subrings.
Easy counter-example: $mathbf Q$ is not finitely generated over $mathbf Z$.
answered May 2 at 20:11
BernardBernard
125k743119
$endgroup$
add a comment |
$begingroup$
The answer is no in general: this would imply that any ring, being finitely generated over itselfn would be finitely generated over any of its subrings.
Easy counter-example: $mathbf Q$ is not finitely generated over $mathbf Z$.
answered May 2 at 20:11
BernardBernard
125k743119
$endgroup$
The answer is no in general: this would imply that any ring, being finitely generated over itselfn would be finitely generated over any of its subrings.
Easy counter-example: $mathbf Q$ is not finitely generated over $mathbf Z$.
answered May 2 at 20:11
BernardBernard
125k743119
edited May 2 at 20:22
answered May 2 at 20:11
BernardBernard
125k743119
answered May 2 at 20:11
BernardBernard
125k743119
answered May 2 at 20:11
BernardBernard
125k743119
125k743119
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add a comment |
$begingroup$
Consider the counterexample of $R=M=mathbbQ$, with $S=mathbbZ$.
answered May 2 at 20:10
Zev ChonolesZev Chonoles
111k16235432
$endgroup$
add a comment |
$begingroup$
Consider the counterexample of $R=M=mathbbQ$, with $S=mathbbZ$.
answered May 2 at 20:10
Zev ChonolesZev Chonoles
111k16235432
$endgroup$
add a comment |
$begingroup$
Consider the counterexample of $R=M=mathbbQ$, with $S=mathbbZ$.
answered May 2 at 20:10
Zev ChonolesZev Chonoles
111k16235432
$endgroup$
Consider the counterexample of $R=M=mathbbQ$, with $S=mathbbZ$.
answered May 2 at 20:10
Zev ChonolesZev Chonoles
111k16235432
answered May 2 at 20:10
Zev ChonolesZev Chonoles
111k16235432
answered May 2 at 20:10
Zev ChonolesZev Chonoles
111k16235432
answered May 2 at 20:10
Zev ChonolesZev Chonoles
111k16235432
111k16235432
add a comment |
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$begingroup$
Take $M=R$ where $R$ is a $S$-algebra that it not finitely generated.
$endgroup$
– Captain Lama
May 2 at 20:09
1
$begingroup$
Take $R=M=mathbbZ[x_1,x_2,x_3,ldots]$ and $S=mathbbZ$.
$endgroup$
– Arturo Magidin
May 2 at 20:10
2
$begingroup$
Your proposed counterexample is wrong: every finitely generated $mathbbZ$-module is also finitely generated as a $2mathbbZ$-module (since it is already finitely generated simply as an abelian group). Note that if you're talking about non-unital modules, the submodule generated by a set is not simply the set of linear combinations with coefficients in the ring, since you can also take integral linear combinations of the generators.
$endgroup$
– Eric Wofsey
May 2 at 20:12