Mixing
Finitely generated modules over a product of domains, where is finite generation/domain used?
The question is to show that given a finite product of (commutative) domains $A_i$, any finitely generated module $M$ over $A=prod_{i = 1}^{n} A_{i}$ naturally decomposes as $bigoplus_{i=1}^{n} M_i$, where each $M_i$ is an $A_i$ module, and the action on each $M_i$ is induced by the projection $Arightarrow A_i$.
The thought is to decompose $M$ as a sum of $1_j M$ submodules, where $1_j$ is the primitive idempotent corresponding to the identity in each $A_i$ factor of $A$. Since we have: $$m=sum_{i=1}^n 1_i m$$ These submodules sum to $M$, and by applying $1_j$ to both sides of this equation, using that $1_i 1_j=0$ if $ineq j$, we get that the sum is direct as $A$ modules. It seems like neither the domain condition on the $A_i$, nor the finite generation of $M$ was needed for this, which makes me think there is an error somewhere.
Is this proof valid?
proof-verification commutative-algebra modules
asked Nov 20 '18 at 6:01
user277182
426212
add a comment |
The question is to show that given a finite product of (commutative) domains $A_i$, any finitely generated module $M$ over $A=prod_{i = 1}^{n} A_{i}$ naturally decomposes as $bigoplus_{i=1}^{n} M_i$, where each $M_i$ is an $A_i$ module, and the action on each $M_i$ is induced by the projection $Arightarrow A_i$.
The thought is to decompose $M$ as a sum of $1_j M$ submodules, where $1_j$ is the primitive idempotent corresponding to the identity in each $A_i$ factor of $A$. Since we have: $$m=sum_{i=1}^n 1_i m$$ These submodules sum to $M$, and by applying $1_j$ to both sides of this equation, using that $1_i 1_j=0$ if $ineq j$, we get that the sum is direct as $A$ modules. It seems like neither the domain condition on the $A_i$, nor the finite generation of $M$ was needed for this, which makes me think there is an error somewhere.
Is this proof valid?
proof-verification commutative-algebra modules
asked Nov 20 '18 at 6:01
user277182
426212
1
No, your proof is fine; you in fact don't need that the $A_i$ are domains, and you don't need that $M$ is finitely generated. You also don't need that the $A_i$ are commutative.
– Qiaochu Yuan
Nov 20 '18 at 8:55
Awesome, thanks :)
– user277182
Nov 20 '18 at 10:17
add a comment |
The question is to show that given a finite product of (commutative) domains $A_i$, any finitely generated module $M$ over $A=prod_{i = 1}^{n} A_{i}$ naturally decomposes as $bigoplus_{i=1}^{n} M_i$, where each $M_i$ is an $A_i$ module, and the action on each $M_i$ is induced by the projection $Arightarrow A_i$.
The thought is to decompose $M$ as a sum of $1_j M$ submodules, where $1_j$ is the primitive idempotent corresponding to the identity in each $A_i$ factor of $A$. Since we have: $$m=sum_{i=1}^n 1_i m$$ These submodules sum to $M$, and by applying $1_j$ to both sides of this equation, using that $1_i 1_j=0$ if $ineq j$, we get that the sum is direct as $A$ modules. It seems like neither the domain condition on the $A_i$, nor the finite generation of $M$ was needed for this, which makes me think there is an error somewhere.
Is this proof valid?
proof-verification commutative-algebra modules
asked Nov 20 '18 at 6:01
user277182
426212
The question is to show that given a finite product of (commutative) domains $A_i$, any finitely generated module $M$ over $A=prod_{i = 1}^{n} A_{i}$ naturally decomposes as $bigoplus_{i=1}^{n} M_i$, where each $M_i$ is an $A_i$ module, and the action on each $M_i$ is induced by the projection $Arightarrow A_i$.
The thought is to decompose $M$ as a sum of $1_j M$ submodules, where $1_j$ is the primitive idempotent corresponding to the identity in each $A_i$ factor of $A$. Since we have: $$m=sum_{i=1}^n 1_i m$$ These submodules sum to $M$, and by applying $1_j$ to both sides of this equation, using that $1_i 1_j=0$ if $ineq j$, we get that the sum is direct as $A$ modules. It seems like neither the domain condition on the $A_i$, nor the finite generation of $M$ was needed for this, which makes me think there is an error somewhere.
Is this proof valid?
proof-verification commutative-algebra modules
proof-verification commutative-algebra modules
asked Nov 20 '18 at 6:01
user277182
426212
asked Nov 20 '18 at 6:01
user277182
426212
asked Nov 20 '18 at 6:01
user277182
426212
asked Nov 20 '18 at 6:01
user277182
426212
asked Nov 20 '18 at 6:01
user277182
426212
426212
1
No, your proof is fine; you in fact don't need that the $A_i$ are domains, and you don't need that $M$ is finitely generated. You also don't need that the $A_i$ are commutative.
– Qiaochu Yuan
Nov 20 '18 at 8:55
Awesome, thanks :)
– user277182
Nov 20 '18 at 10:17
add a comment |
1
No, your proof is fine; you in fact don't need that the $A_i$ are domains, and you don't need that $M$ is finitely generated. You also don't need that the $A_i$ are commutative.
– Qiaochu Yuan
Nov 20 '18 at 8:55
Awesome, thanks :)
– user277182
Nov 20 '18 at 10:17
1
1
No, your proof is fine; you in fact don't need that the $A_i$ are domains, and you don't need that $M$ is finitely generated. You also don't need that the $A_i$ are commutative.
– Qiaochu Yuan
Nov 20 '18 at 8:55
No, your proof is fine; you in fact don't need that the $A_i$ are domains, and you don't need that $M$ is finitely generated. You also don't need that the $A_i$ are commutative.
– Qiaochu Yuan
Nov 20 '18 at 8:55
Awesome, thanks :)
– user277182
Nov 20 '18 at 10:17
Awesome, thanks :)
– user277182
Nov 20 '18 at 10:17
add a comment |
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1
No, your proof is fine; you in fact don't need that the $A_i$ are domains, and you don't need that $M$ is finitely generated. You also don't need that the $A_i$ are commutative.
– Qiaochu Yuan
Nov 20 '18 at 8:55
Awesome, thanks :)
– user277182
Nov 20 '18 at 10:17