Specific example of the completion of a filtered module?
$begingroup$
Let $R$ be a ring, $A = R[x_1, ldots, x_n]$ an $R$-module, and $I = sum_{i=1}^n x_iA = (x_1, ldots, x_n)A$. Then $A supseteq I supseteq I^2 supseteq ldots$ is a filtration of $A$, since it is a decreasing sequence and $I^iI^j = I^{i+j}$. Let $hat{A}$ be the completion of this module with respect to this filtration. I want to show that this completion is equal to $R[[x_1, ldots, x_n]]$. I am fairly new to all of these ideas, so I'm not sure where to begin. Is it best to use the universal mapping property of completions? Or is there a more direct way to proceed? Any small hints are welcome. Thank you.
commutative-algebra
edited Dec 11 '18 at 21:21
user26857
39.4k124183
asked Dec 11 '18 at 9:26
Wyatt GregoryWyatt Gregory
706
$endgroup$
add a comment |
$begingroup$
Let $R$ be a ring, $A = R[x_1, ldots, x_n]$ an $R$-module, and $I = sum_{i=1}^n x_iA = (x_1, ldots, x_n)A$. Then $A supseteq I supseteq I^2 supseteq ldots$ is a filtration of $A$, since it is a decreasing sequence and $I^iI^j = I^{i+j}$. Let $hat{A}$ be the completion of this module with respect to this filtration. I want to show that this completion is equal to $R[[x_1, ldots, x_n]]$. I am fairly new to all of these ideas, so I'm not sure where to begin. Is it best to use the universal mapping property of completions? Or is there a more direct way to proceed? Any small hints are welcome. Thank you.
commutative-algebra
edited Dec 11 '18 at 21:21
user26857
39.4k124183
asked Dec 11 '18 at 9:26
Wyatt GregoryWyatt Gregory
706
$endgroup$
$begingroup$
$A$ is an $R$-algebra considered as $R$-module.
$endgroup$
– user26857
Dec 11 '18 at 21:21
add a comment |
$begingroup$
Let $R$ be a ring, $A = R[x_1, ldots, x_n]$ an $R$-module, and $I = sum_{i=1}^n x_iA = (x_1, ldots, x_n)A$. Then $A supseteq I supseteq I^2 supseteq ldots$ is a filtration of $A$, since it is a decreasing sequence and $I^iI^j = I^{i+j}$. Let $hat{A}$ be the completion of this module with respect to this filtration. I want to show that this completion is equal to $R[[x_1, ldots, x_n]]$. I am fairly new to all of these ideas, so I'm not sure where to begin. Is it best to use the universal mapping property of completions? Or is there a more direct way to proceed? Any small hints are welcome. Thank you.
commutative-algebra
edited Dec 11 '18 at 21:21
user26857
39.4k124183
asked Dec 11 '18 at 9:26
Wyatt GregoryWyatt Gregory
706
$endgroup$
Let $R$ be a ring, $A = R[x_1, ldots, x_n]$ an $R$-module, and $I = sum_{i=1}^n x_iA = (x_1, ldots, x_n)A$. Then $A supseteq I supseteq I^2 supseteq ldots$ is a filtration of $A$, since it is a decreasing sequence and $I^iI^j = I^{i+j}$. Let $hat{A}$ be the completion of this module with respect to this filtration. I want to show that this completion is equal to $R[[x_1, ldots, x_n]]$. I am fairly new to all of these ideas, so I'm not sure where to begin. Is it best to use the universal mapping property of completions? Or is there a more direct way to proceed? Any small hints are welcome. Thank you.
commutative-algebra
commutative-algebra
edited Dec 11 '18 at 21:21
user26857
39.4k124183
asked Dec 11 '18 at 9:26
Wyatt GregoryWyatt Gregory
706
edited Dec 11 '18 at 21:21
user26857
39.4k124183
asked Dec 11 '18 at 9:26
Wyatt GregoryWyatt Gregory
706
edited Dec 11 '18 at 21:21
user26857
39.4k124183
edited Dec 11 '18 at 21:21
user26857
39.4k124183
edited Dec 11 '18 at 21:21
user26857
39.4k124183
39.4k124183
asked Dec 11 '18 at 9:26
Wyatt GregoryWyatt Gregory
706
asked Dec 11 '18 at 9:26
Wyatt GregoryWyatt Gregory
706
asked Dec 11 '18 at 9:26
Wyatt GregoryWyatt Gregory
706
706
$begingroup$
$A$ is an $R$-algebra considered as $R$-module.
$endgroup$
– user26857
Dec 11 '18 at 21:21
add a comment |
$begingroup$
$A$ is an $R$-algebra considered as $R$-module.
$endgroup$
– user26857
Dec 11 '18 at 21:21
$begingroup$
$A$ is an $R$-algebra considered as $R$-module.
$endgroup$
– user26857
Dec 11 '18 at 21:21
$begingroup$
$A$ is an $R$-algebra considered as $R$-module.
$endgroup$
– user26857
Dec 11 '18 at 21:21
add a comment |
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$begingroup$
$A$ is an $R$-algebra considered as $R$-module.
$endgroup$
– user26857
Dec 11 '18 at 21:21