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Does $sum_nge1x_n^n$ converge $mathfrak m$-adically in $K[[x_1,x_2,dots]] ?$
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Let $K$ be a field, let $x_1,x_2,dots$ be indeterminates, and form the $K$-algebra $A:=K[[x_1,x_2,dots]]$.
Recall that $A$ can be defined as the set of expressions of the form $sum_ua_uu$, where $u$ runs over the set monomials in $x_1,x_2,dots$, and each $a_u$ is in $K$, the multiplication being the obvious one.
The subset $mathfrak m$ of $A$ defined by the condition $a_1=0$ is easily seen to be a maximal ideal.
Does the series $sum_nge1x_n^n$ converge $mathfrak m$-adically in $A ?$
Three remarks:
(1) The series $sum_nge1x_n^n$ is $mathfrak m$-adically Cauchy.
(2) The condition that $sum_nge1x_n^n$ converges $mathfrak m$-adically is equivalent to the condition that, for all $n$, the element $sum_ige nx_i^i$ of $A$ is in $mathfrak m^n$. Sadly I'm unable to prove (or disprove) this condition even for $n=2$.
(3) Of course the underlying question is to know if $A$ is $mathfrak m$-adically complete, and I would gladly accept an answer proving that $A$ is not $mathfrak m$-adically complete, even if the convergence of $sum_nge1x_n^n$ is not settled.
abstract-algebra commutative-algebra formal-power-series
asked Jul 26 at 11:06
Pierre-Yves Gaillard
12.7k23179
add a comment |Â
up vote
1
down vote
favorite
Let $K$ be a field, let $x_1,x_2,dots$ be indeterminates, and form the $K$-algebra $A:=K[[x_1,x_2,dots]]$.
Recall that $A$ can be defined as the set of expressions of the form $sum_ua_uu$, where $u$ runs over the set monomials in $x_1,x_2,dots$, and each $a_u$ is in $K$, the multiplication being the obvious one.
The subset $mathfrak m$ of $A$ defined by the condition $a_1=0$ is easily seen to be a maximal ideal.
Does the series $sum_nge1x_n^n$ converge $mathfrak m$-adically in $A ?$
Three remarks:
(1) The series $sum_nge1x_n^n$ is $mathfrak m$-adically Cauchy.
(2) The condition that $sum_nge1x_n^n$ converges $mathfrak m$-adically is equivalent to the condition that, for all $n$, the element $sum_ige nx_i^i$ of $A$ is in $mathfrak m^n$. Sadly I'm unable to prove (or disprove) this condition even for $n=2$.
(3) Of course the underlying question is to know if $A$ is $mathfrak m$-adically complete, and I would gladly accept an answer proving that $A$ is not $mathfrak m$-adically complete, even if the convergence of $sum_nge1x_n^n$ is not settled.
abstract-algebra commutative-algebra formal-power-series
asked Jul 26 at 11:06
Pierre-Yves Gaillard
12.7k23179
Does the expression $sum_ix_i$ represent an element of your set? If so, I have difficulty apprehending the totality. If you had defined your set to be $projlim_nK[[x_1,cdots,x_n]]$, maybe I could get my failing brain around it, but this does not seem to be your intent.
– Lubin
Jul 26 at 15:47
@Lubin - Yes, $K[[x_1,x_2,cdots]]$ can also been defined as $projlim_nK[[x_1,cdots,x_n]]$, and $sum_ix_i$ does represent an element of this ring. I followed Bourbaki's phrasing. Note also that $K[[x_1,x_2,cdots]]$ can be defined as a projective limit in various ways. Thanks for your interest!
– Pierre-Yves Gaillard
Jul 26 at 16:41
Hmm. Then presumably $x_n$ is a convergent sequence with limit zero, but not $(x)$-adically convergent?
– Lubin
Jul 27 at 1:34
@Lubin - Thanks for your comment. Just to make sure: did you notice that the series in the question is $sum_nx_n^n$, not $sum_nx_n$? This being said, I agree with what you wrote. But it seems clear to me that the sequence $(x_n^n)$ tends to $0$ in both topologies.
– Pierre-Yves Gaillard
Jul 27 at 9:50
1
Thanks, I certainly was aware of that, but I used this case to help me understand which ring you were considering. I just think that in dealing with $infty$ly many indeterminates, we need to be extremely careful and scrupulous.
– Lubin
Jul 27 at 13:08
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $K$ be a field, let $x_1,x_2,dots$ be indeterminates, and form the $K$-algebra $A:=K[[x_1,x_2,dots]]$.
Recall that $A$ can be defined as the set of expressions of the form $sum_ua_uu$, where $u$ runs over the set monomials in $x_1,x_2,dots$, and each $a_u$ is in $K$, the multiplication being the obvious one.
The subset $mathfrak m$ of $A$ defined by the condition $a_1=0$ is easily seen to be a maximal ideal.
Does the series $sum_nge1x_n^n$ converge $mathfrak m$-adically in $A ?$
Three remarks:
(1) The series $sum_nge1x_n^n$ is $mathfrak m$-adically Cauchy.
(2) The condition that $sum_nge1x_n^n$ converges $mathfrak m$-adically is equivalent to the condition that, for all $n$, the element $sum_ige nx_i^i$ of $A$ is in $mathfrak m^n$. Sadly I'm unable to prove (or disprove) this condition even for $n=2$.
(3) Of course the underlying question is to know if $A$ is $mathfrak m$-adically complete, and I would gladly accept an answer proving that $A$ is not $mathfrak m$-adically complete, even if the convergence of $sum_nge1x_n^n$ is not settled.
abstract-algebra commutative-algebra formal-power-series
asked Jul 26 at 11:06
Pierre-Yves Gaillard
12.7k23179
Let $K$ be a field, let $x_1,x_2,dots$ be indeterminates, and form the $K$-algebra $A:=K[[x_1,x_2,dots]]$.
Recall that $A$ can be defined as the set of expressions of the form $sum_ua_uu$, where $u$ runs over the set monomials in $x_1,x_2,dots$, and each $a_u$ is in $K$, the multiplication being the obvious one.
The subset $mathfrak m$ of $A$ defined by the condition $a_1=0$ is easily seen to be a maximal ideal.
Does the series $sum_nge1x_n^n$ converge $mathfrak m$-adically in $A ?$
Three remarks:
(1) The series $sum_nge1x_n^n$ is $mathfrak m$-adically Cauchy.
(2) The condition that $sum_nge1x_n^n$ converges $mathfrak m$-adically is equivalent to the condition that, for all $n$, the element $sum_ige nx_i^i$ of $A$ is in $mathfrak m^n$. Sadly I'm unable to prove (or disprove) this condition even for $n=2$.
(3) Of course the underlying question is to know if $A$ is $mathfrak m$-adically complete, and I would gladly accept an answer proving that $A$ is not $mathfrak m$-adically complete, even if the convergence of $sum_nge1x_n^n$ is not settled.
abstract-algebra commutative-algebra formal-power-series
asked Jul 26 at 11:06
Pierre-Yves Gaillard
12.7k23179
asked Jul 26 at 11:06
Pierre-Yves Gaillard
12.7k23179
asked Jul 26 at 11:06
Pierre-Yves Gaillard
12.7k23179
asked Jul 26 at 11:06
Pierre-Yves Gaillard
12.7k23179
12.7k23179
Does the expression $sum_ix_i$ represent an element of your set? If so, I have difficulty apprehending the totality. If you had defined your set to be $projlim_nK[[x_1,cdots,x_n]]$, maybe I could get my failing brain around it, but this does not seem to be your intent.
– Lubin
Jul 26 at 15:47
@Lubin - Yes, $K[[x_1,x_2,cdots]]$ can also been defined as $projlim_nK[[x_1,cdots,x_n]]$, and $sum_ix_i$ does represent an element of this ring. I followed Bourbaki's phrasing. Note also that $K[[x_1,x_2,cdots]]$ can be defined as a projective limit in various ways. Thanks for your interest!
– Pierre-Yves Gaillard
Jul 26 at 16:41
Hmm. Then presumably $x_n$ is a convergent sequence with limit zero, but not $(x)$-adically convergent?
– Lubin
Jul 27 at 1:34
@Lubin - Thanks for your comment. Just to make sure: did you notice that the series in the question is $sum_nx_n^n$, not $sum_nx_n$? This being said, I agree with what you wrote. But it seems clear to me that the sequence $(x_n^n)$ tends to $0$ in both topologies.
– Pierre-Yves Gaillard
Jul 27 at 9:50
1
Thanks, I certainly was aware of that, but I used this case to help me understand which ring you were considering. I just think that in dealing with $infty$ly many indeterminates, we need to be extremely careful and scrupulous.
– Lubin
Jul 27 at 13:08
add a comment |Â
Does the expression $sum_ix_i$ represent an element of your set? If so, I have difficulty apprehending the totality. If you had defined your set to be $projlim_nK[[x_1,cdots,x_n]]$, maybe I could get my failing brain around it, but this does not seem to be your intent.
– Lubin
Jul 26 at 15:47
@Lubin - Yes, $K[[x_1,x_2,cdots]]$ can also been defined as $projlim_nK[[x_1,cdots,x_n]]$, and $sum_ix_i$ does represent an element of this ring. I followed Bourbaki's phrasing. Note also that $K[[x_1,x_2,cdots]]$ can be defined as a projective limit in various ways. Thanks for your interest!
– Pierre-Yves Gaillard
Jul 26 at 16:41
Hmm. Then presumably $x_n$ is a convergent sequence with limit zero, but not $(x)$-adically convergent?
– Lubin
Jul 27 at 1:34
@Lubin - Thanks for your comment. Just to make sure: did you notice that the series in the question is $sum_nx_n^n$, not $sum_nx_n$? This being said, I agree with what you wrote. But it seems clear to me that the sequence $(x_n^n)$ tends to $0$ in both topologies.
– Pierre-Yves Gaillard
Jul 27 at 9:50
1
Thanks, I certainly was aware of that, but I used this case to help me understand which ring you were considering. I just think that in dealing with $infty$ly many indeterminates, we need to be extremely careful and scrupulous.
– Lubin
Jul 27 at 13:08
Does the expression $sum_ix_i$ represent an element of your set? If so, I have difficulty apprehending the totality. If you had defined your set to be $projlim_nK[[x_1,cdots,x_n]]$, maybe I could get my failing brain around it, but this does not seem to be your intent.
– Lubin
Jul 26 at 15:47
Does the expression $sum_ix_i$ represent an element of your set? If so, I have difficulty apprehending the totality. If you had defined your set to be $projlim_nK[[x_1,cdots,x_n]]$, maybe I could get my failing brain around it, but this does not seem to be your intent.
– Lubin
Jul 26 at 15:47
@Lubin - Yes, $K[[x_1,x_2,cdots]]$ can also been defined as $projlim_nK[[x_1,cdots,x_n]]$, and $sum_ix_i$ does represent an element of this ring. I followed Bourbaki's phrasing. Note also that $K[[x_1,x_2,cdots]]$ can be defined as a projective limit in various ways. Thanks for your interest!
– Pierre-Yves Gaillard
Jul 26 at 16:41
@Lubin - Yes, $K[[x_1,x_2,cdots]]$ can also been defined as $projlim_nK[[x_1,cdots,x_n]]$, and $sum_ix_i$ does represent an element of this ring. I followed Bourbaki's phrasing. Note also that $K[[x_1,x_2,cdots]]$ can be defined as a projective limit in various ways. Thanks for your interest!
– Pierre-Yves Gaillard
Jul 26 at 16:41
Hmm. Then presumably $x_n$ is a convergent sequence with limit zero, but not $(x)$-adically convergent?
– Lubin
Jul 27 at 1:34
Hmm. Then presumably $x_n$ is a convergent sequence with limit zero, but not $(x)$-adically convergent?
– Lubin
Jul 27 at 1:34
@Lubin - Thanks for your comment. Just to make sure: did you notice that the series in the question is $sum_nx_n^n$, not $sum_nx_n$? This being said, I agree with what you wrote. But it seems clear to me that the sequence $(x_n^n)$ tends to $0$ in both topologies.
– Pierre-Yves Gaillard
Jul 27 at 9:50
@Lubin - Thanks for your comment. Just to make sure: did you notice that the series in the question is $sum_nx_n^n$, not $sum_nx_n$? This being said, I agree with what you wrote. But it seems clear to me that the sequence $(x_n^n)$ tends to $0$ in both topologies.
– Pierre-Yves Gaillard
Jul 27 at 9:50
1
1
Thanks, I certainly was aware of that, but I used this case to help me understand which ring you were considering. I just think that in dealing with $infty$ly many indeterminates, we need to be extremely careful and scrupulous.
– Lubin
Jul 27 at 13:08
Thanks, I certainly was aware of that, but I used this case to help me understand which ring you were considering. I just think that in dealing with $infty$ly many indeterminates, we need to be extremely careful and scrupulous.
– Lubin
Jul 27 at 13:08
add a comment |Â
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Does the expression $sum_ix_i$ represent an element of your set? If so, I have difficulty apprehending the totality. If you had defined your set to be $projlim_nK[[x_1,cdots,x_n]]$, maybe I could get my failing brain around it, but this does not seem to be your intent.
– Lubin
Jul 26 at 15:47
@Lubin - Yes, $K[[x_1,x_2,cdots]]$ can also been defined as $projlim_nK[[x_1,cdots,x_n]]$, and $sum_ix_i$ does represent an element of this ring. I followed Bourbaki's phrasing. Note also that $K[[x_1,x_2,cdots]]$ can be defined as a projective limit in various ways. Thanks for your interest!
– Pierre-Yves Gaillard
Jul 26 at 16:41
Hmm. Then presumably $x_n$ is a convergent sequence with limit zero, but not $(x)$-adically convergent?
– Lubin
Jul 27 at 1:34
@Lubin - Thanks for your comment. Just to make sure: did you notice that the series in the question is $sum_nx_n^n$, not $sum_nx_n$? This being said, I agree with what you wrote. But it seems clear to me that the sequence $(x_n^n)$ tends to $0$ in both topologies.
– Pierre-Yves Gaillard
Jul 27 at 9:50
1
Thanks, I certainly was aware of that, but I used this case to help me understand which ring you were considering. I just think that in dealing with $infty$ly many indeterminates, we need to be extremely careful and scrupulous.
– Lubin
Jul 27 at 13:08