Maximal ideal in ring of power series
3
$begingroup$
If $R$ is a commutative ring with identity we know that the maximal ideals of the ring of power series over $R$ have the form $M’=(M,x)$ where $M$ is a maximal ideal of $R$. Do you have a counterexample that shows that if $R$ doesn’t have an identity then the theorem doesn’t hold?
I really don’t have any idea where or how to start.
Reference: Burton’s “First course in ring and ideals” page 117 theorem 7-4
abstract-algebra maximal-and-prime-ideals
asked Dec 18 '18 at 19:54
MathIsLoveMathIsLove
214
$endgroup$
|
show 3 more comments
3
$begingroup$
If $R$ is a commutative ring with identity we know that the maximal ideals of the ring of power series over $R$ have the form $M’=(M,x)$ where $M$ is a maximal ideal of $R$. Do you have a counterexample that shows that if $R$ doesn’t have an identity then the theorem doesn’t hold?
I really don’t have any idea where or how to start.
Reference: Burton’s “First course in ring and ideals” page 117 theorem 7-4
abstract-algebra maximal-and-prime-ideals
asked Dec 18 '18 at 19:54
MathIsLoveMathIsLove
214
$endgroup$
$begingroup$
You mean you want a ring $R$ without identity such that $R[[x]]$ has a maximal ideal that's not of that form? In rings without identity, authors usually require more of an ideal to consider it "maximal". For instance, they sometimes require it to be "modular." Is there any such requirement or is it just "a proper ideal maximal among proper ideals."
$endgroup$
– rschwieb
Dec 18 '18 at 20:26
$begingroup$
I am just now considering what $R[x]$ and $R[[x]]$ look like when $R^2={0}$... it seems... interesting, if you define it so that $xin R[x]$
$endgroup$
– rschwieb
Dec 18 '18 at 20:30
$begingroup$
Is this an actual question that was posed or is it something less formal? it's pretty uncommon for algebraists to be interested in such a situation.
$endgroup$
– rschwieb
Dec 18 '18 at 20:37
$begingroup$
I’m only searching a counterexample to this theorem. I thought that searching among ring without identity was easier than searching among noncommutative ring. However I don’t know what it means to be modular.
$endgroup$
– MathIsLove
Dec 18 '18 at 20:42
2
$begingroup$
Yes, for me a maximal ideal is just a maximal ideal among proper ideali. If R isn’t commutative, the maximal ideal is right. For the last question, I mean what you say.
$endgroup$
– MathIsLove
Dec 18 '18 at 21:00
|
show 3 more comments
3
3
3
0
$begingroup$
If $R$ is a commutative ring with identity we know that the maximal ideals of the ring of power series over $R$ have the form $M’=(M,x)$ where $M$ is a maximal ideal of $R$. Do you have a counterexample that shows that if $R$ doesn’t have an identity then the theorem doesn’t hold?
I really don’t have any idea where or how to start.
Reference: Burton’s “First course in ring and ideals” page 117 theorem 7-4
abstract-algebra maximal-and-prime-ideals
asked Dec 18 '18 at 19:54
MathIsLoveMathIsLove
214
$endgroup$
If $R$ is a commutative ring with identity we know that the maximal ideals of the ring of power series over $R$ have the form $M’=(M,x)$ where $M$ is a maximal ideal of $R$. Do you have a counterexample that shows that if $R$ doesn’t have an identity then the theorem doesn’t hold?
I really don’t have any idea where or how to start.
Reference: Burton’s “First course in ring and ideals” page 117 theorem 7-4
abstract-algebra maximal-and-prime-ideals
abstract-algebra maximal-and-prime-ideals
asked Dec 18 '18 at 19:54
MathIsLoveMathIsLove
214
asked Dec 18 '18 at 19:54
MathIsLoveMathIsLove
214
asked Dec 18 '18 at 19:54
MathIsLoveMathIsLove
214
asked Dec 18 '18 at 19:54
MathIsLoveMathIsLove
214
asked Dec 18 '18 at 19:54
MathIsLoveMathIsLove
214
214
$begingroup$
You mean you want a ring $R$ without identity such that $R[[x]]$ has a maximal ideal that's not of that form? In rings without identity, authors usually require more of an ideal to consider it "maximal". For instance, they sometimes require it to be "modular." Is there any such requirement or is it just "a proper ideal maximal among proper ideals."
$endgroup$
– rschwieb
Dec 18 '18 at 20:26
$begingroup$
I am just now considering what $R[x]$ and $R[[x]]$ look like when $R^2={0}$... it seems... interesting, if you define it so that $xin R[x]$
$endgroup$
– rschwieb
Dec 18 '18 at 20:30
$begingroup$
Is this an actual question that was posed or is it something less formal? it's pretty uncommon for algebraists to be interested in such a situation.
$endgroup$
– rschwieb
Dec 18 '18 at 20:37
$begingroup$
I’m only searching a counterexample to this theorem. I thought that searching among ring without identity was easier than searching among noncommutative ring. However I don’t know what it means to be modular.
$endgroup$
– MathIsLove
Dec 18 '18 at 20:42
2
$begingroup$
Yes, for me a maximal ideal is just a maximal ideal among proper ideali. If R isn’t commutative, the maximal ideal is right. For the last question, I mean what you say.
$endgroup$
– MathIsLove
Dec 18 '18 at 21:00
|
show 3 more comments
$begingroup$
You mean you want a ring $R$ without identity such that $R[[x]]$ has a maximal ideal that's not of that form? In rings without identity, authors usually require more of an ideal to consider it "maximal". For instance, they sometimes require it to be "modular." Is there any such requirement or is it just "a proper ideal maximal among proper ideals."
$endgroup$
– rschwieb
Dec 18 '18 at 20:26
$begingroup$
I am just now considering what $R[x]$ and $R[[x]]$ look like when $R^2={0}$... it seems... interesting, if you define it so that $xin R[x]$
$endgroup$
– rschwieb
Dec 18 '18 at 20:30
$begingroup$
Is this an actual question that was posed or is it something less formal? it's pretty uncommon for algebraists to be interested in such a situation.
$endgroup$
– rschwieb
Dec 18 '18 at 20:37
$begingroup$
I’m only searching a counterexample to this theorem. I thought that searching among ring without identity was easier than searching among noncommutative ring. However I don’t know what it means to be modular.
$endgroup$
– MathIsLove
Dec 18 '18 at 20:42
2
$begingroup$
Yes, for me a maximal ideal is just a maximal ideal among proper ideali. If R isn’t commutative, the maximal ideal is right. For the last question, I mean what you say.
$endgroup$
– MathIsLove
Dec 18 '18 at 21:00
$begingroup$
You mean you want a ring $R$ without identity such that $R[[x]]$ has a maximal ideal that's not of that form? In rings without identity, authors usually require more of an ideal to consider it "maximal". For instance, they sometimes require it to be "modular." Is there any such requirement or is it just "a proper ideal maximal among proper ideals."
$endgroup$
– rschwieb
Dec 18 '18 at 20:26
$begingroup$
You mean you want a ring $R$ without identity such that $R[[x]]$ has a maximal ideal that's not of that form? In rings without identity, authors usually require more of an ideal to consider it "maximal". For instance, they sometimes require it to be "modular." Is there any such requirement or is it just "a proper ideal maximal among proper ideals."
$endgroup$
– rschwieb
Dec 18 '18 at 20:26
$begingroup$
I am just now considering what $R[x]$ and $R[[x]]$ look like when $R^2={0}$... it seems... interesting, if you define it so that $xin R[x]$
$endgroup$
– rschwieb
Dec 18 '18 at 20:30
$begingroup$
I am just now considering what $R[x]$ and $R[[x]]$ look like when $R^2={0}$... it seems... interesting, if you define it so that $xin R[x]$
$endgroup$
– rschwieb
Dec 18 '18 at 20:30
$begingroup$
Is this an actual question that was posed or is it something less formal? it's pretty uncommon for algebraists to be interested in such a situation.
$endgroup$
– rschwieb
Dec 18 '18 at 20:37
$begingroup$
Is this an actual question that was posed or is it something less formal? it's pretty uncommon for algebraists to be interested in such a situation.
$endgroup$
– rschwieb
Dec 18 '18 at 20:37
$begingroup$
I’m only searching a counterexample to this theorem. I thought that searching among ring without identity was easier than searching among noncommutative ring. However I don’t know what it means to be modular.
$endgroup$
– MathIsLove
Dec 18 '18 at 20:42
$begingroup$
I’m only searching a counterexample to this theorem. I thought that searching among ring without identity was easier than searching among noncommutative ring. However I don’t know what it means to be modular.
$endgroup$
– MathIsLove
Dec 18 '18 at 20:42
2
2
$begingroup$
Yes, for me a maximal ideal is just a maximal ideal among proper ideali. If R isn’t commutative, the maximal ideal is right. For the last question, I mean what you say.
$endgroup$
– MathIsLove
Dec 18 '18 at 21:00
$begingroup$
Yes, for me a maximal ideal is just a maximal ideal among proper ideali. If R isn’t commutative, the maximal ideal is right. For the last question, I mean what you say.
$endgroup$
– MathIsLove
Dec 18 '18 at 21:00
|
show 3 more comments
1 Answer
1
active
oldest
votes
1
$begingroup$
I don't know for sure if this suits your needs or not, but if $R=2mathbb Z/4mathbb Z$ and $M=(2x)lhd R[[x]]$, then $R[[x]]/Mcong R$ has two elements, so $M$ is maximal (in the sense you specified in the comments.)
answered Dec 18 '18 at 21:14
rschwiebrschwieb
107k12103252
$endgroup$
$begingroup$
I don’t think it is the couterexample I’m searching. Sorry. I think that I’m not able to explain what I want to say in English.
$endgroup$
– MathIsLove
Dec 18 '18 at 21:21
$begingroup$
@MathIsLove What condition did I not satisfy? (I don't mean it rhetorically, I'm asking seriously so I can check if I can improve it.)
$endgroup$
– rschwieb
Dec 18 '18 at 21:22
$begingroup$
I have to search a maximal ideal M’ in the ring of power series over R (R without identity) that doesn’t have the form M’=(M,x) where M is maximal in R. You didn’t showed that it can’t have that form but only that it is maximal.
$endgroup$
– MathIsLove
Dec 18 '18 at 21:26
$begingroup$
@MathIsLove It can't have that form because $xnotin R[[x]]$, in my case. And $(2x)cap R={0}$, so $M$ couldn't be anything but ${0}$.
$endgroup$
– rschwieb
Dec 18 '18 at 21:28
$begingroup$
Oh. It seems to work. Thanks!
$endgroup$
– MathIsLove
Dec 18 '18 at 21:36
add a comment |
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1 Answer
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1
$begingroup$
I don't know for sure if this suits your needs or not, but if $R=2mathbb Z/4mathbb Z$ and $M=(2x)lhd R[[x]]$, then $R[[x]]/Mcong R$ has two elements, so $M$ is maximal (in the sense you specified in the comments.)
answered Dec 18 '18 at 21:14
rschwiebrschwieb
107k12103252
$endgroup$
$begingroup$
I don’t think it is the couterexample I’m searching. Sorry. I think that I’m not able to explain what I want to say in English.
$endgroup$
– MathIsLove
Dec 18 '18 at 21:21
$begingroup$
@MathIsLove What condition did I not satisfy? (I don't mean it rhetorically, I'm asking seriously so I can check if I can improve it.)
$endgroup$
– rschwieb
Dec 18 '18 at 21:22
$begingroup$
I have to search a maximal ideal M’ in the ring of power series over R (R without identity) that doesn’t have the form M’=(M,x) where M is maximal in R. You didn’t showed that it can’t have that form but only that it is maximal.
$endgroup$
– MathIsLove
Dec 18 '18 at 21:26
$begingroup$
@MathIsLove It can't have that form because $xnotin R[[x]]$, in my case. And $(2x)cap R={0}$, so $M$ couldn't be anything but ${0}$.
$endgroup$
– rschwieb
Dec 18 '18 at 21:28
$begingroup$
Oh. It seems to work. Thanks!
$endgroup$
– MathIsLove
Dec 18 '18 at 21:36
add a comment |
1
$begingroup$
I don't know for sure if this suits your needs or not, but if $R=2mathbb Z/4mathbb Z$ and $M=(2x)lhd R[[x]]$, then $R[[x]]/Mcong R$ has two elements, so $M$ is maximal (in the sense you specified in the comments.)
answered Dec 18 '18 at 21:14
rschwiebrschwieb
107k12103252
$endgroup$
$begingroup$
I don’t think it is the couterexample I’m searching. Sorry. I think that I’m not able to explain what I want to say in English.
$endgroup$
– MathIsLove
Dec 18 '18 at 21:21
$begingroup$
@MathIsLove What condition did I not satisfy? (I don't mean it rhetorically, I'm asking seriously so I can check if I can improve it.)
$endgroup$
– rschwieb
Dec 18 '18 at 21:22
$begingroup$
I have to search a maximal ideal M’ in the ring of power series over R (R without identity) that doesn’t have the form M’=(M,x) where M is maximal in R. You didn’t showed that it can’t have that form but only that it is maximal.
$endgroup$
– MathIsLove
Dec 18 '18 at 21:26
$begingroup$
@MathIsLove It can't have that form because $xnotin R[[x]]$, in my case. And $(2x)cap R={0}$, so $M$ couldn't be anything but ${0}$.
$endgroup$
– rschwieb
Dec 18 '18 at 21:28
$begingroup$
Oh. It seems to work. Thanks!
$endgroup$
– MathIsLove
Dec 18 '18 at 21:36
add a comment |
1
1
1
$begingroup$
I don't know for sure if this suits your needs or not, but if $R=2mathbb Z/4mathbb Z$ and $M=(2x)lhd R[[x]]$, then $R[[x]]/Mcong R$ has two elements, so $M$ is maximal (in the sense you specified in the comments.)
answered Dec 18 '18 at 21:14
rschwiebrschwieb
107k12103252
$endgroup$
I don't know for sure if this suits your needs or not, but if $R=2mathbb Z/4mathbb Z$ and $M=(2x)lhd R[[x]]$, then $R[[x]]/Mcong R$ has two elements, so $M$ is maximal (in the sense you specified in the comments.)
answered Dec 18 '18 at 21:14
rschwiebrschwieb
107k12103252
answered Dec 18 '18 at 21:14
rschwiebrschwieb
107k12103252
answered Dec 18 '18 at 21:14
rschwiebrschwieb
107k12103252
answered Dec 18 '18 at 21:14
rschwiebrschwieb
107k12103252
107k12103252
$begingroup$
I don’t think it is the couterexample I’m searching. Sorry. I think that I’m not able to explain what I want to say in English.
$endgroup$
– MathIsLove
Dec 18 '18 at 21:21
$begingroup$
@MathIsLove What condition did I not satisfy? (I don't mean it rhetorically, I'm asking seriously so I can check if I can improve it.)
$endgroup$
– rschwieb
Dec 18 '18 at 21:22
$begingroup$
I have to search a maximal ideal M’ in the ring of power series over R (R without identity) that doesn’t have the form M’=(M,x) where M is maximal in R. You didn’t showed that it can’t have that form but only that it is maximal.
$endgroup$
– MathIsLove
Dec 18 '18 at 21:26
$begingroup$
@MathIsLove It can't have that form because $xnotin R[[x]]$, in my case. And $(2x)cap R={0}$, so $M$ couldn't be anything but ${0}$.
$endgroup$
– rschwieb
Dec 18 '18 at 21:28
$begingroup$
Oh. It seems to work. Thanks!
$endgroup$
– MathIsLove
Dec 18 '18 at 21:36
add a comment |
$begingroup$
I don’t think it is the couterexample I’m searching. Sorry. I think that I’m not able to explain what I want to say in English.
$endgroup$
– MathIsLove
Dec 18 '18 at 21:21
$begingroup$
@MathIsLove What condition did I not satisfy? (I don't mean it rhetorically, I'm asking seriously so I can check if I can improve it.)
$endgroup$
– rschwieb
Dec 18 '18 at 21:22
$begingroup$
I have to search a maximal ideal M’ in the ring of power series over R (R without identity) that doesn’t have the form M’=(M,x) where M is maximal in R. You didn’t showed that it can’t have that form but only that it is maximal.
$endgroup$
– MathIsLove
Dec 18 '18 at 21:26
$begingroup$
@MathIsLove It can't have that form because $xnotin R[[x]]$, in my case. And $(2x)cap R={0}$, so $M$ couldn't be anything but ${0}$.
$endgroup$
– rschwieb
Dec 18 '18 at 21:28
$begingroup$
Oh. It seems to work. Thanks!
$endgroup$
– MathIsLove
Dec 18 '18 at 21:36
$begingroup$
I don’t think it is the couterexample I’m searching. Sorry. I think that I’m not able to explain what I want to say in English.
$endgroup$
– MathIsLove
Dec 18 '18 at 21:21
$begingroup$
I don’t think it is the couterexample I’m searching. Sorry. I think that I’m not able to explain what I want to say in English.
$endgroup$
– MathIsLove
Dec 18 '18 at 21:21
$begingroup$
@MathIsLove What condition did I not satisfy? (I don't mean it rhetorically, I'm asking seriously so I can check if I can improve it.)
$endgroup$
– rschwieb
Dec 18 '18 at 21:22
$begingroup$
@MathIsLove What condition did I not satisfy? (I don't mean it rhetorically, I'm asking seriously so I can check if I can improve it.)
$endgroup$
– rschwieb
Dec 18 '18 at 21:22
$begingroup$
I have to search a maximal ideal M’ in the ring of power series over R (R without identity) that doesn’t have the form M’=(M,x) where M is maximal in R. You didn’t showed that it can’t have that form but only that it is maximal.
$endgroup$
– MathIsLove
Dec 18 '18 at 21:26
$begingroup$
I have to search a maximal ideal M’ in the ring of power series over R (R without identity) that doesn’t have the form M’=(M,x) where M is maximal in R. You didn’t showed that it can’t have that form but only that it is maximal.
$endgroup$
– MathIsLove
Dec 18 '18 at 21:26
$begingroup$
@MathIsLove It can't have that form because $xnotin R[[x]]$, in my case. And $(2x)cap R={0}$, so $M$ couldn't be anything but ${0}$.
$endgroup$
– rschwieb
Dec 18 '18 at 21:28
$begingroup$
@MathIsLove It can't have that form because $xnotin R[[x]]$, in my case. And $(2x)cap R={0}$, so $M$ couldn't be anything but ${0}$.
$endgroup$
– rschwieb
Dec 18 '18 at 21:28
$begingroup$
Oh. It seems to work. Thanks!
$endgroup$
– MathIsLove
Dec 18 '18 at 21:36
$begingroup$
Oh. It seems to work. Thanks!
$endgroup$
– MathIsLove
Dec 18 '18 at 21:36
add a comment |
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$begingroup$
You mean you want a ring $R$ without identity such that $R[[x]]$ has a maximal ideal that's not of that form? In rings without identity, authors usually require more of an ideal to consider it "maximal". For instance, they sometimes require it to be "modular." Is there any such requirement or is it just "a proper ideal maximal among proper ideals."
$endgroup$
– rschwieb
Dec 18 '18 at 20:26
$begingroup$
I am just now considering what $R[x]$ and $R[[x]]$ look like when $R^2={0}$... it seems... interesting, if you define it so that $xin R[x]$
$endgroup$
– rschwieb
Dec 18 '18 at 20:30
$begingroup$
Is this an actual question that was posed or is it something less formal? it's pretty uncommon for algebraists to be interested in such a situation.
$endgroup$
– rschwieb
Dec 18 '18 at 20:37
$begingroup$
I’m only searching a counterexample to this theorem. I thought that searching among ring without identity was easier than searching among noncommutative ring. However I don’t know what it means to be modular.
$endgroup$
– MathIsLove
Dec 18 '18 at 20:42
2
$begingroup$
Yes, for me a maximal ideal is just a maximal ideal among proper ideali. If R isn’t commutative, the maximal ideal is right. For the last question, I mean what you say.
$endgroup$
– MathIsLove
Dec 18 '18 at 21:00