$mathbb{R} ^ mathbb{R}$ is a commutative ring with identity that is neither noetherian nor artinian.
$begingroup$
let $R=mathbb{R}^ mathbb{R}$ (all the functions like $f:mathbb{R} rightarrow mathbb{R}$). For each $f, g in R$ and $a in R$:
$$(f+g)(a):=f(a)+g(a)$$
$$(fg)(a):=f(a)g(a)$$
I want to show that $R$ is a commutative ring with identity which is neither noetherian nor artinian.
abstract-algebra noetherian artinian
asked Jan 8 at 8:07
t.ysnt.ysn
1397
$endgroup$
add a comment |
$begingroup$
let $R=mathbb{R}^ mathbb{R}$ (all the functions like $f:mathbb{R} rightarrow mathbb{R}$). For each $f, g in R$ and $a in R$:
$$(f+g)(a):=f(a)+g(a)$$
$$(fg)(a):=f(a)g(a)$$
I want to show that $R$ is a commutative ring with identity which is neither noetherian nor artinian.
abstract-algebra noetherian artinian
asked Jan 8 at 8:07
t.ysnt.ysn
1397
$endgroup$
4
$begingroup$
Just do it. Where are you stuck?
$endgroup$
– Berci
Jan 8 at 8:11
$begingroup$
I don't know how to show that it's neither noetherian nor atrinian. Don't have any idea about taking the ideals... @Berci
$endgroup$
– t.ysn
Jan 8 at 8:14
1
$begingroup$
Does an approach similar to this one math.stackexchange.com/a/4051/42969 work?
$endgroup$
– Martin R
Jan 8 at 8:18
$begingroup$
Hint: If $A$ is any subset of $mathbb{R}$, then $Z(A)={fin R:f(x)=0,forall xin A}$ is an ideal of $R$; if $Asubsetneq B$, then $Z(A)supsetneq Z(B)$.
$endgroup$
– egreg
Jan 8 at 8:43
$begingroup$
Same strategy as this works.
$endgroup$
– rschwieb
Jan 8 at 14:58
add a comment |
$begingroup$
let $R=mathbb{R}^ mathbb{R}$ (all the functions like $f:mathbb{R} rightarrow mathbb{R}$). For each $f, g in R$ and $a in R$:
$$(f+g)(a):=f(a)+g(a)$$
$$(fg)(a):=f(a)g(a)$$
I want to show that $R$ is a commutative ring with identity which is neither noetherian nor artinian.
abstract-algebra noetherian artinian
asked Jan 8 at 8:07
t.ysnt.ysn
1397
$endgroup$
let $R=mathbb{R}^ mathbb{R}$ (all the functions like $f:mathbb{R} rightarrow mathbb{R}$). For each $f, g in R$ and $a in R$:
$$(f+g)(a):=f(a)+g(a)$$
$$(fg)(a):=f(a)g(a)$$
I want to show that $R$ is a commutative ring with identity which is neither noetherian nor artinian.
abstract-algebra noetherian artinian
abstract-algebra noetherian artinian
asked Jan 8 at 8:07
t.ysnt.ysn
1397
asked Jan 8 at 8:07
t.ysnt.ysn
1397
edited Jan 8 at 8:37
t.ysn
asked Jan 8 at 8:07
t.ysnt.ysn
1397
asked Jan 8 at 8:07
t.ysnt.ysn
1397
asked Jan 8 at 8:07
t.ysnt.ysn
1397
1397
4
$begingroup$
Just do it. Where are you stuck?
$endgroup$
– Berci
Jan 8 at 8:11
$begingroup$
I don't know how to show that it's neither noetherian nor atrinian. Don't have any idea about taking the ideals... @Berci
$endgroup$
– t.ysn
Jan 8 at 8:14
1
$begingroup$
Does an approach similar to this one math.stackexchange.com/a/4051/42969 work?
$endgroup$
– Martin R
Jan 8 at 8:18
$begingroup$
Hint: If $A$ is any subset of $mathbb{R}$, then $Z(A)={fin R:f(x)=0,forall xin A}$ is an ideal of $R$; if $Asubsetneq B$, then $Z(A)supsetneq Z(B)$.
$endgroup$
– egreg
Jan 8 at 8:43
$begingroup$
Same strategy as this works.
$endgroup$
– rschwieb
Jan 8 at 14:58
add a comment |
4
$begingroup$
Just do it. Where are you stuck?
$endgroup$
– Berci
Jan 8 at 8:11
$begingroup$
I don't know how to show that it's neither noetherian nor atrinian. Don't have any idea about taking the ideals... @Berci
$endgroup$
– t.ysn
Jan 8 at 8:14
1
$begingroup$
Does an approach similar to this one math.stackexchange.com/a/4051/42969 work?
$endgroup$
– Martin R
Jan 8 at 8:18
$begingroup$
Hint: If $A$ is any subset of $mathbb{R}$, then $Z(A)={fin R:f(x)=0,forall xin A}$ is an ideal of $R$; if $Asubsetneq B$, then $Z(A)supsetneq Z(B)$.
$endgroup$
– egreg
Jan 8 at 8:43
$begingroup$
Same strategy as this works.
$endgroup$
– rschwieb
Jan 8 at 14:58
4
4
$begingroup$
Just do it. Where are you stuck?
$endgroup$
– Berci
Jan 8 at 8:11
$begingroup$
Just do it. Where are you stuck?
$endgroup$
– Berci
Jan 8 at 8:11
$begingroup$
I don't know how to show that it's neither noetherian nor atrinian. Don't have any idea about taking the ideals... @Berci
$endgroup$
– t.ysn
Jan 8 at 8:14
$begingroup$
I don't know how to show that it's neither noetherian nor atrinian. Don't have any idea about taking the ideals... @Berci
$endgroup$
– t.ysn
Jan 8 at 8:14
1
1
$begingroup$
Does an approach similar to this one math.stackexchange.com/a/4051/42969 work?
$endgroup$
– Martin R
Jan 8 at 8:18
$begingroup$
Does an approach similar to this one math.stackexchange.com/a/4051/42969 work?
$endgroup$
– Martin R
Jan 8 at 8:18
$begingroup$
Hint: If $A$ is any subset of $mathbb{R}$, then $Z(A)={fin R:f(x)=0,forall xin A}$ is an ideal of $R$; if $Asubsetneq B$, then $Z(A)supsetneq Z(B)$.
$endgroup$
– egreg
Jan 8 at 8:43
$begingroup$
Hint: If $A$ is any subset of $mathbb{R}$, then $Z(A)={fin R:f(x)=0,forall xin A}$ is an ideal of $R$; if $Asubsetneq B$, then $Z(A)supsetneq Z(B)$.
$endgroup$
– egreg
Jan 8 at 8:43
$begingroup$
Same strategy as this works.
$endgroup$
– rschwieb
Jan 8 at 14:58
$begingroup$
Same strategy as this works.
$endgroup$
– rschwieb
Jan 8 at 14:58
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let's prove it's neither Noetherian nor Artinian.
To each $A subseteq mathbb{R}$, we can assign a set $varphi(A) subseteq mathbb{R}^mathbb{R}$ as follows:
$$varphi(A) = {f in mathbb{R}^mathbb{R} : forall x in A, f(x) = 0}.$$
Exercise 0. Show that $varphi(A)$ is an ideal for all sets $A subseteq mathbb{R}$.
Conclude that $$varphi : mathcal{P}(mathbb{R}) rightarrow mathcal{P}(mathbb{R}^mathbb{R})$$ can be viewed as a function $$mathcal{P}(mathbb{R}) rightarrow mathrm{Ideal}(mathbb{R}^mathbb{R}).$$
Exercise 1. Show that $varphi : mathcal{P}(mathbb{R}) rightarrow mathrm{Ideal}(mathbb{R}^mathbb{R})$ is injective and order-reversing.
Exercise 2. Find an order-reversing injection $psi : mathbb{Z} rightarrow mathcal{P}(mathbb{R})$.
Conclude that $varphi circ psi : mathbb{Z} rightarrow mathrm{Ideal}(mathbb{R}^mathbb{R})$ is an order-preserving injection.
Conclude that $mathbb{R}^mathbb{R}$ is neither Artinian nor Noetherian.
answered Jan 8 at 9:05
goblingoblin
37.1k1159193
$endgroup$
$begingroup$
I get the distinct sense this website is slowly dying...
$endgroup$
– goblin
Jan 8 at 9:44
$begingroup$
Thank you so much. I'm still a little bit confused but I try to understand it...
$endgroup$
– t.ysn
Jan 8 at 12:58
$begingroup$
@t.ysn, which part is unclear? (I'll answer tomorrow since I'm going to bed now)
$endgroup$
– goblin
Jan 8 at 14:15
$begingroup$
This answer is interesting and OK... but it seems to be going out of its way to go over the posters head! I think a good compromise that keeps it a hint would be to add something like "consider what $phi$ does to the chain of sets of the form $[-1/n, 1/n]$ and $[-n,n]$ for varying $nin mathbb N^+$ ordered by inclusion."
$endgroup$
– rschwieb
Jan 8 at 15:05
$begingroup$
@rschwieb, I take your point, but for me, such hints were always very confusing, perhaps because my reading comprehension skills aren't very strong. I try to offer people a bit more formality than is usual in "hint" answers for this reason. Anyway, why not post your own answer referencing the $varphi$ function I've just defined and then reconceptualizing things in your own terms?
$endgroup$
– goblin
Jan 9 at 8:49
|
show 1 more comment
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Let's prove it's neither Noetherian nor Artinian.
To each $A subseteq mathbb{R}$, we can assign a set $varphi(A) subseteq mathbb{R}^mathbb{R}$ as follows:
$$varphi(A) = {f in mathbb{R}^mathbb{R} : forall x in A, f(x) = 0}.$$
Exercise 0. Show that $varphi(A)$ is an ideal for all sets $A subseteq mathbb{R}$.
Conclude that $$varphi : mathcal{P}(mathbb{R}) rightarrow mathcal{P}(mathbb{R}^mathbb{R})$$ can be viewed as a function $$mathcal{P}(mathbb{R}) rightarrow mathrm{Ideal}(mathbb{R}^mathbb{R}).$$
Exercise 1. Show that $varphi : mathcal{P}(mathbb{R}) rightarrow mathrm{Ideal}(mathbb{R}^mathbb{R})$ is injective and order-reversing.
Exercise 2. Find an order-reversing injection $psi : mathbb{Z} rightarrow mathcal{P}(mathbb{R})$.
Conclude that $varphi circ psi : mathbb{Z} rightarrow mathrm{Ideal}(mathbb{R}^mathbb{R})$ is an order-preserving injection.
Conclude that $mathbb{R}^mathbb{R}$ is neither Artinian nor Noetherian.
answered Jan 8 at 9:05
goblingoblin
37.1k1159193
$endgroup$
$begingroup$
I get the distinct sense this website is slowly dying...
$endgroup$
– goblin
Jan 8 at 9:44
$begingroup$
Thank you so much. I'm still a little bit confused but I try to understand it...
$endgroup$
– t.ysn
Jan 8 at 12:58
$begingroup$
@t.ysn, which part is unclear? (I'll answer tomorrow since I'm going to bed now)
$endgroup$
– goblin
Jan 8 at 14:15
$begingroup$
This answer is interesting and OK... but it seems to be going out of its way to go over the posters head! I think a good compromise that keeps it a hint would be to add something like "consider what $phi$ does to the chain of sets of the form $[-1/n, 1/n]$ and $[-n,n]$ for varying $nin mathbb N^+$ ordered by inclusion."
$endgroup$
– rschwieb
Jan 8 at 15:05
$begingroup$
@rschwieb, I take your point, but for me, such hints were always very confusing, perhaps because my reading comprehension skills aren't very strong. I try to offer people a bit more formality than is usual in "hint" answers for this reason. Anyway, why not post your own answer referencing the $varphi$ function I've just defined and then reconceptualizing things in your own terms?
$endgroup$
– goblin
Jan 9 at 8:49
|
show 1 more comment
$begingroup$
Let's prove it's neither Noetherian nor Artinian.
To each $A subseteq mathbb{R}$, we can assign a set $varphi(A) subseteq mathbb{R}^mathbb{R}$ as follows:
$$varphi(A) = {f in mathbb{R}^mathbb{R} : forall x in A, f(x) = 0}.$$
Exercise 0. Show that $varphi(A)$ is an ideal for all sets $A subseteq mathbb{R}$.
Conclude that $$varphi : mathcal{P}(mathbb{R}) rightarrow mathcal{P}(mathbb{R}^mathbb{R})$$ can be viewed as a function $$mathcal{P}(mathbb{R}) rightarrow mathrm{Ideal}(mathbb{R}^mathbb{R}).$$
Exercise 1. Show that $varphi : mathcal{P}(mathbb{R}) rightarrow mathrm{Ideal}(mathbb{R}^mathbb{R})$ is injective and order-reversing.
Exercise 2. Find an order-reversing injection $psi : mathbb{Z} rightarrow mathcal{P}(mathbb{R})$.
Conclude that $varphi circ psi : mathbb{Z} rightarrow mathrm{Ideal}(mathbb{R}^mathbb{R})$ is an order-preserving injection.
Conclude that $mathbb{R}^mathbb{R}$ is neither Artinian nor Noetherian.
answered Jan 8 at 9:05
goblingoblin
37.1k1159193
$endgroup$
$begingroup$
I get the distinct sense this website is slowly dying...
$endgroup$
– goblin
Jan 8 at 9:44
$begingroup$
Thank you so much. I'm still a little bit confused but I try to understand it...
$endgroup$
– t.ysn
Jan 8 at 12:58
$begingroup$
@t.ysn, which part is unclear? (I'll answer tomorrow since I'm going to bed now)
$endgroup$
– goblin
Jan 8 at 14:15
$begingroup$
This answer is interesting and OK... but it seems to be going out of its way to go over the posters head! I think a good compromise that keeps it a hint would be to add something like "consider what $phi$ does to the chain of sets of the form $[-1/n, 1/n]$ and $[-n,n]$ for varying $nin mathbb N^+$ ordered by inclusion."
$endgroup$
– rschwieb
Jan 8 at 15:05
$begingroup$
@rschwieb, I take your point, but for me, such hints were always very confusing, perhaps because my reading comprehension skills aren't very strong. I try to offer people a bit more formality than is usual in "hint" answers for this reason. Anyway, why not post your own answer referencing the $varphi$ function I've just defined and then reconceptualizing things in your own terms?
$endgroup$
– goblin
Jan 9 at 8:49
|
show 1 more comment
$begingroup$
Let's prove it's neither Noetherian nor Artinian.
To each $A subseteq mathbb{R}$, we can assign a set $varphi(A) subseteq mathbb{R}^mathbb{R}$ as follows:
$$varphi(A) = {f in mathbb{R}^mathbb{R} : forall x in A, f(x) = 0}.$$
Exercise 0. Show that $varphi(A)$ is an ideal for all sets $A subseteq mathbb{R}$.
Conclude that $$varphi : mathcal{P}(mathbb{R}) rightarrow mathcal{P}(mathbb{R}^mathbb{R})$$ can be viewed as a function $$mathcal{P}(mathbb{R}) rightarrow mathrm{Ideal}(mathbb{R}^mathbb{R}).$$
Exercise 1. Show that $varphi : mathcal{P}(mathbb{R}) rightarrow mathrm{Ideal}(mathbb{R}^mathbb{R})$ is injective and order-reversing.
Exercise 2. Find an order-reversing injection $psi : mathbb{Z} rightarrow mathcal{P}(mathbb{R})$.
Conclude that $varphi circ psi : mathbb{Z} rightarrow mathrm{Ideal}(mathbb{R}^mathbb{R})$ is an order-preserving injection.
Conclude that $mathbb{R}^mathbb{R}$ is neither Artinian nor Noetherian.
answered Jan 8 at 9:05
goblingoblin
37.1k1159193
$endgroup$
Let's prove it's neither Noetherian nor Artinian.
To each $A subseteq mathbb{R}$, we can assign a set $varphi(A) subseteq mathbb{R}^mathbb{R}$ as follows:
$$varphi(A) = {f in mathbb{R}^mathbb{R} : forall x in A, f(x) = 0}.$$
Exercise 0. Show that $varphi(A)$ is an ideal for all sets $A subseteq mathbb{R}$.
Conclude that $$varphi : mathcal{P}(mathbb{R}) rightarrow mathcal{P}(mathbb{R}^mathbb{R})$$ can be viewed as a function $$mathcal{P}(mathbb{R}) rightarrow mathrm{Ideal}(mathbb{R}^mathbb{R}).$$
Exercise 1. Show that $varphi : mathcal{P}(mathbb{R}) rightarrow mathrm{Ideal}(mathbb{R}^mathbb{R})$ is injective and order-reversing.
Exercise 2. Find an order-reversing injection $psi : mathbb{Z} rightarrow mathcal{P}(mathbb{R})$.
Conclude that $varphi circ psi : mathbb{Z} rightarrow mathrm{Ideal}(mathbb{R}^mathbb{R})$ is an order-preserving injection.
Conclude that $mathbb{R}^mathbb{R}$ is neither Artinian nor Noetherian.
answered Jan 8 at 9:05
goblingoblin
37.1k1159193
edited Jan 8 at 9:45
answered Jan 8 at 9:05
goblingoblin
37.1k1159193
answered Jan 8 at 9:05
goblingoblin
37.1k1159193
answered Jan 8 at 9:05
goblingoblin
37.1k1159193
37.1k1159193
$begingroup$
I get the distinct sense this website is slowly dying...
$endgroup$
– goblin
Jan 8 at 9:44
$begingroup$
Thank you so much. I'm still a little bit confused but I try to understand it...
$endgroup$
– t.ysn
Jan 8 at 12:58
$begingroup$
@t.ysn, which part is unclear? (I'll answer tomorrow since I'm going to bed now)
$endgroup$
– goblin
Jan 8 at 14:15
$begingroup$
This answer is interesting and OK... but it seems to be going out of its way to go over the posters head! I think a good compromise that keeps it a hint would be to add something like "consider what $phi$ does to the chain of sets of the form $[-1/n, 1/n]$ and $[-n,n]$ for varying $nin mathbb N^+$ ordered by inclusion."
$endgroup$
– rschwieb
Jan 8 at 15:05
$begingroup$
@rschwieb, I take your point, but for me, such hints were always very confusing, perhaps because my reading comprehension skills aren't very strong. I try to offer people a bit more formality than is usual in "hint" answers for this reason. Anyway, why not post your own answer referencing the $varphi$ function I've just defined and then reconceptualizing things in your own terms?
$endgroup$
– goblin
Jan 9 at 8:49
|
show 1 more comment
$begingroup$
I get the distinct sense this website is slowly dying...
$endgroup$
– goblin
Jan 8 at 9:44
$begingroup$
Thank you so much. I'm still a little bit confused but I try to understand it...
$endgroup$
– t.ysn
Jan 8 at 12:58
$begingroup$
@t.ysn, which part is unclear? (I'll answer tomorrow since I'm going to bed now)
$endgroup$
– goblin
Jan 8 at 14:15
$begingroup$
This answer is interesting and OK... but it seems to be going out of its way to go over the posters head! I think a good compromise that keeps it a hint would be to add something like "consider what $phi$ does to the chain of sets of the form $[-1/n, 1/n]$ and $[-n,n]$ for varying $nin mathbb N^+$ ordered by inclusion."
$endgroup$
– rschwieb
Jan 8 at 15:05
$begingroup$
@rschwieb, I take your point, but for me, such hints were always very confusing, perhaps because my reading comprehension skills aren't very strong. I try to offer people a bit more formality than is usual in "hint" answers for this reason. Anyway, why not post your own answer referencing the $varphi$ function I've just defined and then reconceptualizing things in your own terms?
$endgroup$
– goblin
Jan 9 at 8:49
$begingroup$
I get the distinct sense this website is slowly dying...
$endgroup$
– goblin
Jan 8 at 9:44
$begingroup$
I get the distinct sense this website is slowly dying...
$endgroup$
– goblin
Jan 8 at 9:44
$begingroup$
Thank you so much. I'm still a little bit confused but I try to understand it...
$endgroup$
– t.ysn
Jan 8 at 12:58
$begingroup$
Thank you so much. I'm still a little bit confused but I try to understand it...
$endgroup$
– t.ysn
Jan 8 at 12:58
$begingroup$
@t.ysn, which part is unclear? (I'll answer tomorrow since I'm going to bed now)
$endgroup$
– goblin
Jan 8 at 14:15
$begingroup$
@t.ysn, which part is unclear? (I'll answer tomorrow since I'm going to bed now)
$endgroup$
– goblin
Jan 8 at 14:15
$begingroup$
This answer is interesting and OK... but it seems to be going out of its way to go over the posters head! I think a good compromise that keeps it a hint would be to add something like "consider what $phi$ does to the chain of sets of the form $[-1/n, 1/n]$ and $[-n,n]$ for varying $nin mathbb N^+$ ordered by inclusion."
$endgroup$
– rschwieb
Jan 8 at 15:05
$begingroup$
This answer is interesting and OK... but it seems to be going out of its way to go over the posters head! I think a good compromise that keeps it a hint would be to add something like "consider what $phi$ does to the chain of sets of the form $[-1/n, 1/n]$ and $[-n,n]$ for varying $nin mathbb N^+$ ordered by inclusion."
$endgroup$
– rschwieb
Jan 8 at 15:05
$begingroup$
@rschwieb, I take your point, but for me, such hints were always very confusing, perhaps because my reading comprehension skills aren't very strong. I try to offer people a bit more formality than is usual in "hint" answers for this reason. Anyway, why not post your own answer referencing the $varphi$ function I've just defined and then reconceptualizing things in your own terms?
$endgroup$
– goblin
Jan 9 at 8:49
$begingroup$
@rschwieb, I take your point, but for me, such hints were always very confusing, perhaps because my reading comprehension skills aren't very strong. I try to offer people a bit more formality than is usual in "hint" answers for this reason. Anyway, why not post your own answer referencing the $varphi$ function I've just defined and then reconceptualizing things in your own terms?
$endgroup$
– goblin
Jan 9 at 8:49
|
show 1 more comment
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4
$begingroup$
Just do it. Where are you stuck?
$endgroup$
– Berci
Jan 8 at 8:11
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I don't know how to show that it's neither noetherian nor atrinian. Don't have any idea about taking the ideals... @Berci
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– t.ysn
Jan 8 at 8:14
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Does an approach similar to this one math.stackexchange.com/a/4051/42969 work?
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– Martin R
Jan 8 at 8:18
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Hint: If $A$ is any subset of $mathbb{R}$, then $Z(A)={fin R:f(x)=0,forall xin A}$ is an ideal of $R$; if $Asubsetneq B$, then $Z(A)supsetneq Z(B)$.
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– egreg
Jan 8 at 8:43
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Same strategy as this works.
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– rschwieb
Jan 8 at 14:58