$mathbb{R} ^ mathbb{R}$ is a commutative ring with identity that is neither noetherian nor artinian.












3












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















3












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














3












3








3


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 8:37







t.ysn

















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














  • 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










1 Answer
1






active

oldest

votes


















3












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















3












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














3












3








3





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 8 at 9:45

























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


















  • $begingroup$
    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


















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