$mathbb{R} ^ mathbb{R}$ 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
$endgroup$
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
$endgroup$
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
$endgroup$
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
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
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.
$endgroup$
$begingroup$
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
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3065913%2fmathbbr-mathbbr-is-a-commutative-ring-with-identity-that-is-neither-no%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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.
$endgroup$
$begingroup$
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.
$endgroup$
$begingroup$
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.
$endgroup$
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.
edited Jan 8 at 9:45
answered Jan 8 at 9:05
goblingoblin
37k1159193
37k1159193
$begingroup$
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
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3065913%2fmathbbr-mathbbr-is-a-commutative-ring-with-identity-that-is-neither-no%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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