Mixing
Confusion regarding the proof of “Every PID satisfies the Ascending Chain Condition”.
$begingroup$
I refer to this proof of the fact that Principal Ideal Domains satisfy the Ascending Chain Condition.
It says
Let $bigcuplimits_{i=1}^{infty}I_i=(a)$. As $a$ is present in $bigcuplimits_{i=1}^{infty}I_i$, it should be present in some $I_msubseteq bigcuplimits_{i=1}^{infty}I_i$
I don't understand why. For example, in $Bbb{Z},(2^8)subset (2^4)subseteq(2^2)subseteq(2)$. We know $6in (2)$, and $6notin (2^8),(2^4),(2^2)$. Also, $(2)=(2)bigcup(2^2)bigcup(2^4)bigcup(2^8)$. Hence, it is possible that an element is present in $(2)bigcup(2^2)bigcup(2^4)bigcup(2^8)$, but not in any of its sub-ideals ($(2^2),(2^4)$ or $(2^8)$).
You might say this is an example in which the number of ideals is finite, and here we have an infinite number of ideals. However, I don't see why it should be any diferent for an infintie number of ideals. Any clarification would be great!
Thanks in advance!
abstract-algebra principal-ideal-domains
asked Oct 18 '13 at 6:55
fierydemonfierydemon
4,41012155
$endgroup$
add a comment |
$begingroup$
I refer to this proof of the fact that Principal Ideal Domains satisfy the Ascending Chain Condition.
It says
Let $bigcuplimits_{i=1}^{infty}I_i=(a)$. As $a$ is present in $bigcuplimits_{i=1}^{infty}I_i$, it should be present in some $I_msubseteq bigcuplimits_{i=1}^{infty}I_i$
I don't understand why. For example, in $Bbb{Z},(2^8)subset (2^4)subseteq(2^2)subseteq(2)$. We know $6in (2)$, and $6notin (2^8),(2^4),(2^2)$. Also, $(2)=(2)bigcup(2^2)bigcup(2^4)bigcup(2^8)$. Hence, it is possible that an element is present in $(2)bigcup(2^2)bigcup(2^4)bigcup(2^8)$, but not in any of its sub-ideals ($(2^2),(2^4)$ or $(2^8)$).
You might say this is an example in which the number of ideals is finite, and here we have an infinite number of ideals. However, I don't see why it should be any diferent for an infintie number of ideals. Any clarification would be great!
Thanks in advance!
abstract-algebra principal-ideal-domains
asked Oct 18 '13 at 6:55
fierydemonfierydemon
4,41012155
$endgroup$
add a comment |
$begingroup$
I refer to this proof of the fact that Principal Ideal Domains satisfy the Ascending Chain Condition.
It says
Let $bigcuplimits_{i=1}^{infty}I_i=(a)$. As $a$ is present in $bigcuplimits_{i=1}^{infty}I_i$, it should be present in some $I_msubseteq bigcuplimits_{i=1}^{infty}I_i$
I don't understand why. For example, in $Bbb{Z},(2^8)subset (2^4)subseteq(2^2)subseteq(2)$. We know $6in (2)$, and $6notin (2^8),(2^4),(2^2)$. Also, $(2)=(2)bigcup(2^2)bigcup(2^4)bigcup(2^8)$. Hence, it is possible that an element is present in $(2)bigcup(2^2)bigcup(2^4)bigcup(2^8)$, but not in any of its sub-ideals ($(2^2),(2^4)$ or $(2^8)$).
You might say this is an example in which the number of ideals is finite, and here we have an infinite number of ideals. However, I don't see why it should be any diferent for an infintie number of ideals. Any clarification would be great!
Thanks in advance!
abstract-algebra principal-ideal-domains
asked Oct 18 '13 at 6:55
fierydemonfierydemon
4,41012155
$endgroup$
I refer to this proof of the fact that Principal Ideal Domains satisfy the Ascending Chain Condition.
It says
Let $bigcuplimits_{i=1}^{infty}I_i=(a)$. As $a$ is present in $bigcuplimits_{i=1}^{infty}I_i$, it should be present in some $I_msubseteq bigcuplimits_{i=1}^{infty}I_i$
I don't understand why. For example, in $Bbb{Z},(2^8)subset (2^4)subseteq(2^2)subseteq(2)$. We know $6in (2)$, and $6notin (2^8),(2^4),(2^2)$. Also, $(2)=(2)bigcup(2^2)bigcup(2^4)bigcup(2^8)$. Hence, it is possible that an element is present in $(2)bigcup(2^2)bigcup(2^4)bigcup(2^8)$, but not in any of its sub-ideals ($(2^2),(2^4)$ or $(2^8)$).
You might say this is an example in which the number of ideals is finite, and here we have an infinite number of ideals. However, I don't see why it should be any diferent for an infintie number of ideals. Any clarification would be great!
Thanks in advance!
abstract-algebra principal-ideal-domains
abstract-algebra principal-ideal-domains
asked Oct 18 '13 at 6:55
fierydemonfierydemon
4,41012155
asked Oct 18 '13 at 6:55
fierydemonfierydemon
4,41012155
asked Oct 18 '13 at 6:55
fierydemonfierydemon
4,41012155
asked Oct 18 '13 at 6:55
fierydemonfierydemon
4,41012155
asked Oct 18 '13 at 6:55
fierydemonfierydemon
4,41012155
4,41012155
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The claim is that the generator of the union of all the ideals lies in one of the "bottom" ideals - so the claim is that $2$ lies in one of the ideals, which it does: $2 in (2)$. It does not imply that every element of the "top" ideal lies in one of the others.
$endgroup$
add a comment |
$begingroup$
There is no trouble that $2$ is not in the other ideals. That is, in this case $m=1$, or $2in (2)$. It must be in one at least, and that is enough. Recall that if $mathscr C$ is a collection of sets then $xinbigcup mathscr C$, if and only if $xin C$ for at least one set $C$ in $mathscr C$.
answered Oct 18 '13 at 6:58
Pedro Tamaroff♦Pedro Tamaroff
96.6k10153297
$endgroup$
add a comment |
$begingroup$
The statement resembles the statement of Proposition 12.2.14, Algebra by Artin (second edition).
Warm-up
What is confusing about this proof and your quoted bit is that we did not explicitly mention the Axiom of Choice (or Zorn's lemma).
To illustrate the point. Imagine we have constructed an infinite chain of strictly increasing ideals.
$$(a_1) subset (a_2) subset(a_3) subset dots$$
If the chain is indeed infinite then for any arbitrary ideal $a_i$ there exists "a strictly larger" ideal $a_{i+1}$
$$(a_i) subset (a_{i+1})$$
"Strictly larger" for such ordering means there are elements in the set (=our ring R), that are in ideal $(a_{i+1})$, but are NOT in ideal $(a_i)$.
Axiom of Choice
What this axiom is about best described in Wikipedia. For our purposes, with this axiom we are allowed to pick generating elements from each ideal. Equivalently, we are allowed to construct a union of all ideals in our given ring R.
The essence of Axiom of Choice is that we don't care that ring R may contain infinite number of ideals, or even infinite uncountable number of ideals (e.g ideals indexed by reals numbers $mathcal{R}$). Again, for finite sets we may have an explicit rule to help us pick elements from each subset. If the number of subsets is infinite, we do not care about any rules. We simply state union of all ideals still can be made by an axiom. Not always easy to convince intuition, but this is an axiom.
Such union of ideals will be itself an ideal, having some generator (b) provided that ring R is a Principal Ideal Domain.
With such (b) (=bound), [or ideal (a) in your case] we are unable to produce a "strictly larger" ideal in our ring R. There are basically no elements in the set we haven't included in the union already.
Zorn's lemma
In our theorem we have a set of elements in ring R, partially ordered by the symbol of set inclusion. That is, we can put subsets into larger subsets.
Ideals that are arranged into chains represent total ordering. That is, each larger ideal must be "strictly" larger, include additional elements of the set.
All chains of ideals in ring R have an upper bound, the unit ideal $(1)$ (=the whole ring R).
The upper bound necessary exists as an element in the chain of ideals. In other words, each chain of ideals cannot include anything more when it gets to the unit ideal (=include the whole ring).
Zorn's lemma states that ring R therefore has a maximal element. It says, either each chain of ideals has a maximal element in the form of the unit ideal $(1)$ (=includes the whole ring), or the chain ends in some other maximal element. Both options rule out possibility for an infinite strictly increasing chain of ideals.
When we do factoring into irreducible elements, the algorithm does not allow us to move further if at the next step we do not have proper divisors. There are no elements in the set to break our element into a product of proper divisors. The unit ideal $(1)$ at the next step is not an option.
So for Zorn's lemma chains of ideals are not allowed to go till the unit ideal, they have to terminate at some irreducible element before.
Conclusion
There is also a related theorem in Abstract Algebra by Dummit & Foote.
Proposition 7. Every nonzero prime ideal is a PID is a maximal ideal.
Still, in my opinion the main struggling point in the story is what we can do with ideals if their number is infinite, even uncountably infinite.
If we accept the Axiom of Choice (Zorn's lemma), then we can take their union. Infinite number of ideals is not a problem, even if it is hard to imagine intuitively.
answered Jan 7 at 19:00
Mikhail DMikhail D
34325
$endgroup$
add a 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%2f530690%2fconfusion-regarding-the-proof-of-every-pid-satisfies-the-ascending-chain-condit%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The claim is that the generator of the union of all the ideals lies in one of the "bottom" ideals - so the claim is that $2$ lies in one of the ideals, which it does: $2 in (2)$. It does not imply that every element of the "top" ideal lies in one of the others.
$endgroup$
add a comment |
$begingroup$
The claim is that the generator of the union of all the ideals lies in one of the "bottom" ideals - so the claim is that $2$ lies in one of the ideals, which it does: $2 in (2)$. It does not imply that every element of the "top" ideal lies in one of the others.
$endgroup$
add a comment |
$begingroup$
The claim is that the generator of the union of all the ideals lies in one of the "bottom" ideals - so the claim is that $2$ lies in one of the ideals, which it does: $2 in (2)$. It does not imply that every element of the "top" ideal lies in one of the others.
$endgroup$
The claim is that the generator of the union of all the ideals lies in one of the "bottom" ideals - so the claim is that $2$ lies in one of the ideals, which it does: $2 in (2)$. It does not imply that every element of the "top" ideal lies in one of the others.
answered Oct 18 '13 at 6:59
user61527
add a comment |
add a comment |
$begingroup$
There is no trouble that $2$ is not in the other ideals. That is, in this case $m=1$, or $2in (2)$. It must be in one at least, and that is enough. Recall that if $mathscr C$ is a collection of sets then $xinbigcup mathscr C$, if and only if $xin C$ for at least one set $C$ in $mathscr C$.
answered Oct 18 '13 at 6:58
Pedro Tamaroff♦Pedro Tamaroff
96.6k10153297
$endgroup$
add a comment |
$begingroup$
There is no trouble that $2$ is not in the other ideals. That is, in this case $m=1$, or $2in (2)$. It must be in one at least, and that is enough. Recall that if $mathscr C$ is a collection of sets then $xinbigcup mathscr C$, if and only if $xin C$ for at least one set $C$ in $mathscr C$.
answered Oct 18 '13 at 6:58
Pedro Tamaroff♦Pedro Tamaroff
96.6k10153297
$endgroup$
add a comment |
$begingroup$
There is no trouble that $2$ is not in the other ideals. That is, in this case $m=1$, or $2in (2)$. It must be in one at least, and that is enough. Recall that if $mathscr C$ is a collection of sets then $xinbigcup mathscr C$, if and only if $xin C$ for at least one set $C$ in $mathscr C$.
answered Oct 18 '13 at 6:58
Pedro Tamaroff♦Pedro Tamaroff
96.6k10153297
$endgroup$
There is no trouble that $2$ is not in the other ideals. That is, in this case $m=1$, or $2in (2)$. It must be in one at least, and that is enough. Recall that if $mathscr C$ is a collection of sets then $xinbigcup mathscr C$, if and only if $xin C$ for at least one set $C$ in $mathscr C$.
answered Oct 18 '13 at 6:58
Pedro Tamaroff♦Pedro Tamaroff
96.6k10153297
answered Oct 18 '13 at 6:58
Pedro Tamaroff♦Pedro Tamaroff
96.6k10153297
answered Oct 18 '13 at 6:58
Pedro Tamaroff♦Pedro Tamaroff
96.6k10153297
answered Oct 18 '13 at 6:58
Pedro Tamaroff♦Pedro Tamaroff
96.6k10153297
96.6k10153297
add a comment |
add a comment |
$begingroup$
The statement resembles the statement of Proposition 12.2.14, Algebra by Artin (second edition).
Warm-up
What is confusing about this proof and your quoted bit is that we did not explicitly mention the Axiom of Choice (or Zorn's lemma).
To illustrate the point. Imagine we have constructed an infinite chain of strictly increasing ideals.
$$(a_1) subset (a_2) subset(a_3) subset dots$$
If the chain is indeed infinite then for any arbitrary ideal $a_i$ there exists "a strictly larger" ideal $a_{i+1}$
$$(a_i) subset (a_{i+1})$$
"Strictly larger" for such ordering means there are elements in the set (=our ring R), that are in ideal $(a_{i+1})$, but are NOT in ideal $(a_i)$.
Axiom of Choice
What this axiom is about best described in Wikipedia. For our purposes, with this axiom we are allowed to pick generating elements from each ideal. Equivalently, we are allowed to construct a union of all ideals in our given ring R.
The essence of Axiom of Choice is that we don't care that ring R may contain infinite number of ideals, or even infinite uncountable number of ideals (e.g ideals indexed by reals numbers $mathcal{R}$). Again, for finite sets we may have an explicit rule to help us pick elements from each subset. If the number of subsets is infinite, we do not care about any rules. We simply state union of all ideals still can be made by an axiom. Not always easy to convince intuition, but this is an axiom.
Such union of ideals will be itself an ideal, having some generator (b) provided that ring R is a Principal Ideal Domain.
With such (b) (=bound), [or ideal (a) in your case] we are unable to produce a "strictly larger" ideal in our ring R. There are basically no elements in the set we haven't included in the union already.
Zorn's lemma
In our theorem we have a set of elements in ring R, partially ordered by the symbol of set inclusion. That is, we can put subsets into larger subsets.
Ideals that are arranged into chains represent total ordering. That is, each larger ideal must be "strictly" larger, include additional elements of the set.
All chains of ideals in ring R have an upper bound, the unit ideal $(1)$ (=the whole ring R).
The upper bound necessary exists as an element in the chain of ideals. In other words, each chain of ideals cannot include anything more when it gets to the unit ideal (=include the whole ring).
Zorn's lemma states that ring R therefore has a maximal element. It says, either each chain of ideals has a maximal element in the form of the unit ideal $(1)$ (=includes the whole ring), or the chain ends in some other maximal element. Both options rule out possibility for an infinite strictly increasing chain of ideals.
When we do factoring into irreducible elements, the algorithm does not allow us to move further if at the next step we do not have proper divisors. There are no elements in the set to break our element into a product of proper divisors. The unit ideal $(1)$ at the next step is not an option.
So for Zorn's lemma chains of ideals are not allowed to go till the unit ideal, they have to terminate at some irreducible element before.
Conclusion
There is also a related theorem in Abstract Algebra by Dummit & Foote.
Proposition 7. Every nonzero prime ideal is a PID is a maximal ideal.
Still, in my opinion the main struggling point in the story is what we can do with ideals if their number is infinite, even uncountably infinite.
If we accept the Axiom of Choice (Zorn's lemma), then we can take their union. Infinite number of ideals is not a problem, even if it is hard to imagine intuitively.
answered Jan 7 at 19:00
Mikhail DMikhail D
34325
$endgroup$
add a comment |
$begingroup$
The statement resembles the statement of Proposition 12.2.14, Algebra by Artin (second edition).
Warm-up
What is confusing about this proof and your quoted bit is that we did not explicitly mention the Axiom of Choice (or Zorn's lemma).
To illustrate the point. Imagine we have constructed an infinite chain of strictly increasing ideals.
$$(a_1) subset (a_2) subset(a_3) subset dots$$
If the chain is indeed infinite then for any arbitrary ideal $a_i$ there exists "a strictly larger" ideal $a_{i+1}$
$$(a_i) subset (a_{i+1})$$
"Strictly larger" for such ordering means there are elements in the set (=our ring R), that are in ideal $(a_{i+1})$, but are NOT in ideal $(a_i)$.
Axiom of Choice
What this axiom is about best described in Wikipedia. For our purposes, with this axiom we are allowed to pick generating elements from each ideal. Equivalently, we are allowed to construct a union of all ideals in our given ring R.
The essence of Axiom of Choice is that we don't care that ring R may contain infinite number of ideals, or even infinite uncountable number of ideals (e.g ideals indexed by reals numbers $mathcal{R}$). Again, for finite sets we may have an explicit rule to help us pick elements from each subset. If the number of subsets is infinite, we do not care about any rules. We simply state union of all ideals still can be made by an axiom. Not always easy to convince intuition, but this is an axiom.
Such union of ideals will be itself an ideal, having some generator (b) provided that ring R is a Principal Ideal Domain.
With such (b) (=bound), [or ideal (a) in your case] we are unable to produce a "strictly larger" ideal in our ring R. There are basically no elements in the set we haven't included in the union already.
Zorn's lemma
In our theorem we have a set of elements in ring R, partially ordered by the symbol of set inclusion. That is, we can put subsets into larger subsets.
Ideals that are arranged into chains represent total ordering. That is, each larger ideal must be "strictly" larger, include additional elements of the set.
All chains of ideals in ring R have an upper bound, the unit ideal $(1)$ (=the whole ring R).
The upper bound necessary exists as an element in the chain of ideals. In other words, each chain of ideals cannot include anything more when it gets to the unit ideal (=include the whole ring).
Zorn's lemma states that ring R therefore has a maximal element. It says, either each chain of ideals has a maximal element in the form of the unit ideal $(1)$ (=includes the whole ring), or the chain ends in some other maximal element. Both options rule out possibility for an infinite strictly increasing chain of ideals.
When we do factoring into irreducible elements, the algorithm does not allow us to move further if at the next step we do not have proper divisors. There are no elements in the set to break our element into a product of proper divisors. The unit ideal $(1)$ at the next step is not an option.
So for Zorn's lemma chains of ideals are not allowed to go till the unit ideal, they have to terminate at some irreducible element before.
Conclusion
There is also a related theorem in Abstract Algebra by Dummit & Foote.
Proposition 7. Every nonzero prime ideal is a PID is a maximal ideal.
Still, in my opinion the main struggling point in the story is what we can do with ideals if their number is infinite, even uncountably infinite.
If we accept the Axiom of Choice (Zorn's lemma), then we can take their union. Infinite number of ideals is not a problem, even if it is hard to imagine intuitively.
answered Jan 7 at 19:00
Mikhail DMikhail D
34325
$endgroup$
add a comment |
$begingroup$
The statement resembles the statement of Proposition 12.2.14, Algebra by Artin (second edition).
Warm-up
What is confusing about this proof and your quoted bit is that we did not explicitly mention the Axiom of Choice (or Zorn's lemma).
To illustrate the point. Imagine we have constructed an infinite chain of strictly increasing ideals.
$$(a_1) subset (a_2) subset(a_3) subset dots$$
If the chain is indeed infinite then for any arbitrary ideal $a_i$ there exists "a strictly larger" ideal $a_{i+1}$
$$(a_i) subset (a_{i+1})$$
"Strictly larger" for such ordering means there are elements in the set (=our ring R), that are in ideal $(a_{i+1})$, but are NOT in ideal $(a_i)$.
Axiom of Choice
What this axiom is about best described in Wikipedia. For our purposes, with this axiom we are allowed to pick generating elements from each ideal. Equivalently, we are allowed to construct a union of all ideals in our given ring R.
The essence of Axiom of Choice is that we don't care that ring R may contain infinite number of ideals, or even infinite uncountable number of ideals (e.g ideals indexed by reals numbers $mathcal{R}$). Again, for finite sets we may have an explicit rule to help us pick elements from each subset. If the number of subsets is infinite, we do not care about any rules. We simply state union of all ideals still can be made by an axiom. Not always easy to convince intuition, but this is an axiom.
Such union of ideals will be itself an ideal, having some generator (b) provided that ring R is a Principal Ideal Domain.
With such (b) (=bound), [or ideal (a) in your case] we are unable to produce a "strictly larger" ideal in our ring R. There are basically no elements in the set we haven't included in the union already.
Zorn's lemma
In our theorem we have a set of elements in ring R, partially ordered by the symbol of set inclusion. That is, we can put subsets into larger subsets.
Ideals that are arranged into chains represent total ordering. That is, each larger ideal must be "strictly" larger, include additional elements of the set.
All chains of ideals in ring R have an upper bound, the unit ideal $(1)$ (=the whole ring R).
The upper bound necessary exists as an element in the chain of ideals. In other words, each chain of ideals cannot include anything more when it gets to the unit ideal (=include the whole ring).
Zorn's lemma states that ring R therefore has a maximal element. It says, either each chain of ideals has a maximal element in the form of the unit ideal $(1)$ (=includes the whole ring), or the chain ends in some other maximal element. Both options rule out possibility for an infinite strictly increasing chain of ideals.
When we do factoring into irreducible elements, the algorithm does not allow us to move further if at the next step we do not have proper divisors. There are no elements in the set to break our element into a product of proper divisors. The unit ideal $(1)$ at the next step is not an option.
So for Zorn's lemma chains of ideals are not allowed to go till the unit ideal, they have to terminate at some irreducible element before.
Conclusion
There is also a related theorem in Abstract Algebra by Dummit & Foote.
Proposition 7. Every nonzero prime ideal is a PID is a maximal ideal.
Still, in my opinion the main struggling point in the story is what we can do with ideals if their number is infinite, even uncountably infinite.
If we accept the Axiom of Choice (Zorn's lemma), then we can take their union. Infinite number of ideals is not a problem, even if it is hard to imagine intuitively.
answered Jan 7 at 19:00
Mikhail DMikhail D
34325
$endgroup$
The statement resembles the statement of Proposition 12.2.14, Algebra by Artin (second edition).
Warm-up
What is confusing about this proof and your quoted bit is that we did not explicitly mention the Axiom of Choice (or Zorn's lemma).
To illustrate the point. Imagine we have constructed an infinite chain of strictly increasing ideals.
$$(a_1) subset (a_2) subset(a_3) subset dots$$
If the chain is indeed infinite then for any arbitrary ideal $a_i$ there exists "a strictly larger" ideal $a_{i+1}$
$$(a_i) subset (a_{i+1})$$
"Strictly larger" for such ordering means there are elements in the set (=our ring R), that are in ideal $(a_{i+1})$, but are NOT in ideal $(a_i)$.
Axiom of Choice
What this axiom is about best described in Wikipedia. For our purposes, with this axiom we are allowed to pick generating elements from each ideal. Equivalently, we are allowed to construct a union of all ideals in our given ring R.
The essence of Axiom of Choice is that we don't care that ring R may contain infinite number of ideals, or even infinite uncountable number of ideals (e.g ideals indexed by reals numbers $mathcal{R}$). Again, for finite sets we may have an explicit rule to help us pick elements from each subset. If the number of subsets is infinite, we do not care about any rules. We simply state union of all ideals still can be made by an axiom. Not always easy to convince intuition, but this is an axiom.
Such union of ideals will be itself an ideal, having some generator (b) provided that ring R is a Principal Ideal Domain.
With such (b) (=bound), [or ideal (a) in your case] we are unable to produce a "strictly larger" ideal in our ring R. There are basically no elements in the set we haven't included in the union already.
Zorn's lemma
In our theorem we have a set of elements in ring R, partially ordered by the symbol of set inclusion. That is, we can put subsets into larger subsets.
Ideals that are arranged into chains represent total ordering. That is, each larger ideal must be "strictly" larger, include additional elements of the set.
All chains of ideals in ring R have an upper bound, the unit ideal $(1)$ (=the whole ring R).
The upper bound necessary exists as an element in the chain of ideals. In other words, each chain of ideals cannot include anything more when it gets to the unit ideal (=include the whole ring).
Zorn's lemma states that ring R therefore has a maximal element. It says, either each chain of ideals has a maximal element in the form of the unit ideal $(1)$ (=includes the whole ring), or the chain ends in some other maximal element. Both options rule out possibility for an infinite strictly increasing chain of ideals.
When we do factoring into irreducible elements, the algorithm does not allow us to move further if at the next step we do not have proper divisors. There are no elements in the set to break our element into a product of proper divisors. The unit ideal $(1)$ at the next step is not an option.
So for Zorn's lemma chains of ideals are not allowed to go till the unit ideal, they have to terminate at some irreducible element before.
Conclusion
There is also a related theorem in Abstract Algebra by Dummit & Foote.
Proposition 7. Every nonzero prime ideal is a PID is a maximal ideal.
Still, in my opinion the main struggling point in the story is what we can do with ideals if their number is infinite, even uncountably infinite.
If we accept the Axiom of Choice (Zorn's lemma), then we can take their union. Infinite number of ideals is not a problem, even if it is hard to imagine intuitively.
answered Jan 7 at 19:00
Mikhail DMikhail D
34325
answered Jan 7 at 19:00
Mikhail DMikhail D
34325
answered Jan 7 at 19:00
Mikhail DMikhail D
34325
answered Jan 7 at 19:00
Mikhail DMikhail D
34325
34325
add a comment |
add a 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%2f530690%2fconfusion-regarding-the-proof-of-every-pid-satisfies-the-ascending-chain-condit%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
