What does “under inclusion” mean in: $R$ is Noetherian ring $iff$ Every nonempty set of ideals of $R$...
$begingroup$
$R$ is Noetherian ring $iff$ Every nonempty set of ideals of $R$ contains a maximal element under inclusion.
What does the phrase "maximal element $textbf{under inclusion}$" mean? I am having a hard time understanding the context of "under inclusion".
For e.g. assume that we are given any increasing chain of ideals $I_1subset I_2subset cdots$. Since, every set (say $mathcal{S}$) of ideals of $R$ contains a maximal element under inclusion, meaning that if $mathcal{S}={I_1}$, then there exist maximal ideal $M$ which satisfy $I_1 hookrightarrow M$. Similarly, if $mathcal{S}={I_1,I_2}$, then there exist maximal ideal $M$ which satisfy $I_1 hookrightarrow M$ and $I_2 hookrightarrow M$. From here, it is easy to prove that $R$ is Noetherian.
Conversely, Assume that $R$ is Noetherian ring. Let $mathcal{S}$ be any non-empty set of ideals of $R$ with no
maximal element. Since, $mathcal{S} neq emptyset$, let $I_1$ be in $mathcal{S}$.
Now, I have seen in some of the references that "Because $I_1$ is not maximal, we can choose $I_2$ in $mathcal{S}$ with $I_1 subset I_2$ and $I_1 neq I_2$."
Doubt: Why does there exist such $I_2$ in the above statement. (Or you can give some other explanation. I guess this is where "under inclusion" comes to picture.)
abstract-algebra commutative-algebra
asked Dec 21 '18 at 10:45
MUHMUH
410316
$endgroup$
add a comment |
$begingroup$
$R$ is Noetherian ring $iff$ Every nonempty set of ideals of $R$ contains a maximal element under inclusion.
What does the phrase "maximal element $textbf{under inclusion}$" mean? I am having a hard time understanding the context of "under inclusion".
For e.g. assume that we are given any increasing chain of ideals $I_1subset I_2subset cdots$. Since, every set (say $mathcal{S}$) of ideals of $R$ contains a maximal element under inclusion, meaning that if $mathcal{S}={I_1}$, then there exist maximal ideal $M$ which satisfy $I_1 hookrightarrow M$. Similarly, if $mathcal{S}={I_1,I_2}$, then there exist maximal ideal $M$ which satisfy $I_1 hookrightarrow M$ and $I_2 hookrightarrow M$. From here, it is easy to prove that $R$ is Noetherian.
Conversely, Assume that $R$ is Noetherian ring. Let $mathcal{S}$ be any non-empty set of ideals of $R$ with no
maximal element. Since, $mathcal{S} neq emptyset$, let $I_1$ be in $mathcal{S}$.
Now, I have seen in some of the references that "Because $I_1$ is not maximal, we can choose $I_2$ in $mathcal{S}$ with $I_1 subset I_2$ and $I_1 neq I_2$."
Doubt: Why does there exist such $I_2$ in the above statement. (Or you can give some other explanation. I guess this is where "under inclusion" comes to picture.)
abstract-algebra commutative-algebra
asked Dec 21 '18 at 10:45
MUHMUH
410316
$endgroup$
1
$begingroup$
See e.g. the following post : it is the maximal element in a chain of inclusions.
$endgroup$
– Mauro ALLEGRANZA
Dec 21 '18 at 10:55
$begingroup$
See the Poset formed by the set of subsets of a given set (its power set) ordered by inclusion.
$endgroup$
– Mauro ALLEGRANZA
Dec 21 '18 at 10:59
1
$begingroup$
If there did not exist such an $I_2$, then $I_1$ would be maximal in $mathcal{S}$ with respect to subset inclusion, thus contradicting the supposistion.
$endgroup$
– Adam Higgins
Dec 21 '18 at 10:59
$begingroup$
@MauroALLEGRANZA Can we say, equivalently, that If $R$ is Noetherian ring and if the set $mathcal{S}$ contains ideals $I_1,I_2,I_3,cdots$ under inclusion (that is to say that $I_1 subset I_2 subset I_3subset cdots$) then we have to prove that the maximal element exist for this inclusion? (I guess the point is that maximal element makes sense under some "orderedness" which "inclusion" provides here)
$endgroup$
– MUH
Dec 21 '18 at 12:04
add a comment |
$begingroup$
$R$ is Noetherian ring $iff$ Every nonempty set of ideals of $R$ contains a maximal element under inclusion.
What does the phrase "maximal element $textbf{under inclusion}$" mean? I am having a hard time understanding the context of "under inclusion".
For e.g. assume that we are given any increasing chain of ideals $I_1subset I_2subset cdots$. Since, every set (say $mathcal{S}$) of ideals of $R$ contains a maximal element under inclusion, meaning that if $mathcal{S}={I_1}$, then there exist maximal ideal $M$ which satisfy $I_1 hookrightarrow M$. Similarly, if $mathcal{S}={I_1,I_2}$, then there exist maximal ideal $M$ which satisfy $I_1 hookrightarrow M$ and $I_2 hookrightarrow M$. From here, it is easy to prove that $R$ is Noetherian.
Conversely, Assume that $R$ is Noetherian ring. Let $mathcal{S}$ be any non-empty set of ideals of $R$ with no
maximal element. Since, $mathcal{S} neq emptyset$, let $I_1$ be in $mathcal{S}$.
Now, I have seen in some of the references that "Because $I_1$ is not maximal, we can choose $I_2$ in $mathcal{S}$ with $I_1 subset I_2$ and $I_1 neq I_2$."
Doubt: Why does there exist such $I_2$ in the above statement. (Or you can give some other explanation. I guess this is where "under inclusion" comes to picture.)
abstract-algebra commutative-algebra
asked Dec 21 '18 at 10:45
MUHMUH
410316
$endgroup$
$R$ is Noetherian ring $iff$ Every nonempty set of ideals of $R$ contains a maximal element under inclusion.
What does the phrase "maximal element $textbf{under inclusion}$" mean? I am having a hard time understanding the context of "under inclusion".
For e.g. assume that we are given any increasing chain of ideals $I_1subset I_2subset cdots$. Since, every set (say $mathcal{S}$) of ideals of $R$ contains a maximal element under inclusion, meaning that if $mathcal{S}={I_1}$, then there exist maximal ideal $M$ which satisfy $I_1 hookrightarrow M$. Similarly, if $mathcal{S}={I_1,I_2}$, then there exist maximal ideal $M$ which satisfy $I_1 hookrightarrow M$ and $I_2 hookrightarrow M$. From here, it is easy to prove that $R$ is Noetherian.
Conversely, Assume that $R$ is Noetherian ring. Let $mathcal{S}$ be any non-empty set of ideals of $R$ with no
maximal element. Since, $mathcal{S} neq emptyset$, let $I_1$ be in $mathcal{S}$.
Now, I have seen in some of the references that "Because $I_1$ is not maximal, we can choose $I_2$ in $mathcal{S}$ with $I_1 subset I_2$ and $I_1 neq I_2$."
Doubt: Why does there exist such $I_2$ in the above statement. (Or you can give some other explanation. I guess this is where "under inclusion" comes to picture.)
abstract-algebra commutative-algebra
abstract-algebra commutative-algebra
asked Dec 21 '18 at 10:45
MUHMUH
410316
asked Dec 21 '18 at 10:45
MUHMUH
410316
asked Dec 21 '18 at 10:45
MUHMUH
410316
asked Dec 21 '18 at 10:45
MUHMUH
410316
asked Dec 21 '18 at 10:45
MUHMUH
410316
410316
1
$begingroup$
See e.g. the following post : it is the maximal element in a chain of inclusions.
$endgroup$
– Mauro ALLEGRANZA
Dec 21 '18 at 10:55
$begingroup$
See the Poset formed by the set of subsets of a given set (its power set) ordered by inclusion.
$endgroup$
– Mauro ALLEGRANZA
Dec 21 '18 at 10:59
1
$begingroup$
If there did not exist such an $I_2$, then $I_1$ would be maximal in $mathcal{S}$ with respect to subset inclusion, thus contradicting the supposistion.
$endgroup$
– Adam Higgins
Dec 21 '18 at 10:59
$begingroup$
@MauroALLEGRANZA Can we say, equivalently, that If $R$ is Noetherian ring and if the set $mathcal{S}$ contains ideals $I_1,I_2,I_3,cdots$ under inclusion (that is to say that $I_1 subset I_2 subset I_3subset cdots$) then we have to prove that the maximal element exist for this inclusion? (I guess the point is that maximal element makes sense under some "orderedness" which "inclusion" provides here)
$endgroup$
– MUH
Dec 21 '18 at 12:04
add a comment |
1
$begingroup$
See e.g. the following post : it is the maximal element in a chain of inclusions.
$endgroup$
– Mauro ALLEGRANZA
Dec 21 '18 at 10:55
$begingroup$
See the Poset formed by the set of subsets of a given set (its power set) ordered by inclusion.
$endgroup$
– Mauro ALLEGRANZA
Dec 21 '18 at 10:59
1
$begingroup$
If there did not exist such an $I_2$, then $I_1$ would be maximal in $mathcal{S}$ with respect to subset inclusion, thus contradicting the supposistion.
$endgroup$
– Adam Higgins
Dec 21 '18 at 10:59
$begingroup$
@MauroALLEGRANZA Can we say, equivalently, that If $R$ is Noetherian ring and if the set $mathcal{S}$ contains ideals $I_1,I_2,I_3,cdots$ under inclusion (that is to say that $I_1 subset I_2 subset I_3subset cdots$) then we have to prove that the maximal element exist for this inclusion? (I guess the point is that maximal element makes sense under some "orderedness" which "inclusion" provides here)
$endgroup$
– MUH
Dec 21 '18 at 12:04
1
1
$begingroup$
See e.g. the following post : it is the maximal element in a chain of inclusions.
$endgroup$
– Mauro ALLEGRANZA
Dec 21 '18 at 10:55
$begingroup$
See e.g. the following post : it is the maximal element in a chain of inclusions.
$endgroup$
– Mauro ALLEGRANZA
Dec 21 '18 at 10:55
$begingroup$
See the Poset formed by the set of subsets of a given set (its power set) ordered by inclusion.
$endgroup$
– Mauro ALLEGRANZA
Dec 21 '18 at 10:59
$begingroup$
See the Poset formed by the set of subsets of a given set (its power set) ordered by inclusion.
$endgroup$
– Mauro ALLEGRANZA
Dec 21 '18 at 10:59
1
1
$begingroup$
If there did not exist such an $I_2$, then $I_1$ would be maximal in $mathcal{S}$ with respect to subset inclusion, thus contradicting the supposistion.
$endgroup$
– Adam Higgins
Dec 21 '18 at 10:59
$begingroup$
If there did not exist such an $I_2$, then $I_1$ would be maximal in $mathcal{S}$ with respect to subset inclusion, thus contradicting the supposistion.
$endgroup$
– Adam Higgins
Dec 21 '18 at 10:59
$begingroup$
@MauroALLEGRANZA Can we say, equivalently, that If $R$ is Noetherian ring and if the set $mathcal{S}$ contains ideals $I_1,I_2,I_3,cdots$ under inclusion (that is to say that $I_1 subset I_2 subset I_3subset cdots$) then we have to prove that the maximal element exist for this inclusion? (I guess the point is that maximal element makes sense under some "orderedness" which "inclusion" provides here)
$endgroup$
– MUH
Dec 21 '18 at 12:04
$begingroup$
@MauroALLEGRANZA Can we say, equivalently, that If $R$ is Noetherian ring and if the set $mathcal{S}$ contains ideals $I_1,I_2,I_3,cdots$ under inclusion (that is to say that $I_1 subset I_2 subset I_3subset cdots$) then we have to prove that the maximal element exist for this inclusion? (I guess the point is that maximal element makes sense under some "orderedness" which "inclusion" provides here)
$endgroup$
– MUH
Dec 21 '18 at 12:04
add a comment |
1 Answer
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What does the phrase "maximal element under inclusion" mean?
The phrase "maximal element" has no meaning unless some partial order is indicated.
That's what "under inclusion" means: the partial order is the set containment relation $subseteq$.
Now, I have seen in some of the references that "Because $I_1$ is not maximal, we can choose $I_2$ in $mathcal{S}$ with $I_1 subset I_2$ and $I_1 neq I_2$." [...] Doubt: Why does there exist such $I_2$ in the above statement.
Because that is what "not maximal" means. If no such $I_2$ existed, $I_1$ would be maximal. I would encourage you to look at the definition of "maximal" and contemplate its negation, and you will understand.
answered Dec 21 '18 at 14:13
rschwiebrschwieb
107k12102250
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add a comment |
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$begingroup$
What does the phrase "maximal element under inclusion" mean?
The phrase "maximal element" has no meaning unless some partial order is indicated.
That's what "under inclusion" means: the partial order is the set containment relation $subseteq$.
Now, I have seen in some of the references that "Because $I_1$ is not maximal, we can choose $I_2$ in $mathcal{S}$ with $I_1 subset I_2$ and $I_1 neq I_2$." [...] Doubt: Why does there exist such $I_2$ in the above statement.
Because that is what "not maximal" means. If no such $I_2$ existed, $I_1$ would be maximal. I would encourage you to look at the definition of "maximal" and contemplate its negation, and you will understand.
answered Dec 21 '18 at 14:13
rschwiebrschwieb
107k12102250
$endgroup$
add a comment |
$begingroup$
What does the phrase "maximal element under inclusion" mean?
The phrase "maximal element" has no meaning unless some partial order is indicated.
That's what "under inclusion" means: the partial order is the set containment relation $subseteq$.
Now, I have seen in some of the references that "Because $I_1$ is not maximal, we can choose $I_2$ in $mathcal{S}$ with $I_1 subset I_2$ and $I_1 neq I_2$." [...] Doubt: Why does there exist such $I_2$ in the above statement.
Because that is what "not maximal" means. If no such $I_2$ existed, $I_1$ would be maximal. I would encourage you to look at the definition of "maximal" and contemplate its negation, and you will understand.
answered Dec 21 '18 at 14:13
rschwiebrschwieb
107k12102250
$endgroup$
add a comment |
$begingroup$
What does the phrase "maximal element under inclusion" mean?
The phrase "maximal element" has no meaning unless some partial order is indicated.
That's what "under inclusion" means: the partial order is the set containment relation $subseteq$.
Now, I have seen in some of the references that "Because $I_1$ is not maximal, we can choose $I_2$ in $mathcal{S}$ with $I_1 subset I_2$ and $I_1 neq I_2$." [...] Doubt: Why does there exist such $I_2$ in the above statement.
Because that is what "not maximal" means. If no such $I_2$ existed, $I_1$ would be maximal. I would encourage you to look at the definition of "maximal" and contemplate its negation, and you will understand.
answered Dec 21 '18 at 14:13
rschwiebrschwieb
107k12102250
$endgroup$
What does the phrase "maximal element under inclusion" mean?
The phrase "maximal element" has no meaning unless some partial order is indicated.
That's what "under inclusion" means: the partial order is the set containment relation $subseteq$.
Now, I have seen in some of the references that "Because $I_1$ is not maximal, we can choose $I_2$ in $mathcal{S}$ with $I_1 subset I_2$ and $I_1 neq I_2$." [...] Doubt: Why does there exist such $I_2$ in the above statement.
Because that is what "not maximal" means. If no such $I_2$ existed, $I_1$ would be maximal. I would encourage you to look at the definition of "maximal" and contemplate its negation, and you will understand.
answered Dec 21 '18 at 14:13
rschwiebrschwieb
107k12102250
answered Dec 21 '18 at 14:13
rschwiebrschwieb
107k12102250
answered Dec 21 '18 at 14:13
rschwiebrschwieb
107k12102250
answered Dec 21 '18 at 14:13
rschwiebrschwieb
107k12102250
107k12102250
add a comment |
add a comment |
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1
$begingroup$
See e.g. the following post : it is the maximal element in a chain of inclusions.
$endgroup$
– Mauro ALLEGRANZA
Dec 21 '18 at 10:55
$begingroup$
See the Poset formed by the set of subsets of a given set (its power set) ordered by inclusion.
$endgroup$
– Mauro ALLEGRANZA
Dec 21 '18 at 10:59
1
$begingroup$
If there did not exist such an $I_2$, then $I_1$ would be maximal in $mathcal{S}$ with respect to subset inclusion, thus contradicting the supposistion.
$endgroup$
– Adam Higgins
Dec 21 '18 at 10:59
$begingroup$
@MauroALLEGRANZA Can we say, equivalently, that If $R$ is Noetherian ring and if the set $mathcal{S}$ contains ideals $I_1,I_2,I_3,cdots$ under inclusion (that is to say that $I_1 subset I_2 subset I_3subset cdots$) then we have to prove that the maximal element exist for this inclusion? (I guess the point is that maximal element makes sense under some "orderedness" which "inclusion" provides here)
$endgroup$
– MUH
Dec 21 '18 at 12:04