Clarification on Notes about Minimal Prime Ideals over an Ideal $I$ of a Noetherian ring $R$
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Recently in my Algebra course, we defined the minimal prime ideals over an ideal $I$ of a Noetherian ring $R$, and then proved a result about them saying that $sqrt{I}$ is the intersection of minimal primes over $I$. However, this is strange given the way he defines the minimal prime ideals over $I$. Below is the section of the notes, hopefully illustrating why it is confusing for me:
Now suppose $I$ is any ideal of a Noetherian ring R. By (2.13), $sqrt{I} = P_1 cap dotscap P_m$ for some primes $P_i$ such that $P_j notsubset P_i ; forall i neq j$.
Note that if $P$ is prime containing $I$, then:
$prod_{i}P_i leq P_1 cap dots cap P_m = sqrt{I} leq P$, and so: $P_i leq P$
Definition (2.14): The minimal primes over an ideal $I$ of a Noetherian ring $R$ are these primes.
Lemma (2.15): Let $I$ be an ideal of a Noetherian ring $R$. Then $sqrt{I}$ is the intersection of the minimal primers over $I$, and $I$ contains a finite product of them, possibly with repetitions.
This is what I have in my notes. The use of "these primes" in his definition made me think that he was defining the minimal primes to be the $P_1, dots, P_m$ he mentioned earlier when talking about $sqrt{I}$. However, the lemma then seems almost entirely redundant if this is how we define the minimal prime ideals.
Am I supposed to think that minimal primes over $I$ are just prime ideals of $R$ containing $I$ that are minimal with respect to inclusion, AND it can be shown that they have this special property relating to $sqrt{I}$?
I'm just really confused about what this part of my notes means and what the point of this lemma is.
In case it's important, (2.13) says that for a Noetherian ring $R$, a radical ideal is the intersection of finitely many prime ideals.
Also, $sqrt{I} = {{r in R mid r^n in I text{ for some } n}}$ is referred to as a radical ideal. We also defined the Nilradical and Jacobson Radical, but I don't know if these are included when we use the word "radical" here.
abstract-algebra definition
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$begingroup$
Recently in my Algebra course, we defined the minimal prime ideals over an ideal $I$ of a Noetherian ring $R$, and then proved a result about them saying that $sqrt{I}$ is the intersection of minimal primes over $I$. However, this is strange given the way he defines the minimal prime ideals over $I$. Below is the section of the notes, hopefully illustrating why it is confusing for me:
Now suppose $I$ is any ideal of a Noetherian ring R. By (2.13), $sqrt{I} = P_1 cap dotscap P_m$ for some primes $P_i$ such that $P_j notsubset P_i ; forall i neq j$.
Note that if $P$ is prime containing $I$, then:
$prod_{i}P_i leq P_1 cap dots cap P_m = sqrt{I} leq P$, and so: $P_i leq P$
Definition (2.14): The minimal primes over an ideal $I$ of a Noetherian ring $R$ are these primes.
Lemma (2.15): Let $I$ be an ideal of a Noetherian ring $R$. Then $sqrt{I}$ is the intersection of the minimal primers over $I$, and $I$ contains a finite product of them, possibly with repetitions.
This is what I have in my notes. The use of "these primes" in his definition made me think that he was defining the minimal primes to be the $P_1, dots, P_m$ he mentioned earlier when talking about $sqrt{I}$. However, the lemma then seems almost entirely redundant if this is how we define the minimal prime ideals.
Am I supposed to think that minimal primes over $I$ are just prime ideals of $R$ containing $I$ that are minimal with respect to inclusion, AND it can be shown that they have this special property relating to $sqrt{I}$?
I'm just really confused about what this part of my notes means and what the point of this lemma is.
In case it's important, (2.13) says that for a Noetherian ring $R$, a radical ideal is the intersection of finitely many prime ideals.
Also, $sqrt{I} = {{r in R mid r^n in I text{ for some } n}}$ is referred to as a radical ideal. We also defined the Nilradical and Jacobson Radical, but I don't know if these are included when we use the word "radical" here.
abstract-algebra definition
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1
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Yes, (2.14) should have defined minimal primes over $I$ as these primes $P$ containing $I$ that is minimal with respect to inclusion.
$endgroup$
– user10354138
Oct 22 '18 at 15:17
add a comment |
$begingroup$
Recently in my Algebra course, we defined the minimal prime ideals over an ideal $I$ of a Noetherian ring $R$, and then proved a result about them saying that $sqrt{I}$ is the intersection of minimal primes over $I$. However, this is strange given the way he defines the minimal prime ideals over $I$. Below is the section of the notes, hopefully illustrating why it is confusing for me:
Now suppose $I$ is any ideal of a Noetherian ring R. By (2.13), $sqrt{I} = P_1 cap dotscap P_m$ for some primes $P_i$ such that $P_j notsubset P_i ; forall i neq j$.
Note that if $P$ is prime containing $I$, then:
$prod_{i}P_i leq P_1 cap dots cap P_m = sqrt{I} leq P$, and so: $P_i leq P$
Definition (2.14): The minimal primes over an ideal $I$ of a Noetherian ring $R$ are these primes.
Lemma (2.15): Let $I$ be an ideal of a Noetherian ring $R$. Then $sqrt{I}$ is the intersection of the minimal primers over $I$, and $I$ contains a finite product of them, possibly with repetitions.
This is what I have in my notes. The use of "these primes" in his definition made me think that he was defining the minimal primes to be the $P_1, dots, P_m$ he mentioned earlier when talking about $sqrt{I}$. However, the lemma then seems almost entirely redundant if this is how we define the minimal prime ideals.
Am I supposed to think that minimal primes over $I$ are just prime ideals of $R$ containing $I$ that are minimal with respect to inclusion, AND it can be shown that they have this special property relating to $sqrt{I}$?
I'm just really confused about what this part of my notes means and what the point of this lemma is.
In case it's important, (2.13) says that for a Noetherian ring $R$, a radical ideal is the intersection of finitely many prime ideals.
Also, $sqrt{I} = {{r in R mid r^n in I text{ for some } n}}$ is referred to as a radical ideal. We also defined the Nilradical and Jacobson Radical, but I don't know if these are included when we use the word "radical" here.
abstract-algebra definition
$endgroup$
Recently in my Algebra course, we defined the minimal prime ideals over an ideal $I$ of a Noetherian ring $R$, and then proved a result about them saying that $sqrt{I}$ is the intersection of minimal primes over $I$. However, this is strange given the way he defines the minimal prime ideals over $I$. Below is the section of the notes, hopefully illustrating why it is confusing for me:
Now suppose $I$ is any ideal of a Noetherian ring R. By (2.13), $sqrt{I} = P_1 cap dotscap P_m$ for some primes $P_i$ such that $P_j notsubset P_i ; forall i neq j$.
Note that if $P$ is prime containing $I$, then:
$prod_{i}P_i leq P_1 cap dots cap P_m = sqrt{I} leq P$, and so: $P_i leq P$
Definition (2.14): The minimal primes over an ideal $I$ of a Noetherian ring $R$ are these primes.
Lemma (2.15): Let $I$ be an ideal of a Noetherian ring $R$. Then $sqrt{I}$ is the intersection of the minimal primers over $I$, and $I$ contains a finite product of them, possibly with repetitions.
This is what I have in my notes. The use of "these primes" in his definition made me think that he was defining the minimal primes to be the $P_1, dots, P_m$ he mentioned earlier when talking about $sqrt{I}$. However, the lemma then seems almost entirely redundant if this is how we define the minimal prime ideals.
Am I supposed to think that minimal primes over $I$ are just prime ideals of $R$ containing $I$ that are minimal with respect to inclusion, AND it can be shown that they have this special property relating to $sqrt{I}$?
I'm just really confused about what this part of my notes means and what the point of this lemma is.
In case it's important, (2.13) says that for a Noetherian ring $R$, a radical ideal is the intersection of finitely many prime ideals.
Also, $sqrt{I} = {{r in R mid r^n in I text{ for some } n}}$ is referred to as a radical ideal. We also defined the Nilradical and Jacobson Radical, but I don't know if these are included when we use the word "radical" here.
abstract-algebra definition
abstract-algebra definition
asked Oct 22 '18 at 14:52
user366818user366818
961410
961410
1
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Yes, (2.14) should have defined minimal primes over $I$ as these primes $P$ containing $I$ that is minimal with respect to inclusion.
$endgroup$
– user10354138
Oct 22 '18 at 15:17
add a comment |
1
$begingroup$
Yes, (2.14) should have defined minimal primes over $I$ as these primes $P$ containing $I$ that is minimal with respect to inclusion.
$endgroup$
– user10354138
Oct 22 '18 at 15:17
1
1
$begingroup$
Yes, (2.14) should have defined minimal primes over $I$ as these primes $P$ containing $I$ that is minimal with respect to inclusion.
$endgroup$
– user10354138
Oct 22 '18 at 15:17
$begingroup$
Yes, (2.14) should have defined minimal primes over $I$ as these primes $P$ containing $I$ that is minimal with respect to inclusion.
$endgroup$
– user10354138
Oct 22 '18 at 15:17
add a comment |
1 Answer
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To your question:
Am I supposed to think that minimal primes over $I$ are just prime ideals of $R$ containing $I$ that are minimal with respect to inclusion, AND it can be shown that they have this special property relating to $sqrt{I}$?
Yes. I think the point of 2.15 is that we know the radical is the intersection of all prime ideals over $I$, but in fact taking the minimal ones are enough. This follows trivially from your previous discussion.
In fact, you should think of the expression of $sqrt{I}=P_1capcdotscap P_n$ as a special case of primary decomposition, where for radicals, it is actually prime decompositions. Then (treating it as primary decomposition) each $P_i$ is trivially $P_i$-primary, and we assumed that those $P_i$ are distinct. Therefore everyone is minimal, and there are no embedded primes.
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add a comment |
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$begingroup$
To your question:
Am I supposed to think that minimal primes over $I$ are just prime ideals of $R$ containing $I$ that are minimal with respect to inclusion, AND it can be shown that they have this special property relating to $sqrt{I}$?
Yes. I think the point of 2.15 is that we know the radical is the intersection of all prime ideals over $I$, but in fact taking the minimal ones are enough. This follows trivially from your previous discussion.
In fact, you should think of the expression of $sqrt{I}=P_1capcdotscap P_n$ as a special case of primary decomposition, where for radicals, it is actually prime decompositions. Then (treating it as primary decomposition) each $P_i$ is trivially $P_i$-primary, and we assumed that those $P_i$ are distinct. Therefore everyone is minimal, and there are no embedded primes.
$endgroup$
add a comment |
$begingroup$
To your question:
Am I supposed to think that minimal primes over $I$ are just prime ideals of $R$ containing $I$ that are minimal with respect to inclusion, AND it can be shown that they have this special property relating to $sqrt{I}$?
Yes. I think the point of 2.15 is that we know the radical is the intersection of all prime ideals over $I$, but in fact taking the minimal ones are enough. This follows trivially from your previous discussion.
In fact, you should think of the expression of $sqrt{I}=P_1capcdotscap P_n$ as a special case of primary decomposition, where for radicals, it is actually prime decompositions. Then (treating it as primary decomposition) each $P_i$ is trivially $P_i$-primary, and we assumed that those $P_i$ are distinct. Therefore everyone is minimal, and there are no embedded primes.
$endgroup$
add a comment |
$begingroup$
To your question:
Am I supposed to think that minimal primes over $I$ are just prime ideals of $R$ containing $I$ that are minimal with respect to inclusion, AND it can be shown that they have this special property relating to $sqrt{I}$?
Yes. I think the point of 2.15 is that we know the radical is the intersection of all prime ideals over $I$, but in fact taking the minimal ones are enough. This follows trivially from your previous discussion.
In fact, you should think of the expression of $sqrt{I}=P_1capcdotscap P_n$ as a special case of primary decomposition, where for radicals, it is actually prime decompositions. Then (treating it as primary decomposition) each $P_i$ is trivially $P_i$-primary, and we assumed that those $P_i$ are distinct. Therefore everyone is minimal, and there are no embedded primes.
$endgroup$
To your question:
Am I supposed to think that minimal primes over $I$ are just prime ideals of $R$ containing $I$ that are minimal with respect to inclusion, AND it can be shown that they have this special property relating to $sqrt{I}$?
Yes. I think the point of 2.15 is that we know the radical is the intersection of all prime ideals over $I$, but in fact taking the minimal ones are enough. This follows trivially from your previous discussion.
In fact, you should think of the expression of $sqrt{I}=P_1capcdotscap P_n$ as a special case of primary decomposition, where for radicals, it is actually prime decompositions. Then (treating it as primary decomposition) each $P_i$ is trivially $P_i$-primary, and we assumed that those $P_i$ are distinct. Therefore everyone is minimal, and there are no embedded primes.
answered Dec 14 '18 at 9:25
Matt KellerMatt Keller
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Yes, (2.14) should have defined minimal primes over $I$ as these primes $P$ containing $I$ that is minimal with respect to inclusion.
$endgroup$
– user10354138
Oct 22 '18 at 15:17