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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), $\sqrt{I} = P_1 \cap \dots\cap P_m$ for some primes $P_i$ such that $P_j \not\subset P_i \; \forall i \neq j$.

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.

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    $\begingroup$ 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
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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 $\sqrt{I}=P_1\cap\cdots\cap P_n$ 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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