Csdrhrt

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.

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.