Irredundant primary decomposition of radical from irredundant primary decomposition under certain conditions

Question

Let $R$ be a noetherian ring, and $\mathfrak{a}$ a proper ideal of $R$. Let all the associated primes of $R/\mathfrak{a}$ ($Ass\left(R/\mathfrak{a}\right)=\{\mathfrak{p}_{1},\dots,\mathfrak{p}_{r}\}$) be minimal and let $$ \mathfrak{a}=\mathfrak{q}_{1}\cap \dots\cap \mathfrak{q}_{r} $$ be the irredundant primary decomposition of $\mathfrak{a}$ (with $\mathfrak{q}_{i}$ a $\mathfrak{p}_{i}$-primary ideal, $i\in\{1,\dots,r\}$). Let us take "radicals" in both sides, using the properties of radical of an ideal, to obtain: $$ \sqrt{\mathfrak{a}}=\mathfrak{p}_{1}\cap \dots\cap \mathfrak{p}_{r}. $$ My question is: Is this last primary decomposition irredundant in my conditions?

I would have to show, according to my definition of irredundant decomposition, that:

1. $\not\exists S\subsetneq \{1,\dots,r\}$ such that $\sqrt{a}=\bigcap_{i\in S}\mathfrak{p}_{i}$.

2. $\mathfrak{p}_{i}\neq \mathfrak{p}_{j}$ for all $i,j\in\{1,\dots,r\}$, $i\neq j$.

We have that 2. is evident for the latter arguments. I would have to show 1. For that, let suppose that there exists $S\subsetneq \{1,\dots,r\}$ such that $\sqrt{a}=\bigcap_{i\in S}\mathfrak{p}_{i}$. In this case, can I arrive to some contradiction with the initial hypothesis?

Any hint will be highly appreciated. Thanks in advanced.

Answer 1

Let's suppose that there exists $S\subsetneq \{1,\dots,r\}$ such that $\sqrt{\mathfrak{a}}=\bigcap_{i∈S}\mathfrak{p}_{i}$. Then, there exists $j\in\{1,\dots,r\}\setminus S$ such that $\mathfrak{p}_{j}\supseteq \bigcap_{i\in S} \mathfrak{p}_{i}$ (as we would have that $\bigcap_{i\in S} \mathfrak{p}_{i}=\left(\bigcap_{i\in S} \mathfrak{p}_{i}\right)\cap \mathfrak{p}_{j}$). As the ideals involved are prime ideals, $\mathfrak{p}_{j}\supseteq \bigcap_{i\in S} \mathfrak{p}_{i}$ implies that there exists $i_{0}\in S$ such that $\mathfrak{p}_{j}\supseteq \mathfrak{p}_{i_{0}}$; I state this clearer:

If $R$ is a ring, $\mathfrak{p}\in Spec(R)$ and $\mathfrak{p}\supseteq \mathfrak{a}_{1}\cap\dots\cap\mathfrak{a}_{m}$ (with $\mathfrak{a}_{1},\dots,\mathfrak{a}_{m}$ ideals of $R$), then, there exists $i\in\{1,\dots,m\}$ such that $\mathfrak{a}_{i}\subseteq \mathfrak{p}$.

Proof: If there doesn't exist $i\in\{1,\dots,m\}$ such that $\mathfrak{a}_{i}\subseteq \mathfrak{p}$, then, for all $i\in\{1,\dots,m\}$, there exist $x_{i}\in\mathfrak{a}_{i}\setminus\mathfrak{p}$. Then $x=\prod_{i=1}^{m}x_{i}\in \prod_{i=1}^{m} \mathfrak{a}_{i}\subseteq \bigcap_{i=1}^{m} \mathfrak{a}_{i}\subseteq \mathfrak{p}$, and this is a contradiction with the fact that $\mathfrak{p}$ is prime.

Then, we would have to options:

• $\mathfrak{p}_{j}=\mathfrak{p}_{i_{0}}$, that contradicts the fact that $\mathfrak{q}_{1}\cap \dots\cap\mathfrak{q}_{r}$ is an irredundant decomposition.

• $\mathfrak{p}_{j}\supsetneq \mathfrak{p}_{i_{0}}$, that implies that $\mathfrak{p}_{j}$ wouldn't be a minimal prime over $\mathfrak{a}$, contradicting our initial hypothesis.

So, we conclude that there doesn't exist $S\subsetneq \{1,\dots,r\}$ such that $\sqrt{\mathfrak{a}}=\bigcap_{i∈S}\mathfrak{p}_{i}$, as we were willing to prove.