1

Let $I$ be a proper ideal of a commutative Noetherian ring $R$. Let $J:=\bigcap_{n=1}^\infty I^n$. Then, is it true that $J \subseteq P$ for some minimal prime $P$ of $R$?

By prime avoidance, I am equivalently asking: Is $J$ necessarily contained in the union of all the (finitely many) minimal primes of $R$?

By Krull-intersection Theorem, I know this to be true if either
(1) $I$ is inside the Jacobson radical of $R$, or
(2) $R$ is an integral domain.
(In both these cases, the intersection is in fact $0$.)

But I don't know what happens in general.

Please help.

user26857
  • 52,094
strat
  • 329

2 Answers2

3

It is well known that $$\bigcap_{i=1}^\infty I^n=\{x\in R\mid\exists a\in I\text{ such that }x=ax\} $$ (See here.)

Let $\mathrm{Min}(R)=\{P_1,\dots,P_m\}$. If $J\not\subseteq P_i$ for all $i=1,\dots,m$, there exists $x_i\in J$ such that $x_i\notin P_i$ for all $i=1,\dots,m$. Let $a_i\in I$ such that $x_i(1-a_i)=0$. It follows that $1-a_i\in P_i$ for all $i=1,\dots,m$. Then $(1-a_1)\cdots(1-a_m)$ belongs to the product of all minimal prime ideals, hence it belongs to the intersection of all minimal prime ideals, so it is nilpotent. Now write $(1-a_1)\cdots(1-a_m)=1-a$ with $a\in I$. Since $1-a$ is nilpotent, $a=1-(1-a)$ is invertible, so $I=R$, a contradiction.

user26857
  • 52,094
1

Let $P$ be a minimal prime of $R$, let $K=I+P$, and suppose that $K$ is a proper ideal of $R$. Then, since $R/P$ is a Noetherian domain, we know $O:=\bigcap_{n\in\mathbb{N}}(K/P)^n$ is trivial in $R/P$. But for any $a\in J$, we have $a+P\in O$, whence this would imply $J\leqslant P$, as needed.

So, enumerating the minimal primes of $R$ as $P_1,\dots,P_n$, we have that, if $J$ is not contained in any minimal prime, then $I+P_i=R$ for all $i\in[n]$. For each $i$, fix $a_i\in I$ and $p_i\in P_i$ with $a_i+p_i=1$; then $1-p_1\dots p_n=(a_1+p_1)\dots(a_n+p_n)-p_1\dots p_n\in I$. But $p:=p_1\dots p_n$ is an element of $\bigcap_{i\in[n]}P_i$, so that $1-p$ is a unit, contradicting that $I$ is proper.

So indeed $J$ must be contained a minimal prime.

Atticus Stonestrom
  • 11,426
  • 1
    Thanks! An alternative way to see $I+P$ has to be a proper ideal for some minimal prime $P$ is as follows: As $I$ is proper, so $I\subseteq \mathfrak m$ for some maximal ideal $\mathfrak m$ of $R$. Now $\mathfrak m$ contains a minimal prime of $R$, say $P$. Hence, $I+P\subseteq \mathfrak m$ is a proper ideal – strat May 22 '22 at 02:53
  • I have also recently asked this question https://math.stackexchange.com/questions/4455927/mathfrak-mr-mathfrak-m-primary-ideal-is-the-localization-of-some-mathf in which you might be interested – strat May 22 '22 at 06:38
  • @strat you're right, that is a much cleaner way of showing it! :) – Atticus Stonestrom May 22 '22 at 08:31