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Let $R$ be a Noetherian commutative ring. Show that $$\bigcap_{n \in \mathbb N} \bigcap_{\mathfrak p \in \mathrm{Spec}(R)}\mathfrak p^n = \{0\}.$$

I definitely want to use Krull intersection theorem as $\bigcap_{p \in Spec(R)}p^n $ is inside the jacobson radical. So I want to find some ideal $J$ in the jacobson radical such that for every $n$, there exists $n_i$ such that $\bigcap_{p \in Spec(R)}p^n \subset J^{n_i} $, then the result applies. Right know I am thinking of letting $J$ be the nipotent radical, but I'm not sure if this is valid. Any suggestion?

user26857
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The One
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    You can also view the intersection in every localization of $R$ (where you can apply Krull!) – Aphelli Feb 04 '22 at 08:53
  • @Mindlack I believe localization only commutes with finite intersection. How can I apply localization when the intersection is infinite – The One Feb 04 '22 at 22:23
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    Sure, it doesn’t per se commute, but there’s still a useful inclusion working! – Aphelli Feb 04 '22 at 22:42

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In this answer it is mentioned the following useful result: if $x\in\bigcap_{n=1}^{\infty}I^n$, then $x\in xI$, that is, there is $a\in I$ such that $(1-a)x=0$. One uses this result for a prime ideal $P$, and get an element outside of $P$ which multiplied by $x$ gives zero. When this happens for all primes we can conclude that $x=0$.

user26857
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You could also think about this in terms of primary decompositions.

Since $R$ is Noetherian, $0$ has a primary decomposition $0 = \bigcap Q_i$.

Since $P_i = \sqrt{Q_i}$ is finitely generated, there exists $n_i$ such that $P_i^{n_i} \subseteq Q_i$. Therefore $\bigcap P_i^{n_i} = 0$

Badam Baplan
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