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The Nullstellensatz says that any ideal $I \subset \mathbb{C}[x_1,\dots,x_n]$ has the property that

$\sqrt{I} = \bigcap_{\text{maximal } \mathfrak{m} \supset I} \mathfrak{m}$

Is it also true that we can write any ideal as above as

$I = \bigcap_{\text{maximal } \mathfrak{m} \supset I} \mathfrak{q}_{\mathfrak{m}}$

for certain $\mathfrak{m}$-primary ideals $\mathfrak{q}_{\mathfrak{m}}$? Note that I'm only allowing a single primary ideal for each maximal ideal $\mathfrak{m}$; if I allowed countably many then the problem would be trivial.

I suppose my question applies to any Jacobson ring instead of $\mathbb{C}[x_1,\dots,x_n]$ .

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  • $\begingroup$ Primary decomposition reduces the problem to $I$ primary, but I don't know how to go on from there. $\endgroup$
    – Sasha
    Aug 24, 2022 at 14:12

1 Answer 1

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The answer is yes for any excellent Hilbert ring, hence for $\mathbb{C}[x_1, \dots, x_n]$.

This follows from:

Eisenbud and Hochster's Theorem [Corollary 2, 1]. Let $A$ be a ring finitely generated over a field or over the integers. Let $I$ be an ideal of $A$. Then $I$ can be written as $$I = \bigcap_{\mathfrak{m} \in \mathcal{N}} (\mathfrak{m}^e + I)$$ where $\mathcal{N}$ is the set of maximal ideals of $A$ containing $I$ and $e$ is an integer depending on the degree of nilpotency of $A/I$.

According to the remarks following [Corollary 2, 1], it suffices that $A$ be an excellent Hilbert ring so that the above theorem holds. Furthermore, a concrete value for $e$ can be obtained from a primary decomposition of $I$.

Note. Observe that in any commutative ring $R$ with identity, an ideal which contains some power of a maximal ideal $\mathfrak{m}$ of $R$ is $\mathfrak{m}$-primary.


[1] D. Eisenbud and M. Hochster, "A Nullstellensatz with Nilpotents and Zariski’s Main Lemma on Holomorphic Functions", 1979.

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