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It is well-known that any ideal in a Dedekind domain can be generated by at most two elements. However, already for Noetherian domains of dimension 2, it is easy to construct examples of ideals that require arbitrariry large finite numbers of elements to generate.

Nevertheless, I believe I can prove the following theorem.

Let $R$ be a commutative Noetherian ring of finite Krull dimension $d$. Given an ideal $I\subset R$, we denote by $r(I)$ the radical of $I$, i.e., the ideal formed by all the elements $a\in R$ for which there exists a positive integer $n$ such that $a^n\in I$.

Let $I\subset R$ be an ideal such that $r(I)=I$. Then there exist $d+1$ elements $a_0$, $\dots$, $a_d\in I$ such that $r(a_0,\dots,a_d)=I$, where $(a_0,\dots,a_d)$ denotes the ideal generated by $a_0$, $\dots$, $a_d$ in $R$.

This assertion can be restated as follows: let $\operatorname{Spec} R$ be a Noetherian affine scheme of Krull dimension $d$ and $W\subset\operatorname{Spec}R$ be an open subscheme. Then $W$ can be presented as the union of at most $d+1$ principal affine open subschemes of $\operatorname{Spec} R$.

If true, this certainly must be well-known. Is there any reference?

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2 Answers

up vote 8 down vote accepted

This is a classical result. See, for instance, Iyengar S. et al., Twenty-Four Hours of Local Cohomology, Remark 9.14.

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This answers it. Thank you! –  Leonid Positselski Nov 13 '12 at 2:09
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Fyi: the result holds for non-Noetherian commutative rings as well (Heitmann, 1984).

And for Noetherian $R$ of dimension $\le n$, the following sharpening holds: every ideal of $R[X]$ has the same radical as a suitable ideal generated by $n+1$ elements (Eisenbud-Evans-Storch). Cf. http://hlombardi.free.fr/publis/nilregular.pdf.

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