Radical generation of ideals in Noetherian rings

Question

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 arbitrarily 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?

Answer 1

This is a classical result. See, for instance the book of Iyengar, Leuschke, Leikin, Miller, Miller, Singh, Walther, Twenty-Four Hours of Local Cohomology, Remark 9.14.

Answer 2

Fyi: the result holds for non-Noetherian commutative rings as well, provided $I$ is finitely generated (Heitmann, R. Generating non_Noetherian modules efficiently, Michigan Math. J. 31 2, 1984).

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

Added June 2017: as the result for the non-Noetherian case doesn't appear to be folklore, I include a simple proof here.

Theorem: if $R$ is a commutative ring of finite Krull dimension $\le d$, and $a,b_{0},\ldots,b_{d}\in R$, there exist $x_{0},\ldots,x_{d}\in R$ such that $\sqrt{(a,b_{0},\ldots,b_{d})} = \sqrt{(b_{0}+ax_{0},\ldots,b_{d}+ax_{d})}$.

Proof: induction on $d$. When $d=-1$, then $R=0$ is the trivial ring, so that $a=0$, and the ideal $\sqrt{(a)}=0$ is generated by the empty set (consisting of $d+1=0$ generators).

Now let $d\ge 0$, and put $\mathfrak{a}=(b_{d})+\sqrt{0}:b_{d}$, where $\sqrt{0}:b_{d}=\{x\in R\, | \, xb_{d}$ is nilpotent$\}$. If $\mathfrak{p}$ is a minimal prime ideal of $R$, and $b_{d}\in\mathfrak{p}$, then $b_{d}\in\mathfrak{p}R_{\mathfrak{p}}$. As the latter is the only prime ideal of $R_{\mathfrak{p}}$, hence its nilradical, it follows that $b_{d}$ is nilpotent in $R_{\mathfrak{p}}$, say $b_{d}\,^{n}=0$. So there exists an $s\in R-\mathfrak{p}$ such that $s.b_{d}\,^{n}=0$ in $R$. Then $s.b_{d}$ is nilpotent in $R$, hence $s\in \sqrt{0}:b_{d}$. But as $s$ is not in $\mathfrak{p}$, we find $\mathfrak{a}\nsubseteq\mathfrak{p}$. Hence $\mathfrak{a}$ is not contained in any minimal prime ideal of $R$. So dim$(R/\mathfrak{a})$ < dim$(R)$, and therefore dim$(R/\mathfrak{a})\le d-1$.

By the induction hypothesis, $\sqrt{(a,b_{0},\ldots,b_{d-1})} = \sqrt{(b_{0}+ax_{0},\ldots,b_{d-1}+ax_{d-1})}$ in $R/\mathfrak{a}$ for suitable $x_{0},\ldots,x_{d-1}\in R$ (where for ease of notation we simply write $x$ for $x$ mod $\mathfrak{a}$ when $x\in R$). Let $\mathfrak{b}$ denote the ideal $(b_{0}+ax_{0},\ldots,b_{d-1}+ax_{d-1})$ of $R$. Then $\sqrt{(a,b_{0},\ldots,b_{d-1})} =\sqrt{\mathfrak{b}}$ in $R/\mathfrak{a}$, so that $a\in\sqrt{\mathfrak{b}}$ in $R/\mathfrak{a}$. By definition of $\mathfrak{a}$, there exist $x_{d},r\in R$ such that $x_{d}b_{d}\in\sqrt{0}$ and $a-rb_{d}-x_{d}\in\sqrt{\mathfrak{b}}$ in $R$. So $a\in\sqrt{\mathfrak{b}+Rb_{d}+Rx_{d}}$, hence also $a\in\sqrt{\mathfrak{b}+Rb_{d}+Rax_{d}}$.

Note that, in any commutative ring $A$, $\sqrt{(u,v)}=\sqrt{(u+v,uv)}$ for all $u,v\in A$. Indeed, we have $u^{2}=u(u+v)-uv$ and similarly for $v^{2}$. If, moreover, $uv\in\sqrt{0_{A}}$, then $\sqrt{(u,v)}=\sqrt{(u+v)}$.

In our case, this gives $\sqrt{Rb_{d}+Rax_{d}}=\sqrt{R(b_{d}+ax_{d})}$, so that $a\in\sqrt{\mathfrak{b}+R(b_{d}+ax_{d})}$. Writing $\mathfrak{c}:=\mathfrak{b}+R(b_{d}+ax_{d})=(b_{0}+ax_{0},\ldots,b_{d}+ax_{d})$, one has $a\in\sqrt{\mathfrak{c}}$, and therefore the $b_{i}$ are also in $\sqrt{\mathfrak{c}}$ (for $0\le i\le d$). Thus $\sqrt{(a,b_{0},\ldots,b_{d})}\subseteq\sqrt{\mathfrak{c}}$. And since the converse inclusion obviously holds as well, we are done. $\Box$.

The radical of any finitely generated ideal $I$ of $R$ will therefore equal the radical of some suitable ideal generated by $d+1$ elements - by induction on the number of generators of $I$.

Answer 3

As to the Noetherian case, here's an elementary proof of the Eisenbud-Evans-Storch theorem. The theorem yields that every algebraic set in the affine space $\mathbb{A}_{k}^{d}$ over a field $k$ is the intersection of $d$ or fewer hypersurfaces, improving the old result of Kroneckers's that needed $d+1$ or less.

Theorem: If $R$ is a Noetherian commutative ring of finite Krull dimension $d$, then every ideal of $R[X]$ (a ring of dimension $d+1$) has the same radical as a suitable ideal generated by $\le d+1$ elements.

Proof: since dim$(R)=$dim$(R/\sqrt{0})$, and for ideals $\mathfrak{a},\mathfrak{b}$ of $R$, we have $\sqrt{\mathfrak{a}}=\sqrt{\mathfrak{b}}$ in $R$ iff $\sqrt{\overline{\mathfrak{a}}}=\sqrt{\overline{\mathfrak{b}}}$ in $R/\sqrt{0}$, we can assume $R$ is reduced (i.e., $\sqrt{0}=0$).

Induction on $d$. The case $d=-1$ is trivial, for then $R=0=R[X]$, the trivial ring. So let $d\ge 0$.

Let Min$(R)=\{\mathfrak{p_{1}},\ldots,\mathfrak{p_{n}}\}$ be the set of minimal prime ideals of $R$, and $S=R-\bigcup$Min$(R)$ the set of non zero dividers of $R$. (Briefly, if $x\in R$ is a zero divisor, say $xy=0\ne y$, then if $x\notin\mathfrak{p_{i}}$ for all $1\le i\le n$, we have $y\in\mathfrak{p_{i}}$ for all $i$, so $y\in\bigcap$Min$(R)=\sqrt{0}=0$, contradiction. Conversely, if $x\in\mathfrak{p_{i}}=:\mathfrak{p}$, then $x\in\mathfrak{p}R_{\mathfrak{p}}$, the unique prime ideal of $R_{\mathfrak{p}}$, hence its nilradical, so $sx^{m}=0$ in $R$ for some $s\in R-\mathfrak{p}$ and positive integer $m$. But then $s.x$ is nilpotent in $R$, hence vanishes, so $x$ is a zero divisor.)

$T:=S^{-1}R$ is zero dimensional and reduced, so it must be a finite product of fields. (Indeed, one has Max$(T)=$ Min$(T)=\{\mathfrak{p_{1}}T,\ldots,\mathfrak{p_{n}}T\}$, and the natural map $T\to T/\mathfrak{p_{1}}T\times\ldots\times T/\mathfrak{p_{n}}T$ is surjective by the Chinese Remainder Theorem, and injective since $T$ is reduced.) It is easy to see that this implies that every ideal of $T[X]$ is principal.

Now let $\mathfrak{a}=(f_{1},\ldots,f_{m})$ be any ideal of $R[X]$. It follows that the ideal $\mathfrak{a}T[X]$ of $T[X]$ is principal, say $\mathfrak{a}T[X]=gT[X]$ for some $g\in R[X]$. Clearing sufficiently many denominators, there exists an element $s\in S$ such that $s.gR[X]\subseteq\mathfrak{a}$ and $s.\mathfrak{a}\subseteq gR[X]$ in $R[X]$.

We set $A:=R/(s)$. It has dimension $<d$, because $s$ is not in any minimal prime ideal of $R$. By the induction hypothesis, $\sqrt{\mathfrak{a}A[X]}=\sqrt{(\overline{g_{1}},\ldots,\overline{g_{d}})}$ for certain $g_{j}\in R[X]$ (with $1\le j\le d)$, where the $\overline{g_{j}}$ denote their images in $A[X]$. This means that $\sqrt{(s,f_{1},\ldots,f_{m})}=\sqrt{(s,g_{1},\ldots,g_{d})}$ in $R[X]$.

Put $a_{0}:=sg\in\mathfrak{a}$, and take $u\in\mathbb{N}$ large enough so that for all $1\le j\le d$ we have $g_{j}^{u}=r_{j}s+a_{j}$ with $r_{j}\in R[X]$ and $a_{j}\in\mathfrak{a}$. We now put $\mathfrak{b}:=(a_{0},\ldots,a_{d})\subseteq\mathfrak{a}$. Then $\sqrt{\mathfrak{b}}\subseteq\sqrt{\mathfrak{a}}$, while for each $1\le i\le m$ one has $\sqrt{(f_{i})}\subseteq\sqrt{(s,g_{1},\ldots,g_{d})}=\sqrt{(s,a_{1},\ldots,a_{d})}$. Hence, as is easily seen, also $\sqrt{(f_{i})}\subseteq\sqrt{(sf_{i},a_{1},\ldots,a_{d})}$. But $\sqrt{(sf_{i})}\subseteq\sqrt{(g)}$, that is, $\sqrt{(s)}\cap\sqrt{(f_{i})}\subseteq\sqrt{(g)}$, so that $\sqrt{(s)}\cap\sqrt{(f_{i})}\subseteq\sqrt{(s)}\cap\sqrt{(g)}=\sqrt{(sg)} =\sqrt{(a_{0})}$, and therefore $\sqrt{(sf_{i})}\subseteq\sqrt{(a_{0})}$. It follows that $\sqrt{(f_{i})}\subseteq\sqrt{(a_{0},a_{1},\ldots,a_{d})}=\sqrt{\mathfrak{b}}$. Hence $\sqrt{\mathfrak{a}}\subseteq\sqrt{\mathfrak{b}}$, so $\sqrt{\mathfrak{a}}=\sqrt{\mathfrak{b}}$, $\Box$.