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 $I$.