I am currently reading Algebraic Geometry and Commutative Algebra by S. Bosch and I have troubles proving the following Theorem:
Let $R$ be a Noetherian local ring with maximal ideal $m$. Then there exists an $m$-primary ideal in $R$ generated by $d=\operatorname{dim}R$ elements, but no such ideal that is generated by less than $d$ elements
The author writes, the proof follows by Krull's Dimension Theorem and the following Lemma
Lemma. Let $R$ be a Noetherian ring and $I$ an ideal in $R$ of height $\operatorname{ht}I=r$. Assume there are elements $a_1,\ldots,a_{s-1}$ such that $\operatorname{ht}(a_1,\ldots,a_{s-1})=s-1$ for some $s\in\mathbb{N}$ such that $1\leq s \leq r$. Then there exists an element $a_s\in I$ such that $\operatorname{ht}(a_1,\ldots,a_s)=s$.
So my question is, how dow I know, that in the situation of the Theorem, there always exist $a_1,\ldots,a_s\in m$ for some $s\in\mathbb{N}$ such that $1\leq s\leq d$ and $\operatorname{ht}(a_1,\ldots,a_s)=s$? Given such $a_i$ one can apply the above Lemma and the rest is clear. Thanks in advance.
EDIT: I saw in this thread, basically the same question was asked. In the comments it was pointed out that in Proposition 11.13 from Atiyah and Macdonald there would be a proof. However, the proof uses induction and I feel like the induction basis, which is skipped, is precisely my question. So I am still unsure how to prove the above.
Best Answer
My guess is that the base step of the induction is $s = 1$. Then the set $\{a_1,\dots, a_{s-1}\}$ vacuously equals $\varnothing$, so the it generated is the zero ideal $(0)$. Furthermore, $\mathrm{ht}(0) = \dim R_{(0)} = 0$ since $R_{(0)}$ is a field.