Lemma 10. Let $R$ be a Noetherian ring and $\mathfrak a$ an ideal in $R$ of height $\mathrm{ht}\,\mathfrak a = r$. Assume there are elements $a_1, a_2, \dots, a_{s-1}\in \mathfrak a$ such that $\mathrm{ht}(a_1, a_2, \dots, a_{s-1})= s-1$ for some $s\in\mathbb{N}$ where $1 \le s \le r$. Then there exists an element $a_s\in\mathfrak a$ such that $\mathrm{ht}\, (a_1, a_2, \dots, a_s)= s$.
Proof. Consider elements $a_1, a_2, \dots, a_{s-1}\in \mathfrak a$, $1 \le s \le r$, such that $\mathrm{ht}(a_1, a_2, \dots, a_{s-1})= s-1$, and let $\mathfrak p_1,\mathfrak p_2,\dots,\mathfrak p_n\subset R$ be the minimal prime divisors of $(a_1, a_2, \dots, a_{s-1})$. Then, using Krull’s Dimension Theorem, we have $\mathrm{ht}\,\mathfrak p_i\le s-1<r$ for all $i$. Since $\mathrm{ht}\,\mathfrak a=r$, this implies a $\mathfrak a\not\subset \mathfrak p_i$ for all $i$ and, hence, $\mathfrak a\,\not \subset \cup_{i=1}^n\,\mathfrak p_i$ by 1.3/7. Thus, choosing $a_s\in \mathfrak a-\cup_{i=1}^n\,\mathfrak p_i$, we get $\mathrm{ht}(a_1, a_2, \dots, a_s)\ge s$ and, in fact $\mathrm{ht}\,(a_1, a_2, \dots, a_s)= s$ by Krull’s Dimension Theorem.
This Lemma comes from Siegfried Bosch's book Algebraic Geometry and Commutative Algebra, section 2.4, Lemma 10, page 77. I don't understand why we obtain $\mathrm{ht}(a_1, a_2, \dots, a_s)\ge s$ from $\mathrm{ht}(a_1, a_2, \dots, a_{s-1})= s-1$. I can only get $\mathrm{ht}(a_1, a_2, \dots, a_s)\ge s-1$ due to $(a_1, a_2, \dots, a_{s-1})\subset (a_1, a_2, \dots, a_{s})$.
Could you show me the proof of this point? Thanks very much!
$\operatorname{ht}(a_1,\cdots, a_s)=\min \operatorname{ht} \mathfrak{p}$ where $\mathfrak{p}$ is a prime ideal containing $(a_1,\cdots, a_s).$
If $\operatorname{ht}(a_1,\cdots, a_s)<s$, then there is a prime ideal $\mathfrak{p}$ containing $(a_1,\cdots, a_s)$ such that $\operatorname{ht} \mathfrak{p}\leq s-1<s$.
Hence $\mathfrak{p}$ is one of the $\mathfrak{p}_i$'s. But that implies $(a_1,\cdots, a_s)\subset \mathfrak{p}\subset \cup_i \mathfrak{p}_i$, hence $a_s\in \cup_i \mathfrak{p}_i$, a contradiction. Thus, $\operatorname{ht}(a_1,\cdots, a_s)\geq s$.