Prove that $\operatorname {ht}(p/a)\leq \operatorname {ht}(p)\leq \operatorname {ht}(p/a)+n$

Question

In the Milne's book A Primer of Commutative Algebra, pg. 100, there's a proof that $\operatorname {ht}(p/a)\leq \operatorname {ht}(p)\leq \operatorname {ht}(p/a)+n$. I understand the first inequality, but I'm lost in the second inequality, when it says Then p is a minimal prime of $(a_{1},..,a_{d},b_{1},...,b_{n})$, and hence has height d+n. I don't know how it concludes that. It says that $p/a$ is a minimal prime ideal of $(a'_{1},..,a'_{d})$ and $a$ is generated by $b_{1},...,b_{n}$. How this implies the first? I'd like somebody help me please. Thanks.

Answer 1

The book says "and hence has height $\le d+n$", and this follows form the Krull's Principal Ideal Theorem; see Theorem 21.6 in Milne's book.

If you want to prove that $\mathfrak p$ is minimal over $(a_{1},..,a_{d},b_{1},...,b_{n})$, then suppose the contrary, and pick a prime ideal $\mathfrak p'$ such that $(a_{1},..,a_{d},b_{1},...,b_{n})\subseteq \mathfrak p'\subsetneq \mathfrak p.$ Then $$(a'_{1},..,a'_{d})=(a_{1},..,a_{d},b_{1},...,b_{n})/(b_{1},...,b_{n})\subseteq \mathfrak p'/(b_{1},...,b_{n})\subsetneq \mathfrak p/(b_{1},...,b_{n}),$$ a contradiction.