Prove that $mathrm{ht}(P/Ra)=mathrm{ht}(P) -1$

Lemma 1: Let $R$ be a Noetherian ring and $(a) subset Iin mathrm{Spec(R)}$ such that $mathrm{ht}(I) = 0$. Then $a$ is a divisor of $0$.

Proof: Consider the localized ring $R_{I}$. Clearly, $R_{I}$ is a Noetherian ring and $mathrm{Spec}(R_{I}) = {I':= I_I}$ (since $I$ has height $0$ and there is a bijection between $mathrm{Spec}(R_I)$ and ${Ain mathrm{Spec}(R); Acap (Rsetminus I) = emptyset}$) we can conclude that $R_I$ is an Artinian ring. Theorefore $I'$ is nilpotent. Thus, there exists $sin Rsetminus I$ and $ninmathbb{R}$, such that $s a^{n} = 0$. Implying that $a$ is a $0$ divisor.

---

Suppose by reductio ad absurdum that $mathrm{ht}(P/(a)) = mathrm{ht}(P) = n$. Then there there exists a saturated chain of prime ideals in $R/(a)$ such that $$Q_0 subsetneq Q_1 subsetneq ldotssubsetneq Q_{n-1}subsetneq P/(a). $$

It is well known that there is a bijection between the set $mathrm{Spec}(R/(a))$ and ${Iin mathrm{Spec}(R); (a)subset I}$. Then $forall iin{0,1,ldots,n-1}$, there exists a unique $P_iin mathrm{Spec}(R)$, such that $(a)subset P_i$ and $P_i/(a) = Q_i$. Thus, we can define the following chain of prime ideals in $R$ $$P_0 subsetneq P_1 subsetneq ldotssubsetneq P_{n-1}subset P. $$ Since $mathrm{ht}(P)=n$ the above chain is saturated. So $mathrm{ht}(P_0) = 0$ and $(a)subset P_0$. Using Lemma 1, $a$ is a zero divisor, which is a contradiction. Therefore $mathrm{ht}(P/(a))<mathrm{ht}(P)$.

Corollary 15.15, of the same book, states that the following inequality holds $$mathrm{ht}(P/(a)) leq mathrm{ht}(P)leq mathrm{ht}(P/(a)) +1, $$ so, $$left|mathrm{ht}(P) - mathrm{ht}(P/(a)) right| leq 1. $$

We already have proved that $ 0<left|mathrm{ht}(P) - mathrm{ht}(P/(a)) right|$. Therefore $$mathrm{ht}(P) = mathrm{ht}(P/(a)) +1.$$

$$hspace{10cm}square $$