maximal chains of graded prime ideals

In Theorem 1.5.8 [Bruns,Herzog - Cohen-Macaulay-Rings] it is proved that for a noetherian graded ring $R$, a finitely generated graded $R$-module $M$ and any chain $\mathfrak{p}_0 \subsetneq \mathfrak{p}_1 \subsetneq \dots \subsetneq \mathfrak{p}_d$ of prime ideals with $\mathfrak{p}_i \in \operatorname{Supp}(M)$ and $d = \dim M_{\mathfrak{p}_d}$ we can replace all but the last prime by homogeneous primes.

In the second case of the induction step of the proof we assume that all primes but $\mathfrak{p}_{d-1}$ are homogeneous (even $\mathfrak{p}_d$). Then we take a homogeneous element $x \in \mathfrak{p}_d \setminus \mathfrak{p}_{d-2}$, and consider a minimal prime $\mathfrak{q}$ over $\mathfrak{p}_{d-2} + (x)$. Now the authors claim

Since $\operatorname{height}(\mathfrak{p}_d/\mathfrak{q}) = 1$, it is impossible for $\mathfrak{q}$ to equal $\mathfrak{p}_d$.

This statement really confuses me. I can see that $\mathfrak{p}_d \neq \mathfrak{q}$ since $\operatorname{height}(\mathfrak{q}/\mathfrak{p}_{d-2})=1$ by Krull's principal ideal theorem, so the assumption $\mathfrak{p}_d = \mathfrak{q}$ leads to a contradiction as there is a prime lying properly between $\mathfrak{p}_d$ and $\mathfrak{p}_{d-2}$. But I have no idea how to show $\operatorname{height}(\mathfrak{p}_d/\mathfrak{q}) = 1$ without proving $\mathfrak{p}_d \neq \mathfrak{q}$ first.

What could the authors have meant?

Do we really need the assumption that $M$ is finitely generated?

• @user26857: You are probably right. I just wondered if there is a simpler argument here since the authors don't even mention that they are using some other theorem, and Krull's is a pretty heavy one. – Dune Aug 8 '14 at 16:29