Dimension of $R/\mathfrak{p}$ is less than dimension of $R$?

Question

I have the following exercise:

Let $R$ be a domain of finite dimension $r$, and $\mathfrak{p}$ a nonzero prime. Prove that $\text{dim}(R/\mathfrak{p}) < r$.

What I don't understand is, the chain of prime ideals with maximal length doesn't have to contain the prime $\mathfrak{p}$, so how can we conclude that $\text{dim}(R/\mathfrak{p}) < r$?

Thanks for your help in advance.

Answer 1

No, it doesn’t. The point is that any chain of primes in $R / \mathfrak p$ lifts to a chain of primes in $R$ above $\mathfrak p$. But since $R$ is an integral domain, you can at least extend this chain below by $(0)$, since $\mathfrak p ≠ 0$.

So, no matter how long a chain of primes is in $R / \mathfrak p$, you can always lift and extend it to a chain in $R$, making any chain of primes in $R$ of maximal length, whatever it may be, longer than any such chain in $R/\mathfrak p$.

Answer 2

That's the point! As $R$ is a domain, we conclude $\langle0\rangle$ is a prime ideal of $R$. So $\dim R=r$ means that there is a maximal chain like $\langle0\rangle=P_0\subsetneq P_1\subsetneq\cdots\subsetneq P_r$ such that $\forall 0\leq i\leq r$ we have $P_i \in \operatorname{Spec}(R)$. By definition we have: $$\dim\dfrac{R}{P}=\max\{s\in\mathbb{N}: \left< 0 \right>=Q_0\subsetneq Q_1\subsetneq\cdots\subsetneq Q_s\quad\text{such that for each } i,Q_i\in\operatorname{Spec}(\dfrac{R}{P})\}.$$ But we know that $$\operatorname{Spec}(\dfrac{R}{P})=\{Q\in\operatorname{Spec}(R) : P\subseteq Q\}.$$ As $P\neq0$, so $s\leq r-1$.