If $p$ and $q$ are prime ideals in a ring such that $p\subsetneq q$ is ht$(p)<$ht$(q)$ If we have a containment of prime ideals in a  commutative ring with $1$ is the "larger" prime ideal necessarily of the higher height? 
 A: Remember, the height of a prime ideal $\mathfrak{p}$ is equal to the number of strict inclusions in a maximal chain of primes ending in $\mathfrak{p}$. So, if $\mathfrak{p}$ is strictly contained in $\mathfrak{q}$, then a maximal chain of primes ending in $\mathfrak{q}$ must necessarily have more inclusions - so, yes (provided that we are in a Noetherian ring, thus there are no infinite chains of prime ideals).
Let $\mathfrak{p}_0 \subsetneq \mathfrak{p}_1 \subsetneq \ldots \subsetneq \mathfrak{p}_n = \mathfrak{p}$ be a maximal chain of primes with respect to inclusion (so that $ht(\mathfrak{p})=n$), then this chain can be extended to $\mathfrak{p}_0 \subsetneq \mathfrak{p}_1 \subsetneq \ldots \subsetneq \mathfrak{p}_n \subsetneq \mathfrak{q}$, which has one more inclusion than the original chain. Since $ht(\mathfrak{q})$ depends on the maximum length of a chain of primes, it may only increase from $n+1$ (since perhaps our extended chain is not of maximal length).
A: Without restricting to finite heights, the answer is no; we could have $\mathfrak{p}\subsetneq\mathfrak{q}$ but $\operatorname{ht}(\mathfrak{p})=\operatorname{ht}(\mathfrak{q})=\infty$. For example, in $R=\mathbb{C}[x_1,x_2,\ldots]$, the polynomial ring in infinitely many variables over $\mathbb{C}$, the prime ideal $\mathfrak{p}=(x_2,x_3,\ldots)$ is strictly contained inside $\mathfrak{q}=(x_1,x_2,\ldots)$, but both $\mathfrak{p}$ and $\mathfrak{q}$ contain infinitely long chains of strict inclusions of prime ideals, e.g.
$$\mathfrak{q}\supsetneq\mathfrak{p}\supsetneq(x_3,x_4,\ldots)\supsetneq(x_4,x_5,\ldots)\supsetneq\cdots$$
so $\operatorname{ht}(\mathfrak{p})=\operatorname{ht}(\mathfrak{q})=\infty$.
