Let $a\subset p$ be prime ideals in a noetherian ring $A$. I want to prove that height of $p/a$ is less or equal than the height of $p$.
The hint to prove that is that exists a bijection between ideals that contains $a$ and ideals of $A/a$, but $p/a$ is not a prime necessarily, so I don't know how to proceed. Thanks.
Suppose that $a$ and $p$ are prime ideal of $A$ and $a\subset p$ then $p/a$ is prime. Let $f:A\rightarrow A/a$ be the projection map remark $f^{-1}(p/a)=p$: let $z\in f^{-1}(p/a)$, there exists $z'\in p$ such that $f(z')=f(z)$ thus $f(z-z')=0$, $z-z'\in a\subset p$ we deduce $z\in p$.
Let $x,y\in A/a$ such that $xy\in p/a$, write $x=f(u), y=f(v)$, $uv\in f^{-1}(p/a)=p$ since $p$ is a prime, $u\in p$ or $v\in p$ this implies that $f(u)=x\in p/a$ or $f(v)=y\in p/a$. We deduce that $p/a$ is a prime ideal of $A/a$.
if $q$ is a prime of $A/a$, $f^{-1}(q)$ is a prime of $A$. Let $u,v\in A$, $uv\in f^{-1}(q), f(u)f(v)\in q$ since $q$ is a prime $f(u)\in q$ or $f(v)\in q$. Suppose that $f(u)\in q$ there exists $u'\in f^{-1}(q)$ such that $f(u)=f(u')$ this implies $u-u'\in a \subset f^{-1}(q)$ thus $u\in f^{-1}(q)$.
Let $p_1\subset....\subset p_n\subset p/a$ a chain of distinct primes $f^{-1}(p_1)....\subset f^{-1}(p_n)\subset p$ is a chain of distinct primes thus the height of $p/a$ is inferior to the height of $p$.