Let be $D$ a domain, $\mathfrak{a,b,p}\subsetneq D$ ideals with $\mathfrak{ab}=\lambda D\, , \lambda\in D\setminus\{0\}$ and $\mathfrak{p}$ prime maximal.
Show that $\mathfrak{ap\subsetneq a}$
Is easy to show $\mathfrak{ap\subset a}$ but I can't get $\mathfrak{ap\neq a}$
EDIT Forget write $\mathfrak{p}$ is maximal.
Edit: I dropped the hypothesis that $D$ has a unity, but the following argument still assumes $D$ commutative (and doesn't use the hypothesis that $\mathfrak{p}$ is prime).
If you multiply both sides by $\mathfrak{b}$ we see that it is enough to prove $$ \lambda \mathfrak{p} \subsetneq \lambda D. $$ Now, since $\mathfrak{p} \subsetneq D$ we can find some $d \in D \setminus \mathfrak{p}$ and since $\mathfrak{p}$ is an ideal (so $0 \in \mathfrak{p}$) we know that $d \neq 0$. Then suppose that $\lambda D = \lambda \mathfrak{p}$. In particular, we can find a $p \in \mathfrak{p}$ such that $$ \lambda d = \lambda p. $$ Since $D$ is a domain we can cancel out $\lambda$ obtaining $$ d = p \in \mathfrak{p} $$ which is absurd because we explicitly chose $d \notin \mathfrak{p}$.