Let $R$ be a commutative ring and $\mathfrak{p}\subseteq\mathfrak{q}$ be two prime ideals. We send $R_\mathfrak{q}\to R_\mathfrak{p},r/s\mapsto r/s$. Is this map always injective?
I was considering like this: let $r/s=0$ in $R_\mathfrak{p}$, i.e. there exists $u\in R\setminus\mathfrak{p}$ such that $ur=0$. If it is injective, then the existent $v\in R\setminus\mathfrak{q}$ such that $vr=0$ is a must. But this can not be guaranteed since $R\setminus\mathfrak{q}$ is smaller than $R\setminus\mathfrak{p}$.
Then I was thinking about some counter-examples. First, the counter-examples cannot be integral domains, otherwise $r=0$. I tried $\mathbb{Z}_n[X]$ for $n$ composite which did not work either. So this is equivalent to find a case that a commutative ring has two prime ideals $\mathfrak{p}\subseteq\mathfrak{q}$, and for some element $a\in\mathfrak{p}$, Ann$(a)\subseteq\mathfrak{q}$.
No. For instance, let $k$ be a field and $R=k[x,y]/(xy)$, $\mathfrak{p}=(x)$, and $\mathfrak{q}=(x,y)$. Then $x$ is nonzero in $R_\mathfrak{q}$ (its annihilator is $(y)\subset\mathfrak{q}$), but it is zero in $R_\mathfrak{p}$ since it is annihilated by $y\in R\setminus\mathfrak{p}$.