Localisation of a primary submodule either stays primary or becomes the whole module

Put $N=M\mathrm{/}Q$. Since $Q$ is $\mathfrak{p}$-primary, [L1] gives $n\ge 1$ with ${\mathfrak{p}}^{n}N=0$. If $S\cap \mathfrak{p}\mathrm{\ne }\mathrm{\varnothing }$, choose $s\in S\cap \mathfrak{p}$. Then $s^{n}N=0$, so in the localization $S^{-1}N$ the unit $s\mathrm{/}1$ annihilates every element. Hence $S^{-1}N=0$, and [L3] shows $(S^{-1}M)\mathrm{/}(S^{-1}Q)=0$, that is, $S^{-1}Q=S^{-1}M$.

Assume now that $S\cap \mathfrak{p}=\mathrm{\varnothing }$. Since $Q$ is $\mathfrak{p}$-primary, the quotient $N$ has ${\mathrm{Ass}}_R(N)=\{\mathfrak{p}\}$ by [L1]. Fact [L2] therefore gives ${\mathrm{Ass}}_{S^{-1}R}(S^{-1}N)=\{S^{-1}\mathfrak{p}\}\mathrm{.}$ Also ${\mathfrak{p}}^{n}N=0$ from [L1], so $(S^{-1}\mathfrak{p}{)}^n{S}^{-1}N=0$.

Let $u=a\mathrm{/}s\in S^{-1}R$ with $u\mathrm{\notin }{S}^{-1}\mathfrak{p}$. Then $a\mathrm{\notin }\mathfrak{p}$, so multiplication by $a$ on $N$ is injective by [L1]. Because $s\mathrm{/}1$ is a unit, multiplication by $u=(a\mathrm{/}1)(s\mathrm{/}1{)}^{-1}$ on $S^{-1}N$ is also injective. The module $S^{-1}M$ is finitely generated over the Noetherian ring $S^{-1}R$: if $m_1,\dots ,m_t$ generate $M$, then $m_1\mathrm{/}1,\dots ,m_t\mathrm{/}1$ generate $S^{-1}M$. Hence [L1], applied over $S^{-1}R$, shows via [L3] that $S^{-1}Q$ is $S^{-1}\mathfrak{p}$-primary in $S^{-1}M$.

Steps 1.1 and 2.1 prove the two localization alternatives.