The radicals in a minimal primary decomposition are exactly the associated primes of the quotient

Let $\delta :M\mathrm{/}N⟶{⨁}_{i=1}^{r}M\mathrm{/}{Q}_i$ be the diagonal map. Its kernel is zero, because $m+N$ maps to zero exactly when $m\in Q_i$ for every $i$, that is, when $m\in N$. Thus $\delta$ is injective. By [L3] and the direct-sum part of [L2], ${\mathrm{Ass}}_R\text{\hspace{0.17em}⁣}\left({⨁}_{i=1}^{r}M\mathrm{/}{Q}_i\right)=\{{\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_r\}\mathrm{.}$ Since $M\mathrm{/}N$ is a submodule of that direct sum, the left-inclusion part of [L2] gives ${\mathrm{Ass}}_R(M\mathrm{/}N)\subseteq \{{\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_r\}\mathrm{.}$

Fix $i$. Put $I_i={\bigcap }_{j\mathrm{\ne }i}{Q}_j$. By [L1], the decomposition is irredundant, so $I_i⊈Q_i$ and therefore $I_i\mathrm{/}N\mathrm{\ne }0$. The map $I_i\mathrm{/}N⟶M\mathrm{/}{Q}_i,m+N⟼m+Q_i$ is injective because its kernel is $(I_i\cap Q_i)\mathrm{/}N=N\mathrm{/}N=0$. Hence $I_i\mathrm{/}N$ is a nonzero submodule of $M\mathrm{/}{Q}_i$. By [L2] and [L3], ${\mathrm{Ass}}_R(I_i\mathrm{/}N)\subseteq {\mathrm{Ass}}_R(M\mathrm{/}{Q}_i)=\{{\mathfrak{p}}_i\}\mathrm{.}$ Fact [L4] makes ${\mathrm{Ass}}_R(I_i\mathrm{/}N)$ nonempty, so ${\mathrm{Ass}}_R(I_i\mathrm{/}N)=\{{\mathfrak{p}}_i\}$. Applying [L2] again to the inclusion $I_i\mathrm{/}N\hookrightarrow M\mathrm{/}N$ yields ${\mathfrak{p}}_i\in {\mathrm{Ass}}_R(M\mathrm{/}N)$.

Step 1.2 shows every ${\mathfrak{p}}_i$ belongs to ${\mathrm{Ass}}_R(M\mathrm{/}N)$, and step 1.1 gives the reverse inclusion. Therefore ${\mathrm{Ass}}_R(M\mathrm{/}N)=\{{\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_r\}$.