A contradiction to $\textrm{Ass}(M) \subset \textrm{Ass}(N)$ when $M \subset N$?

Question

I am learning about the associated primes of module and just proved a result that $\textrm{Ass}(R /p) = \{ p\}$ for prime $p$ using the fact that $R/p$ is integral domain and therefore has no zero divisors except $p$. But then I remembered that $\textrm{Ass}(M) \subset \textrm{Ass}(N)$ if $M \subset N$. I wonder if for two primes $p \subset q$, we have $R/q \subset R/p$, how is it the case that $$\{q\} = \textrm{Ass}(R/q) \subset \textrm{Ass}(R/p) = \{p\}\ ?$$ Isn't this a contradiction?

Answer 1

If $\mathfrak{p} \subsetneq \mathfrak{q}$, then $R/\mathfrak{q}$ is not contained in $R/\mathfrak{p}$ (think on $\mathbb{Z}$ and $\mathbb{Z}/3\mathbb{Z}$ for instance). Rather, $R/\mathfrak{p} \to R/\mathfrak{q}$ is surjective, but the claimed inclusion does not hold. The closest we can hope for is $\operatorname{Ass}(R/\mathfrak{p}) \subseteq \operatorname{Ass}(R/\mathfrak{q}) \cup \operatorname{Ass}(\mathfrak{q}/\mathfrak{p})$ (for instance, by this lemma), but this does not lead to a contradiction. In fact, we can check that $\operatorname{Ass}(\mathfrak{q}/\mathfrak{p})=\{\mathfrak{p}\}$, just as you already show.