Lemma 10.63.14. Let $R$ be a ring. Let $I$ be an ideal. Let $M$ be an $R/I$-module. Via the canonical injection $\mathop{\mathrm{Spec}}(R/I) \to \mathop{\mathrm{Spec}}(R)$ we have $\text{Ass}_{R/I}(M) = \text{Ass}_ R(M)$.

**Proof.**
Omitted.
$\square$

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