Let $M$ be a module over a ring $R$.
Let $\operatorname{Ass}(M)$ be the set of annihilator ideals $\operatorname{Ann}(x)$, which are prime, so
$\operatorname{Ass}(M) = \{\operatorname{Ann}(x) \mid \operatorname{Ann}(x)\text{ is prime}, x \in M\}.$
Recall that $\operatorname{Ann}(x) = \{r \in R \mid rx=0\}$.
If $M_1$ and $M_2$ are two modules, I wish to prove that $\operatorname{Ass}(M_1 \oplus M_2) = \operatorname{Ass}(M_1) \cup \operatorname{Ass}(M_2),$ where $\oplus$ is direct sum and $\cup$ is ordinary union of sets.
I need to do this by considering an element of the left hand side and show it is in the right hand side, so nothing fancy. The direction from right to left is easy, since for any $m_1 \in M_1$ I have $\operatorname{Ann}(m_1) = \operatorname{Ann}(m_1,0)$, but the other direction causes me trouble.