Associated primes of a sum of modules 
Let $M$ be a module with $M_1$ and $M_2$ submodules such that their sum (not necessarily a direct sum) is $M$. Is it true in full generality that $\text{Ass}(M) = \text{Ass}(M_1) \cup \text{Ass}(M_2)$? If so prove, if not, provide a counterexample.

I believe the statement is false. The fact that $\text{Ass}(M_i) \subset M$ is obvious, and I know it holds for a direct sum, however I was hoping someone could clarify with a counter example. Thanks is advance.
 A: Mariano's answer proves that if $A$ is a commutative ring and if $M_1,M_2$ are $A$-modules, then $\text{Ass}_{A}(M_1\oplus M_2)=\text{Ass}_{A}(M_1)\cup\text{Ass}_{A}(M_2)$. However, when $M$ is an $A$-module and when there exist $A$-submodules $M_1,M_2$ of $M$ such that $M=M_1+M_2$, then we cannot conclude in general that $\text{Ass}_{A}(M)=\text{Ass}_{A}(M_1)\cup\text{Ass}_{A}(M_2)$. (However, we can conclude in general that $\text{Ass}_{A}(M_1)\cup\text{Ass}_{A}(M_2)\subseteq\text{Ass}_{A}(M)$.)
Let me give a counterexample:
Exercise: Let $M=\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}$ be a $\mathbb{Z}$-module and let $M_1=\{(n,0):n\in\mathbb{Z}\}$, $M_2=\{(n,\overline{n}):n\in\mathbb{Z}\}$ (where $\overline{n}$ is the residue of $n$ modulo $2$). 
(a) Prove that $M_1$ and $M_2$ are $\mathbb{Z}$-submodules of $M$. 
(b) Prove that $M=M_1+M_2$. 
(c) Prove that $M_1\cong \mathbb{Z}\cong M_2$ as $\mathbb{Z}$-modules. In particular, deduce that $\text{Ass}_{\mathbb{Z}}(M_1)=\{0\}=\text{Ass}_{\mathbb{Z}}(M_2)$. 
(d) Prove that $\text{Ass}_{\mathbb{Z}}(M)=\{0,2\}$. (Hint: Mariano's answer is relevant.)
(e) Note that we have a counterexample to your claim.
I hope this helps! 
A: $\newcommand\ass{\operatorname{Ass}}$Show that whenever you have a short exact sequence $$0\to M\to N\to P\to0$$ then $$\ass M\subseteq\ass N\subseteq\ass M\cup\ass P.$$ Then what you want follows from the existence of the two obvious exact sequences $$0\to M_1\to M_1\oplus M_2\to M_2\to0$$ and $$0\to M_2\to M_1\oplus M_2\to M_1\to0$$ 
