Today I came across a problem in my notes which is:
Let $R$ is Commutative and unital ring and $N,N'$ are submodules of M which is an R-module. Then show that $$Ann(N+N')=Ann(N) \cap Ann(N'),$$ where $Ann (N)$ denotes annihilator of N.
I know that for $x\in R$
If $x\in Ann(N+N')\implies x(N+N')=xN+xN'=0$ but how can I show that $xN=0=xN'$
Can anyone please help me?
To say $x \in Ann(N + N')$ is to say that $x$ annihilates every element of the sum $N + N'$ of the modules $N$ and $N'$, i.e., that whenever $n \in N$ and $n' \in N'$, then $x(n + n') = 0$. In particular, if $x \in Ann(N + N')$, taking $n \in N$ and $n'=0$, you have $xn = 0$, and, taking $n = 0$ and $n' \in N'$, you have $xn' = 0$, so $x \in Ann(N)$ and $x \in Ann(N')$. This shows that $$Ann(N + N') \subseteq Ann(N) \cap Ann(N').$$ Conversely if $x \in Ann(N) \cap Ann(N')$, and $n \in N$ and $n' \in N'$, you have $x(n + n') = xn + xn' = 0$, so $$Ann(N) \cap Ann(N') \subseteq Ann(N + N').$$ From this we can conclude $$ Ann(N + N') = Ann(N) \cap Ann(N'). $$ However, $Ann(N) \cap Ann(N')$ is a proper subset of $Ann(N) + Ann(N')$ in general (so, as you realised, you made a typo in your notes). For an example, where $$Ann(N) \cap Ann(N') \neq Ann(N) + Ann(N'), $$ take $R = \Bbb{Z} \times \Bbb{Z}$, and $N$ and $N'$ to be the ideals generated by $(1, 0)$ and $(0, 1)$. Then $Ann(N) = N'$ and $Ann(N') = N$, so $Ann(N) \cap Ann(N') = \{0\}$, but $Ann(N) + Ann(N') = R$.
EDIT: I've left my answer to the original question (with the typo) more or less unchanged as it answers the corrected question and as the example arising from the typo may be of interest.