# Let $N$, $K$ be sub-modules of $M$ with $I=\mathrm{Ann}(N)$, $J=\mathrm{Ann}(K)$. Show $I+J$ is a proper subset of $\mathrm{Ann}(N \cap K)$.

Let $N$ and $K$ be sub-modules of $M$ with $I=\operatorname{Ann}(N)$ and $J=\operatorname{Ann}(K)$. Show that $I+J$ is a proper subset of $\operatorname{Ann}(N \cap K)$.

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When $N=K$, for example, there's equality $I+J= \text{Ann}(N\cap K)$. Perhaps the word "proper" is redundant? – Ofir Dec 31 '12 at 10:34

For any $i\in I$ we have that $in=0$ for all $n\in N$. For any $j\in J$ we have that $jk=0$ for all $k\in K$. Now take $i+j$. Look at $x\in N\cap K$. $ix=0$ because $x\in N$. $jx=0$ because $x\in K$. Thus $(i+j)x=ix+jx=0$ so $i+j\in \text{Ann}(N\cap K)$ so $I+J\leqslant \text{Ann}(N\cap K)$.
I don't see why it should be true in general that $I+J$ should be proper. Aside from the obvious case with $N=K$, we have for example from where $N$ is a submodule of $K$ (where $I+J$ is sometimes proper but not always).
Or take the ring $R = \mathbb{Z}/6 \mathbb{Z}.$ Take $M = R,$ take $N$ to be the ideal $2R$ and $K$ to be the ideal $3R.$ Then $I +J$ contains the unit $5,$ so is all of $R.$ – Geoff Robinson Dec 31 '12 at 10:01