Do we have $ U\cap L + W=(U+W)\cap (L+W)$ for $L$ a lattice and $U$, $W$ in a vector space $V$?

Question

Let $V$ be a finite dimensional $\mathbb{R}$-vector space, and let $L$ be a $\mathbb{Z}$-lattice of full rank in $V$. Let also $U$ and $W$ be two subspaces of $V$.

We have an inclusion of $\mathbb{Z}$-modules: $$ U\cap L + W \subseteq (U+W)\cap (L+W). $$ Is this an equality ?

Many thanks !

Answer 1

Let $V=\mathbb{R}^2$, $U=\mathbb{R}\times 0$, and $W=0\times\mathbb{R}$, and let $L$ be the subgroup of $V$ generated by $(0,1)$ and $(\pi,\pi)$. Then we have $U\cap L=\{(0,0)\}$, so that $(U\cap L)+W=W=0\times\mathbb{R}$. On the other hand, we have $U+W=V$, and $L+W=\mathbb{Z}\pi\times\mathbb{R}$, so that $(U+W)\cap(L+W)=\mathbb{Z}\pi\times\mathbb{R}$.