# Dual of intersections (lattices)

In this paper ( http://cseweb.ucsd.edu/classes/wi10/cse206a/lec2.pdf ), it says that for two sub-lattices $$L_1$$ and $$L_2$$ of $$\mathbb{Z}^n$$, it is easy to see that: $$\widehat{L_1\cap L_2}=\widehat{L}_1+\widehat{L}_2$$ where $$\widehat{L}$$ is the dual of lattice $$L$$ and $$+"$$ stands for the linear combinations of lattices.

Well, the relation $$\widehat{L}_1+\widehat{L}_2\subset \widehat{L_1\cap L_2}$$ is easy, but I can not verify that : $$\widehat{L_1\cap L_2}\subset\widehat{L}_1+\widehat{L}_2$$. How can I check it? That is not a homework, I just want to profoundly understand all parts of the lecture note.

Thanks.

Let us use the notations given in http://cseweb.ucsd.edu/classes/wi10/cse206a/lec2.pdf. Let $$B$$ and $$B'$$ be two bases and $$D$$ and $$D'$$ be dual bases of the $$B$$ and $$B'$$ respectively, i.e., $$D^TB = I, D'^TB' = I$$. We want to show that $$\widehat{\mathcal{L}(B) \cap \mathcal{L}(B')} = \mathcal{L}(D|D')$$ which is equivalent to $$\mathcal{L}(B) \cap \mathcal{L}(B') = \widehat{\mathcal{L}(D|D')}$$. For the equality to be true, it is necessary that $$span(B) = span(B')$$. So, we will consider $$B, B' \in \mathbb{Z}^{n \times m}$$.
Let us show that $$\mathcal{L}(B) \cap \mathcal{L}(B') \subseteq \widehat{\mathcal{L}(D|D')}$$. Let $$v \in \mathcal{L}(B) \cap \mathcal{L}(B')$$, i.e., $$v = Bz = B'z'$$ where $$z, z' \in \mathbb{Z}^m$$. For any $$w \in \mathcal{L}(D|D')$$, i.e., $$w = Dy + D'y'$$ where $$y, y' \in \mathbb{Z}^m$$, we have $$\ =\ + \ \in \ \mathbb{Z}$$ which implies $$v \in \widehat{\mathcal{L}(D|D')}$$. Since, this is true for every $$v \in \mathcal{L}(B) \cap \mathcal{L}(B')$$, we have $$\mathcal{L}(B) \cap \mathcal{L}(B') \subseteq \widehat{\mathcal{L}(D|D')}$$.
Let us now show that $$\widehat{\mathcal{L}(D|D')} \subseteq \mathcal{L}(B) \cap \mathcal{L}(B')$$. From the definition of dual lattice, we have $$span(D) = span(B)$$ and $$span(D') = span(B')$$ which implies $$span(D) = span(D') = span(D|D')$$. So, for any $$v \in \widehat{\mathcal{L}(D|D')}$$, we can write $$v = Bc = B'c'$$ where $$c, c' \in \mathbb{Q}^m$$. If $$c \notin \mathbb{Z}^m$$, then there exist an $$i$$ such that $$c_i \notin \mathbb{Z}$$. This implies that for $$w = De_i \in \mathcal{L}(D|D')$$, $$ = = c_i \notin \mathbb{Z}$$ But, this is contrary to the assumption that $$v \in \widehat{\mathcal{L}(D|D')}$$. Hence, $$c \in \mathbb{Z}^m$$ and using similar argument, we have $$c' \in \mathbb{Z}^m$$. Therefore, $$\widehat{\mathcal{L}(D|D')} \subseteq \mathcal{L}(B) \cap \mathcal{L}(B')$$.