# Is $\Lambda(q)^{\bot}(A) = \{y \in \mathbb{Z}^m: Ay = b \mod q\}$ also a lattice?

Given a matrix $$\bf{A} \in \mathbb{Z}_q^{n \times m}$$, then $$\Lambda_q^{\bot}(A) = \{y \in \mathbb{Z}^m: Ay = 0 \mod q\}$$ is a q-ary lattice. I am wondering whether $$\Lambda_q^{\bot}(A) = \{y \in \mathbb{Z}^m: Ay = b \mod q\}$$ for a given $$b \neq \vec{0}$$ is a lattice, too?

I would assume no, since it does not hold that: $$A(y + y') = b$$, for two $$y, y'$$ where $$Ay = b$$ and $$Ay' = b$$. If this is correct, is there any other way to construct a lattice that contains all solutions to the equation $$Ay = b$$?

If $$b \neq 0 \ mod \ q$$, you are right. Another way to argue is that a lattice must always contain $$0$$, but $$A0 \neq b\ mod\ q$$. From the definition of a lattice $$\mathcal{L}$$, if $$y \in \mathcal{L}$$ then $$zy \in \mathcal{L}$$ for any $$z \in \mathbb{Z}$$. So, if you want a lattice that contains all the solutions to the equation $$Ay = b\ mod \ q$$, then it will immediately contain all the solutions to the equation $$Ay = zb\ mod \ q$$ where $$z \in \mathbb{Z}$$. So, the simplest lattice I can come up with is $$\mathcal{L} = \{y \in \mathbb{Z}^m : Ay = zb\ mod \ q, z \in \{0,1, \dots, q-1\}\}$$