Rings and Lattices From Silverman and Tate Rational Points on Elliptic Curves, Exercise 6.16.

Let $L \subset \mathbb{C}$ be a lattice. Define a set of complex numbers $R_L$ by $$R_L = \{ c \in \mathbb{C} : cL \subset L \} .$$
  Prove $R_L$ is  a ring. 

$L$ is a lattice, so it is of the form $\omega_1 \mathbb{Z} + \omega_2 \mathbb{Z}$, where $\{ \omega_1, \omega_2 \}$ are a basis for $\mathbb{C}$.
Then, if $cL \subset L$ that means that $c\omega_1 \mathbb{Z} + c\omega_2 \mathbb{Z} = \omega_1 \mathbb{Z} + \omega_2 \mathbb{Z}.$
Then,  if $c , d \in R_L$, $(c+d)L = (c+d)\omega_1 \mathbb{Z} + (c+d)\omega_2 \mathbb{Z} = cL + dL$ by properties of vectors.
It is then straightforward proving the other axioms in order to show that this is a ring.

Is the definition of lattice I used correct? Doesn't that mean $c$ has to be an integer?

 A: The lattice might be rank 1 (i.e. $L = \omega\mathbf{Z}$) or rank 0. It doesn't feel like you should need to pick a generating set for $L$ though. You can write the proof like this: if $cL \subseteq L$ and $dL \subseteq L$ then


*

*$c(dL) \subseteq cL \subseteq L$. 

*$(c + d)L \subseteq cL + dL \subseteq L + L = L$.


(1.) doesn't require any lattice properties of $L$. (2.) only requires that $L + L \subseteq L$ which is true because $(L,+)$ is a group.

Doesn't that mean c has to be an integer?

No. For example if $L = \mathbf{Z}[i]$ then $cL \subseteq L$ for any $c \in L$ (in fact $R_L = L$). I think you're confusing yourself by choosing a generating set for $L$.

Is the definition of lattice I used correct? 

The definition of lattice that I normally use is that a lattice is a discrete subgroup of $(\mathbf{C},+)$—or, more generally, $(\mathbf{R}^n,+)$—meaning that $L \cap B_r(0)$ is finite where $B_r(0)$ is a ball of any radius $r$ around $0$. One can show that lattices are finitely generated, torsion-free $\mathbf{Z}$ modules of rank $\le \dim \mathbf{C} = 2$.
