Let $n \geq 4$ be an integer divisible by 4. Let $E_0 = \{v \in \mathbb{Z}^n \, :\, \sum_{i =0}^n{v_i} \equiv 0 \pmod{2}\}$. Let $e = (1/2, \dots, 1/2) \in \mathbb{Q}^n$ and let $E = \mathbb{Z}e + E_0$ (in $\mathbb{Q}^n$).
Since $E_0$ has index $2$ in $\mathbb{Z}^n$, the $\mathbb{Z}$-module $E$ is free of rank $n$ and it is easy to check that $E_0$ has index $2$ in $E$. Also $E$ is clearly finitely generated. These three facts alone do not suffice to conclude that $E$ is free of rank $n$ in general, as the example $\mathbb{Z}^n \hookrightarrow \mathbb{Z}^n \oplus \mathbb{Z}/2 \mathbb{Z}$ shows.
Is there an easy way to show that $E$ is free of rank $n$? (The example is taken from chapter V in Serre's Course in Arithmetic, by the way). Is there an obvious $\mathbb{Z}$-basis for $E$ that one can write down?
First note that $u_i=e_1+e_i$ for $1\leq i\leq n$ is a set of generators for $E_0$. For if $v\in E_0$ then \begin{align} v&=\sum_{i=1}^nv_ie_i\\ &=-\frac 12\left(\sum_{i=1}^nv_i\right)2e_1+\sum_{i=1}^nv_i(e_1+e_i)\\ &=-\frac 12\left(\sum_{i=1}^nv_i\right)u_1+\sum_{i=1}^nv_iu_i \end{align} But for even $n$ we have $$u_2=2e+\frac{n-2}2u_1-\sum_{i=3}^nu_i$$ hence $u_1,e,u_i$ for $3\leq i\leq n$ generates $E$. This is also a basis for $E$, for if \begin{align} 0 &=a_1u_1+a_2e+\sum_{i=3}^na_iu_i\\ &=\left(2a_1+\frac{a_2}2+\sum_{i=3}^na_i\right)e_1+\frac{a_2}2e_2+\sum_{i=3}^n\left(\frac{a_2}2+a_i\right)e_i \end{align} implies $a_2=0$, hence $a_i=-\frac{a_2}2=0$ for $i\geq 3$ and $a_1=-\frac{a_2}4-\frac 12\sum_{i=3}^na_i=0$.