$\mathbb Z[1/2]$ is not finitely generated? 
$\mathbb Z[1/2]$ is not finitely generated ?

Maybe I misunderstood, what finitely generated means. Here it says, we need finitely many elements and I think $1$ and $1/2$ suffices as generators.
But here, at the end of Proposition $5.1.4$, from the fact that $\mathbb Z[1/2]$ is not finitely generated it is concluded that $1/2$ is not integral.
 A: I believe you are confusing finitely generated modules with finitely generated algebras. In a module we only have addition and scalar multiplication by the base ring, which in this case is $\mathbb Z$. $\frac1 4$ is not in the submodule generated by $1$ and $\frac 1 2$ because $\frac 1 4$ is not an integer linear combination of these. To see that the module is not finitely generated, note that any finite set of elements will have a largest denominator, and hence elements with larger denominators will not be in the submodule generated by the finite set of elements.
A: As a $\Bbb{Z}$-algebra, $\Bbb{Z}[1/2]$ is finitely generated, but it is not a finitely generated $\Bbb{Z}$-module!
In fact, if $A=\left\{ \frac{a_i}{2^i} \right\}_{i=1}^N$ is finite subset of $\Bbb{Z}[1/2]$, the $\Bbb{Z}$-submodule generated by $A$ does not contain powers of $\frac{1}{2}$ bigger than $\frac{1}{2^N}$.
A: $\mathbb{Z}[\frac{1}{2}]$ is a finitely-generated $\mathbb{Z}$-algebra, but not a finitely generated $\mathbb{Z}$-module.
