Let $D$ be a commutative domain with multiplicative identity $1$ and assume that the additive group $D$ is finitely generated. Prove the characteristic of $D$ equals $0$ iff $(D, +)$ is a free abelian group.
I'm honestly not even really understanding the difference between a group being finitely generated versus a group being free. For a free group with generators $x_1,...,x_n$ the idea is that all the elements $\phi \in G$ can be written as $\phi = k_1x_1 + \cdots + k_n x_n$. In a finitely generated group, we can all the elements of the group $G$ as a finite product of the elements of our generator set and their inverses. When both are abelian aren't these two notions the same?
The underlying group of $D$ is finitely generated, meaning that it is isomorphic to $$\mathbb Z^n\oplus \mathbb Z_{p_1^{n_1}}\oplus \dots\oplus \mathbb Z_{p_r^{n_r}},$$ where $p_1,\dots,p_r$ are prime. Then consider $x:=(0,1,0,\dots ,0)$; $p_1^{n_1}x=0$, $x\neq 0$, and so (being a domain) $D$ has characteristic $p_1$. Hence, if the characteristic is $0$, $D\cong \mathbb Z^n$.
It's easy to see that conversely if $D\cong \mathbb Z^n$ as groups, its characteristic is $0$.