Example of a ring with non-finitely generated unit group? The well known Dirichlet's unit theorem states that the unit group of a maximal order in a quadratic number field is finitely generated of rank blah blah blah. I think it's pretty naive to expect a most radical generalization to hold, namely:

Generalized Dirichlet unit theorem. The unit group of an associative unital ring with finitely generated additive group is finitely generated.

Is there any concrete counterexample?
 A: At Lee Mosher's request, here is how Borel-Harish-Chandra is applied, which is the bulk of the argument (with some additional work in the case with torsion as described in Aurel's post).
Let $A$ be a finitely generated (associative unital) ring whose underlying additive group is free abelian of finite rank. From left multiplication we get a representation of $A$ on $B=A\otimes_\mathbf{Z}\mathbf{R}$, and thus an injective ring homomorphism $i:A\to M=\mathrm{End}(B)\simeq M_d(\mathbf{R})$, where $\mathrm{End}$ means as real vector space; it maps into $E=E(\mathbf{R})$, the set of endomorphisms as algebra. 
Inside $E\times E$, consider the set $G$ of pairs $(x,y)$ such that $xy=1$. This is a Zariski closed submonoid of $E\times E$, and can be identified to the automorphism group of $B$; it lies inside $M_{2d}(\mathbf{R})$. For $x\in A^\times$, consider $j(x)=(x,x^{1})$. Then the group homomorphism $j$ maps injectively $A^\times$ into $G(\mathbf{Z})$. We claim that this is surjective. Indeed, consider $(x,y)\in E(\mathbf{Z})\times E(\mathbf{Z})$ with $xy=1$. Then $x$ preserves the lattice of integral points, i.e., $x$ induces an automorphism of $A=B(\mathbf{Z})$. [Note: we had to introduce $E\times E$ instead of $E$ to describe $G$ as a closed subgroup, which is necessary to apply Borel-Harish-Chandra.]
Thus we have $A\simeq G(\mathbf{Z})$. Borel and Harish-Chandra precisely proved (Annals of Math, 1962) that $G(\mathbf{Z})$ is finitely generated (and even finitely presented, as mentioned in Borel's ICM proceeding of 1962) for every $\mathbf{Q}$-defined subgroup of $\mathrm{GL}_k$. 
[Remark: $G(\mathbf{Z})$ is possibly not a lattice in $G(\mathbf{R})$; however $G(\mathbf{Z})=H(\mathbf{Z})$ for some $\mathbf{Q}$-defined normal subgroup $H$ of $G$, such that $G^0/H^0$ is a $\mathbf{Q}$-split torus, and $H(\mathbf{Z})$ is a lattice in $H(\mathbf{R})$.]
A: Let $A$ be a ring that is finitely generated as an abelian group. Let $I$ be the subgroup of torsion elements. By finite generation, $I$ is finite. On the other hand, $I$ is a two-sided ideal of $A$, and $A/I$ is a ring that is torsion-free as an abelian group. By Borel and Harish-Chandra, $(A/I)^\times$ is finitely-generated. Since $I$ is finite, the map $A^\times \to (A/I)^\times$ is an isogeny (i.e. has finite kernel and cokernel, see Theorem 1.3 in this paper), and therefore $A^\times$ is finitely generated.
