Is the unit group of any finitely generated reduced $\Bbb Z$ algebra finitely generated?

Solution 1:

"I think the case that $A$ is finite as $\mathbf Z$-module is always true".
Yes, it's true, it's even presented as "a generalization of the unit theorem" in §4.7 of P. Samuel's booklet on ANT. The particular case that $A$ is an integer domain is easy, because then, as you said, $A$ would be an order of some number field in characteristic $0$, or $A$ would be finite in non zero characteristic.

But what worries me is your hint (which I can't quite grasp) at the finite number of minimal primes of $A$ to reduce to that particular case, whereas Samuel feels obliged to go on with a technical inductive proof on the nilradical $N$ of $A$. More precisely, the induction bears on the exponent $s$ such that $N^s = (0)$. The starting step is $s=0$, i.e. $A$ is a reduced ring in which ($0$) is the intersection of finitely many prime ideals $P_i$'s, and so $A^*$ injects into the direct product of the $(A/P_i)^*$'s, which are of finite type according to the particular case. Next assume $s>1$ and consider the natural map $\phi : A \to A/N^{s-1}$. By the induction hypothesis $\operatorname{Im}\phi$ is finitely generated, and Samuel shows that $\ker\phi = 1+N^{s-1}$ and that the latter group is finitely generated.

Finally, your original question, with the additional assumption that $A$ is reduced, has an affirmative answer, see P. Samuel, "A propos du théorème des unités", Bull. Sc. Math., 90 (1966), 89-96).