# uncountable principal ideal domain with few units

Does there exist an uncountable principal ideal domain with only countably many units (or, even, with only finitely many units) ?
The answer would be yes, if only unique factorization domain was required, for instance the ring of polynomials with uncountably many indeterminates over a finite field.

• Very strange. I wonder if some version of the following construction works: start with, say, $D_0 = \mathbb{Q}[x, y]$. At each step, adjoin generators of all ideals that aren't principal in some reasonably nice way. Iterate uncountably many times...? Sep 2, 2020 at 20:20
• @Qiaochu Yuan: How should this iteration be working for uncountably many times? $\mathbb{Q}[x, y]$ as well as the of its ideals is countable ...
– Ulli
Sep 3, 2020 at 7:04
• At each step in the construction you've added new elements that produce new ideals that may not be principal. In any case, this isn't a big problem, if you accidentally end up with a PID before the construction becomes uncountable you just add a new generator! But I have no idea how to control the result of this, I just couldn't rule out something like this working when I tried to write a proof. Sep 3, 2020 at 7:06
• @Qiaochu: yes, I also had in mind a transfinite induction up to $\omega_1$. But I think it's rather difficult to control the PID requirement. I got your point that you do not care about the successor case, but I think the limit case might even be harder, if you do not include some very special precautions at each successor step. But I would be happy to see it working!
– Ulli
Sep 3, 2020 at 7:30

Start with any finite or countable field $$K$$. Let $$D$$ be a polynomial ring over $$K$$ with uncountably many indeterminates. Then $$D$$ is a unique factorization domain (but, of course, not a PID).
Now, the essential argument is Theorem 4.5 in this paper of Heinzer and Roitman. With the notation in this paper, $$I_\omega(D)$$ is a PID, which contains $$D$$ as a subring and has the same units as $$D$$, hence $$K \setminus \{0\}$$.