UFD with infinitely many units and infinitely many non-units

Question

It is known that a UFD which is not a field either has

• infinitely many units, or
• infinitely many non-associated prime elements.

I assume that if the first condition is satisfied, nothing can be said about the second one. What if we add the condition that the set of non-units is also infinite?

To state the question clearly: If a UFD has infinitely many units and infinitely many non-units, can we conclude that the UFD has infinitely many non-associated prime elements?

Answer 1

Counter-example:

Let $K$ be an infinite field. The ring $K[\mkern-1.5mu[X]\mkern-1.5mu]$ has infinitely many units, since a formal power series is invertible if & only if its constant term is non-zero. On the other hand, it has a single prime element, namely $X$.

Answer 2

Look at the ring $A=\{ a + b\sqrt 2 \mid a,b\in\mathbf{Z}\}$. This is a Euclidean domain, hence PID, hence UFD. The subset of elements with $a=0$ will be non-units, which is infinite.

The number $u=3+2\sqrt2$ is a unit as its inverse is $3-2\sqrt2$, which is in $A$. Then all powers $u^n$ which are infinite in number are also units.

The ring $A$ is Euclidean follows from the norm function $N(a+b\sqrt 2) = |a^2-2b^2|$, and division algorithm yielding a remainder of smaller norm. You can imitate usual proof for infinite number of primes here.