Examples of non-Euclidean domains which have a universal side divisor

Solution 1:

Take $R$ to be the following subring of $\mathbb{Q}[x]$: $$ R = \{ P\in \mathbb{Q}[x] \ | \ P(0)\in \mathbb{Z} \}.$$

Then $2$ is a universal side divisor. Indeed, the only units of $R$ are $\pm 1$, and an element $P$ of $R$ is divisible by $2$ if and only if $P(0)$ is even, so $P$ or $P-1$ is divisible by $2$.

However, $R$ is not Euclidean. In fact, it is not even principal: the ideal $$ I = \{ P\in R \ | \ P(0)=0 \}$$ is not principal. Indeed, if it was principal and generated by $Q$, then $Q$ would have to be of degree $1$, since $x\in I$. Moreover, the only polynomials of degree $1$ in $I$ would be integer multiples of $Q$, so some polynomials of degree $1$ would be missing in $I$, a contradiction.

Hence $R$ is a non-Euclidean domain containing a universal side divisor.

Edit: In fact, $R$ is not even a UFD, since $x$ is divisible by infinitely many pairwise non-associated irreducible elements (for any prime number $p$, $x = p \cdot \frac{x}{p}$). This implies that $R$ cannot be principal nor Euclidean.

Solution 2:

I claim that $\pi:=11-3 \sqrt{10}$ is a universal side divisor in $\mathbb{Z}[\sqrt{10}]$. Since $\mathbb{Z}[\sqrt{10}]$ is not a UFD, it can't be euclidean.

Verification The element $\pi$ has norm $11^2-3^2 \cdot 10 = 31$, so $\mathbb{Z}[\sqrt{10}]/\pi$ is the field $\mathbb{F}_{31}$. In this quotient, $\sqrt{10} \equiv 11/3 \equiv 14$. Let $u$ be the unit $3+\sqrt{10}$, so $u \equiv 3+14 \equiv 17 \bmod \pi$. We want to show that every element of $\mathbb{Z}[\sqrt{10}]$ is either $0 \bmod \pi$ or congruent to a power of $u$ modulo $\pi$. In other words, it is enough to show that $u$ is a generator of the unit group of $\mathbb{Z}[\sqrt{10}]/\pi$ or, concretely, that $17$ is a generator of $\mathbb{F}_{31}^{\times}$. As it happens this is true.

Let's also check that $\mathbb{Z}[\sqrt{10}]$ is not a UFD: We have $\sqrt{10}^2 = 2 \cdot 5$ so, if we had unique factorization, we'd have to have $2$ split as $\alpha \beta$ with $N(\alpha) = N(\beta) = \pm 2$. But $x^2-10 y^2 = \pm 2$ is impossible modulo $5$.

Method I wanted an infinite unit group, so there would be many units to chose from. The easiest way to get that was a real quadratic field. I wanted it NOT to be a PID, so that it wouldn't be Euclidean. So I took the first real quadratic field with class number bigger than $1$.

I wanted the element $\pi$ to be a prime so that the unit group modulo $\pi$ would be cyclic, so I took the first prime in $\mathbb{Z}[\sqrt{10}]$.

Then I checked, and I won! Even if the first prime hadn't worked, we expect that any non-perfect power is a primitive root infinitely often, so I'd expect to win eventually.