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I've encountered several proofs that $A = \mathbb{Z}[\frac{1 +i\sqrt{19}}{2}]$ is indeed a PID that is not euclidean. I'm quite confused with this one, which seems to be nonetheless using similar arguments. We consider that $A$ is euclidean, therefore it features a euclidean function $\nu$. Let $x \neq 0$ non-unit such that $\nu(x)$ is minimal, and let $\bar{y} \neq 0$ an element of the quotient space $A/(x)$. By definition of $x$, it is clear that if $y = qx + r$, then $r$ must be invertible, so $r = 1$ or $r = -1$ (because $\nu(r) < \nu(x)$).

From that we deduce that either $A \simeq \mathbb{F}_2$ or $A \simeq \mathbb{F}_3$. I'm not totally confident with this first fact. Indeed, it seems the only possible remainder (i.e. elements of $A/(x)$) are $0, 1, -1$. Either everyone of them is possible, and $A \simeq \mathbb{F}_3$, or only two of them occur ($0$ and $1$ for instance) and $A \simeq \mathbb{F}_2$. But when I tried writing actual proofs, I ended up proving the hypothesis (that $A$ is euclidian) is false.

(1) If $A \simeq \mathbb{F}_2$, then $\bar{y}$ is a root of $T^2 + T + 1 \in A[T]$.

(2) If $A \simeq \mathbb{F}_3$ then $\bar{y}$ is a root of $T^2 - T - 1 \in A[T]$.

Both statements are allegedly contradictory, which seems wrong to me. ($\bar{y} \neq 0$, then $\bar{y} = 1 \in \mathbb{F}_2$, and clearly (1) is not satisfied.)

user26857
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Hermès
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    (It's written Euclidean, even though it is pronounced euclidian) ---Well Wikipedia writes "Euclidean (or, less commonly, Euclidian)" so maybe both are fine. Excuse this remark then. – Myself Dec 04 '15 at 13:22

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In fact, $A/(x)\simeq\mathbb F_2$ or $A/(x)\simeq\mathbb F_3$. Now use that $\alpha^2-\alpha+5=0$ in $A$, where $\alpha=\frac{1 +i\sqrt{19}}{2}$, and then reduce the equation modulo $(x)$. In $\mathbb F_2$ this leads to $\alpha^2+\alpha+1=0$, and in $\mathbb F_3$ to $\alpha^2-\alpha-1=0$. But there are no elements in these two finite fields which satisfy the given equations.

user26857
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