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Is $R=\mathbb Z [(1+\sqrt{-19})/2]$ a Euclidean domain?

Its Voronoi region seems relatively small, but its hard to have intuition about division with remainder. I predict it is not since the norm of $(1+\sqrt{-19})/2$ is $\sqrt5>1$.

Here is my attempt at a proof: I show that an Euclidean Domain $D$ (with not every nonzero element a unit) has a non-unit element $p$ such that $\forall x\in D$ $p|{x}$ or $p|{x-u}$ for some unit $u\in D$. I now want to reason by trying to show the conditions of the contrapositive are true for $R$. In other words, that $\nexists p\in R$ such that $p|{x}$ or $p|{x-u}\ \ \forall x\in R$ for $p$ nonunital and for some unit $u\in D$.

I'm having difficulty showing this. Any help appreciated.

Matt R
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    This is a hard nut, imo: read the last example in 8.1 (and the whole section, in fact) of Dummit&Foote's "Algebra": this ring is a PID yet not an Euclidean domain since it does have "universal side divisors". – DonAntonio Dec 08 '13 at 04:52
  • @DonAntonio OK. I have it handy...thanks for the tip. – Matt R Dec 08 '13 at 05:04
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    Here is a nice sketch of proof, I may fulfill the details on demand http://www.maths.qmul.ac.uk/~raw/MTH5100/PIDnotED.pdf – Stephen Dedalus Mar 09 '14 at 01:10

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