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I cite from Wikipedia: Commutative rings ⊃ integral domains ⊃ integrally closed domains ⊃ unique factorization domains ⊃ principal ideal domains ⊃ Euclidean domains ⊃ fields

All of these properties have a well-known geometric interpretation, except perhaps "(norm) Euclidean domain". Do you know one?

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    $\begingroup$ Related: mathoverflow.net/questions/15354/… $\endgroup$ – Andrew Stout Oct 17 '13 at 16:15
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    $\begingroup$ A related question: What is the geometric content of Bezout's identity? $\endgroup$ – Matthias Ludewig Oct 17 '13 at 16:35
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    $\begingroup$ this appears to me to be a duplicate question with a duplicate accepted answer. the paper quoted below is exactly the same as the paper quoted in the link I posted. although the answer here is more precise perhaps they should be merged. $\endgroup$ – Andrew Stout Oct 18 '13 at 12:37
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It is a theorem of Pierre Samuel that if $R$ is normal and of finite type over a field $k$, then the following are equivalent:

1) The map taking $x$ to ($1$ plus the $k$-dimension of $R/(x)$) is a degree map that makes $R$ euclidean.

2) $R$ is a PID and there exists a fractional ideal whose divisor has odd degree at infinity.

The relevant paper is from the Journal of Algebra, 1971.

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Roughly speaking near every element of the field of fractions of a norm euclidean domain you have an element of the euclidean domain. More explicitly for every $q$ in the field of fractions you have an element $m$ in the euclidean domain such that $N(q-m)<1$.

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    $\begingroup$ To cement this image, it helps to recall how one typically proves that the Gaussian integers form a Euclidean domain: the key is to show that the ball of radius $1$ (for the algebraic norm on $\mathbb{Q}[i]$) about any point in $\mathbb{Q}[i]$ contains a lattice point in $\mathbb{Z}[i]$, since the furthest you can get from a lattice point is $\sqrt{2}/2$. This really is a thoroughgoingly geometric image, isn't it? $\endgroup$ – Todd Trimble Oct 17 '13 at 17:26
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    $\begingroup$ @ToddTrimble I get the feeling the OP was asking for "geometric" in the sense of $\operatorname{Spec}$. For instance, a reduced ring $A$ is an integral domain if and only if $\operatorname{Spec} A$ is irreducible. $\endgroup$ – Zhen Lin Oct 17 '13 at 20:36
  • $\begingroup$ @ZhenLin Actually, I also got that feeling. But Yazdegerd III's response still seems worthwhile to me. $\endgroup$ – Todd Trimble Oct 17 '13 at 20:43
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Although this isn't "if and only if", one property of Euclidean domains $R$ which is not true of all PID's is that the special Whitehead group $SK_1(R)$ is trivial.

This is Theorem 2.3.2 of Rosenberg's book "Algebraic K-theory and its applications".

I leave it to others to decide whether this property is geometric.

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  • $\begingroup$ This is just a fancy way of saying that euclidean domains, well, admit an euclidean algorithm (applied to the entries of a matrix). :) I don't see any geometric content. $\endgroup$ – Martin Brandenburg Oct 24 '13 at 23:03

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