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[parent] proof that a Euclidean domain is a PID (Result)

Let $ D$ be a Euclidean domain, and let $ \mathfrak{a} \subseteq D$ be a nonzero ideal. We show that $ \mathfrak{a}$ is principal. Let

$\displaystyle A = \{\nu(x) : x \in \mathfrak{a}, x \neq 0\} $
be the set of Euclidean valuations of the non-zero elements of $ \mathfrak{a}$. Since $ A$ is a non-empty set of non-negative integers, it has a minimum $ m$. Choose $ d\in \mathfrak{a}$ such that $ \nu(d) = m$. Claim that $ \mathfrak{a} = (d)$. Clearly $ (d) \subseteq \mathfrak{a}$. To see the reverse inclusion, choose $ x\in \mathfrak{a}$. Since $ D$ is a Euclidean domain, there exist elements $ y,r\in D$ such that
$\displaystyle x = yd + r $
with $ \nu(r) < \nu(d)$ or $ r = 0$. Since $ r \in \mathfrak{a}$ and $ \nu(d)$ is minimal in $ A$, we must have $ r = 0$. Thus $ d \lvert x$ and $ x\in(d)$.



"proof that a Euclidean domain is a PID" is owned by rm50. [ full author list (2) | owner history (1) ]
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See Also: PID, UFD, integral domain, Euclidean valuation


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Cross-references: minimal, inclusion, integers, Euclidean valuations, ideal, Euclidean domain

This is version 4 of proof that a Euclidean domain is a PID, born on 2002-06-03, modified 2008-03-23.
Object id is 3015, canonical name is ProofThatAnEuclideanDomainIsAPID.
Accessed 2939 times total.

Classification:
AMS MSC: 13F07 (Commutative rings and algebras :: Arithmetic rings and other special rings :: Euclidean rings and generalizations)
Pending Errata and Addenda
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