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[parent] uniqueness of division algorithm in Euclidean domain (Theorem)

Theorem. Let $a,\,b$ be non-zero elements of a Euclidean domain $D$ with the Euclidean valuation $\nu$ . The incomplete quotient $q$ and the remainder $r$ of the division algorithm $$a = qb+r \quad \mbox{where} \quad r = 0 \;\; \mbox{or}\;\; \nu(r) < \nu(b)$$ are unique if and only if

$\displaystyle \nu(a+b) \leqq \max\{\nu(a),\,\nu(b)\}.$ (1)

Proof. Assume first (1) for the elements $a,\,b$ of $D$ . If we had
\begin{align*}\begin{cases}a = qb+r \quad \mbox{with} \quad r = 0 \;\;\;\;\lor\;... ...d \mbox{with} \quad r' = 0 \;\;\lor\;\; \nu(r') < \nu(b) \end{cases}\end{align*}    

and $r' \neq r$ , $q' \neq q$ , then the properties of the Euclidean valuation and the assumption yield the chain of inequalities $$\nu(b) \leqq \nu((q'-q)b) = \nu(r'-r) \leqq \max\{\nu(r'),\,\nu(-r)\} < \nu(b)$$ which is impossible. We must infer that $r'-r = 0$ or $q'-q = 0$ . But these two conditions are equivalent. Thus the division algorithm is unique.

Conversely, assume that (1) is not true for non-zero elements $a,\,b$ of $D$ , i.e. $$\nu(a+b) > \max\{\nu(a),\,\nu(b)\}.$$ Then we obtain two repsesentations $$b = 0(a+b)+b = 1(a+b)-a$$ where $\nu(b) < \nu(a+b)$ and $\nu(-a) = \nu(a) < \nu(a+b)$ . Thus the incomplete quotient and the remainder are not unique.




"uniqueness of division algorithm in Euclidean domain" is owned by pahio.
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See Also: Krull valuation, quotient

Also defines:  incomplete quotient

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Cross-references: conversely, inequalities, proof, division algorithm, remainder, Euclidean valuation, Euclidean domain, theorem

This is version 4 of uniqueness of division algorithm in Euclidean domain, born on 2008-03-06, modified 2008-03-08.
Object id is 10366, canonical name is UniquenessOfDivisionAlgorithmInEuclideanDomain.
Accessed 1070 times total.

Classification:
AMS MSC13F07 (Commutative rings and algebras :: Arithmetic rings and other special rings :: Euclidean rings and generalizations)

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