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单词 ProofOfBezoutsTheorem
释义

proof of Bezout’s Theorem


Let D be an integral domain with an Euclidean valuation. Let a,bD not both 0. Let (a,b)={ax+by|x,yD}. (a,b) is an ideal in D{0}. We choose d(a,b) such that μ(d) is the smallest positive value. Then (a,b) is generated by d and has the property d|a and d|b. Two elements x and y in D are associateMathworldPlanetmath if and only if μ(x)=μ(y). So d is unique up to a unit in D. Hence d is the greatest common divisorMathworldPlanetmathPlanetmath of a and b.

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