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[parent] properties of a gcd domain (Result)

Let $ D$ be a gcd domain. For any $ a\in D$, denote $ [a]$ the set of all elements in $ D$ that are associates of $ a$, $ \operatorname{GCD}(a,b)$ the set of all gcd's of elements $ a$ and $ b$ in $ D$, and any $ S\subseteq D$, $ mS:=\lbrace ms\mid s\in S\rbrace$. Then

  1. $ \operatorname{GCD}(a,b)=[a]$ iff $ a\mid b$.
  2. $ \operatorname{GCD}(ma,mb)=m\operatorname{GCD}(a,b)$.
  3. If $ \operatorname{GCD}(ab,c)=[1]$, then $ \operatorname{GCD}(a,c)=[1]$
  4. If $ \operatorname{GCD}(a,b)=[1]$ and $ \operatorname{GCD}(a,c)=[1]$, then $ \operatorname{GCD}(a,bc)=[1]$.
  5. If $ \operatorname{GCD}(a,b)=[1]$ and $ a\mid bc$, then $ a\mid c$.
Proof. To aid in the proof of these properties, let us denote, for $ a\in D$ and $ S\subseteq D$, $ a\vert S$ to mean that every element of $ S$ is divisible by $ a$, and $ S\vert a$ to mean that every element in $ S$ divides $ a$. We take the following four steps:
  1. One direction is obvious from the definition. So now suppose $ a\mid b$. Then $ a\mid\operatorname{GCD}(a,b)$. But by definition, $ \operatorname{GCD}(a,b)\mid a$, so $ [a]=\operatorname{GCD}(a,b)$.
  2. Suppose $ d \in \operatorname{GCD}(a,b)$. We want to show that $ md\in \operatorname{GCD}(ma,mb)$. In other words, if $ x\in \operatorname{GCD}(ma,mb)$, we want to show that $ md$ and $ x$ are associates. By assumption, $ d\mid a$ and $ d\mid b$, so $ md\mid ma$ and $ md\mid mb$, which implies that $ md\mid x$. Write $ x=mn$ for some $ n\in D$. Then $ mn\mid ma$ and $ mn\mid mb$ imply that $ n\mid a$ and $ n\mid b$, and therefore $ n\mid d$ since $ d$ is a gcd of $ a$ and $ b$. As a result, $ mn\mid md$, or $ x\mid md$, showing that $ x$ and $ md$ are associates.

    On the other hand, if $ x\in \operatorname{GCD}(ma,mb)$, we want to show that $ x=md$ for some $ d\in \operatorname{GCD}(a,b)$. Pick some $ y\in \operatorname{GCD}(a,b)$, then $ my \in \operatorname{GCD}(ma,mb)$ by the previous paragraph. Therefore $ x$ and $ my$ are associates. Write $ x=mye$ for some unit $ e\in D$. Setting $ d=ye$ gives us the desired gcd of $ a$ and $ b$.

  3. If $ d\mid a$ and $ d\mid c$, then $ d\mid ab$ and $ d\mid c$. So $ d\mid\operatorname{GCD}(ab,c)=[1]$, hence $ d$ is a unit and the result follows.
  4. Suppose $ d\mid a$ and $ d\mid bc$. Then $ d\mid ab$ and $ d\mid bc$ and hence $ d\mid\operatorname{GCD}(ab,bc)=b\operatorname{GCD}(a,c)=[b]$. But $ d\mid a$ also, so $ d\mid\operatorname{GCD}(a,b)=[1]$ and $ d$ is a unit.
  5. $ \operatorname{GCD}(a,b)=[1]$ implies $ [c]=\operatorname{GCD}(ac,bc)$. Now, $ a\mid ac$ and by assumption, $ a\mid bc$. Therefore, $ a\mid\operatorname{GCD}(ac,bc)=[c]$.
$ \qedsymbol$



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Cross-references: unit, implies, obvious, divides, divisible, mean, properties, proof, iff, gcd's, associates, gcd domain
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This is version 1 of properties of a gcd domain, born on 2008-08-12.
Object id is 10936, canonical name is PropertiesOfAGcdDomain.
Accessed 270 times total.

Classification:
AMS MSC13G05 (Commutative rings and algebras :: Integral domains)

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