653: For Unique Factorization Domain and Finite Subset, if Greatest Common Divisors of Each Pair Subset of Subset Are Unit Associates, Greatest Common Divisors of Subset Are Unit Associates, but Not Vice Versa

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description/proof of that for unique factorization domain and finite subset, if greatest common divisors of each pair subset of subset are unit associates, greatest common divisors of subset are unit associates, but not vice versa

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Topics

About: ring

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Proof

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Starting Context

• The reader knows a definition of unique factorization domain.
• The reader knows a definition of greatest common divisors of subset of commutative ring.
• The reader knows a definition of associates of element of commutative ring.
• The reader knows a definition of representatives set of quotient set.
• The reader admits the proposition that for any unique factorization domain, the method of getting the greatest common divisors of any finite subset by factorizing each element of the subset with the representatives set of the associates quotient set works.

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Target Context

• The reader will have a description and a proof of the proposition that for any unique factorization domain and any finite subset with more than 1 elements, if the greatest common divisors of each pair subset of the subset are the unit associates, the greatest common divisors of the subset are the unit associates, but not vice versa.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$R$: $\in \{\text{ the unique factorization domains }\}$
$U$: $=\{\text{ the units of }R\}$
$I$: $=\{\text{ the irreducible elements of }R\}$
$R/Asc$: $=\text{ the quotient set by the associates equivalence relation }$
$f$: $:R/Asc\to R$ such that $\mathrm{\forall }p\in R/Asc,f(p)\in p$
$\overline{R/Asc-f}$: $=\text{ the representatives set of }R/Asc\text{ by }f$
$S$: $=\{p_1,...,p_n\}$, $\in \{\text{ the finite subsets of }R\}$, with more than 1 elements
$gcd(S)$:
//

Statements:
(
$\mathrm{\forall }{p}_j,p_k\in S\text{ such that }{p}_j\ne p_k(gcd(\{p_j,p_k\})=Asc(1))$
$⟹$
$gcd(S)=Asc(1)$
)
$\wedge$
(
$gcd(S)=Asc(1)$
$\mathrm{\neg }⟹$
)
$\mathrm{\forall }{p}_j,p_k\in S(gcd(\{p_j,p_k\})=Asc(1))$
//

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2: Natural Language Description

For any unique factorization domain, $R$, the set of the units of $R$, $U$, the set of the irreducible elements of $R$, $I$, the quotient set by the associates equivalence relation, $R/Asc$, any map, $f:R/Asc\to R$, such that for each $p\in R/Asc$, $f(p)\in p$, the representatives set of $R/Asc$ by $f$, $\overline{R/Asc-f}$, and any finite subset, $S=\{p_1,...,p_n\}\subseteq R$, with more than 1 elements, if $\mathrm{\forall }{p}_j,p_k\in S\text{ such that }{p}_j\ne p_k(gcd(\{p_j,p_k\})=Asc(1))$, $gcd(S)=Asc(1)$, but not vice versa.

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3: Proof

As this proof is based on the proposition that for any unique factorization domain, the method of getting the greatest common divisors of any finite subset by factorizing each element of the subset with the representatives set of the associates quotient set works, that proposition is supposed to be understood.

Let $I^{\prime }=\{i_1,...,i_l\}$ be the set of some irreducible elements cited in that proposition. The point is that the factorizations of the elements of $S$ with respect to $I^{\prime }$ are unique, because we have chosen the fixed representatives of the associates equivalence classes.

Let us suppose that $\mathrm{\forall }{p}_j,p_k\in S(gcd(\{p_j,p_k\})=Asc(1))$.

While $p_j=u_j{i}_1^{c_{j,1}}...i_l^{c_{j,l}}$, if $0<c_{j,k}$, $c_{s,k}=0$ for each $s\ne j$, because otherwise, $i_k$ would be a common divisor of $\{p_j,p_s\}$ and $1$ would not be any greatest common divisor of $\{p_j,p_s\}$, because $1=q{i}_k$ would not hold, because that would mean that $i_k$ was a unit.

So, $(m_1,...,m_l)=(min(\{c_{j,1}|j\in \{1,...,n\}\}),...,min(\{c_{j,l}|j\in \{1,...,n\}\}))=(0,...,0)$.

So, $d:=i_1^{m_1}...i_l^{m_l}=1$.

So, $gcd(S)=Asc(1)$.

Let us suppose that $gcd(S)=Asc(1)$.

That means that $(m_1,...,m_l)=(min(\{c_{j,1}|j\in \{1,...,n\}\}),...,min(\{c_{j,l}|j\in \{1,...,n\}\}))=(0,...,0)$.

But $min(\{c_{j,s}|j\in \{1,...,n\}\})=0$ does not imply that $min(\{c_{j,s},c_{k,s}\})=0$ for each pair, $j,k$. So, $m_s$ may be positive, which would imply that $d\ne 1$.

As a counterexample, $R=\mathbb{Z}$; $S=\{1,2,4\}$: $gcd(S)=Asc(1)$, but $gcd(\{2,4\})=Asc(2)$.

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References

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