657: Principal Integral Domain Is Greatest Common Divisors Domain, and for 2 Elements, Each of Greatest Common Divisors Is One by Which Sum of Principal Ideals by 2 Elements Is Principal Ideal

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description/proof of that principal integral domain is greatest common divisors domain, and for 2 elements, each of greatest common divisors is one by which sum of principal ideals by 2 elements is principal ideal

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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 principal integral domain.
• The reader knows a definition of greatest common divisors domain.
• The reader admits the proposition that for any ring and any finite number of ideals, the sum of the ideals is an ideal.
• The reader admits the proposition that for any integral domain and any subset, if the greatest common divisors of the subset exist, they are the associates of a greatest common divisor.

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Target Context

• The reader will have a description and a proof of the proposition that any principal integral domain is a greatest common divisors domain, and for each 2 elements on the principal integral domain, each of the greatest common divisors of the 2 elements is a one by which the sum of the principal ideals by the 2 elements is the principal ideal.

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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:
$R$: $\in \{\text{ the principal integral domains }\}$
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Statements:
$R\in \{\text{ the greatest common divisors domains }\}$
$\wedge$
$\mathrm{\forall }\{p_1,p_2\}\subseteq R(\mathrm{\forall }d\in gcd(\{p_1,p_2\})(dR=p_{1}R+p_{2}R))$
//

As an immediate corollary, $\mathrm{\forall }\{p_1,p_2\}\subseteq R(gcd(\{p_1,p_2\})\subseteq p_{1}R+p_{2}R)$.

As another corollary, $\mathrm{\forall }\{p_1,p_2\}\subseteq R(gcd(\{p_1,p_2\})=Asc(1)⟺1\in p_{1}R+p_{2}R⟺p_{1}R+p_{2}R=R)$.

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2: Natural Language Description

For any principal integral domain, $R$, $R$ is a greatest common divisors domain, and for each subset, $\{p_1,p_2\}\subseteq R$, $\mathrm{\forall }\{p_1,p_2\}\subseteq R(\mathrm{\forall }d\in gcd(\{p_1,p_2\})(dR=p_{1}R+p_{2}R))$. As an immediate corollary, $\mathrm{\forall }\{p_1,p_2\}\subseteq R(gcd(\{p_1,p_2\})\subseteq p_{1}R+p_{2}R)$. As another corollary, $\mathrm{\forall }\{p_1,p_2\}\subseteq R(gcd(\{p_1,p_2\})=Asc(1)⟺1\in p_{1}R+p_{2}R⟺p_{1}R+p_{2}R=R)$.

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3: Proof

$p_{1}R+p_{2}R$ is an ideal, by the proposition that for any ring and any finite number of ideals, the sum of the ideals is an ideal, while each $p_{j}R$ is a principal ideal.

As $R$ is a principal integral domain, there is a $d\in R$ such that $p_{1}R+p_{2}R=dR$.

$p_j\in dR$, because $p_1=p_{1}1+p_{2}0\in p_{1}R+p_{2}R=dR$ and likewise for $p_2$, and $p_j=r_{j}d$ for a $r_j\in R$, so, $d$ is a common divisor of $\{p_1,p_2\}$.

For each common divisor of $\{p_1,p_2\}$, $d^{\prime }$, $p_j=q_j^{\prime }{d}^{\prime }$, but as $d=1d\in dR=p_{1}R+p_{2}R$, $d=p_1{r}_1+p_2{r}_2=q_1^{\prime }{d}^{\prime }{r}_1+q_2^{\prime }{d}^{\prime }{r}_2=(q_1^{\prime }{r}_1+q_2^{\prime }{r}_2)d^{\prime }$, which means that $d$ is a greatest common divisor.

So, $gcd(\{p_1,p_2\})\ne \mathrm{\varnothing }$ and $R$ is a greatest common divisors domain.

For each $d^{\prime }\in gcd(\{p_1,p_2\})$, $d^{\prime }=ud$ for a unit, $u$, by the proposition that for any integral domain and any subset, if the greatest common divisors of the subset exist, they are the associates of a greatest common divisor. So, $d^{\prime }R=dR$, because for each $p\in d^{\prime }R$, $p=d^{\prime }r=udr=dur\in dR$; for each $p\in dR$, $p=dr=u^{-1}{d}^{\prime }r=d^{\prime }{u}^{-1}r\in d^{\prime }R$. So, $d^{\prime }R=dR=p_{1}R+p_{2}R$.

For each $d\in gcd(\{p_1,p_2\}$, $d=d1\in dR=p_{1}R+p_{2}R$.

So, $gcd(\{p_1,p_2\})\subseteq p_{1}R+p_{2}R$.

Let us suppose that $gcd(\{p_1,p_2\})=Asc(1)$.

$1\in Asc(1)=gcd(\{p_1,p_2\})\subseteq p_{1}R+p_{2}R$.

Let us suppose that $1\in p_{1}R+p_{2}R$.

$1\in p_{1}R+p_{2}R=dR$, which implies that $1=rd$ for an $r\in R$, which implies that $d$ is a unit, so, $gcd(\{p_1,p_2\})=Asc(1)$, by the proposition that for any integral domain and any subset, if the greatest common divisors of the subset exist, they are the associates of a greatest common divisor.

$1\in p_{1}R+p_{2}R⟺p_{1}R+p_{2}R=R$ is a common fact that comes from that $p_{1}R+p_{2}R$ is an ideal: if $1\in I$ for any ideal, $I$, of $R$, $I=R$, because $R=R1\subseteq RI=I$ while $I\subseteq R$; if $I=R$, $1\in R=I$.

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References

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