660: For Principal Integral Domain and Finite Subset, Sum of Principal Ideals by Elements of Subset Is Principal Ideal by Any of Greatest Common Divisors of Subset

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description/proof of that for principal integral domain and finite subset, sum of principal ideals by elements of subset is principal ideal by any of greatest common divisors of subset

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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 of subset of commutative ring.
• 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.
• The reader admits the proposition that for any integral domain, if any principal ideal by any element is also by any another element, the elements are associates with each other, and the principal ideal is by any associate.

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Target Context

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

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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:
$R$: $\in \{\text{ the principal integral domains }\}$
$S$: $=\{p_1,...,p_n\}\subseteq R$
$p_{1}R+...+p_{n}R$:
$gcd(S)$:
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Statements:
$p_{1}R+...+p_{n}R\in \{\text{ the principal ideals of }R\}$
$\wedge$
$\mathrm{\forall }d\in gcd(S)(p_{1}R+...+p_{n}R=dR)$
//

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2: Natural Language Description

For any principal integral domain, $R$, and any subset, $S:=\{p_1,...,p_n\}\subseteq R$, $p_{1}R+...+p_{n}R$ is the principal ideal by any of $gcd(S)$.

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3: Proof

Each $p_{j}R$ is an (principal) ideal, and $p_{1}R+...+p_{n}R$ is an ideal, by the proposition that for any ring and any finite number of ideals, the sum of the ideals is an ideal.

$p_{1}R+...+p_{n}R$ is a principal ideal, because $R$ is a principal integral domain. So, there is a $d\in R$ such that $p_{1}R+...+p_{n}R=dR$.

$p_j=p_{j}1\in p_{1}R+...+p_{n}R=dR$, which implies that $p_j=q_{j}d$. So, $d$ is a common divisor of $S$.

For each common divisor of $S$, $d^{\prime }$, $p_j=q_j^{\prime }{d}^{\prime }$. As $d=d1\in dR=p_{1}R+...+p_{n}R$, $d=p_1{r}_1+...+p_n{r}_n=q_1^{\prime }{d}^{\prime }{r}_1+...+q_n^{\prime }{d}^{\prime }{r}_n=(q_1^{\prime }{r}_1+...+q_n^{\prime }{r}_n)d^{\prime }$. So, $d$ is a greatest common divisor of $S$.

For each $d^{\prime }\in gcd(S)$, $d^{\prime }=ud$ 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. $d^{\prime }R=dR$, by the proposition that for any integral domain, if any principal ideal by any element is also by any another element, the elements are associates with each other, and the principal ideal is by any associate. So, $d^{\prime }R=dR=p_{1}R+...+p_{n}R$.

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References

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