2024-06-30

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

Topics


About: ring

The table of contents of this article


Starting Context



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.

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.


Main Body


1: Structured Description


Here is the rules of Structured Description.

Entities:
\(R\): \(\in \{\text{ the principal integral domains }\}\)
//

Statements:
\(R \in \{\text{ the greatest common divisors domains }\}\)
\(\land\)
\(\forall \{p_1, p_2\} \subseteq R (\forall d \in gcd (\{p_1, p_2\}) (d R = p_1 R + p_2 R))\)
//

As an immediate corollary, \(\forall \{p_1, p_2\} \subseteq R (gcd (\{p_1, p_2\}) \subseteq p_1 R + p_2 R)\).

As another corollary, \(\forall \{p_1, p_2\} \subseteq R (gcd (\{p_1, p_2\}) = Asc (1) \iff 1 \in p_1 R + p_2 R \iff p_1 R + p_2 R = R)\).


2: Natural Language Description


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


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 \(p_1 R + p_2 R = d R\).

\(p_j \in d R\), because \(p_1 = p_1 1 + p_2 0 \in p_1 R + p_2 R = d R\) 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 \(\{p_1, p_2\}\), \(d'\), \(p_j = q'_j d'\), but as \(d = 1 d \in d R = p_1 R + p_2 R\), \(d = p_1 r_1 + p_2 r_2 = q'_1 d' r_1 + q'_2 d' r_2 = (q'_1 r_1 + q'_2 r_2) d'\), which means that \(d\) is a greatest common divisor.

So, \(gcd (\{p_1, p_2\}) \neq \emptyset\) and \(R\) is a greatest common divisors domain.

For each \(d' \in gcd (\{p_1, p_2\})\), \(d' = u d\) 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, \(d' R = d R\), because for each \(p \in d' R\), \(p = d' r = u d r = d u r \in d R\); for each \(p \in d R\), \(p = d r = u^{-1} d' r = d' u^{-1} r \in d' R\). So, \(d' R = d R = p_1 R + p_2 R\).

For each \(d \in gcd (\{p_1, p_2\}\), \(d = d 1 \in d R = p_1 R + p_2 R\).

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 \(1 \in p_1 R + p_2 R\).

\(1 \in p_1 R + p_2 R = d R\), which implies that \(1 = r d\) 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.

\(1 \in p_1 R + p_2 R \iff p_1 R + p_2 R = R\) 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 \(R = R 1 \subseteq R I = I\) while \(I \subseteq R\); if \(I = R\), \(1 \in R = I\).


References


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