description/proof of that for commutative ring, if each elements pair has greatest common divisor, each finite subset has greatest common divisor, which can be gotten sequentially
Topics
About: ring
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Note
- 4: Proof
Starting Context
- The reader knows a definition of ring.
- The reader knows a definition of greatest common divisors of subset of commutative ring.
- 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.
Target Context
- The reader will have a description and a proof of the proposition that for any commutative ring, if each elements pair has a greatest common divisor, each finite subset has a greatest common divisor, which can be gotten sequentially.
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:
\(R\): \(\in \{\text{ the commutative rings }\}\)
//
Statements:
\(\forall \{p_1, p_2\} \subseteq R (gcd (\{p_1, p_2\}) \neq \emptyset)\)
\(\implies\)
(
\(\forall S = \{p_1, ..., p_n\} \in \{\text{ the finite subsets of R }\} (gcd (S) \neq \emptyset)\)
\(\land\)
\(gcd (\{p_1, gcd (\{p_2, gcd (\{p_3, ... gcd (\{p_{n - 1}, p_n\}) ... \})\})\}) \subseteq gcd (\{p_1, ..., p_n\})\)
)
//
\(gcd (\{p_1, gcd (\{p_2, gcd (\{p_3, ... gcd (\{p_{n - 1}, p_n\}) ... \})\})\}) \subseteq gcd (\{p_1, ..., p_n\})\) means that if we take a greatest common divisor of \(\{p_{n - 1}, p_{n}\}\) as \(d_{n - 1}\), then, take a greatest common divisor of \(\{p_{n - 2}, d_{n - 1}\}\) as \(d_{n - 2}\), then, ..., and so on, then a greatest common divisor of \(\{p_1, d_2\}\) is a common divisor of \(S\).
This proposition does not claim \(gcd (\{p_1, gcd (\{p_2, gcd (\{p_3, ... gcd (\{p_{n - 1}, p_n\}) ... \})\})\}) = gcd (\{p_1, ..., p_n\})\), but when \(R\) is an integral domain, it holds, because while there is a \(d \in gcd (\{p_1, gcd (\{p_2, gcd (\{p_3, ... gcd (\{p_{n - 1}, p_n\}) ... \})\})\})\), for each \(d' \in gcd (\{p_1, ..., p_n\})\), \(d' = u d\), 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, and \(d' \in gcd (\{p_1, gcd (\{p_2, gcd (\{p_3, ... gcd (\{p_{n - 1}, p_n\}) ... \})\})\})\), 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.
2: Natural Language Description
For any commutative ring, \(R\), if for each \(\{p_1, p_2\} \subseteq R\), \(gcd (\{p_1, p_2\}) \neq \emptyset)\), for each finite subset, \(S = \{p_1, ..., p_n\} \subseteq R\), \(gcd (S) \neq \emptyset\), and \(gcd (\{p_1, gcd (\{p_2, gcd (\{p_3, ... gcd (\{p_{n - 1}, p_n\}) ... \})\})\}) \subseteq gcd (\{p_1, ..., p_n\})\).
3: Note
This proposition does not suppose that \(R\) is a greatest common divisors domain (\(R\) can be a non-integral-domain), but when \(R\) is an integral domain, \(R\) is a greatest common divisors domain, and this proposition holds for any greatest common divisors domain.
4: Proof
Let us prove it inductively with respect to \(n\).
For \(n = 2\), it obviously holds.
Let us suppose that it holds through \(n = n'\).
Let us think of \(S = \{p_1, ..., p_{n' + 1}\}\).
There is a \(d \in gcd (\{p_2, ..., p_{n' + 1}\})\). \(p_j = q_j d\) for each \(j \in \{2, ..., n' + 1\}\).
There is a \(d' \in gcd (\{g_1, d\})\). \(p_1 = q'_1 d'\) and \(d = q' d'\).
\(p_j = q_j q' d'\) for each \(j \in \{2, ..., n' + 1\}\). So, \(d'\) is a common divisor of \(S = \{p_1, ..., p_{n' + 1}\}\).
Let \(d''\) be any common divisor of \(S\). \(p_1 = q''_1 d''\) and \(p_j = q''_j d''\) for each \(j \in \{2, ..., n' + 1\}\). \(d''\) is a common divisor of \(\{p_2, ..., p_{n' + 1}\}\). So, \(d = q'' d''\).
So, \(d''\) is a common divisor of \(\{p_1, d\}\). So, \(d' = q''' d''\).
So, \(d'\) is a greatest common divisor of \(S\).
So, by the induction principle, each \(S\) for each \(n\) has a greatest common divisor.
We have seen that \(gcd (\{p_1, gcd (\{p_2, ..., p_n\})\}) \subseteq gcd (\{p_1, ..., p_n\})\).
As \(gcd (\{p_2, gcd (\{p_3, ..., p_n\})\}) \subseteq gcd (\{p_2, ..., p_n\})\) and so on, \(gcd (\{p_1, gcd (\{p_2, gcd (\{p_3, ..., p_n\})\})\}) \subseteq gcd (\{p_1, ..., p_n\})\) and so on, and after all, \(gcd (\{p_1, gcd (\{p_2, gcd (\{p_3, ... gcd (\{p_{n - 1}, p_n\}) ... \})\})\}) \subseteq gcd (\{p_1, ..., p_n\})\).
