description/proof of that for ring and set of subrings, intersection of set is subring
Topics
About: ring
The table of contents of this article
Starting Context
- The reader knows a definition of ring.
- The reader knows a definition of intersection of set.
- The reader admits the proposition that for any group, the intersection of any possibly uncountable number of subgroups of the group is a subgroup of the group.
Target Context
- The reader will have a description and a proof of the proposition that for any ring and any set of subrings, the intersection of the set is a subring.
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:
\(R\): \(\in \{\text{ the rings }\}\)
\(B\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{R_\beta \vert \beta \in B\}\): \(R_\beta \in \{\text{ the subrings of } R\}\)
//
Statements:
\(\cap \{R_\beta \vert \beta \in B\} \in \{\text{ the subrings of } R\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(\cap \{R_\beta \vert \beta \in B\}\) is an Abelian group under the addition; Step 2: see that \(\cap \{R_\beta \vert \beta \in B\}\) is a monoid under the multiplication; Step 3: see that the multiplication is distributive with respect to the addition.
Step 1:
Let us see that \(\cap \{R_\beta \vert \beta \in B\}\) is an Abelian group under the addition.
\(\cap \{R_\beta \vert \beta \in B\}\) is a subgroup of \(R\) under the addition, by the proposition that for any group, the intersection of any possibly uncountable number of subgroups of the group is a subgroup of the group.
\(\cap \{R_\beta \vert \beta \in B\}\) is an Abelian group, because for each \(r_1, r_2 \in \cap \{R_\beta \vert \beta \in B\}\), \(r_1 + r_2 = r_2 + r_1\), because \(r_1, r_2 \in R\) and \(r_1 + r_2 = r_2 + r_1\) on \(R\).
Step 2:
Let us see that \(\cap \{R_\beta \vert \beta \in B\}\) is a monoid under the multiplication.
\(\cap \{R_\beta \vert \beta \in B\}\) is closed under the multiplication, because for each \(r_1, r_2 \in \cap \{R_\beta \vert \beta \in B\}\), \(r_1, r_2 \in R_\beta\) for each \(\beta\), so, \(r_1 r_2 \in R_\beta\) for each \(\beta\), and so, \(r_1 r_2 \in \cap \{R_\beta \vert \beta \in B\}\).
For each \(r_1, r_2, r_3 \in \cap \{R_\beta \vert \beta \in B\}\), \((r_1 r_2) r_3 = r_1 (r_2 r_3)\), because it is so on the ambient \(R\).
\(1 \in \cap \{R_\beta \vert \beta \in B\}\), because \(1 \in R_\beta\) for each \(\beta\), and for each \(r \in \cap \{R_\beta \vert \beta \in B\}\), \(1 r = r 1 = r\), because it is so on the ambient \(R\).
Step 3:
Let us see that the multiplication is distributive with respect to the addition.
For each \(r_1, r_2, r_3 \in \cap \{R_\beta \vert \beta \in B\}\), \(r_1 (r_2 + r_3) = (r_1 r_2) + (r_1 r_3)\) and \((r_1 + r_2) r_3 = (r_1 r_3) + (r_2 r_3)\), because it is so on the ambient \(R\).
