description/proof of that for ring and set of subfields, intersection of set is subfield
Topics
About: ring
About: field
The table of contents of this article
Starting Context
- The reader knows a definition of ring.
- The reader knows a definition of field.
- The reader knows a definition of intersection of set.
- The reader admits the proposition that for any ring and any set of subrings, the intersection of the set is a subring.
- The reader admits the proposition that for any ring, if an element has an inverse, the inverse is unique.
Target Context
- The reader will have a description and a proof of the proposition that for any ring and any set of subfields, the intersection of the set is a subfield.
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:
\(R\): \(\in \{\text{ the rings }\}\)
\(B\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{F_\beta \vert \beta \in B\}\): \(F_\beta \in \{\text{ the subfields of } R\}\)
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Statements:
\(\cap \{F_\beta \vert \beta \in B\} \in \{\text{ the subfields of } R\}\)
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2: Note
A point is that \(R\) does not need to be any field.
But \(R\) is allowed to be a field for this proposition.
3: Proof
Whole Strategy: Step 1: see that \(\cap \{F_\beta \vert \beta \in B\}\) is a subring of \(R\); Step 2: see that \(\cap \{F_\beta \vert \beta \in B\}\) is a commutative ring; Step 3: see that each element of \(\cap \{F_\beta \vert \beta \in B\}\) has an inverse.
Step 1:
\(\cap \{F_\beta \vert \beta \in B\}\) is a subring of \(R\), by the proposition that for any ring and any set of subrings, the intersection of the set is a subring: each \(F_\beta\) is a subring of \(R\).
Step 2:
Let us see that \(\cap \{F_\beta \vert \beta \in B\}\) is a commutative ring.
For each \(r_1, r_2 \in \cap \{F_\beta \vert \beta \in B\}\), \(r_1, r_2 \in F_\beta\) for each \(\beta\), so, \(r_1 r_2 = r_2 r_1\), because \(F_\beta\) is a field.
Step 3:
Let us see that each element of \(\cap \{F_\beta \vert \beta \in B\}\) has an inverse.
For each \(r \in \cap \{F_\beta \vert \beta \in B\}\), \(r \in F_\beta\) for each \(\beta\), so, the inverse, \(r_\beta\) (we cannot rule out the possibility that it depends on \(\beta\), at this point), is on \(F_\beta\) for each \(\beta\), but the inverse is an inverse also on \(R\), because \(r r_\beta = r_\beta r = 1\) holds on \(R\), and it is unique on \(R\), by the proposition that for any ring, if an element has an inverse, the inverse is unique; so, the \(r_\beta\) s are in fact the same, which can be denoted as \(r^{-1}\), so, \(r^{-1} \in F_\beta\) for each \(\beta\) and \(r^{-1} \in \cap \{F_\beta \vert \beta \in B\}\).
