978: For Ring and Set of Subfields, Intersection of Set Is Subfield

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description/proof of that for ring and set of subfields, intersection of set is subfield

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Topics

About: ring
About: field

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note
• 3: Proof

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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.

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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.

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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:
$R$: $\in \{\text{ the rings }\}$
$B$: $\in \{\text{ the possibly uncountable index sets }\}$
$\{F_{\beta }|\beta \in B\}$: $F_{\beta }\in \{\text{ the subfields of }R\}$
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Statements:
$\cap \{F_{\beta }|\beta \in B\}\in \{\text{ the subfields of }R\}$
//

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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.

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3: Proof

Whole Strategy: Step 1: see that $\cap \{F_{\beta }|\beta \in B\}$ is a subring of $R$; Step 2: see that $\cap \{F_{\beta }|\beta \in B\}$ is a commutative ring; Step 3: see that each element of $\cap \{F_{\beta }|\beta \in B\}$ has an inverse.

Step 1:

$\cap \{F_{\beta }|\beta \in B\}$ 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 $\cap \{F_{\beta }|\beta \in B\}$ is a commutative ring.

For each $r_1,r_2\in \cap \{F_{\beta }|\beta \in B\}$, $r_1,r_2\in F_{\beta }$ 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 $\cap \{F_{\beta }|\beta \in B\}$ has an inverse.

For each $r\in \cap \{F_{\beta }|\beta \in B\}$, $r\in F_{\beta }$ 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 $r^{-1}\in \cap \{F_{\beta }|\beta \in B\}$.

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References

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