976: For Ring and Set of Subrings, Intersection of Set Is Subring

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description/proof of that for ring and set of subrings, intersection of set is subring

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Topics

About: ring

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The table of contents of this article

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

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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.

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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 subrings, the intersection of the set is a subring.

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

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2: Proof

Whole Strategy: Step 1: see that $\cap \{R_{\beta }|\beta \in B\}$ is an Abelian group under the addition; Step 2: see that $\cap \{R_{\beta }|\beta \in B\}$ 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 $\cap \{R_{\beta }|\beta \in B\}$ is an Abelian group under the addition.

$\cap \{R_{\beta }|\beta \in B\}$ 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.

$\cap \{R_{\beta }|\beta \in B\}$ is an Abelian group, because for each $r_1,r_2\in \cap \{R_{\beta }|\beta \in B\}$, $r_1+r_2=r_2+r_1$, because $r_1,r_2\in R$ and $r_1+r_2=r_2+r_1$ on $R$.

Step 2:

Let us see that $\cap \{R_{\beta }|\beta \in B\}$ is a monoid under the multiplication.

$\cap \{R_{\beta }|\beta \in B\}$ is closed under the multiplication, because for each $r_1,r_2\in \cap \{R_{\beta }|\beta \in B\}$, $r_1,r_2\in R_{\beta }$ for each $\beta$, so, $r_1{r}_2\in R_{\beta }$ for each $\beta$, and so, $r_1{r}_2\in \cap \{R_{\beta }|\beta \in B\}$.

For each $r_1,r_2,r_3\in \cap \{R_{\beta }|\beta \in B\}$, $(r_1{r}_2)r_3=r_1(r_2{r}_3)$, because it is so on the ambient $R$.

$1\in \cap \{R_{\beta }|\beta \in B\}$, because $1\in R_{\beta }$ for each $\beta$, and for each $r\in \cap \{R_{\beta }|\beta \in B\}$, $1r=r1=r$, 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 $r_1,r_2,r_3\in \cap \{R_{\beta }|\beta \in B\}$, $r_1(r_2+r_3)=(r_1{r}_2)+(r_1{r}_3)$ and $(r_1+r_2)r_3=(r_1{r}_3)+(r_2{r}_3)$, because it is so on the ambient $R$.

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References

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