656: For Ring and Finite Number of Ideals, Sum of Ideals Is Ideal

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description/proof of that for ring and finite number of ideals, sum of ideals is ideal

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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: Natural Language Description
• 3: Proof

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Starting Context

• The reader knows a definition of ring.
• The reader knows a definition of ideal of ring.
• The reader admits the proposition that for any group, the product of any finite number of normal subgroups is commutative and is a normal subgroup.

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Target Context

• The reader will have a description and a proof of the proposition that for any ring and any finite number of ideals, the sum of the ideals is an ideal.

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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:
$R$: $\in \{\text{ the rings }\}$
$\{I_{l,1},...,I_{l,n}\}$: $\subseteq \{\text{ the left ideals of }R\}$
$\{I_{r,1},...,I_{r,n}\}$: $\subseteq \{\text{ the right ideals of }R\}$
$\{I_{b,1},...,I_{b,n}\}$: $\subseteq \{\text{ the both-sided ideals of }R\}$
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Statements:
$I_{l,1}+...+I_{l,n}\in \{\text{ the left ideals of }R\}$
$\wedge$
$I_{r,1}+...+I_{r,n}\in \{\text{ the right ideals of }R\}$
$\wedge$
$I_{b,1}+...+I_{b,n}\in \{\text{ the both-sided ideals of }R\}$.
//

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2: Natural Language Description

For any ring, $R$, any left ideals of $R$, $I_{l,1},...,I_{l,n}$, any right ideals of $R$, $I_{r,1},...,I_{r,n}$, and any both-sided ideals of $R$, $I_{b,1},...,I_{b,n}$, $I_{l,1}+...+I_{l,n}$ is a left ideal of $R$, $I_{r,1}+...+I_{r,n}$ is a right ideal of $R$, and $I_{b,1}+...+I_{b,n}$ is a both-sided ideal of $R$.

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3: Proof

$I_{l,1}+...+I_{l,n}$ is an additive subgroup, by the proposition that for any group, the product of any finite number of normal subgroups is commutative and is a normal subgroup: $R$ is an Abelian group with respect to the addition; $I_{l,j}$ is a normal subgroup of $R$, because any subgroup of any Abelian group is a normal subgroup; $I_{l,1}+...+I_{l,n}$ is in fact a product mentioned in that proposition (while the notation uses "+", it is $I_{l,1}...I_{l,n}$ with the multiplication notation).

$R(I_{l,1}+...+I_{l,n})=I_{l,1}+...+I_{l,n}$?

Let us see that $R(I_{l,1}+...+I_{l,n})=R{I}_{l,1}+...+R{I}_{l,n}$. For each $p\in R(I_{l,1}+...+I_{l,n})$, $p=r(p_1+...+p_n)=r{p}_1+...+r{p}_n\in R{I}_{l,1}+...+R{I}_{l,n}$. For each $p\in R{I}_{l,1}+...+R{I}_{l,n}$, $p=r_1{p}_1+...+r_n{p}_n$, but as $R{I}_{l,j}=I_{l_j}$, $r_j{p}_j=p_j^{\prime }=1p_j^{\prime }$, so, $p=r_1{p}_1+...+r_n{p}_n=1p_1^{\prime }+...+1p_n^{\prime }=1(p_1^{\prime }+...+p_n^{\prime })\in R(I_{l,1}+...+I_{l,n})$.

So, $R(I_{l,1}+...+I_{l,n})=R{I}_{l,1}+...+R{I}_{l,n}=I_{l,1}+...+I_{l,n}$.

So, $I_{l,1}+...+I_{l,n}$ is a left ideal of $R$.

$I_{r,1}+...+I_{r,n}$ is an additive subgroup, by the proposition that for any group, the product of any finite number of normal subgroups is commutative and is a normal subgroup.

$(I_{r,1}+...+I_{r,n})R=I_{r,1}+...+I_{r,n}$?

Let us see that $(I_{r,1}+...+I_{r,n})R=I_{r,1}R+...+I_{r,n}R$. For each $p\in (I_{r,1}+...+I_{r,n})R$, $p=(p_1+...+p_n)r=p_{1}r+...+p_{n}r\in I_{r,1}R+...+I_{r,n}R$. For each $p\in I_{r,1}R+...+I_{r,n}R$, $p=p_1{r}_1+...+p_n{r}_n$, but as $I_{r,j}R=I_{r_j}$, $p_j{r}_j=p_j^{\prime }=p_j^{\prime }1$, so, $p=p_1{r}_1+...+p_n{r}_n=p_1^{\prime }1+...+p_n^{\prime }1=(p_1^{\prime }+...+p_n^{\prime })1\in (I_{r,1}+...+I_{r,n})R$.

So, $(I_{r,1}+...+I_{r,n})R=I_{r,1}R+...+I_{r,n}R=I_{r,1}+...+I_{r,n}$.

So, $I_{r,1}+...+I_{r,n}$ is a right ideal of $R$.

As $I_{b,j}$ is a left ideal, $I_{b,1}+...+I_{b,n}$ is a left ideal by the above argument. So, $I_{b,1}+...+I_{b,n}$ is an additive subgroup of $R$, and $R(I_{b,1}+...+I_{b,n})=I_{b,1}+...+I_{b,n}$.

As $I_{b,j}$ is a right ideal, $I_{b,1}+...+I_{b,n}$ is a right ideal by the above argument. So, $(I_{b,1}+...+I_{b,n})R=I_{b,1}+...+I_{b,n}$.

So, $I_{b,1}+...+I_{b,n}$ is a both-sided ideal of $R$.

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References

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