description/proof of that for ring and finite number of ideals, sum of ideals is ideal
Topics
About: ring
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
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.
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.
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:
\(R\): \(\in \{\text{ the rings }\}\)
\(\{I_{l, 1}, ..., I_{l, n}\}\): \(\subseteq \{\text{ the left ideals of } R\}\)
\(\{I_{r, 1}, ..., I_{r, n}\}\): \(\subseteq \{\text{ the right ideals of } R\}\)
\(\{I_{b, 1}, ..., I_{b, n}\}\): \(\subseteq \{\text{ the both-sided ideals of } R\}\)
//
Statements:
\(I_{l, 1} + ... + I_{l, n} \in \{\text{ the left ideals of } R\}\)
\(\land\)
\(I_{r, 1} + ... + I_{r, n} \in \{\text{ the right ideals of } R\}\)
\(\land\)
\(I_{b, 1} + ... + I_{b, n} \in \{\text{ the both-sided ideals of } R\}\).
//
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 \(R\).
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 \(R I_{l, j} = I_{l_j}\), \(r_j p_j = p'_j = 1 p'_j\), so, \(p = r_1 p_1 + ... + r_n p_n = 1 p'_1 + ... + 1 p'_n = 1 (p'_1 + ... + p'_n) \in R (I_{l, 1} + ... + I_{l, n})\).
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 \(I_{r, j} R = I_{r_j}\), \(p_j r_j = p'_j = p'_j 1\), so, \(p = p_1 r_1 + ... + p_n r_n = p'_1 1 + ... + p'_n 1 = (p'_1 + ... + p'_n) 1 \in (I_{r, 1} + ... + I_{r, n}) R\).
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 \(R\).
