generated ideal of a union and sum

1. Proposition

Let \(R\) be a ring, \(A_i \subseteq R\). Then

\begin{align*} (\bigcup A_i) = \sum (A_i) \end{align*}

2. Proof

2.1. \(\subseteq\)

Let \(a \in \sum (A_i)\), then there exists \(r_j\) and \(b_{j,i} \in (A_i)\) such that

\begin{align*} \sum_{j=1}^{n} r_j b_{j,i} = a \end{align*}

Furthremore, each \(b_{j,i}\) is of the form

\begin{align*} \sum_{l=1}^{m} r_l a_{l,j,i} \end{align*}

for \(r_l= \in R, a_{l,j,i} \in A_i\) Therefore \(a\) is also a linear kombination of elements from \(\bigcup A_i\)

2.2. \(\supseteq\)

Let \(a = \sum_{j=1}^{n} r_i a_j \in (\bigcup A_i)\), then each \(a_j\) is an element in an \(A_i\) for suitable \(i \in I\) Therefore \(a \in \sum (A_i)\)