Proposition 4

Let $A$ be a ring and $(a_\lambda)_{\lambda \in L}$ a family of left ideals of $A$. The left ideal generated by $\bigcup_{\lambda \in L} a_\lambda$ consists of the sums $\sum_{\lambda \in L} y_\lambda$ where $(y_\lambda)_{\lambda \in L}$ is a family with finite support such that $y_\lambda \in a_\lambda$ for all $\lambda \in L$.

Let $a$ be the set of sums $\sum_{\lambda \in L} y_\lambda$ with $y_\lambda \in a_\lambda$ for all $\lambda \in L$. The formulae $\sum_{\lambda \in L} x_\lambda + \sum_{\lambda \in L} y_\lambda = \sum_{\lambda \in L} (x_\lambda + y_\lambda)$ and $a \cdot \sum_{\lambda \in L} x_\lambda = \sum_{\lambda \in L} a x_\lambda$ shows that $a$ is a left ideal of $A$. Let $\lambda \in L$ and $x \in a_\lambda$; write $y_\lambda = x$ and $y_\mu = 0$ for $\mu \neq \lambda$; then $x = \sum_{\lambda \in L} y_\lambda$, whence $x \in a$ and finally $a_\lambda \subset a$. If a left ideal $a'$ contains $a_\lambda$ for all $\lambda \in L$, it obviously contains $a$ and hence $a$ is generated by $\bigcup_{\lambda \in L} a_\lambda$.

The ideal $a$ generated by $\bigcup_{\lambda \in L} a_\lambda$ is called the sum of the left ideals $a_\lambda$ and is denoted by $\sum_{\lambda \in L} a_\lambda$ (cf. II, § 1, no. 7). In particular, the sum $a_1 + a_2$ of the two left ideals consists of the sums $a_1 + a_2$ where $a_1 \in a_1$ and $a_2 \in a_2$.