Let \(R\) be a ring, let \(I\le R\) be a left ideal, let \(A\) be a left \(R\)-module, and let \(S\subseteq A\) be nonempty. Define \[ IS=\left\{\sum_{i=1}^n r_i a_i:n\ge1,\ r_i\in I,\ a_i\in S\right\}. \] Show that \(IS\) is a submodule of \(A\).
Solution
Since \(S\) is nonempty, choose \(a\in S\). Then \[ 0=0\cdot a\in IS, \] so \(IS\) is nonempty.
Let \[ x=\sum_{i=1}^n r_i a_i, \qquad y=\sum_{j=1}^m s_j b_j \] be elements of \(IS\). Concatenating the two sums gives \[ x+y =\sum_{i=1}^n r_i a_i+ \sum_{j=1}^m s_j b_j\in IS. \]
If \[ x=\sum_i r_i a_i, \] then, because \(I\) is an additive subgroup, \[ -x=\sum_i (-r_i)a_i\in IS. \] Hence \(IS\) is an additive subgroup of \(A\).
Finally, for \(r\in R\), \[ rx =r\sum_i r_i a_i =\sum_i (rr_i)a_i. \] Since \(I\) is a left ideal, \[ rr_i\in I. \] Therefore \[ rx\in IS. \] Thus \(IS\) is closed under the \(R\)-action and is a submodule of \(A\).