Let \(I\) be a left ideal of a ring \(R\), and let \(A\) be an \(R\)-module.
- Show that if \(S\) is a nonempty subset of \(A\), then \[ IS \coloneqq \left\{ \sum_{i=1}^n r_i a_i \mid n\in \mathbb{N}^*; r_i \in I; a_i \in S \right\} \] is a submodule of \(A\).
Note that if \(S = \{a\}\), then \(IS = Ia = \{ra \mid r\in I\}\).
- If \(I\) is a two-sided ideal, then \(A/IA\) is an \(R/I\) module with the action of \(R/I\) given by \[(r+I)(a+IA) = ra + IA.\]
Solution
1. If \(S\subseteq A\) is nonempty, then \(IS\) is an \(R\)-submodule of \(A\).
Choose \(a\in S\). Since \(0\in I\), \[ 0=0a\in IS, \] so \(IS\) is nonempty.
Let \[ x=\sum_{i=1}^m r_i a_i, \qquad y=\sum_{j=1}^n s_j b_j \] belong to \(IS\), with \(r_i,s_j\in I\) and \(a_i,b_j\in S\). Then \[ x+y=\sum_{i=1}^m r_i a_i+\sum_{j=1}^n s_j b_j\in IS, \] and \[ -x=\sum_{i=1}^m(-r_i)a_i\in IS \] because \(I\) is an additive subgroup of \(R\).
Finally, for \(t\in R\), \[ tx=\sum_{i=1}^m (tr_i)a_i. \] Since \(I\) is a left ideal, \(tr_i\in I\) for every \(i\), so \(tx\in IS\). Thus \(IS\) is closed under addition, additive inverses, and scalar multiplication, hence is an \(R\)-submodule of \(A\).
2. If \(I\) is two-sided, then the formula\[ (r+I)(a+IA)=ra+IA \]is well-defined.
By 1 with \(S=A\), the subset \(IA\) is an \(R\)-submodule of \(A\).
Suppose \(r-r'\in I\) and \(a-a'\in IA\). Then \[ ra-r'a'=r(a-a')+(r-r')a'. \] The first term belongs to \(IA\) because \(IA\) is a submodule. The second belongs to \(IA\) because \(r-r'\in I\) and \(a'\in A\). Hence \[ ra+IA=r'a'+IA, \] so the value depends only on the cosets \(r+I\) and \(a+IA\).
3. The formula in 2 makes \(A/IA\) into an \(R/I\)-module.
Because \(I\) is two-sided, \(R/I\) is a ring. For \(r,s\in R\) and \(a,b\in A\), the module laws follow from those in \(A\): \[ (r+I)((a+IA)+(b+IA)) =r(a+b)+IA =(ra+IA)+(rb+IA), \] \[ ((r+I)+(s+I))(a+IA) =(r+s)a+IA =(ra+IA)+(sa+IA), \] and \[ (r+I)((s+I)(a+IA)) =r(sa)+IA =(rs)a+IA =((r+I)(s+I))(a+IA). \] Thus \(A/IA\) is an \(R/I\)-module with the stated action.