Let \(R\) be a commutative ring with identity, let \(I\trianglelefteq R\) be an ideal, and let \(A\) be an \(R\)-module. Prove that \(A/IA\) is naturally an \(R/I\)-module via \[ (r+I)(a+IA)=ra+IA. \]
Solution
1. The displayed scalar action is well-defined.
Suppose \[ r+I=r'+I, \qquad a+IA=a'+IA. \] Then \(r-r'\in I\) and \(a-a'\in IA\). We compute \[ ra-r'a' =r(a-a')+(r-r')a'. \] Because \(IA\) is an \(R\)-submodule of \(A\), the first term lies in \(IA\). Since \(r-r'\in I\), the second term lies in \(IA\) by definition. Hence \[ ra+IA=r'a'+IA, \] so the action depends only on the cosets.
2. Scalar addition distributes over the module addition.
For \(r,s\in R\) and \(a\in A\), \[ ((r+I)+(s+I))(a+IA) =(r+s)a+IA =ra+sa+IA, \] which equals \[ (r+I)(a+IA)+(s+I)(a+IA). \]
3. Scalars distribute over addition in \(A/IA\).
For \(r\in R\) and \(a,b\in A\), \[ (r+I)((a+IA)+(b+IA)) =r(a+b)+IA =ra+rb+IA, \] which is \[ (r+I)(a+IA)+(r+I)(b+IA). \]
4. Scalar multiplication is associative.
For \(r,s\in R\) and \(a\in A\), \[ ((r+I)(s+I))(a+IA) =(rs)a+IA =r(sa)+IA =(r+I)((s+I)(a+IA)). \]
5. The identity coset acts as the identity.
For \(a\in A\), \[ (1+I)(a+IA)=1a+IA=a+IA. \]
6. Therefore \(A/IA\) is an \(R/I\)-module.
The quotient \(A/IA\) is already an abelian group. By 1 the scalar action is well-defined, and <1>2–<1>5 verify the module axioms.