Let \(R\) be a principal ideal domain,\(A\) a unitary left \(R\)-module, and \(p\in R\) a prime (and thus irreducible) element. Define \[\begin{aligned} pA &\coloneqq \{ pa \mid a\in A\} \\ A[p] &\coloneqq \{ a\in A \mid pa = 0\}.\end{aligned}\]
Show the following:
\(R/(p)\) is a field.
\(pA\) and \(A[p]\) are submodules of \(A\).
\(A/pA\) is a vector space over \(R/(p)\), with \[(r + (p))(a + pA) = ra + pA.\]
\(A[p]\) is a vector space over \(R/(p)\) with \[(r + (p))a = ra.\]
Solution
1. The quotient \(R/(p)\) is a field.
In a principal ideal domain, every irreducible element is prime and the principal ideal generated by a prime element is maximal. Hence \((p)\) is a maximal ideal of \(R\), so the quotient ring \(R/(p)\) is a field.
2. The subsets \(pA\) and \(A[p]\) are \(R\)-submodules of \(A\).
The map \[ \mu_p:A\longrightarrow A, \qquad a\longmapsto pa \] is \(R\)-linear because \(R\) is commutative. Its image is \(pA\) and its kernel is \(A[p]\). Images and kernels of module homomorphisms are submodules.
3. The formula\[ (r+(p))(a+pA)=ra+pA \]makes \(A/pA\) into an \(R/(p)\)-vector space.
By 2, \(pA\) is a submodule. Since \((p)A=pA\), part (2) of P-457V6, applied to the two-sided ideal \((p)\), gives an \(R/(p)\)-module structure on \(A/pA\) by the stated formula. By 1 the scalar ring \(R/(p)\) is a field, so this module is a vector space.
4. The formula\[ (r+(p))a=ra \]makes \(A[p]\) into an \(R/(p)\)-vector space.
First check well-definedness in the scalar coset. If \(r-r'\in(p)\), write \(r-r'=tp\). For \(a\in A[p]\), \[ (r-r')a=t(pa)=0, \] so \(ra=r'a\).
For \(a\in A[p]\) and \(r\in R\), \[ p(ra)=r(pa)=0, \] so \(ra\in A[p]\). Each module axiom for \(A[p]\) over \(R/(p)\), evaluated on coset representatives, is the corresponding axiom for \(A\) over \(R\). Thus \(A[p]\) is an \(R/(p)\)-module, and hence, by 1, an \(R/(p)\)-vector space.