Exercise 4 Hungerford's Algebra IV.3. Following is from Hungerford's book of Algebra, Exercise 4 page 189 :

Let R be a principal ideal domain, A a unitary left R-module, and p \in R a. prime (= irreducible). Let pA = {pa | a \in A} and A[p] = {a \in A | pa = 0}.
(c) A/pA is a vector space over R/(p), with (r + (p))(a + pA) = ra + pA.
(d) A[p] is a vector space over R/(p), with (r + (p))a = ra.

I could solve Items (a) and (b). For Items (c) and (d) (quoted above) My first idea is to show that for (c) : (r+(p))(a + pA)   = ra +pA is well-defined and for (d) : for r+(p) \in R/(p) and a \in A[p] : (r+(p))a    = ra, but I don't know if this is a right track to build on them the solutions.
 A: The four items in Exercise 4 page 189 Hungerford's Algebra book can be solved essentially by applying the definitions.
For item (c):  We have two operations in $A/pA$

*

*for all $a+ pA$ and  $b + pA \in A/pA$   ,  $(a+ pA) +  (b + pA) = a+b +pA$

*for all $r+(p) \in R/(p)$ and $a + pA \in A/pA$ , $(r+(p))(a + pA)   = ra +pA$.

It is easy to prove that those two operations are well-defined.  That means:

*

*if  $a+ pA = a'+ pA$  and $b + pA = b' + pA$ then $a+b +pA = a'+b' +pA$

*if $r+(p)=r'+(p)$ and $a+ pA = a'+ pA$ then $ra +pA = r'a' +pA$.

Then, check that those two operations satisfy the axioms of vector space.
For item (d): The operations are:

*

*if $a, b \in A[p]$, then $a+b$  is just the addition in $A$.

*for all $r+(p) \in R/(p)$ and $a \in A[p]$ ,  $(r+(p))a    = ra$.

It is easy to prove that the operations are well-defined:

*

*if $a, b \in A[p]$, then $pa=pb=0$. So $p(a+b)= pa+pb=0$. So $a+b \in A[p]$.

*$r+(p)=r'+(p)$ and $a \in A[p]$, we have that $r-r' \in (p)$. So there is $k \in R$ such that $r-r'=kp$. So, $ra-r'a = (r-r')a = kpa = 0$. So $ra=r'a$.

Then, check that those two operations satisfy the axioms of vector space.
