Quotient, regular modules and submodules PID. Let $R$ be a Principal Ideal Domain and $I\neq \{0\}$ an ideal of $R$.
Are the $R$-submodules of $R/I$ the same as the $R/I$-submodules of $R/I$?
If the action of $R/I$ as $R$-module is
$r(r'+I) = rr' + I$. Then it coincides with the product in the Quotient Ring $R/I$ as regular module:
$(r+I)(r'+I) = rr' + I$
So are the submodules in the form:
$J/I$ with $J \supseteq I$ submodule of $R$ ? Both modules have the same (finite) length?
 A: Yes the $R$-submodules of $R/I$ are indeed the same as the $R/I$ submodules. The reason ist that the action $R\times R/I \to R/I$ factorises over $R/I \times R/I$.
More generally it is always true that given an $R/I$-module $M$ we can make $M$ into an $R$-module by defining the multiplication to be
$$R \times M \to R/I \times M \to M$$
where the second arrow is the $R/I$-multiplication. This even generalizes to the following: Given any rings $A,B$ and a ringhom. $\phi: A\to B$ we can make every $B$-module $N$ into an $A$-module by setting $a\cdot n := \phi(a)\cdot n$. But this is the same as defining the module structure to be given by
$$ A \times N \xrightarrow{\phi \times id} B \times N \to N$$
where again the second arrow is just the $B$-multiplication on $N$. This process is called restriction of scalars. On the other hand given an $A$-module $M$ and if we consider $B$ as an $A$-module by restriction of scalars we can form the $B$-module $M\otimes_A B$ where the $B$-multiplication is induced by the map
$$B\times (M\times B) \to M\otimes_AB,\quad (b,(m,b'))\mapsto m\otimes bb'.$$
This process is called extension of scalars. In this situation we have bijections between the following sets
$$Hom_A(M,N)=Hom_B(M\otimes_A B,N)$$
where $M$ is an $A$-module, $N$ is a $B$-module and in the LHS we regard $N$ as an $A$-module. This shows that restriction and extension are adjoint.
