The following question came up on a recent exam that I took.
Let $R$ be a PID. Let $I$ and $J$ be ideals in $R$ such that $\operatorname {Hom}_{R}(R/I, R/J) = 0$; that is, there is only the zero map between the quotient rings. Prove that $R = I + J$.
Proposed Proof:
Consider the maps, \begin{eqnarray} \pi_{0} : R/I \longrightarrow R/J, \quad & \pi_{0}(r + I) = 0 + J, \\ \pi_{1} : R \longrightarrow R/I, \quad & \pi_{1}(r) = r + I. \end{eqnarray} Composing the maps, we have that,
\begin{equation} \pi_{2} = \pi_{0} \circ \pi_{1} : R \longrightarrow R/J, \quad \pi_{2}(r) = 0 + J. \end{equation}
But this means that the projection of element in $R$ is the ideal $J$ in $R$. By the Lattice Isomorphism Theorem, we must have that $R = J$. Since $J$ is contained in $I + J$, we then must have that $R = I + J$.
I am not so sure about the proposed proof. For example, I have not used the fact the $R$ is a PID. Please comment if you think the proof is wrong. Also, please do share the correct proof to this question.
This quesion was for 10 points, so I am trying to see if I have at least some parts correct.
Since $R$ is a PID, there exists $a,b,d\in R$ such that $I=Ra$, $J=Rb$ and $I+J=Rd$. Then $d$ is a gcd of $a$ and $b$, hence $a=d\bar a$ and $b=d\bar b$ for some $\bar a,\bar b\in R$. We have $I+J=R$ if and only if $d$ is a unit. Consider the $R$-module homomorphism \begin{align} &\varphi:R\to R/J& &x\mapsto \bar b x+J \end{align} Then $\varphi(a)=a\bar b+J=\bar ab+J=0$, hence $\varphi$ give rise to an $R$-module homomorphism $\bar\varphi:R/I\to R/J$. By assumption, $\bar\varphi=0$ so that $0=\bar\varphi(1)=\bar b+J$ so that $\bar b\in J$.
Consequently, there exists $u\in R$ such that $\bar b=ub=u\bar bd$ from which $ud=1$, that's $d$ is a unit, hence $I+J=R$.