$M/(I_1\cap I_2)M$ is a free $R/(I_1\cap I_2)$-module of rank $1$. Let $R$ be a unital Noetherian ring, $M$ a finitely generated $R$-module, and $I_1$,$I_2$ two ideals in $R$ such that $M/I_jM$ is free $(R/I_j)$-module of rank $1$ for $j=1,2$. Assume that the group of units of $R/(I_1\cap I_2)$ is mapped surjectively onto that of $R/(I_1+I_2)$. Show that $M/(I_1\cap I_2)M$ is a free $R/(I_1\cap I_2)$-module of rank $1$.
I am not sure how to use the surjectivity between the units. I thought about using the short exact sequence$$0 \longrightarrow R/(I_1\cap I_2)\stackrel{f} \longrightarrow R/I_1\oplus R/I_2\stackrel{g}{\longrightarrow}R/(I_1+I_2) \longrightarrow 0$$but I don't know how to proceed.
 A: By passing to quotient we may assume $I_1 \cap I_2 = 0$. Denote the canonical projection from $M$ to $M/I_i M$ by $p_i$.
We firstly show that there exists $m \in M$ such that $p_i(m)$ are basis for $M/I_i M$. From the exact sequence
\begin{equation}
0 \to M/(I_1 M\cap I_2 M) \to M/I_1 M \oplus M/I_2 M \to M/(I_1 + I_2)M \to 0
\end{equation}
it suffices to find $m_i$ representing basis of $M/I_i M$ such that $m_1 - m_2 \in I_1 M + I_2 M$. Take basis $m_i$, so we have
\begin{equation}
m_2 - r_1 m_1 \in I_1 M, \quad m_1 - r_2 m_2 \in I_2 M
\end{equation}
for some $r_1,r_2 \in R$. Now It follows that
\begin{equation}
(1-r_1r_2) m_1 \in (I_1 + I_2) M.
\end{equation}
Thus $r_1 r_2$ is a unit in $R/(I_1+I_2)$ and then $r_1 \equiv u \mod I_1 +I_2$ where $u$
is a unit in $R$. Now $m_2 - u m_1 \in (I_1 + I_2)M$, change $m_1$ to $um_1$ we get the desired basis.
From the five lemma we deduce that
\begin{equation}
M/(I_1 M \cap I_2 M) \simeq R.
\end{equation}
But then $M \simeq R \oplus N$ since $M$ admits an epimorphism to $M/(I_1 M \cap I_2 M)$. However, $N/I_1 N = N/I_2 N = 0$, which implies $N = I_1 N = I_2 N = I_1 I_2 N = 0$.
