Prove that $J \cap R(1 - e) \neq \{0\}$ Let $R$ be a commutative unitary ring and I and J be two maximal ideals of $R$ such that $I \neq J$ and $I \cap J = Re$ where $e \in R$ is idempotent. 
Prove that $J \cap R(1 - e) \neq \{0\}$. 
I have no clue. I tried to work by contradiction to make use of the second isomorphism theorem. I said that since $e$ is idempotent, $R = Re \oplus R(1 - e)$. Also $J + R(1-e)$ is an ideal of $R$ containing $J$ and $\neq J$ so $R = J \oplus R(1-e)$, so $J \cong Re = I \cap J$. I don't see how this can be made useful. 
Help is much appreciated. 
 A: Assume the intersection to be zero.
We have
$$R/J = (J+R(1-e))/J \cong R(1-e)/(J \cap R(1-e)) = R(1-e) \cong R/Re,$$
but $Re = I \cap J$ is not maximal.
A: Because of $I\cap J = Re$, $I \neq J$, and $J$ maximal, we know that $Re \subsetneq J$. Let $\phi:R \to Re\oplus R(1-e)$ be the canonical isomorphism, and set $J' = \phi(J)$. 
Take an element $(a, b) \in J'$ such that $b \neq 0$ (possible because $J \not \subseteq Re$). Because of $Re \subset J$, we also have $(a, 0)\in J'$, so $(a, b) - (a, 0) = (0, b) \in J'$. But $(0,b) \in \{0\}\oplus R(1-e)$, so $\phi^{-1}(0,b)$ is both an element of $J$ and of $R(1-e)$.
A: Maybe this is the most "down-to-earth" proof:
As $1-e \notin J$ and $J$ is maximal, we have some $r \in R$ with $1-r(1-e) \in J$. Multiply with $1-e$ to obtain $(1-r)(1-e) \in J$, hence $(1-r)(1-e) \in J \cap R(1-e)$.
If this element is non-zero, we are done. If it is zero, we have $1-r \in Ann(1-e)=Re$. Replace $r$ by $r+j$ for some $j \in J \setminus Re$. Then $1-(r+j) \notin Re=Ann(1-e)$, but we still have $(1-(r+j))(1-e) \in J$, so this is a non-zero element of the intersection.
