Lets $A,B$ be two rings. Show that $M\times N$ is a maximal ideal of $A\times B \iff$ $M\times N$ is the form
$I \times B$ or $A \times J$, where $I$ is a maximal ideal of $A$ and $J$ a is a maximal ideal of $B$.
$\textbf{My Attempt:}$
$(\Leftarrow)$
Suppose that $I$ is a maximal ideal of $A$ and let $L\times B$ an ideal of $A\times B $ such that $I \times B \subset L \times B$ and $I\times B \neq L \times B$. Then there is $l \in L$ such that $l \notin I$.
Since $I$ is maximal ideal of $A$ we have $I+ (l) = A$, so $L = A$. Therefore $L \times B = A \times B$ and $I \times B$ is maximal ideal of $A \times B$.
The same argument goes for the other coordinate.
But I cannot see how to solve the otherside. Can you help me?
If $M \times N$ is a maximal ideal of $A \times B$, then $\dfrac{A \times B}{M \times N} \cong \dfrac{A}{M}\times\dfrac{B}{N}$ is a field. An element of the form $(0 + M,b + N) \in \dfrac{A}{M}\times\dfrac{B}{N}$, where $b \not\in N$, does not have an inverse unless $\dfrac{A}{M} \cong \{0\}$ and $\dfrac{B}{N}$ is a field.