$A=\{(3x,y)\mid x,y\in \mathbb Z\}$ is a maximal ideal of $\mathbb Z \oplus \mathbb Z$ I've a question from gallian which states:  
Show that $A=\{(3x,y)\mid x,y\in \mathbb Z\}$ is a maximal ideal of $\mathbb Z \oplus \mathbb Z$.Generalize.What happens if $3x$ is replaced by $4x$...  
can anyone explain me the strategy to show it is a maximal ideal (I just know the basic definition of maximal ideal)...  
thanks in advance for any help..
 A: Let $B$ be an ideal such that $A$ is a proper subset of $B$. We show that $B$ is the full ring.
Let $(s,t)$ be in $B$ but not in $A$. Then $3$ does not divide $s$. Note that $(s,0)$ is in $B$ but not in $A$. The remainder when we divide $s$ by $3$ is $1$ or $2$. Or, better, $s$ is congruent to $1$ or $-1$ modulo $3$.
If the remainder is $1$, then $s=1+3q$ for some $q$, so $(1,0)$ is in $B$, and therefore $(w,y)$ is in $B$ for every $w$ and $y$, meaning that $B$ is the whole ring.
A similar argument deals with the case $s$ congruent to $-1$ nodulo $3$. Or just work with $-s$.   
In general, the set of all $(px,y)$, where $p$ is prime, is a maximal ideal. The set of pairs $(4x,y)$ is not, since the set of pairs $(2x,y)$ is a larger proper ideal. 
Remarks: $1.$ For proving the result for all primes, one can use the fact that if $a$ and $b$ are relatively prime, then there exist integers $u$ and $v$ such that $au+bv=1$. 
$2.$ Soon you will have more conceptual tools to deal with such problems.  
