Vtyjkyuk

Problem

Show that $A=x,y in Z$ is a maximal ideal . Generalize the result.

Proof

$A$ is an ideal

For any$ (a,b) in Z+Z$ and $(3x,y) in A,
(a,b)(3x,y) in Z$. Also it is easy to see $A$ is a subring.

$A$ is a maximal ideal

Consider a maximal ideal $B$ which contains $A$. Now$ (c,d) in B$ such that $gcd(3,c)=1. 1=3alpha+cbeta$. Then
$(1,1)=(3alpha+cbeta,1-d+d)$ . Hence $(1,1) in B$. So $A$ is maximal.

Doubt

1.What is the generalization of the result?

$A$=$(nx,y)

2.Is my proof ok?

With respect to your proof, the idea is correct (in fact I find it quite original), but to be honest, it is terribly written. Let me rewrite it appropriately:

As you have said, it is easy to check that $A$ is indeed an ideal (another way to argue is to observe that $A=3mathbbZ times mathbbZ$ is a product of two ideals of $mathbbZ$). Let $B subset mathbbZ times mathbbZ$ a maximal ideal containing $A$. Arguing by contradiction, we suppose that the inclusion is strict; that is, there exists $(c,d) in B$ such that $textgcd(3,c)=1$. By Bézout's Identity, there exist $alpha, beta in mathbbZ$ such that
$$3alpha+cbeta=1.$$
Hence,
$$(1,1)=(3alpha,1-d)+(beta,1)(c,d) in B,$$
against $B$ being proper. Therefore, $A=B$ and $A$ is indeed a maximal ideal.

We now come to consider the question of the generalization. For every $n geq 0$ we denote
$$A_n:=(nx,y) / x,y in mathbbZ,$$
and observe that in fact
$$A_n=nmathbbZ times mathbbZ.$$
Therefore, $A_n$ is a product of ideals of $mathbbZ$ and hence, an ideal of $mathbbZ times mathbbZ$. Then:
$$fracmathbbZ times mathbbZA_n=fracmathbbZ times mathbbZnmathbbZ times mathbbZ simeq fracmathbbZnmathbbZ times fracmathbbZmathbbZ = fracmathbbZnmathbbZ times 0+mathbbZ simeq fracmathbbZnmathbbZ.$$
Thus,
$$A_n text is a maximal ideal iff mathbbZ times mathbbZ/A_n text is a field iff mathbbZ/nmathbbZ text is a field iff n text is a prime number.$$
This also solves the original problem for $n=3$.