Proving an ideal is maximal

Question

Let p be a prime. show that A = {(px,y) : x,y $\in$ $\mathbb Z$ } is a maximal ideal of $\mathbb Z$ x $\mathbb Z$.

I am having trouble showing that A is maximal. To show A is an ideal, first note that $\mathbb Z$ x $\mathbb Z$ is a commutative ring. Let (px,y) $\in$ A and let (a,b) $\in$ $\mathbb Z$ x $\mathbb Z$. Then (px,y)(a,b) = (pxa,yb) $\in$ A. Thus A is an ideal (Is this sufficient?). To show A is maximal, I think I could either show that ($\mathbb Z$ x $\mathbb Z$)/A is a field or show that any ideal containing A must equal $\mathbb Z$ x $\mathbb Z$, perhaps by showing a unit must be in in an ideal containing A?. The main trouble I am having is seeing what the cosets from ($\mathbb Z$ x $\mathbb Z$)/A look like. Do I consider possible remainders for each coordinate? Any hints as to either approach or an idea for a different approach are appreciated.

Answer 1

What is the kernel of the homomorphism $f: \mathbb{Z}\times \mathbb{Z} \to\mathbb{Z}/p\mathbb{Z}$ given by $f(s,t) = s\pmod{p}$?

Answer 2

An ideal $I$ in a commutative ring $R$ is maximal iff $\frac{R}{I}$ is a field.

In this case it is easy to see $\frac{\mathbb Z\times \mathbb Z}{A}$ is isomorphic to $\mathbb Z_p\times \{0\}$, and $\mathbb Z_p$ is a field.