TITLE: Prove $M$ is a Maximal Ideal in $\Bbb Z\times \Bbb Z$
QUESTION [6 upvotes]: A problem from introduction to abstract algebra by Hungerford.
It asks:
If $p$ is a prime integer, prove that $M$ is a maximal ideal in $\mathbb Z \times \mathbb Z$, where $M =\{(pa,b)\mid a,b\in \mathbb Z\}$
Though, we know that an ideal is maximal iff $\frac{\mathbb Z \times \mathbb Z}{M}$ is a field. Isn't $\frac{\mathbb Z \times \mathbb Z}{M} = (\mathbb Z_{p}, \mathbb Z_{b})$? And since $b$ can be any integer, say $6$, then this is not a field.  And so $M$ must not be maximal?
Can someone please show what I am misunderstanding?  Thanks.

REPLY [2 votes]: Add an ordered pair $(c,d)$, where $c$ is not a multiple of $p$. Using the Bézout Identity we find that $(1,k)$ is in the resulting ideal for some $k$.  But $(0,k)$ is in the ideal, so $(1,0)$ is.  Since $(0,1)$ is in the ideal, we get all of $\mathbb{Z}\times \mathbb{Z}$.