Prove $M$ is a Maximal Ideal in $\Bbb Z\times \Bbb Z$

Question

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.

Answer 1

You're doing well, but this isn't right: $\frac{\mathbb Z \times \mathbb Z}{M} = (\mathbb Z_{p}, \mathbb Z_{b})$

It's true that everything in the left side of the pairs is of the form $pa$ for $a$ ranging over $\Bbb Z$, and those are exactly the elements of $(p)\lhd \Bbb Z$, so the left part is indeed $\Bbb Z_p$. But in the right hand side, $b$ can be anything, including $1$!

So $\Bbb Z_b$ is not right but $\frac{\mathbb Z \times \mathbb Z}{M} = (\mathbb Z_{p}, ?)$ (you suddenly realize what it actually should be in the comments below...)

Answer 2

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 $\mathbb{Z}\times \mathbb{Z}$.