Maximal ideal in $\mathbb{Z} \times \mathbb{Z}$

Question

While trying to find a maximal ideal in $\mathbb{Z} \times \mathbb{Z}$, I ran into something that seems contradictory.

If we let $J = \{(a,a):a \in \mathbb{Z}\}$ then $J \cong \mathbb{Z}$ given the by isomorphism $(a,a) \mapsto a$.

Then (at least it seems to me): $$J/(7,7)J \cong \mathbb{Z}/7\mathbb{Z}$$

This is a field, which implies $(7,7)J = \{(7a,7a):a \in \mathbb{Z}\}$ is a maximal ideal.

But $(7,7)J \subsetneq \{(7a, 7b): a,b \in \mathbb{Z}\}$ which is a counterexample to the claim that $(7,7)J$ is maximal.

Where have I gone wrong?

Answer 1

What you have written is correct: $J/(7,7)J$ is isomorphic to $\Bbb Z/7\Bbb Z$ but what I think you're doing is you're confusing $J/(7,7)J$ with $\Bbb Z^2/(7,7) J$. $(7,7)J$ is maximal in $J$. It is not maximal in $\Bbb Z^2$ as you have demonstrated.

Answer 2

I suggest thinking about it like this: $\Bbb Z\times\Bbb Z$ is generated by $(1,0)$ and $(1,1)$. When you mod out by $(7,7)$, you are essentially giving $(1,1)$ an order of $7$. Hence, you get $\Bbb Z\times(\Bbb Z/7\Bbb Z)$.