Prove or disprove the maximality of the following ideals

Question

$1.$ $I= \{(a,b) : 7\mid a, 7\mid b\} \subseteq \mathbb{Z}\bigoplus \mathbb{Z}$

$2.$ $I= \{a+bi : 7\mid a, 7\mid b\} \subseteq \mathbb{Z}[i].$

$3.$ $I=\{(a,b) : 9\mid a, 9\mid b\} \subseteq \mathbb{Z}\bigoplus \mathbb{Z}$

I'm not looking for a higher-level mathematics approach; i'm not there yet. I just need an "elementary" approach somewhat similar to what I've done below.

$1.$

Consider the ideal $I= \{(a,b) : 7\mid a, 7\mid b\}\subseteq \mathbb{Z}\bigoplus\mathbb{Z}.$ Since $(1,1)\not\in I, I\neq \mathbb{Z}\bigoplus\mathbb{Z}.$ We will show that this is not maximal. Let $J=\{(a,b) : 7 \mid a\}\subseteq \mathbb{Z}\bigoplus\mathbb{Z}.$ For all $(a,b)\in I,7\mid a,$ so $(a,b)\in J.$ Thus $I\subseteq J.$ However, $(7,1)\not\in I$ but $(7,1)\in J$, so $I\subsetneq J.$ We now show that $J$ is an ideal. Let $(a,b),(c,d)\in J.$ Then $7\mid a,7\mid c$ and $b,d\in\mathbb{Z}.$ Thus $(a,b)-(c,d)=(a-c,b-d).$ Since $7\mid a$ and $7\mid c,7\mid (a-c)$ (since $a=7k_1, c=7k_2,k_1,k_2\in\mathbb{Z},$ and $a-c = 7(k_1-k_2), k_1-k_2\in\mathbb{Z}$). As well, $b-d\in\mathbb{Z}$ so $(a,b)-(c,d)\in J.$ Now let $(x,y)\in \mathbb{Z}\bigoplus\mathbb{Z}.$ Then $(a,b)\cdot (x,y) = (ax,by).$ Since $7\mid a, 7\mid ax$. Since $by\in \mathbb{Z},$ we have that $(a,b)(x,y)\in J\;\forall (x,y)\in\mathbb{Z}\bigoplus\mathbb{Z}.$ Thus $J$ is an ideal. Since $I$ is a proper subset of $J\subseteq \mathbb{Z}\bigoplus\mathbb{Z}, I$ is not maximal.

$2.$

Consider the ideal $I=\{a+bi : 7\mid a , 7\mid b\}\subseteq \mathbb{Z}[i].$ Since $1+i \not\in I, I\subsetneq \mathbb{Z}[i].$ We will show that this is maximal. Let $J$ be an ideal of $\mathbb{Z}[i]$ such that $I\subsetneq J\subseteq \mathbb{Z}[i].$ We will show that $J-\mathbb{Z}[i].$ Let $a\in J\backslash I.$ Then $a=(x,y)$ for some integers $x,y$ such that $7\nmid x$ or $7\nmid y.$

$3.$ Will a similar proof to the first proof suffice? (I'm really unsure about this and I'm just asking this one to clarify my understanding)

Answer 1

The abstract algebra is not elementary. Then in my opinion, the necessary and sufficient condition for the ideal to be maximal in a ring is a standard algebra knowledge. :)

1. The ideal $I_1=\{n\in\Bbb Z\colon 7|n\}$ is maximal in $\Bbb Z$ since $\Bbb Z/I_1$ is isomorphic to $\Bbb Z_7$, wnich is a field. Then $\Bbb Z\oplus \Bbb Z/I$ is isomorphic to $\Bbb Z_7\oplus \Bbb Z_7$, but this is not the field (there are nontrivial zero divisors). Then $I$ is not maximal.