I need to prove that in the ring $6\mathbb{Z} = \left\{x \in \mathbb{Z} \mid x = 6q, q \in \mathbb{Z}\right\}$ the subset $12\mathbb{Z}$ is a maximal ideal but not a prime ideal.
I first wanted to prove it is a maximal ideal. Suppose there would exist an ideal $J$ such that $12\mathbb{Z} \subset J \subset 6 \mathbb{Z}$. I want to prove that $6\mathbb{Z} = J$. Let $x \in 6\mathbb{Z} \setminus J$. Then $x$ is a multiple of $6$, and so $2x \in 12\mathbb{Z} \subset J$. But how can I conclude from this that $x \in J$ ?
I didn't find the other part either, i.e. showing that $12\mathbb{Z}$ is not a prime ideal. We need to show there exist $x, y \in 6 \mathbb{Z}$ such that $xy \in 12\mathbb{Z}$, but $x \notin 12\mathbb{Z}$ and $y \notin 12\mathbb{Z}$. I wanted to pick $x = 2$ and $y = 6$, but I see that $x \notin 6\mathbb{Z}$.
Help with this problem is appreciated.
Later Edit:
Definition: A ring $R, + , \cdot$ is a non-empty set on which two binary operations are defined, such that:
(i) R, + is a commutative group.
(ii) $\cdot$ is associative.
(iii) $\cdot$ is distributive with respect to $+$.
Definition: Let $R$ be a commutative ring.
(i) A prime ideal of $R$ is an ideal $I$ of $R$ such that $I \neq R$ and $$\forall x, y \in R: x \cdot y \in I \Rightarrow x \in I \ \text{or} \ y \in I. $$ (ii) A maximal ideal of $R$ is an ideal $I$ of $R$ such that $I \neq R$ and such that there exists no ideal $J$ of $R$ with $I \subset J \subset R$ (here the symbol $\subset$ means a strict inclusion).
Note that $6\times 6=36\in 12\Bbb Z$ but $6\notin 12\Bbb Z$.
Regarding your proof :
Let $12\Bbb Z\subset J\subset 6\Bbb Z$,since $6\Bbb Z$is a PID so $J=\langle a\rangle$.
Then $12\in \langle a\rangle\implies 12=na; n\in \Bbb Z$.
Since $a\in 6\Bbb Z\implies a=6k$
So $12=6kn\implies kn=2\implies k=2 $or $k=1$
If $k=2\implies a=12$;If $k=1\implies a=6$
So either $J=12\Bbb Z$ or $J=6\Bbb Z$