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).

  • $\begingroup$ Not that $n\Bbb Z\cong\Bbb Z$. $\endgroup$ – Zelos Malum Nov 26 '16 at 10:53
  • $\begingroup$ $6\mathbb Z$ is not a ring. $\endgroup$ – Georges Elencwajg Nov 26 '16 at 11:45
  • $\begingroup$ Apparently you are dealing with rngs, rings without a multiplicative unit. I am not familiar with the definitions of maximal ideal and prime ideal in this context, and I think I am not the only one. It would be helpful to cite the precise definitions used. $\endgroup$ – Marc van Leeuwen Nov 26 '16 at 11:45
  • $\begingroup$ Yes it is. In the axiom's of a ring, the identity element for multiplication is not necessary. I'm using the most general definition. $\endgroup$ – Kamil Nov 26 '16 at 11:46
  • $\begingroup$ Nothing in a set of axioms is necessary; if you drop some axiom you get a more general (but not necessarily more interesting) notion. So we agree this question can only be about rngs, with no multiplicative unit required. But we need to know the definitions used. For instance one can define a maximal ideal to be maximal element among the ideals that do not contain a unit for multiplication; that is OK in rings, but with this definition $12\Bbb Z$ is not a maximal ideal in $6\Bbb Z$ (because $6\Bbb Z$ itself is). $\endgroup$ – Marc van Leeuwen Nov 26 '16 at 11:51

The quotient ring $$ 6\mathbb{Z}/12\mathbb{Z} $$ is simple as an additive group, so the ideal is maximal, but it is not a domain. Its elements are $0+12\mathbb{Z}$ and $6+12\mathbb{Z}$; compute the multiplication table.


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 $J=6\Bbb Z$

  • 1
    $\begingroup$ $2\notin6\Bbb Z$ $\endgroup$ – Marc van Leeuwen Nov 26 '16 at 10:47
  • 1
    $\begingroup$ Even to those who do not suppose rings must have a multiplicative unit$~1$, I think the notion of domain does require this (and in addition $1\neq0$). I would certainly not call $6\Bbb Z$ a PID. $\endgroup$ – Marc van Leeuwen Nov 26 '16 at 11:57

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