0

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).

Kamil
  • 5,139
  • Not that $n\Bbb Z\cong\Bbb Z$. – Zelos Malum Nov 26 '16 at 10:53
  • 2
    $6\mathbb Z$ is not a ring. – Georges Elencwajg Nov 26 '16 at 11:45
  • 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. – Marc van Leeuwen Nov 26 '16 at 11:45
  • 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. – Kamil Nov 26 '16 at 11:46
  • 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). – Marc van Leeuwen Nov 26 '16 at 11:51
  • In reply to the "later edit": my question was about the definitions of maximal and prime ideals. – Marc van Leeuwen Nov 26 '16 at 11:59
  • @GeorgesElencwajg It is a ring, a ring does not need a multiplicative identity. – Zelos Malum Nov 26 '16 at 14:16
  • @Zelos: yes a ring does have an identity. At least that's what Artin, Atiyah, Bourbaki, Grothendieck, Jacobson, Lang, Matsumura, Serre and all contemporary algebra textbooks I have browsed claim. Who exactly calls rings (instead of pseudo-rings or rngs) structures without identity elements? – Georges Elencwajg Nov 26 '16 at 18:42
  • They do not, some do, some don't. Rngs was, snd forever is a joke – Zelos Malum Nov 26 '16 at 18:44
  • dear @Zelos, I asked you a precise question which generalities like "some do, some don't" don't really answer. So please tell me who defines a ring as not necessarily having a unit and does that potential person have the expertise of those in my list? Can you quote one article in a top journal like Annals of Mathematics or Inventiones Mathematicae or Publications de l'IHES where rings are allowed to not have an identity? As for Rngs you may consider it a joke, but the joker was Jacobson, one of the great algebraists the 20th century. – Georges Elencwajg Nov 26 '16 at 19:27
  • Every course (5 of them) do not require identity – Zelos Malum Nov 26 '16 at 19:28

2 Answers2

2

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.

egreg
  • 238,574
0

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 $J=6\Bbb Z$

Learnmore
  • 31,062