In my book, the definition of a maximal ideal is as follows:

Let $R$ be a commutative ring. A maximal ideal of $R$ is an ideal $I$ such that:

  1. $I \neq R$.
  2. There exists no ideal $J$ of $R$ such that $I \subsetneq J \subsetneq R$.

When trying to prove the following statement:

$I$ is a maximal ideal of $\mathbb{Z} \implies I = (p)$ with $p$ a prime number.

I came up with an error in my reasoning, but do not understand at all why it is an error. Because all ideals of $\mathbb{Z}$ are of the form $(n)$ with $n \in \mathbb{Z}$, I know that $I = (n)$. So I tried formulating the definition of maximal ideal as follows:

$\forall n' \in \mathbb{Z}: (n') = \mathbb{Z} \ $ or $ \ (n') \subset (n)$.

But this must obviously be false because then $n$ would be a divisor of all $n' \in \mathbb{Z}$, so that $n = 1$ and thus $(n) = \mathbb{Z}$. This cannot be the case since $I$ is a maximal ideal. Where did my reasoning go wrong ?

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    $\begingroup$ Given two ideals in $\mathbb Z$ it is not true that one must be included in the other. $\endgroup$ – Mark Bennet Jan 24 at 14:54

Someone has proposed that you thought that the ideals of $\mathbb Z$ are linearly ordered, but I have a feeling that the mistake was actually simpler than that, and "upstream" so to speak. I think it is just an error in quantification.

It should read

“for all $n’$, if $(n’)\supseteq (n)$ it follows that A or B.”

and not

“for all $n’$, it follows that A or B.”

A stands for $(n')=(n)$ and B stands for $(n')=R$, but I wanted to disguise that the conditions themselves are not important. What was important was the missing "if."

The second one implies that all ideals of $R$ are comparable to $(n)$ (which can be exaggerated to "all ideals are in a chain" as the other solution thought.)

To connect this with your original definition, which was correct:

There exists no ideal $J$ of $R$ such that $I\subsetneq J \subsetneq R$

This would be logically equivalent to

"If $I\subseteq J \subseteq R$, then $J=I$ or $J=R$."

The misstep you wrote looked like this:

"If $J \subseteq R$, then $J=I$ or $J=R$."

  • $\begingroup$ I don't understand why the IF has to be there, since $(n) \subset (n')$ reduces the second condition from $(n') \subset (n)$ to $(n') = (n)$, so it's already fine ? $\endgroup$ – Einsteinwasmyfather Jan 24 at 15:42
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    $\begingroup$ @Einsteinwasmyfather As the example of $\mathbb Z$ shows, not all $(n')$ are comparable to $(n)$. For example, $n'=9$ and $n=2$ shows that $(n)\nsubseteq (n')$ and $(n')\neq \mathbb Z$ can happen! Therefore you don't really mean "for all $n'$" you mean "for all $(n')$ which contain $(n)$" so that you are only talking about relevant ideals. $\endgroup$ – rschwieb Jan 24 at 15:44
  • $\begingroup$ Actually I misspoke. I should have said the second thing reads like "All ideals are $M$ or $R$" and the first one just says "All ideals containing $M$ are $M$ or $R$", which is what you want. $\endgroup$ – rschwieb Jan 24 at 16:00

You seem to be assuming that all the ideals of $\mathbb Z$ are linearly ordered, but they're not.

  • $\begingroup$ Do you mean that the statement "for all ideals $J : \neg(I \subsetneq J \subsetneq R)$" is not equivalent to "for all ideals $J : J = R \ $ or $\ J \subset I$" ? $\endgroup$ – Einsteinwasmyfather Jan 24 at 15:06
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    $\begingroup$ @Einsteinwasmyfather Those are indeed not equivalent, as rschwieb points out there is a third possibility, that J is simply not comparable to I. $\endgroup$ – Slade Jan 24 at 16:17

If you follow the definition of maximal ideals, then for $\mathbb{Z}$, you have that $(n)$ is a maximal ideal if and only if

  • $(n)\ne\mathbb{Z}$;
  • there is no ideal $(m)$ of $\mathbb{Z}$ such that $(n)\subsetneq (m)\subsetneq\mathbb{R}$.

The second condition does not imply that for every $m\in\mathbb{Z}$ (and $m\ne 1$), $(m)\subset (n)$.

For two integers $m$ and $n$, exactly one of the following is true: $$ m<n,\quad m=n,\quad m>n $$ Hence if $m\ge n$ is not true, you must have $m<n$.

However, for two ideals $(m)$ $(n)$, it is possible that none of the following is true $$ (m)\subset(n),\quad (m)=(n),\quad (m)\supset(n) $$

Consider for instance the example $m=3$ and $n=5$.


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