# Confusion about the definition of maximal ideal

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 ?

• Given two ideals in $\mathbb Z$ it is not true that one must be included in the other. – 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.

“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 $$J=R$$."

• 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 ? – Einsteinwasmyfather Jan 24 at 15:42
• @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. – rschwieb Jan 24 at 15:44
• 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. – 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.

• 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 $\ J \subset I$" ? – Einsteinwasmyfather Jan 24 at 15:06
• @Einsteinwasmyfather Those are indeed not equivalent, as rschwieb points out there is a third possibility, that J is simply not comparable to I. – 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: $$mn$$ Hence if $$m\ge n$$ is not true, you must have $$m.

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 $$n=5$$.