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I want to prove that a noetherian ring $R \neq \{0\}$ contains at least one maximal ideal.

My idea is to consider $\langle 0 \rangle$ and $\langle 1 \rangle$: If there is no ideal $I$ with $\langle 0 \rangle \subsetneq I \subsetneq\langle 1 \rangle$ then $\langle 0 \rangle$ is maximal. Otherwise for each infinite chain $\langle 0 \rangle \subset I_1 \subset \cdots$ there exists $i \in \mathbb{N}$ such that for all $j>i$, $I_i = I_j$. Then $I_i$ is maximal.

Is that correct?

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There are various equivalent ways to define a Noetherian ring, one of which is precisely what you have used. What you say is correct, for, if there were no maximal ideal, one could get a strictly-ascending-never-ending sequence of proper ideals in which case $R$ wouldn't have been Noetherian! –  Host-website-on-iPage Jul 11 '12 at 13:24
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So it is not necessary to use Zorn's lemma in noetherian rings other than in general rings? –  joachim Jul 11 '12 at 13:26
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Note that if your argument (as corrected by Seth Baldwin) does not use Zorn's lemma, it does use a choice process (infinitely often) to construct a chain that contradicts the Noetherian property. I'm no expert at this, but my guess would be that you can do without the general axiom of choice, but you need something like dependent choice. –  Marc van Leeuwen Jul 11 '12 at 13:38
    
Does my argument use the axiom of choice? –  Seth Jul 11 '12 at 13:44
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Yes, or more precisely it uses the weaker version called "dependent choice". We've discussed these issues a lot on MO; see mathoverflow.net/questions/7025 , mathoverflow.net/questions/98549 , mathoverflow.net/questions/27163 –  David Speyer Jul 11 '12 at 13:45

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up vote 8 down vote accepted

You have the right idea but your proof isn't correct as stated. It isn't true that each stabilizing chain ends in a maximal ideal. For example, you could just find a Noetherian ring where $0$ isn't maximal and just repeat $0 \subset 0 \subset 0 \subset \dots$

What you should do is define this process:

Start with $0$. If $0$ is maximal then define $I_i = 0$ for $i \geq 1$. Otherwise, find some proper ideal $I_1$ that contains $0$ and put it in the list.

$0 \subset I_1$

If $I_1$ is maximal let $I_i = I_1$ for $i \geq 2$. Otherwise find some proper ideal $I_2$ that contains $I_1$ and put it in the list.

Inductively we have $0 \subset I_1 \subset I_2 \subset \dots \subset I_n \subset \dots$

By the Noetherian property, this chain stabilizes. The ideal it stabilizes to is maximal, or else by construction we would have chosen an ideal properly containing it to succeed it in the chain.

Also note that by a Zorn's lemma argument, every non zero ring has a maximal ideal. They key here is that for Noetherian rings you don't need Zorn's lemma.

EDIT: I was informed that this argument uses the axiom of dependent choice so I will rewrite it here to make this clear.

The axiom of dependent choice states that for any nonempty set $X$ and any entire binary relation $T$ on $X$, there exists a sequence $(x_n)$ such that for all $n \geq 0$, $x_n T x_{n+1}$. A binary relation $T$ on $X$ is entire if for all $x \in X$, there exists $y \in X$ such that $xTy$.

Let $X$ be the set of all proper ideals of a nonzero noetherian ring $R$. Since $R \neq 0$, $X$ is nonempty. Consider the binary relation "<" of strict inclusion. If $R$ has no maximal ideals, then "<" is entire. So by the axiom of dependent choice, if $R$ has no maximal ideals, then we may choose a sequence $(x_n)$ such that

$x_1 < x_2 < \dots < x_n < \dots$

This contradicts the Noetherian property. Hence $R$ has a maximal ideal.

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You need the Axiom of Dependant Choice, which is independant from ZF. –  Martin Brandenburg Jul 11 '12 at 13:50
    
Thanks, I rewrote the argument to make it clear how the axiom of dependant choice is used. –  Seth Jul 11 '12 at 14:13

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