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I want to proof that a noetherian ring $R \neq \{0\}$ contains at least one maximal ideal. My idea is to consinder $\langle 0 \rangle$ 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 $\langle 0 \rangle \subset I_1 \subset \dots$ 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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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 stabalizes. The ideal it stabalizes 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 dependant choice so I will rewrite it here to make this clear. The axiom of dependant choice states that for any nonempty set $X$ and any entire binary operation $T$ on $X$, there exists a sequence $(x_n)$ such that for all $n \geq 0$, $x_n T x_{n+1}$. A binary operation $T$ 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 operation "<" 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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