# A zero-dimensional ring is Noetherian?

Proposition: Let $A$ be a non-zero ring that is not a field. Suppose $A$ is zero dimensional. Then it is Noetherian.

Proof: Let $p$ be a prime ideal of $A$. If $p$ is not maximal, then $p \subsetneq m$ for a maximal ideal $m$. Hence $dim A \ge 1$, contradiction. Hence every prime ideal is maximal. Consequently, every increasing (or decreasing) chain of prime ideals will be trivial in the sense that it will have length $0$. Hence the ring is Noetherian (and Artinian).

Is there a problem with the above proof?

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Yes. The Noetherian condition states that every increasing sequence of (not necessarily prime) ideals stabilizes. –  Alex Becker Jan 8 at 23:13
Just because every increasing chain of primes is trivial doesn't mean every ideal is finitely generated... –  Dylan Wilson Jan 8 at 23:13
As stated in the comments, your proof fails because the Noetherian condition refers to arbitrary ideals, not prime ones. In fact, the proposition you are trying to prove is not true. Several counterexamples are given here, the simplest being $\mathbb Q[x_1,x_2,\ldots]/I$ where $I$ is the ideal generated by $x_ix_j$ for $1\le i\le j$.