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  "body": "\n## Statement\n\nLet $R$ be a [[commutative ring|0GYGHKX834]].\nThen $R$ is [[Noetherian|0MVH4Y1CRZ]] if and only if every [[prime ideal|06HC6HK0Z8]] of $R$ is [[finitely generated|030TZ0DQWT]].\n\n## Remarks\n\nOne direction is clear, since in a Noetherian ring every ideal is finitely generated.\nFor the other, suppose that some ideal is not finitely generated.\nBy Zorn's lemma there is an ideal $I$ that is maximal among the ideals that are not finitely generated.\nLet $a, b \\notin I$ with $ab \\in I$.\nThen $I + (a)$ and the ideal quotient $(I : a) = \\set{r \\in R : ra \\in I}$ are strictly larger than $I$, hence finitely generated.\nOne checks that $I$ would then be finitely generated, which is impossible; so $I$ is prime.\nThis prime ideal is not finitely generated, against the hypothesis.\nFor example, to show that $\\Z[x]$ is Noetherian it would be enough to check its prime ideals, although [[Hilbert's basis theorem|0YKA9J3XEE]] gives this more directly.\n\n## Explanations\n\n### L1 human\n\nTo check that every ideal of a commutative ring is finitely generated, it is enough to check the prime ideals.\nThe prime ideals are often easier to list, so this saves work.\n\n### L2 human\n\nThe proof is a model of a common method: an ideal maximal for not having a property is often prime.\nThe same method shows that a commutative ring in which every prime ideal is principal is a principal ideal ring.\nThe theorem is a test for the [[Noetherian|0670T59HGE]] property, for example for rings of power series and rings of integers.\n",
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