Lemma 10.31.1. Any finitely generated ring over a Noetherian ring is Noetherian. Any localization of a Noetherian ring is Noetherian.

** Noetherian property is stable by passage to finite type extension and localization. **

**Proof.**
The statement on localizations follows from the fact that any ideal $J \subset S^{-1}R$ is of the form $I \cdot S^{-1}R$. Any quotient $R/I$ of a Noetherian ring $R$ is Noetherian because any ideal $\overline{J} \subset R/I$ is of the form $J/I$ for some ideal $I \subset J \subset R$. Thus it suffices to show that if $R$ is Noetherian so is $R[X]$. Suppose $J_1 \subset J_2 \subset \ldots $ is an ascending chain of ideals in $R[X]$. Consider the ideals $I_{i, d}$ defined as the ideal of elements of $R$ which occur as leading coefficients of degree $d$ polynomials in $J_ i$. Clearly $I_{i, d} \subset I_{i', d'}$ whenever $i \leq i'$ and $d \leq d'$. By the ascending chain condition in $R$ there are at most finitely many distinct ideals among all of the $I_{i, d}$. (Hint: Any infinite set of elements of $\mathbf{N} \times \mathbf{N}$ contains an increasing infinite sequence.) Take $i_0$ so large that $I_{i, d} = I_{i_0, d}$ for all $i \geq i_0$ and all $d$. Suppose $f \in J_ i$ for some $i \geq i_0$. By induction on the degree $d = \deg (f)$ we show that $f \in J_{i_0}$. Namely, there exists a $g\in J_{i_0}$ whose degree is $d$ and which has the same leading coefficient as $f$. By induction $f - g \in J_{i_0}$ and we win.
$\square$

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## Comments (1)

Comment #911 by Matthieu Romagny on