Which constructions preserve the Noetherian condition, and which do not

What is proved above. Starting from a Noetherian commutative ring $R$, each of the following is again Noetherian:

• every quotient $R\mathrm{/}I$ and every localisation $S^{-1}R$ (Every quotient and every localisation of a Noetherian ring is Noetherian);
• the polynomial ring $R[x]$ (Hilbert basis theorem: if $R$ is Noetherian then $R[x]$ is Noetherian) and, by induction, $R[x_1,\dots ,x_n]$ for each $n\in \mathbb{N}$ (If $R$ is Noetherian then $R[x_1,\dots ,x_n]$ is Noetherian for every $n\in \mathbb{N}$);
• every commutative $R$-algebra of finite type (Every algebra of finite type over a Noetherian ring is a Noetherian ring);
• every module-finite commutative $R$-algebra, and every ring between $R$ and such an algebra (A module-finite algebra over a Noetherian ring is a Noetherian ring, and so is every ring between the two);
• the product $R\times S$ with $S$ Noetherian (A product of two Noetherian rings is Noetherian);
• a subring $R_0\subseteq R$ that admits an $R_0$-linear retraction $R\to R_0$ fixing $R_0$ pointwise (A subring that admits a module retraction from a Noetherian ring is Noetherian).

What fails, and why it is worth saying. Every entry above supplies some map or some finiteness relating the new ring to $R$. Two natural-looking weakenings supply neither, and both fail.

An arbitrary subring. Being an additive subgroup closed under multiplication gives no way to pull a generating set of an ideal of the subring back from the larger ring, and the conclusion is false: a subring of a Noetherian ring need not be Noetherian. The companion examples page of this pair works a witness inside a polynomial ring in two variables, where the failing ideal is visibly not finitely generated. The retraction hypothesis is exactly what is missing: it is a map back, and with it the argument runs.

Infinitely many indeterminates. The polynomial-ring statement is proved by an induction on the number of indeterminates, and that induction has no limit stage: it covers a finite list and no more. The companion examples page of this pair carries the witness, a ring of polynomials in countably many indeterminates in which the ideals generated by initial segments of the variables form a strictly ascending chain. The same remark applies to the product statement, which is proved for two factors and extends by iteration to a finite list; a product indexed by an infinite set is not among the constructions A product of two Noetherian rings is Noetherian speaks about.

A hypothesis that is not needed anywhere above. No result above assumes that $R$ is an integral domain, that $R$ is nonzero, or that its ideals are principal. The zero ring is Noetherian and is admitted throughout, and rings with zero divisors are admitted in the Hilbert basis argument in particular, which never multiplies two leading coefficients together.