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The kernel of $x\mapsto t^3$, $y\mapsto t^4$, $z\mapsto t^5$ is generated by $y^2-xz$, $yz-x^3$, $z^2-x^{2}y$. Reducing these three relations leaves a unique normal form $A_0(x)+A_1(x)y+A_2(x)z$: its images have distinct exponents modulo three. The curve ring is therefore finite free of rank three over $k[x]$ and has dimension one. In a three-variable polynomial ring its ideal has height two, but its three independent quadratic initial forms force at least three generators. This is a concrete failure of generation by codimension many equations.

The nilradical of a commutative ring is its ideal of nilpotent elements, equivalently the radical of an ideal applied to its zero ideal.

For every commutative ring $R$,Every nilpotent element lies in every prime ideal. If $a$ is not nilpotent, an ideal maximal among those disjoint from $\{1,a,a^2,\dots \}$ is a prime ideal avoiding $a$, by the Zorn lemma and the maximality argument.

In $R=k[t_1,t_2,\dots ]/(t_1^2,t_2^2,\dots )$, the nilradical is generated by all ${\bar{t}}_i$. Every element uses finitely many variables and is nilpotent, while each square-free product ${\bar{t}}_1\cdots {\bar{t}}_j$ is nonzero. Thus this nilradical is not a nilpotent ideal. A Noetherian ring cannot have this behavior: finitely many nilpotent generators give a nilpotent ideal.

A formal power series ring $R[[x]]$ is Noetherian if and only if $R$ is Noetherian. In the forward direction, use the quotient ring $R[[x]]/(x)\cong R$. In the reverse direction, for an ideal $I$ consider the ascending sequence of ideals of $R$ formed by the coefficient of $x^n$ in $I\cap x^{n}R[[x]]$. Stabilization and finite generation give a finite list of series that generate $I$ by successive coefficient cancellation.

Every unital $k$-subalgebra $A\subseteq k[T,T^{-1}]$ is a Noetherian ring. If $A\subseteq k[T]$ or $k[T^{-1}]$, a nonconstant element $f\in A$ makes that ambient polynomial ring a finitely generated module over $k[f]$. Otherwise $A$ contains an $f$ with both positive and negative exponents, and both $T$ and $T^{-1}$ are integral elements over $k[f]$. In either case $A$ is a submodule of a finite module over the Noetherian ring $k[f]$, and every ideal of $A$ is finitely generated over $k[f]$, hence over $A$.