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No. Take $R=k[t]$ over a field $k$, and inside $R[x]=k[t,x]$ takeThis is a subring containing $R$: its elements have arbitrary constant coefficient in $k[t]$ as polynomials in $x$, and every positive-degree coefficient is divisible by $t$; that condition is preserved by sums and products. The ring $R$ is Noetherian. Indeed, a nonzero ideal of $k[t]$ has an element $g$ of least degree, and division by $g$ shows that every element of the ideal is a multiple of $g$. Thus $k[t]$ is a principal ideal domain, and its ideals are finitely generated.

In $S$, let $J_n=(tx,t{x}^2,\dots ,t{x}^n{)}_S$. These form an ascending chain. We show that $t{x}^{n+1}\notin J_n$. If it belonged, we could writeCancel $t$ in the integral domain $k[t,x]$, then apply the evaluation homomorphism $t\mapsto 0$. Every $s_j\in S$ becomes a scalar in $k$, so this would givewhich is impossible by comparison of degrees. Hence $J_n⊊J_{n+1}$ for every $n$, andThis is a constant-plus-ideal non-Noetherian subring construction: the restricted coefficients prevent a finite list of positive powers of $x$ from generating all the required powers.

The coefficient condition says precisely that every nonconstant monomial has positive powers of both variables. ThusIf $c+XYf$ and $d+XYg$ lie in $R$, so do their sum and their productso $R$ is a subring of $K[X,Y]$.

For $n\ge 1$, letThen $I_n\subseteq I_{n+1}$. The containment is strict because $X{Y}^{n+1}\notin I_n$: multiplying a generator $X{Y}^j$ by an element of $R$ produces either a scalar multiple of $X{Y}^j$, or terms divisible by $X^2$. It cannot produce the monomial $X{Y}^{n+1}$ when $j\le n$. Henceis a strictly increasing ideal chain. The constant-plus-ideal non-Noetherian subring therefore satisfies