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We prove the needed Hilbert basis theorem step rather than assuming it. Suppose $A$ is a commutative Noetherian ring and $I\subseteq A[X]$ is an ideal. The set $J$ of leading coefficients of elements of $I$, together with zero, is an ideal of $A$: align degrees with powers of $X$ before addition, and use scalar multiplication. Choose finitely many generators of $J$, represented by polynomials $f_1,\dots ,f_s\in I$, and let $d$ be the maximum of their degrees. If $J=0$ then $I=0$ and there is nothing to prove.

For each $0\le j<d$, let $J_j$ consist of the coefficients of $X^j$ of elements of $I$ of degree at most $j$. This is also an ideal of $A$; choose finitely many generating coefficients and representing polynomials $g_{j,l}\in I$ of degree at most $j$. The finite collection of all $f_i$ and $g_{j,l}$ generates $I$: for any $h\in I$ of degree $m\ge d$, express its leading coefficient using those of $f_i$ and subtract the matching combination of $X^{m-\mathrm{deg}{f}_i}{f}_i$ to lower degree. When the degree becomes $j<d$, use the representatives of $J_j$ to lower it again. Induction on degree eventually leaves zero.

A field is Noetherian because its only ideals are zero and the whole field. Repeatedly applying the proved step gives every ideal of $\boxed{F[X_1,\dots ,X_n]}$ finitely generated.

The coefficient restriction in the second ring says preciselyIt contains one and is closed under addition, additive inverses and multiplication: $(a+XYp)(b+XYq)=ab+XY(aq+bp+XYpq)$. To show it is not Noetherian, consider the ideals in $R$They form an ascending chain. In any ${\sum }_{j=1}^d{r}_{j}X{Y}^j$ with $r_j\in R$, the coefficient of $X^1$ is a polynomial in $Y$ of degree at most $d$, since the nonconstant part of $r_j$ introduces at least $X^2$. It cannot equal $X{Y}^{d+1}$; hence $I_d⊊I_{d+1}$. If the union ideal were finitely generated, all its generators would lie in one $I_d$, forcing the chain to stabilize. This contradiction proves $\boxed{R\text{ is not Noetherian}}$, the constant-plus-ideal non-Noetherian subring example.

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