Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 2 / 11G (Groups, Rings and Modules)

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.