Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 101 / 4

A ring is a Noetherian ring if it satisfies the ascending chain condition on ideals: every chaineventually becomes constant. Equivalently, every ideal has a finite generating set of an ideal. To see this equivalence, if an ideal were not finitely generated, one could repeatedly choose an element outside the ideal generated by the preceding choices, producing a strictly ascending chain. Conversely, the union of an ascending chain is an ideal; if it is generated by finitely many elements, all of them lie in one term of the chain, and every later term equals that term.

A ring is an Artinian ring if it satisfies the descending chain condition on ideals: every chaineventually becomes constant. These are stabilization conditions on all ideals, not merely on principal ideals.

Yes: $R[[x]]$ is Noetherian. We prove Noetherianity of a formal power series ring directly, using only the equivalence in part (a). Let $A=R[[x]]$, a formal power series ring, and fix an ideal $I\subseteq A$. For $n\ge 0$, let $C_n$ be the set of coefficients of $x^n$ in elements of $I\cap x^{n}A$. It is an ideal of $R$, because coefficient extraction is additive and respects multiplication by constants. Multiplication by $x$ givesSince $R$ is Noetherian, there is an $N$ with $C_n=C_N$ for all $n\ge N$. Each $C_n$, for $0\le n\le N$, has a finite generating set of an ideal $a_{n1},\dots ,a_{n{r}_n}$. Choose $f_{nj}\in I\cap x^{n}A$ whose coefficient of $x^n$ is $a_{nj}$. If $C_n=0$, use no generators at that index.

We claim that these finitely many $f_{nj}$ generate $I$ over $A$. Given $f\in I$, start with $g_0=f$ and recursively construct $g_n\in I\cap x^{n}A$. For $n<N$, express the coefficient of $x^n$ in $g_n$ as ${\sum }_j{c}_{nj}{a}_{nj}$, with $c_{nj}\in R$, and subtract ${\sum }_j{c}_{nj}{f}_{nj}$ to obtain $g_{n+1}\in I\cap x^{n+1}A$. For $n\ge N$, use $C_n=C_N$ and subtractto cancel that coefficient. Every step involves a finite sum and leaves a residual still in $I$.

For each $j$, define the formal power seriesThe residual after the $n$th cancellation lies in $x^{n+1}A$, so comparison of each coefficient gives the exact identityThe right side is a finite $A$-linear combination of the chosen generators. No assertion that arbitrary ideals are closed under limits is being used: the equality is verified coefficient by coefficient, and the multipliers $h_j$ belong to $A$. Thus every ideal of $A$ is finitely generated, proving

Yes. Taking the constant coefficient defines a surjective ring homomorphismIts kernel of a ring homomorphism is $(x)$, so the first isomorphism theorem for rings gives $R[[x]]/(x)\cong R$.

We prove the needed preservation result. If $A$ is a Noetherian ring and $J\subseteq A$ is an ideal, any ascending chain of ideals in $A/J$ lifts under the quotient ring homomorphism to an ascending chain of ideals in $A$. The lifted chain stabilizes; surjectivity makes the original chain stabilize as well. Hence every quotient ring of a Noetherian ring is Noetherian.

Applying this with $A=R[[x]]$ and $J=(x)$ gives

No. For example, take $R=k$, any field. It is an Artinian ring, since its only ideals are zero and itself. In the formal power series ring $k[[x]]$, however,is a strictly descending chain of ideals. The inclusion is strict because $x^n$ has a nonzero coefficient of $x^n$, whereas every element of $(x^{n+1})$ has that coefficient zero. Thus $k[[x]]$ fails the descending chain condition and is not Artinian.

The same argument works for any nonzero coefficient ring $R$: the coefficient $1$ in $x^n$ cannot be produced by a multiple of $x^{n+1}$. Therefore $R[[x]]$ is never Artinian when $R\ne 0$. The zero ring is the harmless exception.

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.