Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 85 / 2 / Solution

The variable $t$ is central. To prove the leading-coefficient proof for a central polynomial extension, let $I$ be any right ideal of $R[t]$ and setZero leading coefficients are allowed. Addition and right multiplication by constants show that $L_d$ is a right ideal. Multiplication by $t$ gives $L_d\subseteq L_{d+1}$. Since $R$ is a right Noetherian ring, this chain stabilizes, say at $L_N$, and each $L_d$ for $d\le N$ has finitely many generators $a_{dj}$. Choose $f_{dj}\in I$ of degree at most $d$ lifting these coefficients.

These finitely many polynomial lifts generate $I$ on the right. Indeed, for $f\in I$ of degree $m$, let $d=\min (m,N)$. Write its leading coefficient as ${\sum }_j{a}_{dj}{r}_j$. Thenbelongs to $I$ and has degree less than $m$, because $L_m=L_d$ when $m>N$. Induction on degree reduces $f$ to a right linear combination of the chosen lifts. This proves finite generation of every right ideal, and henceCentrality matters: it is what permits the displayed leading-coefficient cancellation without twisting the coefficients.