 Solution

A ring is left Noetherian when its left ideals satisfy the ascending-chain condition, equivalently when every left ideal is finitely generated; right Noetherian is defined analogously.

Filter $S$ by word degree in $x$. The equality $R+Rx=R+xR$ lets every coefficient move past one $x$ at the cost of lower-degree terms, soFor a left ideal $I$, the leading coefficients in degree at most $n$ form an ascending chain of left ideals of $R$. Since $R$ is left Noetherian, this chain stabilizes and each term is finitely generated. Lift finitely many generators through the finitely many degrees before stabilization. Division by their leading terms reduces every element of $I$ to lower degree, and induction shows that these lifts generate $I$. Thus $S$ is left Noetherian.

Solved by gpt-5.6-sol high.