Question

Prove that every finite ring is noetherian.

Answer 1

Solution:

The statement on localizations follows from the fact that any ideal $J \subset S^{-1} R$ is of the form $I \cdot S^{-1} R$ . Any quotient $R / I$ of a Noetherian ring R is Noetherian because any ideal $\bar{J} \subset R / I$ is of the form $J / I$ for some ideal $I \subset J \subset R$ . Thus it suffices to show that if R is Noetherian so is R[X] . Suppose $J_{1} \subset J_{2} \subset \ldots$ is an ascending chain of ideals in R[X]. Consider the ideals $I_{i, d}$ defined as the ideal of elements of R which occur as leading coefficients of degree d polynomials in $J_{i}$ . Clearly $I_{i, d} \subset I_{i^{\prime}, d^{\prime}}$ whenever $i \leq i^{\prime}$ and $d \leq d^{\prime}$ . By the ascending chain condition in R there are at most finitely many distinct ideals among all of the $I_{i, d}$ . (Hint: Any infinite set of elements of $\mathbf{N} \times \mathbf{N}$ contains an increasing infinite sequence.) Take $i_{0}$ so large that $I_{i, d}=I_{i_{0}, d} \forall i \geq i_{0}$ and all d. Suppose $f \in J_{i}$ for some $i \geq i_{0}$ . By induction on the degree $d=\operatorname{deg}(f)$ we show that $f \in J_{i_{0}}$ . Namely, there exists a $g \in J_{i_{0}}$ whose degree is d and which has the same leading coefficient as f. By induction $f-g \in J_{i_{0}}$ and we proved.