Question #202611

Prove that every finite ring is noetherian.


1
Expert's answer
2021-06-04T05:10:43-0400

Solution:

The statement on localizations follows from the fact that any ideal JS1RJ \subset S^{-1} R is of the form IS1RI \cdot S^{-1} R . Any quotient R/IR / I of a Noetherian ring R is Noetherian because any ideal JˉR/I\bar{J} \subset R / I is of the form J/IJ / I for some ideal IJRI \subset J \subset R . Thus it suffices to show that if R is Noetherian so is R[X] . Suppose J1J2J_{1} \subset J_{2} \subset \ldots is an ascending chain of ideals in R[X]. Consider the ideals Ii,dI_{i, d} defined as the ideal of elements of R which occur as leading coefficients of degree d polynomials in JiJ_{i} . Clearly Ii,dIi,dI_{i, d} \subset I_{i^{\prime}, d^{\prime}} whenever iii \leq i^{\prime} and ddd \leq d^{\prime} . By the ascending chain condition in R there are at most finitely many distinct ideals among all of the Ii,dI_{i, d} . (Hint: Any infinite set of elements of N×N\mathbf{N} \times \mathbf{N} contains an increasing infinite sequence.) Take i0i_{0} so large that Ii,d=Ii0,dii0I_{i, d}=I_{i_{0}, d} \forall i \geq i_{0} and all d. Suppose fJif \in J_{i} for some ii0i \geq i_{0} . By induction on the degree d=deg(f)d=\operatorname{deg}(f) we show that fJi0f \in J_{i_{0}} . Namely, there exists a gJi0g \in J_{i_{0}} whose degree is d and which has the same leading coefficient as f. By induction fgJi0f-g \in J_{i_{0}} and we proved.


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