Prove that every finite ring is noetherian.
Solution:
The statement on localizations follows from the fact that any ideal is of the form . Any quotient of a Noetherian ring R is Noetherian because any ideal is of the form for some ideal . Thus it suffices to show that if R is Noetherian so is R[X] . Suppose is an ascending chain of ideals in R[X]. Consider the ideals defined as the ideal of elements of R which occur as leading coefficients of degree d polynomials in . Clearly whenever and . By the ascending chain condition in R there are at most finitely many distinct ideals among all of the . (Hint: Any infinite set of elements of contains an increasing infinite sequence.) Take so large that and all d. Suppose for some . By induction on the degree we show that . Namely, there exists a whose degree is d and which has the same leading coefficient as f. By induction and we proved.
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