Expert answers to "Why the ring S−1R is Noetherian if S..."

ordanlarroque1xm

Answered question

2022-03-08

Why the ring

S^{- 1} R

is Noetherian if S is multiplicative?

oighreacha12

Beginner2022-03-09Added 2 answers

Step 1
Assume the contrary that the following ascending chain of ideals in

S^{- 1} R

does not stop

S^{- 1} I_{1} \subseteq S^{- 1} I_{2} \subseteq S^{- 1} I_{3} \subseteq S^{- 1} I_{4} \subseteq \dots .

So there is not any integer n such that

S^{- 1} I_{n} = S^{- 1} I_{n + 1}

. Hence we find the following ascending chain of ideals in R which does not stop.

I_{1} \subseteq I_{2} \subseteq I_{3} \subseteq I_{4} \dots .

This is a contradiction.

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