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hello how is the krull intersection theorem if ring is not noetherian?

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What exactly are you asking here? If only "can the Noetherian hypothesis be dropped entirely so that the conclusion still holds", then the answer is no and an example has been given below. (Alternate example: take the ring of Puiseux series or any valuation ring with divisible value group.) One would hope that any algebra text which covers KIT would discuss this. (Unfortunately I doubt they all do, but I hope so...) Or do you want to know something more? – Pete L. Clark May 19 '11 at 16:27

It fails. For example, let $R$ be the ring of germs of smooth real functions at the origin, and let $I$ be the ideal generated by $x$. Now take the function $f(x) = e^{-1/x^2}$. The Taylor expansion of $f$ at $0$ is $0$, so $f$ belongs to $\bigcap_{n=0}^{\infty} I^n$, but $f$ is clearly not the zero function.

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Just a comment: since germs are being considered, $R$ is local and the conclusion of KIT would be that the intersection is zero. However, the same function in the ring of all smooth functions from $\mathbb{R}$ to $\mathbb{R}$ still contradicts the more general form of KIT which has as its conclusion $f \in fI$. If for all $x \in \mathbb{R}$ we had $f(x) = f(x) x g(x)$, then for all $x \neq 0$, $1 = x g(x)$, contradicting the fact that $g$ is smooth at $0$. – Pete L. Clark May 19 '11 at 17:03

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