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

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. 

