Krulls intersection theorem without Noetherian

Question

Krulls intersection theorem states

Let $R$ be a noetherian integral domain and $I\subset R$ an proper ideal. Then $\cap_{n>0}I^n=0$.

What are some simple counterexamples if we forget the fact that $R$ is noetherian or an integral domain?

score 2 · Accepted Answer · answered Mar 30 '18 at 16:12

2

How about $R$ is the ring of $C^\infty$ maps from $\Bbb R$ to $\Bbb R$ and $I$ be those elements of $R$ with $f(0)=0$? Then a function with vanishing Taylor series at the origin will lie in $\bigcap_n I^n$.

answered Mar 30 '18 at 16:12

Angina Seng

158,341

The function $f(x)=e^{-\frac{1}{x^2}}$ for $x\neq 0$ and $f(x)=0$ if $x=0$ would work, correct? Also what is the structure of the $k$'th power of $I$? Is that the elements of $R$ with $f^{(i)}(0)=0$ for alle $0\leq i \leq k$? – njlieta May 03 '18 at 14:36
1

That certainly works @Jta – Angina Seng May 03 '18 at 16:56

score 2 · Answer 2 · answered Mar 30 '18 at 16:18

2

$\mathbb{Z}/6\mathbb{Z}$ is noetherian but not an integral domain. Take the proper ideal $I=\{0, 2, 4\}$.

answered Mar 30 '18 at 16:18

Billy

5,224

Krulls intersection theorem without Noetherian

2 Answers2