Exercise A0.V.1.04

Proposition

The ring $R$ of continuous functions $[0,1] \to \mathbb{R}$ is not Noetherian.

Proof

Let $I_n$ denote the ideal of $R$ containing functions that vanish on the interval $[0, n^{-1}].$ It is clear that for all $n > 0,$ $I_n$ is indeed an ideal.

If a function vanishes on $[0, n^{-1}],$ it certainly also vanishes on $[0, (n+1)^{-1}],$ since $[0, (n+1)^{-1}] \subseteq [0, n^{-1}]$. Thus we have an infinite ascending chain of ideals.

For each $n,$ we have a function $s_n(x)$ that is $0$ on $[0, n^{-1}]$ and is $x - n^{-1}$ elsewhere. It is continuous, since it is made by gluing two continuous functions at an overlap point.

This function doesn’t vanish on any of the larger intervals, and so is not present in any of the previous ideals. Hence, the chain never stabilizes. We conclude that $R$ is not Noetherian.

$\blacksquare$