The ring of all continuously n-times differentiable functions is not an integral domain for all n?

Question

Let $\mathbb{Z^+}$ be the set of all non-negative integers. For each $n\in \mathbb{Z^+}$ let $S_{n} =C^{n}([0,1], \mathbb{R})$ be the ring of all continuously n-times differentiable functions where $S_{0}=C([0,1],\mathbb{R})$. So what can be the possible choice for n for which $S_{n}$ fails to be an integral domain? Obviously not 1 and 2..

Thanks for any help!

Answer 1

Let

$$f_n = \begin{cases} 0 &\text{if }x \in [0,1/2], \\ (x-1/2)^{n+1} &\text{if }x\in [1/2,1].\end{cases}$$

and $g_n(x) = f(-x+1/2)$. Then $f_n, g_n \in C^n([0,1],\mathbb R)$ and $f_n g_n =0$.

So $S_n$ fails to be an integral domain for all $n\in \mathbb Z^+$.

Answer 2

It fails for every $n$, even $n=\infty$, you might want to have a look at the bump function.