# About the existence of an isomorphism of algebras between the algebras of $\mathcal{C}^0$ and $\mathcal{C}^1$ functions from $[0, 1]$ to $\mathbb{R}$

Let $$\mathcal{C}^0 ( [0, 1], \mathbb{R})$$ denote the set of continuous functions from $$[0, 1]$$ to $$\mathbb{R}$$ and $$\mathcal{C}^1 ( [0, 1], \mathbb{R})$$ denote the set of class $$\mathcal{C}^1$$ functions from $$[0, 1]$$ to $$\mathbb{R}$$, both of these sets are algebras over the field $$\mathbb{R}$$.
The question is whether we can find an isomorphism of algebras:
$$\Phi:\mathcal{C}^0 ( [0, 1], \mathbb{R})\longrightarrow\mathcal{C}^1 ( [0, 1], \mathbb{R})$$
it seems, well at least to me, that such a cheesy isomorphism cannot exist, however I couldn't prove it.
I tried the obvious choice, which is proof by contradiction, and tried taking the inverse of such a function which would map differentiable functions to continuous ones, considering that one set already contains the other, however no contradiction seemed to arise.

• I'd consider the maximal ideals of each. – Angina Seng Jun 27 '20 at 17:59
• Maximal ideals of both are $V(x_0) = \{f \in A: f(x_0) = 0\}$, where $x_0$ is some point in $[0, 1]$, and $A$ is either $C^0([0, 1])$ or $C^1([0, 1])$. So, I don't think it helps much. – xyzzyz Jun 27 '20 at 18:08

No such isomorphism exists. A quick way to prove it is to observe that every element of $$\mathcal{C}^0 ( [0, 1], \mathbb{R})$$ has a cube root, but not every element of $$\mathcal{C}^1 ( [0, 1], \mathbb{R})$$ has a cube root (for instance, the function $$f(x)=x$$ does not since $$x\mapsto \sqrt[3]{x}$$ is not differentiable at $$0$$).