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.
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$\begingroup$ I'd consider the maximal ideals of each. $\endgroup$ – Angina Seng Jun 27 '20 at 17:59
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$\begingroup$ 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. $\endgroup$ – xyzzyz Jun 27 '20 at 18:08
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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 $0$).
answered Jun 27 '20 at 22:21
