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Let's denote ring of all continuous functions and differentiable functions from $\mathbb{R}$ to $\mathbb{R}$ by $C(\mathbb{R})$ and $D(\mathbb{R})$, respectively. I want to know whether these rings are isomorphic or not.
From the first part of this answer it is clear that if there exists a ring monomorphism which sends 1 to itself then it is identity on the constant functions. But I can not find anything else. Please help. Thank you.
Suppose they are isomorphic. Then consider the identity function $Id$ which belongs to $D(\mathbb{R})$. Let the pre image of $Id$ is $f$ in $C(\mathbb{R})$. Now if $f\in C(\mathbb{R})$ then $f^{1/3}\in C(\mathbb{R})$. Then by properties of isomorphism image of $f^{1/3}$ is $Id^{1/3}$. But $Id^{1/3}$ is not differentiable at $0$ and so it is not in $D(\mathbb{R})$. Hence contradiction!!
I might be wrong, but in case by $D(\mathbb{R})$ you mean $C^{\infty}(\mathbb{R})$, I think we can arrive at a contradiction. Suppose we have such an isomorphism from $C(\mathbb{R})$ to $C^{\infty}(\mathbb{R})$, call it $\varphi$. We define a derivation on $C(\mathbb{R})$ as below :
$\forall f\in C(\mathbb{R})$, $\delta(f)=\varphi^{-1}\biggl[\biggl(\varphi(f)\biggr)'\biggr]$.
By the standard properties of differentiation, one can check that $\delta$ indeed defines a derivation on $C(\mathbb{R})$. However, $C(\mathbb{R})$ admits only one derivation, namely the trivial derivation. Thus, $Im(\varphi)$ only consists of constant functions, and we have a contradiction.
