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.


2 Answers 2


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!!

  • $\begingroup$ I completely forgot about this question! Nice, +1. $\endgroup$
    – Hmm.
    Jul 2, 2016 at 15:16
  • $\begingroup$ Phrased differently: every element of $C(\Bbb R)$ is a cube whilst not every element of $D(\Bbb R)$ is (e.g. $x$). $\endgroup$
    – anon
    Jul 2, 2016 at 15:38

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.

  • 1
    $\begingroup$ How do we know that $C(\mathbb R)$ admits just one trivial derivation? $\endgroup$ Jun 8, 2016 at 12:36
  • 3
    $\begingroup$ @ Jon Warneke, see here for a proof- ncatlab.org/nlab/show/derivation#DerOfContFuncts $\endgroup$
    – Hmm.
    Jun 8, 2016 at 12:37
  • $\begingroup$ it is more equivalent to saying that by the definition and properties of polynomials, $\varphi(\sum_k c_k x^k) = \sum_k c_k (x+a)^k$ so the only ones isomorphisms of such function rings containing the polynomials is $f(x) \mapsto f(x+a)$ ? $\endgroup$
    – reuns
    Jun 8, 2016 at 12:52

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