Is the Kähler differential of a continuous function ring trivial? Suppose $A=C^0(\mathbb R)$ is the ring of real-valued continuous functions on $\mathbb R$. Is it true that, the Kähler differential $\Omega_{A/\mathbb R}$ trivial? In other words, suppose that $d\colon A\to M$ is a $\mathbb R$-linear map, where $M$ is a $A$-module, such that $d(1)=0$ and $d(ab)=adb+bda$ for $a,b\in A$, is it always true that $d$ is a zero map?
It's known that any derivative $D\colon A\to A$ is trivial. The key idea is that $Df=2\sqrt fD(\sqrt f)$ and that if $f(x)=0$ then $Df(x)=0$. In general, this argument shows that $d(A)\subseteq\mathfrak m_xM$ for all $x\in\mathbb R$, where $\mathfrak m_x$ be the maximal ideal of functions $f\in A$ such that $f(x)=0$.
Any ideas? Thanks!
 A: No, $\Omega_{A/\mathbb{R}}\neq 0$. If $\mathbb{R}\subset K$, where $K$ is a field of positive transcendence degree over $\mathbb{R}$, then there exists non-zero $\mathbb{R}$ derivations $D:K\to K$. If we can find a quotient of $A$ to such a $K$, then by composing, we get a non-zero derivation $D:A\to K$ and thus $\Omega_{A/\mathbb{R}}\neq 0$. Such examples of $A\to K$ are well known and if you need one, I can write up one for you.  Here is such an example, as you wanted. For $n\in\mathbb{Z}$ define $f_n(x)=0$ for $x<n$ and $f_n(x)=x-n$ for $x\geq n$. Then $f_n$ is continuous and let $I$ be the ideal generated by all the $f_n$'s. Then $I$ is a proper ideal, since if not, we can write $1=\sum_ia_{n_i}f_{n_i}$ for some finitely many $n_i\in\mathbb{Z}$. But this is impossible, since the right side vanishes for any $x<\min\{n_i\}$. So, $I$ is contained in a maximal ideal, say $M$ and I claim that $A/M=K$ has positive transcendence degree over the reals. I will leave you to check that the image of $e^x$ has this property, by checking that if both $e^x-a, e^{2x}-a$, for any real number $a$ along with $I$ generates the unit ideal. 
