Showing $d\pi$ is non-zero in module of Kähler differentials While reading about Kähler differentials I came across the seemingly innocent statement that $d\pi \in \Omega_{\mathbb{R/Q}}$ in non-zero. This is kind of "obvious" as if $\alpha$ is algebraic then there is a simple way of showing $d\alpha=0$ which fails if $\alpha$ is transcendental.
To actually prove $d\pi \neq 0$ though seems quite tricky, I imagine we must use the universal property of $\Omega_{\mathbb{R/Q}}$ somewhere, but how I have no idea.
Any ideas would be much appreciated.
 A: The general fact you're looking for is that if $E/F$ is a field extension and $t_1,\cdots,t_n$ is a transcendence basis for $E$ over $F$ (not necessarily finite), then $dt_i$ are a basis for $\Omega_{E/F}$. All the facts necessary for this should be in your favorite commutative algebra textbook, and I'll sketch the proof below:


*

*$\Omega_{k[x_1,\cdots,x_n]/k}$ is free on $dx_i$. This follows from Yoneda and the product rule: we can arbitrarily assign any value to all the $dx_i$ and that determines a derivation. Note that there is no reason to restrict to finite $n$ here.

*$\Omega$ commutes with localization. This lets us pass from $k[x_1,\cdots,x_n]/k$ to $k(x_1,\cdots,x_n)/k$, and see that $\Omega$ is still free on the $dx_i$.

*$\Omega$ plays well with algebraic field extensions, ie if $E/F/k$ is a tower of field extensions with $E/F$ algebraic, then $E\otimes_k \Omega_{F/k} \to \Omega_{E/k}$ is an isomorphism. This can be proven from the first exact sequence: one treats the case of a finite extension using the theorem of the primitive element, and then uses the fact that every algebraic field extension is the colimit of it's finite subextensions to get the result. 

*So if we set up a tower $\Bbb R/\Bbb Q^{tr}/\Bbb Q$ where $\Bbb Q^{tr}$ is a maximal purely transcendental subextension of $\Bbb R/\Bbb Q$ chosen so that $\pi$ is in it's transcendence basis over $\Bbb Q$, we see the result we were after.
