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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.

md2perpe
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MadChickenMan
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  • I'm totally new to this topic, so my question might be stupid. When I read the definition I see nothing that hinders the derivation $d$ to be just the zero map, i.e. $ds=0$ for all $s \in S.$ In this case you would have $d\pi=0.$ But I guess that you are considering non-trivial cases or that there is something more in the definition that I am missing. – md2perpe Aug 24 '18 at 15:21
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    Well $\Omega_{\mathbb{R/Q}}$ is defined by taking the free $\mathbb{R}$-module with generators just formal symbols $d\alpha$ (one for each real number $\alpha$), then taking the quotient by the submodule with generators such as $d(st)-sdt-tds$ (as is described in the link you gave). So to say $d\pi \neq 0$ is to say $d\pi$ does not lie in this submodule, which is hard to show directly, but by the universal property it is enough to show we can find an $\mathbb{R}$-module $M$ with derivation $\delta$ say such that $\delta \pi \neq 0$, though this is still hard to do, hence my question. – MadChickenMan Aug 24 '18 at 15:32

1 Answers1

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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:

  1. $\Omega_{k[x_1,\cdots,x_n]/k}$ 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.

  2. $\Omega$ 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 $dx_i$.

  3. $\Omega$ 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.

  4. 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.

KReiser
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