$HH_1(\mathbb{C})=0$ over $\mathbb{C}$, but $HH_1(\mathbb{C})\neq 0$ over $\mathbb{Q}$

Question

I'm reading Loday's Cyclic homology and I got stuck on the following example:

Hochschild homology of $k$-algebra $A$ depends on a ground ring $k$. As an example $HH_1(\mathbb{C})=0$ over $\mathbb{C}$ (a result which is by the way true in general for $A=k$) but $HH_1(\mathbb{C})\neq 0$ over $\mathbb{Q}$.

I tried to understand this via the following proposition: If $A$ is commutative $k$-algebra with unity, then $HH_1(A)\cong \Omega^1_{A|k}$, where $\Omega^1_{A|k}$ is $A$-module of Kähler differentials over $k$ (module generated by $k$-linear symbols $da, a\in A$ obeying Leibniz rule)

Using this fact I easily obtain an alternative proof for $HH_1(\mathbb{C})=0$ over $\mathbb{C}$ as $dz=0$ for any $z\in \mathbb{C}$ (we have in general $d \lambda=0$ for $\lambda \in k$), but I can't see why exactly $HH_1(\mathbb{C})\neq 0$ over $\mathbb{Q}$.

Hint, or other ideas?

Answer 1

From the "cotangent exact sequence" for $\Omega^1$ (or general properties of derivations) you can compute $\Omega^1(\mathbb{C}/\mathbb{Q})$ by the following steps:

1. if $E/F$ and $F/k$ are field extensions with $E/F$ algebraic and separable (e.g. $E/F$ algebraic and $char(k)=0$), then $$\Omega^1(E/k)\cong E\otimes_F\Omega^1(F/k)$$

2. If $A=k[\{x_i\}_{i\in I}]$ is the polynomial on some variables, then $$\Omega^1(A/k)\cong \oplus_{i\in I}A dx_i$$ is a free A-module, with generators $\{d x_i\}_{i\in I}$

3. If $S\subset A$ is a multiplicative set of a $k$-algebra $A$, then the differentials localize: $$\Omega^1(A[S^{-1}]/k)\cong A[S^{-1}]\otimes_A\Omega^1(A/k)$$

4. Now take $C/Q$ a field extension with $char(Q)=0$, let $X\subset C$ be a maximal set of $Q$-algebraically independent elements in $C$, that is, a set of algebraically independent elements $Q(X)/Q$ is algebraic. Then, from the extensions $Q\subseteq Q(X)\subseteq C$ and the above items you get $$ \Omega^1(C/Q)\cong C\otimes_{Q(X)}(\oplus_{i\in I}Q(X)dx_i) \cong \oplus_{i\in I}Cdx_i $$ In particular,in characteristic zero, $\dim_C(\Omega^1(C/Q))$= the cardinal of a maximal set of $Q$-algebraically independent elements of $C$. Hence, an alternative proof of the fact (least in characteristic zero) that the trascendence degree is well-defined.

In the case $C=\mathbb C$, you have a non-zero element $d\pi$, or also $de$. But since we don't know if $\{\pi,e\}$ is algebraically independent or not, we don't know if $d\pi$ and $de$ are linearly independent over $\mathbb C$ in $\Omega^1(\mathbb C/\mathbb Q)$, or not :\

On the other hand, $di=0$ because $0=d(1)=d(-i^2)=-2idi$