Let $k$ be a field and $K_0(Var_k)$ the Grothendieck ring of varieties over $k$. This is the ring generated by isomorphism classes of varieties over $k$ with multiplication given by

$$ [X \times_k Y] = [X][Y] $$

and the relation that $[X] = [X \setminus Z] + [Z]$ for any closed subvariety $Z \subset X$.

It is well known that if $\pi :X \to Y$ is a Zariski local fibration with fiber $F$, then $[X] = [F][Y]$.

I assume this result is false if $\pi$ is only etale locally a fibration. Is there a counterexample?

  • 5
    $\begingroup$ A very simple concrete counterexample is $\mathbb{C}^*\to \mathbb{C}^*$ by $z\mapsto z^2$. $\endgroup$ – Jim Bryan Mar 4 '15 at 16:02

Yes, even in the quotient ring of $K_0(Var_k)$ by the class $L$ of $\mathbb{A}^1$. By Larsen-Lunts this quotient is the free abelian group over classes of varieties up to stable birational equivalence. Thus it suffices to find a $\mathbb{P}^1$-bundle $P\rightarrow X$ where $P$ is rational and $X$ is not stably rational (so $P$ is zero in the quotient, but $X$ is not). This is provided by the Artin-Mumford example : its Brauer group is $\mathbb{Z}/2$, hence it admits a nontrivial $\mathbb{P}^1$-bundle; this $\mathbb{P}^1$-bundle can be constructed explicitely and turns out to be rational, see §9 in A. Beauville, Variétés rationnelles et unirationnelles, Algebraic Geometry - Open problems (Proc. Ravello 1982), LN 997, 16-33; Springer-Verlag (1983).


To clarify what's happening, let us introduce the etale Grothendieck ring varieties $K^{et}(Var/k)$ by imposing the scissor congruence relation AND the relation $[X] = [F][Y]$ for every finite etale covering $X \to Y$ with (finite) fiber $F$.

Let us prove that at least when $k$ has characteristic zero, we have $K^{et}(Var/k)$ isomorphic to $\mathbb{Z}$ via the Euler characteristic with compact supports.

The proof goes in three steps. First we show by induction on dimension and using Noether Normalization that every element of $K^{et}(Var/k)$ is represented by a polynomial $P(\mathbf{L})$ in the Lefschetz class. The key observation is that any $n$-dimensional variety $X$ has an open subset $U$ which is a finite etale covering of an open subset of a projective space $U' \subset \mathbf P^n$.

Then we show that $\mathbf{L} = 1$; this follows using the $2:1$ covering $\mathbf G_m \to \mathbf G_m$, $z \mapsto z^2$ as in Jim Bryan's comment. Thus every element in $K^{et}(Var/k)$ is represented by an integer.

Finally, since the Euler characteristic respects etale coverings, it gives a well-defined homomorphism $K^{et}(Var/k) \to \mathbb Z$, and we just showed that this is an isomorphism!


A variant of abx's example, that also uses Larsen-Lunts's theorem: in characteristic zero, it follows from this theorem that two (projective, smooth, connected) curves have the same class in the Grothendieck group of varieties if and only if they are isomorphic. It thus suffices to take an étale covering of such curves $f\colon Y\to X$ where $Y$ and $X$ are not isomorphic.

It is natural to try adding the relation $[Y]=n[X]$ when $f$ is such an étale cover of degree $n$. However (in characteristic zero), the corresponding quotient of the Grothendieck group of varieties is isomorphic to $\mathbf Z$, corresponding to the Euler characteristic.


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