@Bjorn: You are right of course, I misremembered. (In my defence, an example of non-cancellation also gives examples of zero-divisors in the Grothendieck ring, the tricky thing is to get an example over $\mathbb Q$.)

I think you may be thinking of the case of a finite morphism $\pi\colon X \to Y$ where $X$ is smooth. In that case the singularities of $Y$ are indeed quotient singularities. Not necessarily cyclic quotient singularities however; an $E_8$-singularity for instance is the quotient of $\mathbb C^2$ by the icosahedral group which is not cyclic.

There are many more normal surface singularities than cyclic quotient singularities or for that matter rational singularities (which is defined by the condition of the second paragraph). You must have misinterpreted something. To take just one example: The cone point of the affine cone of an elliptic curve (embedded in the projective plane say) is not a rational singularity.

It is false also in characteristic $0$ though there it is true for elliptic curves (see Angelo's reply). It is a question of finding an example of non-cancellation for projective modules over a suitable ring and then mirror it for abelian varieties. See Poonen, "The Grothendieck ring of varieties is not a domain" for an example. (Bjorn's example is somewhat involved as he wants an example over $\mathbb Q$. An example over $\mathbb C$ is easier to construct.)

Note that just adding additive inverses to the semiring does <em>not</em> give the usual Grothendieck group. In fact the semiring is, by the Krull-Remak-Schmidt theorem, the free abeliam semigroup on the (isomorphism classes of) indecomposable vector bundles. Hence its group completion is just the free abelian group on the same vector bundles. Hence it is, as Angelo pointed out for the semiring, truly huge. Except in the $1$-dimensional case when the only indecomposable vector bundles are the line bundles.

Sorry, the last $(\mathbb Z/p)^n$ should be $(\mathbb Z_p)^n$.

This is maybe too tautological a characterisation: The complex is flag and there is a total order on the vertices such that if $x<y<z$ and there is one edge between $x$ and $y$ and one between $y$ and $z$, then there is an edge between $x$ and $z$. (In one direction it uses the fact that a partial order can be extended to a total one.)

This looks to be in contradiction with the result that says that the invariant ring is not Cohen-Macaulay as a polynomial ring is certainly Cohen-Macaulay. Also according to the Math Review the condition is that the subspace $\{gv-v\}$ be $1$-dimensional not of codimension $1$. (The space of covariants is the quotient of $V$ by that space.)

[ cont'd ] The Lie algebra is classified by a representation of the Galois group of the base field in $(\mathbb Z/p)^n$. On the other hand $U_{dp}$ is classified by a representation of the Galois group of the base field in $(\mathbb Z_p)^n$. Many representations in $(\mathbb Z/p)^n$ do not lift to a representation in $(\mathbb Z/p)^n$.

[ I had to split this up into two pieces because of size. ] Are you saying that one can construct an $U_{dp}$ from a restricted Lie algebra? At least if you want it canonically as a Hopf algebra this seems fishy. In the case of a restricted Lie algebra which over an algebraic closure is the Lie algebra of a torus (of dimension $n$) we should have the following situation:

@Marc: Specifying in characteristic $2$ what $[x,x]/2$ is means exactly to specify a restricted Lie algebra structure so it is not directly related the definition of an ordinary Lie algebra. @Theo: I agree that a downside of setting $[x,x]=0$ is that we do not get multilinear axioms. On the other hand without it we can not embed the Lie algebra in an associative algebra, i.e., the map to the enveloping algebra vanishes on $[x,x]$.