# Double points in the Grothendieck ring

Let $$K_0(V_k)$$ be the Grothendieck ring of $$k$$-varieties, and consider the scheme $$X = \mathrm{Spec}(k[x]/(x^2))$$.

I understand that this scheme has one point, but I am missing the fact that in $$K_0(V_k)$$, the class of $$X$$ is $$[X] = 1$$ (which is something I read at this site in the context of Grothendieck classes of quadrics).

Why is that? Why is $$[X]$$ the same equivalence class as $$[\mathrm{Spec}(k[x]/(x))]$$?

• $[X]=[X_{\mathrm{red}}]$ for any $k$-scheme $X$, just because the Grothendieck ring is defined by saying that for any closed subscheme $Z$ of $X$ we have $[Z]+[X-Z]=[X]$, including for $Z=X_{\mathrm{red}}$. Commented Mar 10, 2020 at 18:09
• Sure: $\mathrm{Spec}\, k[x]/x$ is a closed subscheme of $\mathrm{Spec}\, k[x]/x^2$ (corresponding to the surjection $k[x]/x^2\to k[x]/x$, and thus definitionally $[\mathrm{Spec}\, k[x]/x]=[\mathrm{Spec}\, k[x]/x^2]$ in the Grothendieck ring. Then, since for any $k$-scheme $X$ we have $X\times_k k \cong X$ we have that $[\mathrm{Spec}\,k]$ is the identity in the Grothendieck ring. Commented Mar 10, 2020 at 18:43
• @DevlinMallory: And by "definitionally," you mean that the complement of $\mathrm{Spec}(k[x]/(x^2))$ in $\mathrm{Spec}(k[x]/(x))$ is empty ?
– THC
Commented Mar 10, 2020 at 19:01
• Yes, that's exactly it (and more generally the same is true for the reduced subscheme $X_{\mathrm{red}}$ of any scheme $X$: the complement is empty, so the classes of $X$ and $X_{\mathrm{red}}$ agree in the Grothendieck ring). Commented Mar 10, 2020 at 19:11
• Maybe you want to ask whether there is some "enriched" Grothendieck ring mapping onto $K_0(V_k)$, which takes into account non-reducedness (so that the class of "$X-X_{\mathrm{red}}$", in some sense, would be meaningful and possibly nonzero...) I don't know how to formalize this properly.
– YCor
Commented Mar 11, 2020 at 11:44

An answer has already been given in the comments: the definition of $$K_0(V_k)$$ includes the relations $$[X]=[X_\text{red}]$$ so fat points are equal to points by definition. The intuition is that $$K_0(V_k)$$ was primarily conceived to study phenomena involving classical topological invariants (Euler characteristic of the underlying topological space, Hodge structure, etc), or point counts over finite fields, which are all insensitive to the scheme structure anyway.

There is however a theory which encodes richer data, though it (at least a priori) depends on more input than just the scheme structure of $$X$$, and is only defined for certain kinds of schemes. This is the notion of a motivic vanishing cycle attached to a pair $$(M,f)$$ where $$f\colon M\to k$$ is a regular function on a smooth variety $$M$$. The intuition is that this should be thought of as a motivic invariant attached to the degeneracy locus $$X=\{df=0\}$$, thought of as a subscheme of $$M$$, which is sensitive to the scheme structure of $$X$$. In your example, we would consider the pair $$(M,f)=({\mathbb A}^1, x^3)$$ which indeed has the double point $$X$$ as degeneracy locus. This motivic vanishing cycle takes values in a ring which is larger than $$K_0(V_k)$$, remembering also monodromy data about the function; in the example at hand, you would need to worry about order-three monodromy. But there is still an Euler characteristic map to the integers, which returns the answer $$2$$ for this example, indicating that indeed we are talking about a double point.

The motivic vanishing cycle was defined by Denef–Loeser, and the interpretation I gave above arose in works of Joyce and Kontsevich–Soibelman in connection with motivic Donaldson–Thomas theory. A starting reference might be On motivic vanishing cycles of critical loci, arXiv:1305.6428, by Bussi–Joyce–Meinhardt.

• This is a neat idea. Is there any sense in which this is independent of the way in which you describe $X$ as a critical locus? Here is an example that is analogous to what I'm looking for. If $X$ is a critical locus, it carries a symmetric perfect obstruction theory, which is enough to define $[X]_{Virt} \in A_0(X)$ independent of the choice of action functional. Does something similar hold for the class in $K(Var)$? Commented Mar 11, 2020 at 17:12
• Actually, I see that the paper you mention contains at least one result along these lines (Theorem 3.2, if anyone else is interested) Commented Mar 11, 2020 at 18:54

This answer addresses both the original question and the poignant comment by YCor.

Such a concept of an "enriched" Grothendieck ring mapping onto $$K_0(V_k)$$ has already been formalized, pun intended, by H. Schoutens around 2009–2012 in two preprints, found here: Schemic Grothendieck rings and motivic rationality and The yoga of schemic Grothendieck rings, a topos-theoretical approach.

One of the main innovative ideas here is to use the functor of points perspective of Algebraic Geometry to pick up a non-empty compliment $$X^{\circ} \setminus X_\text{red}^{\circ}$$ where $$Y^{\circ}$$ denotes the functor of points from finitely generated $$k$$-algebras to sets given by a separated scheme $$Y$$ of finite type over an algebraically closed field $$k$$.

One can define several interesting motivic sites on the category of separated schemes of finite type over a field as different natural collections of sieves which happen to form a distributive lattices (so that scissor relations hold and fiber products work as multiplication).

The most interesting of these sites is the formal site $$\mathbf{Form}_k$$ whose Grothendieck ring I will denote by $$\mathbf{H}$$. It is the right concept in my view as it takes into account thickenings of arbitrary order. Of note, in $$\mathbf{H}$$,

$$[\mathbb{P}_k^1] = \mathbb{L} + \widehat{\mathbb{L}}$$

where $$\widehat{\mathbb{L}}$$ denotes the class of the formal sieve given by the functor of points of the completed affine line at the origin.

Note further that there is a natural surjective ring homomorphism $$\mathbf{H} \to K_0(V_k)$$ given by sending the class of a formal sieve $$[\mathcal{X}]$$ to $$[\mathcal{X}(k)]$$ and if you complete these rings along the dimensional filtration, you obtain continuous ring homomorphisms of topological rings (cf., Theorem 2.1 and Lemma 2.4 of my paper here: On the auto Igusa-zeta function of an algebraic curve).

Note then that it is immediate that $$[X^{\circ}]$$ is sent to $$[X_\text{red}]$$. It is slightly less immediate but true that $$[(\widehat{X}_Y)^{\circ}]$$ is sent to $$[Y_\text{red}]$$ so that in particular $$\widehat{\mathbb{L}} \mapsto 1$$ under this ring homomorphism.

2. Relatedly, it contains a highly nontrivial subring $$\mathbf{H}_0$$ formed by all formal motifs dimension less than or equal to $$o$$ which gets mapped onto $$\mathbb{Z} \subset K_0(V_k)$$.
3. It is very rigid: I suspect that $$[X^{\circ}] = \mathbb{L}^n \implies X \cong \mathbb{A}_k^n$$.
4. The possibility of some type of rationality of the motivic generating series of the form: $$P_m : = \sum_{n=0}^{\infty} [(\underline{\operatorname{Hom}}(D_n, D_m))^{\circ}]\mathbb{L}^{-r_n}t^n$$ where $$D_i : = \operatorname{spec}(k[t]/t^{i+1})$$ and $$r_n = n - \lceil \frac{n}{m}\rceil+1$$ such that $$P_m \mapsto (1-t)^{-1}$$.
5. Relatedly, the possible rationality of $$A := \sum_{n=0}^{\infty} [(\underline{\operatorname{Hom}}(D_n, D_n))^{\circ}]\mathbb{L}^{-n}t^n$$.
6. Is $$\mathbf{H}$$ an integral domain?
7. If the answer to item 6 is negative, then is $$\mathbb{L}$$ a zero divisor in $$\mathbf{H}$$?