Here is my second question on constructible sets, now on Grothendieck rings. Let $K_0(Sch_k)$ be the Grothendieck ring of schemes over $k$. I have read that if $S$ is a constructible set in a projective $k$-space $\mathbf{P}$ (or in a scheme over $k$), then $S$ has a well-defined class in $K_0(Sch_k)$. Is there an easy way to see this ? Also, if $C$ is closed in $S$, does one know that $[S] = [C] + [S \setminus C]$ ? Thanks again !


1 Answer 1


The "scissor relation" $[S]=[C]+[S\setminus C]$ in your last question holds by definition of $K_0(Sch_k)$. As for your first question, a constructible subset $V\subset X$ of a $k$-scheme $X$ can be expressed as a finite disjoint union of locally closed subschemes of $X$, say $$ V=\coprod_{i=1}^n Z_i. $$ Each $Z_i$ has a well-defined class $[Z_i]\in K_0(Sch_k)$ and one can put $$[V]:=\sum_{i=1}^n[Z_i].$$ To check this is independent upon the chosen decomposition, assume $$V=\coprod_{j=1}^m W_j.$$ Observe that $$ \begin{align} [Z_i]&=[Z_i\cap V]=\sum_{j=1}^m[Z_i\cap W_j]\qquad\textrm{for }1\leq i\leq n,\\ [W_j]&=[W_j\cap V]=\sum_{i=1}^n[W_j\cap Z_i]\qquad\textrm{for }1\leq j\leq m. \end{align} $$ Then $\sum_i[Z_i]=\sum_i\sum_j[Z_i\cap W_j]=\sum_j\sum_i[Z_i\cap W_j]=\sum_j[W_j]$.

  • $\begingroup$ Thanks ! Is there some way of expressing that $[S] = [S']$ if $S$ and $S'$ are "isomorphic constructible sets," just as for schemes ? $\endgroup$
    – THC
    Apr 26, 2017 at 11:57
  • $\begingroup$ I guess one way to do it would be to find appropriate decompositions $S = \coprod_i Z_i$ and $S' = \coprod_jZ_j'$ as above, indexed over the same index set $I$ and such that each $Z_i$ is isomorphic to $Z_i'$. $\endgroup$
    – THC
    Apr 26, 2017 at 13:22
  • $\begingroup$ not sure what you mean by "isomorphic constructible sets". If you find decompositions like in your last comment, then certainly $[S]=[S']$. The converse (stating: if two varieties have the same class then they are piecewise isomorphic) is the "cut and paste conjecture" by Larsen and Lunts, which is false in general. $\endgroup$ Apr 26, 2017 at 13:46
  • $\begingroup$ By the way, for obtaining the first formula, is it needed that the scheme is Noetherian ? (Is this a necessary condition ?) $\endgroup$
    – THC
    May 3, 2017 at 15:44
  • $\begingroup$ No, I do not think that is needed. $\endgroup$ May 14, 2017 at 9:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.