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 welldefined 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 !
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 welldefined 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 '17 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 '17 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. $\endgroup$ – Andrea Ricolfi Apr 26 '17 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 '17 at 15:44
