Let $\mathbf{Sch}_k$ be the category of $k$schemes of finite type, and let $K_0(\mathbf{Sch}_k)$ be the Grothendieck ring of $k$schemes. Let $\mathbb{Z}[\mathbb{L}]$ the subring generated by the virtual Lefschetz motive $\mathbb{L} := [\mathbb{A}^1_k]$. Is every element of $\mathbb{Z}[\mathbb{L}]$ the class of a $k$scheme (or at least a $k$constructible set) ? It might be convenient to recall that $K_0(\mathbf{Sch}_k)$ is generated by isomorphism classes of $k$schemes $[X]$, and that $$[X] = [X \setminus C] + [C]$$ whenever $C$ is a closed subscheme in $X$; the product is given by $$[X] \cdot [Y] = [X \times Y].$$

3$\begingroup$ Not $1$, surely. $\endgroup$ – Will Sawin May 4 '17 at 15:02

$\begingroup$ Do you know of a general property on the form of the elements of $\mathbb{Z}[\mathbb{L}]$ so that the answer would be "yes" ? (Say, with $k$ a finite prime field.) Thanks ! $\endgroup$ – THC May 4 '17 at 15:12

$\begingroup$ I think the leading term must be positive. $\endgroup$ – Will Sawin May 4 '17 at 15:13

$\begingroup$ Or maybe "slightly" more restrictive: if $P(\mathbb{L}) \in \mathbb{Z}[\mathbb{L}]$ and $k = \mathbb{F}_p$, then $P(p^n) > 0$ for all positive integers $n$. What do you think ? $\endgroup$ – THC May 4 '17 at 15:59

$\begingroup$ I was thinking over an infinite field. I bet you can extract a proof from arxiv.org/pdf/0907.0488.pdf $\endgroup$ – Will Sawin May 4 '17 at 20:36
Here is a proof of the fact that over an infinite field $f(\mathbb{L})$ is a class of a scheme if and only if the leading term of $f$ is positive, as suggested by Will Sawin.
Example. For $f(t)=a_nt^n+a_{n1}t^{n1}+\dots+a_0$ with $a_n>0$ throw away from affine space of dimension $n$ planes $\mathbb{A}^i$ in amount $a_i$ for each negative $a_i$ such that these planes are pairwise disojoint(this is possible due to the infinitness of the base field). Then take the disjoint union of the resulting complement with $a_n1$ copies $\mathbb{A}^n$ and with $a_j$ copies of $\mathbb{A}^j$ for each positive $a_j$. The variety which we get clearly equals to $f(\mathbb{L})$ in $K_0$(one can construct an irreducible example by doing blowups instead of disjoint unions).
All other polynomials do not give a class of a scheme. 1) First, consider the case $\mathrm{char}\, k=0$ and reduce by a Lefschetz principle type argument to the case $k=\mathbb{C}$. Assume that $[f(\mathbb{L})]=[X]$ and $a_n<0$. This identity is a linear combination of a finite number of standard identities. There exists a subfield $F$ of $k$, finitely generated over $\mathbb{Q}$ such that all schemes involved in these identities are defined over $F$. Thus, we have $[p(\mathbb{L})]=[X_F]$ already in $K_0(Sch_F)$. Then embedding $F$ into $\mathbb{C}$ and base changing to $\mathbb{C}$ we arrive at the case $k=\mathbb{C}$.
Over $\mathbb{C}$ we can use the following motivic measure(strictly speaking, this formula defines it only on $K_0(Var_{\mathbb{C}})$ but if I got the definition of $K_0(Sch)$ right, the relations imply $[X]=[X_{red}]$  there is still something to be said about notseparatedness, but let's omit that): $$\mu([X])=\sum_{k,p,q}(1)^k\dim H^k_c(X,\mathbb{C})^{p,q}u^pv^q\in \mathbb{Z}[u,v]$$ where by $H^k_c(X,\mathbb{C})^{p,q}$ I mean the piece of weight $(p,q)$ in the MHS on $H^k_c(X)$. For an irreducible variety of dimension $n$ the leading term of $\mu([X])$ is $u^nv^n$ so it is at least positive for arbitrary varirty. But for $f(\mathbb{L})$ this motivic measure gives $f(uv)$, qed.
2)The case $\mathrm{char}\, k=p>0$. Exactly as in the first case, we may assume that we have a counterexample defined over a ring which is a finite extension of $\mathbb{F}_p[t_1,\dots,t_n]$(say, Noether normalization). Specializing at some values of $t_i$ we arrive at the case $k=\mathbb{F_q}$ where we can use the motivic measure suggested by THC: $$\mu_n([X])=\#X(\mathbb{F}_{q^n})$$ and for $f(\mathbb{L})$ with negative leading coefficient there exists a big enough $n$ such that $\mu_n(f(\mathbb{L}))<0$.
For finite fields everything seems more delicate  a necessary condition is $f(q^{nm})\geq f(q^n)\geq 0$ for any $n,m$.

1$\begingroup$ One can also prove the characteristic zero case by a spreading out method, as you do for the characteristic p case. $\endgroup$ – Will Sawin May 6 '17 at 5:09