Let $\mathcal{Q}$ be an irreducible quadric in $\mathbb{P}^n(k)$, with $n \geq 2$ and $k$ a finite field. Let $K_0(V_k)$ be the Grothendieck ring of $k$varieties. It is well known (it appears) that the class $[\mathcal{Q}]$ in $K_0(V_k)$ is contained in $\mathbb{Z}[\mathbb{L}]$, where $\mathbb{L} = [\mathbb{A}^1(k)]$. My question is: what is an easy (elementary) way to prove this rigorously ? The more proofs the better !!

3$\begingroup$ Is that true even when $\mathcal{Q}$ has no $k$rational point? I thought that the counting function was determined by the class in the Grothendieck ring . . . $\endgroup$ – Jason Starr Sep 6 '17 at 17:56

6$\begingroup$ @Jason Starr: Any positive dimensional quadric hypersurface over a finite field has a rational point; this follows from the ChevalleyWarning theorem. $\endgroup$ – Daniel Loughran Sep 6 '17 at 20:04

2$\begingroup$ Of course, you are correct. That was silly of me. $\endgroup$ – Jason Starr Sep 6 '17 at 22:03
Edit. Following Remy van Dobben de Bruyn's excellent suggestion, I clarified the use of "irreducible quadrics of dimension $0$."
Daniel Loughran's observation about ChevalleyWarning is the key to one solution of this problem. Let $k$ be a finite field, or any quasialgebraically closed field, or even just a field with $u$invariant $\leq 2$. This means that every quadric $\mathcal{Q}_n$ in $\mathbb{P}^n$ with $n\geq 2$ has a $k$rational point.
Proposition. For every quadric hypersurface $\mathcal{Q}_n$ in $\mathbb{P}^n$, $n\geq 1$, there exists a field extension $K/k$ of degree $\leq 2$ (possibly degree $1$, i.e., $K$ equals $k$) such that the class $[\mathcal{Q}_n]$ is in the $\mathbb{Z}[\mathbb{L}]$module generated by $1$ and the class $[\text{Spec}(K)]$ (this is an Artin motive).
Proof. This is proved by induction on $n$. The base case is when $n$ is equal to $1$. Then $\mathcal{Q}_1$ is a finite, affine $k$scheme. If the $k$scheme is isomorphic to $\text{Spec}(k\times k)$, then $[\mathcal{Q}_1]$ equals $2$. If $\mathcal{Q}_1$ is isomorphic to $\text{Spec}(k[\epsilon]/\langle \epsilon^2 \rangle)$, then $[\mathcal{Q}_1]$ equals $1$. If $\mathcal{Q}_1$ is neither of these, then $\mathcal{Q}_1$ is isomorphic to $\text{Spec}(K)$ for a quadratic field extension $K/k$. Thus $[\mathcal{Q}_1]$ equals $[\text{Spec}(K)]$. So the proposition holds for $n$ equal to $1$.
By way of induction, assume that $n\geq 2$, and assume that the result is proved for all quadric hypersurfaces of dimension smaller than $n1$.
First, consider the case when $\mathcal{Q}_n$ is everywhere nonreduced. If the characteristic is different from $2$, then $\mathcal{Q}_n$ is the zero scheme of the square of a linear equation, i.e., a hyperplane with multiplicity $2$. In this case, $$[\mathcal{Q}_n] = [\mathbb{P}^{n1}_k] = 1+\mathbb{L}+ \dots + \mathbb{L}^{n1}.$$ Thus, without loss of generality, assume that $\mathcal{Q}_n$ is not a double plane.
If the characteristic equals $2$, since the $u$invariant is $\leq 2$, there is only one other possibility with $\mathcal{Q}_n$ everywhere nonreduced: $\mathcal{Q}_n$ might be the zero scheme of a linear combination of precisely two squares of linear equations. In this case, the common zero locus $\Lambda_{n2}$ of the $2$linear equations is a linear space of dimension $n2$. Moreover, $\mathcal{Q}_n$ is a cone with vertex $\Lambda_{n2}$ over an everywhere nonreduced quadric $\mathcal{Q}_1'$ in $\mathbb{P}^1$. Similarly, if $\mathcal{Q}_n$ is reduced yet singular, then the singular locus is a linear space $\Lambda_m$ of some dimension $m$, and $\mathcal{Q}_n$ is a cone with vertex $\Lambda_m$ over a smooth quadric hypersurface $\mathcal{Q}'_{nm1}$ in $\mathbb{P}^{nm1}$. In both of these cases, the reduced case and the nonreduced case, since $\mathcal{Q}_n$ is a cone, we have an identity in the Grothendieck ring of varieties, $$[\mathcal{Q}_n] = [\Lambda_m]+ [\mathcal{Q}_n\setminus \Lambda_m] = $$ $$[\Lambda_m] + [\mathbb{A}^{m+1}_k]\cdot [\mathcal{Q}'_{nm1}] = $$ $$(1+\mathbb{L} + \dots + \mathbb{L}^m) + \mathbb{L}^{m+1}\cdot [\mathcal{Q}'_{nm1}].$$ Therefore $[\mathcal{Q}_n]$ is a $\mathbb{Z}[\mathbb{L}]$linear combination of the class $1$ and the class $[\mathcal{Q}'_{nm1}]$. By the induction hypothesis, there exists a field extension $K/k$ of degree $\leq 2$ such that $[\mathcal{Q}'_{nm1}]$ is in the $\mathbb{Z}[\mathbb{L}]$submodule generated by $1$ and $[\text{Spec}(K)].$ Thus, from the identity above, also $[\mathcal{Q}_n]$ is in this submodule. So the result is proved in this case.
Next assume that $\mathcal{Q}_n$ in $\mathbb{P}^n$ is smooth with $n\geq 3.$ Since $k$ has $u$invariant $\leq 2$, there exists a $k$rational point $p$ in $\mathcal{Q}_n$. Projection away from $p$ is birational, with the tangent hyperplane section $Y = H_p\cap \mathcal{Q}$ being equal to a cone over a quadric $\mathcal{Q}''_{n2}$ in a hyperplane $\Lambda_{n2}$ inside $\mathbb{P}^{n1}$. In fact, after base change to $\overline{k}$, it is straightforward to compute that $\mathcal{Q}''_{n2}$ is smooth and integral. Thus, again we have an identity in the Grothendieck ring, $$[\mathcal{Q}_n] = [\mathbb{P}^{n1}\setminus \Lambda_{n2}] + [Y] = $$ $$(\mathbb{L}^{n1}) + \left( 1 + [\mathcal{Q}''_{n2}]\cdot \mathbb{L} \right) = $$ $$ 1 + \mathbb{L}^{n1} + [\mathcal{Q}''_{n2}]\cdot \mathbb{L}.$$ By the induction hypothesis, there exists a field extension $K/k$ of degree $\leq 2$ such that $[\mathcal{Q}''_{n2}]$ is in the $\mathbb{Z}[\mathbb{L}]$submodule generated by $1$ and $[\text{Spec}(K)]$. Thus, from the identity above, also $[\mathcal{Q}_n]$ is in this submodule. So the result is proved in this case.
Finally, for a smooth quadric hypersurface $\mathcal{Q}_2$ in $\mathbb{P}^2$, the argument is almost the same as in the previous paragraph. The only difference is that $Y$ is already a $k$point with multiplicity $2$. So that gives the identity, $$[\mathcal{Q}_2] = [\mathbb{P}^{1}\setminus \Lambda_{0}] + [Y] = $$ $$\mathbb{L} + 1.$$ For $K/k$ equal to the identity field extension, we again conclude that $[\mathcal{Q}_2]$ is in the $\mathbb{Z}[\mathbb{L}]$submodule generated by $1$ and $[\text{Spec}(K)].$ Thus, in every case, there exists a field extension $K/k$ of degree $\leq 2$ such that $[\mathcal{Q}_n]$ is in the $\mathbb{Z}[\mathbb{L}]$submodule generated by $1$ and $[\text{Spec}(K)].$ Therefore the proposition is proved by induction on $n$. QED
Note. It seems that this argument proves that for every field $k$ with $u$invariant $\leq r$, for every $n\geq 1$, for every quadric hypersurface $\mathcal{Q}_n$ in $\mathbb{P}^n$, $n\geq r$, there exists a quadric hypersurface $\mathcal{Q}'_m$ with $m\leq r1$ such that $[\mathcal{Q}_n]$ is in the $\mathbb{Z}[\mathbb{L}]$submodule generated by $1$ and $[\mathcal{Q}'_m]$.

1$\begingroup$ To be clear, an irreducible quadric in $\mathbb P^1_k$ means the 'Artin motive' $\operatorname{Spec} \mathbb F_{q^2}$ if $k = \mathbb F_q$. (I first thought it was a typo and should have said $\mathbb P^2_k$; this comment is in case someone else is confused as well.) $\endgroup$ – R. van Dobben de Bruyn Sep 7 '17 at 3:47

$\begingroup$ @R.vanDobbendeBruyn. That is a good point. I edited the answer to make this more clear. $\endgroup$ – Jason Starr Sep 7 '17 at 13:14

$\begingroup$ @Jason Starr: is there a reason why you work with affine schemes instead of Projschemes in the first part of your proof (case of dimension $1$) ? Could you explain why this works ? $\endgroup$ – THC Jan 9 '18 at 16:25

$\begingroup$ @THC. I might not be properly understanding your question. Every quadric hypersurface in $\mathbb{P}^1$ is automatically finite, and thus also affine. $\endgroup$ – Jason Starr Jan 9 '18 at 18:11