As I’ve mentioned before, the number of squarefree monic polynomials of degree n in F_q[t] is exactly q^n – q^{n-1}.
I explained in the earlier post how to interpret this fact in terms of the cohomology of the braid group. But one can also ask whether this identity has a motivic interpretation. Namely: let U be the variety over Q parametrizing monic squarefree polynomials of degree d. So U is a nice open subvariety of affine n-space. Now the identity of point-counts above suggests the question:
Question: Is there an identity [U] = [A^n] – [A^{n-1}] in the ring of motives K_0(Var/Q)?
I asked Loeser, who seemed to feel the answer was likely yes, and pointed out to me that one could also ask whether the two classes were identical in the localization K_0(Var/Q)[1/L], where L is the class of A^1. Are these questions different? That is, is there any nontrivial kernel in the natural map K_0(Var/Q) -> K_0(Var/Q)[1/L]? This too is apparently unknown.
Here, I’ll start you off by giving a positive answer in the easy case n=2! Then the monic polynomials are parametrized by A^2, where (b,c) corresponds to the polynomial x^2 + bx + c. The non-squarefree locus (i.e. the locus of vanishing of the discriminant) consists of solutions to b^2 – 4c = 0; the projection to c is an isomorphism to A^1 over Q. So in this case the identity is indeed correct.
Update: I totally forgot that Mike Zieve sent me a one-line argument a few months back for the identity |U(F_q)| = q^n – q^{n-1} which is in fact a proof of the motivic identity as well! Here it is, in my paraphrase.
Write U_e for the subvariety of U consisting of degree-d polynomials of the form a(x)b(x)^2, with a,b monic, a squarefree, and b of degree e. Then U is the union of U_e as e ranges from 1 to d/2. Note that the factorisation as ab^2 is unique; i.e, U_e is naturally identified with {monic squarefree polynomials of degree d-2e} x {monic polynomials of degree e.}
Now let V be the space of all polynomials (not necessarily monic) of degree d-2, so that [V] = [A^{n-1}] – [A^{n-2}]. Let V_e be the space of polynomials which factor as c(x)d(x)^2, with d(x) having degree e-1. Then V is the union of V_e as e ranges from 1 to d/2.
Now there is a map from U_e to V_e which sends a(x)b(x)^2 to a(x)(b(x) – b(0))^2, and one checks that this induces an isomorphism between V_e x A^1 and U_e, done.
But actually, now that I think of it, Mike’s observation allows you to get the motivic identity even without writing down the map above: if we write 
Quomodocumque 
I think the following gives the argument in K_0(Var_C). I’m not sure what it would take to soup it up to get an argument over Q.
The space of monic polynomials is given by C^n = Sym^n(C) where we think of elements on the right as unordered n-tuple of the roots of the polynomial. The discriminant locus pulls back to a union of hyperplanes under the quotient map C^n — > Sym^n(C). Now if I stratify those hyperplanes according to how many hyperplanes each point lies on, then the symmetric group acts transitively on the components in the stratification of a given dimension. The upshot is that the discriminant locus in Sym^n(C) has a stratification whose strata are isomorphic to the strata of a single hyperplane and hence in the ring of motives, the discriminant locus is equal to C^{n-1}.
I don’t know if I totally buy this, Jim — I feel like your strata on the discriminant locus are still going to be _quotients_ by some finite stablizer group of the strata in the hyperplane upstairs.
You are right Jordan, the strata will be quotients by stabilizers. Hmm, serves me right for firing this off 2 minutes after thinking of it. Let me see if I can fix my argument and get back to you.
Perhaps the following works:
The discriminant locus D, i.e. the complement of U in A^n, has a natural stratification parametrized by partitions of n (except n = 1 + 1 + …+ 1 which corresponds to U). The normalisation of the closure of each stratum is isomorphic to an affine space of dimension the number of terms in the corresponding partition. By induction on the dimension one sees that each stratum is in the subring of K_0(Var/k) generated by A^1, where is k any field, and there is a universal formula for this, not depending on k. (To see this one should consider the quotients of A^{n_1+n_2+…+n_k} by the natural action of S_{n_1}x … xS_{n_k} together with their natural stratifications). In particular, D is also in this subring and hence also U. The number of points of U over a finite field with q elements is therefore given by a polynomial in q whose coefficients determine the representation of U as an element of the subring of K_0(Var/k) generated by A^1. Plugging in the formula for this polynomial that you gave above then tells us that [U] = [A^n] – [A^{n-1}].
Of course this is a cheap proof and perhaps not what you wanted. There should be a purely combinatorial proof, but such a proof might require a lot more work.
This seems right to me! And indeed the pure thought “by induction it’s polynomial in [A^1]” argument is in the end the same as the explicit induction argument I just added to the post.
A natural generalisation of your question would be to ask whether the closure of any stratum is equal to an affine space in K_0(Var_k); this seems quite plausible.
Naf, that is a great question, which Ravi Vakil and I also wondered about. Definitely plausible, and the evidence looks good. But the answer is no! The closure of the 1,1,2,2,3 stratum in Sym^9 is =L^5-L^2+L in the Grothendieck ring (that’s the first counterexample). We investigate this phenomena more thoroughly in an upcoming paper “Discriminants in the Grothendieck ring” (which also includes a few different, and different from above, proofs of Jordan’s original puzzle).
In even further response to naf’s question, a recent paper of Almousa and myself http://arxiv.org/pdf/1210.0472v1.pdf addresses the question of for which strata the closure is equal to an affine space in K_0(Var_k).