Weil Conjectures for nonprojective algebraic varieties

If we replace projective variety with algebraic variety in the statement of the Weil conjectures what happens? To me it seems the statement still makes sense. But is it still true?

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Downvoted for having a profoundly uninformative title. –  JSE Nov 25 '09 at 2:37
Title edited. Better now? –  David Speyer Nov 25 '09 at 4:40

Correctly restated, the conjectures hold for any variety $V$ (not necessarily complete or nonsingular) over a finite field $k$.

Dwork proved that the zeta function $Z(V,t)$ of $V$ is a rational function of $t$.

Grothendieck (et al.) expressed $Z(V,t)$ as the alternating product of the characteristic polynomials of the Frobenius map $F$ acting on the etale cohomology groups of $V$ with compact support.

Deligne showed (Weil II) that for each positive integer $i$ and each eigenvalue $a$ of $F$ acting on the $i$th etale cohomology group of $V$ with compact support, there exists an integer $j\leq i$ such that all the complex conjugates of $a$ have absolute value $q^{j/2}$ where $q=|k|$.

When $V$ is nonsingular and complete, these statements together with Poincare duality, give Weil's original conjectures.

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What does "correctly restated" mean? –  John McCarthy Nov 25 '09 at 13:21
He "correctly restated" them in his answer. –  David Zureick-Brown Nov 25 '09 at 17:30
Weil conjectures can even be extended to Artin stacks of finite type over a finite field, if correctly restated. –  shenghao Dec 19 '09 at 6:30
Would the coefficients still have integer coefficients (rather than just rational)? –  Makhalan Duff Oct 24 '11 at 18:49
As usually stated, most of the statements break. Try looking at $\mathbb{A}^1$ minus a point.