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It is well known that Grothendieck had a different idea than Deligne about how one should go about proving the Riemann hypothesis for finite fields. However, since Grothendieck's desired proof never came to fruition, I find it hard to look up what his proposed proof was.

It is clear from texts on the subject that he wanted to use motives in some way or other, and that he wanted to prove the standard conjectures first. Given the standard conjectures, is there an easy proof the Riemann hypothesis over finite fields? What is it? Did he want other things to be true also? What is the sketch he had in mind -- can you write a proof of the Riemann hypothesis with some conjectural black boxes like the standard conjectures?

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Maybe you have already read Grothendieck's own (short) article on standard conjectures. There you may find the 'proof' of the Weil conjecture. –  shenghao Aug 14 '11 at 18:03

2 Answers 2

The standard conjectures imply directly that the category of motives over the finite field $\mathbb{F}_q$ is a polarizable (hence semisimple) Tannakian category. Using only that, we have the following result.

PROPOSITION: Let $X$ be a motive of weight $m$ over $ \mathbb{F}_q$, and let $\alpha\mapsto\alpha^t$ be the involution of $\mathrm{End}(X)$ defined by a Weil form $ \varphi$. The following statements hold for the Frobenius endomorphism $\pi =\pi_X$ of $X$:

(a) $\pi\cdot\pi^t=q^m$; hence $\mathbb{Q}[\pi ]$ is stable under the involution $\alpha\mapsto\alpha^t$;

(b) $\mathbb{Q}[\pi ]\subset\mathrm{End}(X)$ is a product of fields;

(c) for every homomorphism $\rho\:\mathbb{Q}[\pi ]\rightarrow \mathbb{C}$, $\rho (\pi^t)=\iota (\rho\pi )$, and $|\rho\pi |=q^{m/2}$. ($\iota$ is complex conjugation)

PROOF: (a) By definition, $\varphi$ is a morphism $X\otimes X\to T^{\otimes (-m)}$ ($T$ is the Tate object). It is invariant under $\pi$, and so $$\varphi (\pi x,\pi y)=\pi (\varphi (x,y))=q^m\varphi (x,y)=\varphi (x,q^my).$$ But $\varphi (\pi x,\pi y)=\varphi (x,\pi^t\pi y)$, and because $ \varphi$ is nondegenerate, this implies that $\pi^t\cdot\pi =q^m$. Therefore $\mathbb{Q}[\pi ]$ is stable under $ \alpha\mapsto\alpha^t$, and we obtain (a).

(b) Let $R$ be a commutative subalgebra of $\mathrm{End}(X)$ stable under $\alpha\mapsto\alpha^t$, and let $r$ be a nonzero element of $R$. Then $ s=rr^t\neq 0$ because $\mathrm{Tr}(rr^t)>0$. As $s^t=s$, $\mathrm{Tr}(s^2)=\mathrm{Tr}(ss^t)>0$, and so $s^2\neq 0$. Similarly $s^4\neq 0$, and so on, which implies that $s$ is not nilpotent, and so neither is $r$. Thus $R$ is a finite-dimensional commutative $\mathbb{Q}$-algebra without nonzero nilpotents, and the only such algebras are products of fields.

(c) In an abuse of notation, we set $\mathbb{R}[\pi ]=\mathbb{R}\otimes_{ \mathbb{Q}}\mathbb{Q}[\pi ]$. As in (b), this is a product of fields stable under $\alpha\mapsto\alpha^t$. This involution permutes the maximal ideals of $\mathbb{R}[\pi ]$ and, correspondingly, the factors of $ \mathbb{R}[\pi ]$. If the permutation were not the identity, then $\alpha\mapsto\alpha^ t$ would not be a positive involution. Therefore each factor of $\mathbb{R}[\pi ]$ is stable under the involution. The only involution of $\mathbb{R}$ is the identity map (= complex conjugation), and the only positive involution of $\mathbb{C}$ is complex conjugation. Therefore we obtain the first statement of (c), and the second then follows from (a).

This (conjectural) proof of the Riemann hypothesis for motives is very close to Weil's original proof for abelian varieties (Weil 1940).

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Look at the article:

Kleiman, S. L. Algebraic cycles and the Weil conjectures. Dix esposés sur la cohomologie des schémas, pp. 359–386. North-Holland, Amsterdam; Masson, Paris, 1968

or

Kleiman, Steven L. The standard conjectures. Motives (Seattle, WA, 1991), 3–20, Proc. Sympos. Pure Math., 55, Part 1, Amer. Math. Soc., Providence, RI, 1994.

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A good online summary: [link text][1] [1]: sbseminar.wordpress.com/2010/05/03/… –  Denis Chaperon de Lauzières Aug 14 '11 at 17:09
    
I don't have access to the first paper, but the second seems quite readable and fun! –  James D. Taylor Aug 14 '11 at 17:15

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