Talk:Weil conjectures

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This article has comments.

Functional equation[edit]

I think that, in the second version of the functional equation, the $T$ on the left hand side should be replaced by $T^{-1}$. —Preceding unsigned comment added by 84.167.241.29 (talk) 09:44, 14 March 2011 (UTC)

I think that you're right. Thanks! Ozob (talk) 01:52, 17 March 2011 (UTC)

Move[edit]

I moved the page here because it is a group of conjectures. LittleDan 17:26 12 Jun 2003 (UTC)

I think the conjectures was made mainly by taniyama. you should mention the fact it is now calld shimura-taniyama conjectures. http://www.fact-index.com/t/ta/taniyama_shimura_theorem.html

No, you are mistaken about that. These Weil conjectures were made by Weil around 1949. The Taniyama-Shimura conjecture was made about the time of the conference in Nikko, in 1954.

Charles Matthews 16:29, 20 Sep 2004 (UTC)

Who?[edit]

Who proved these? Deligne? Others? Many others? Deligne seems to have gotten a Fields medal for this or at least something related.

Dwork and Deligne are responsible for the proofs themselves, Weil gets credit for the conjectures, and Grothendieck is responsible for creating the theory of etale cohomology which was used by Deligne in his proofs. - Gauge 20:45, 10 July 2005 (UTC)
Some credit goes also to Mike Artin, I think. Charles Matthews 21:16, 10 July 2005 (UTC)
Per Allyn Jackson's article [1] on page 1203, it lists B. Dwork's solution of part 1 as being from 1959 (versus 1960 in the article text), and it lists A. Grothendieck as the solver of part 2. It also mentions Grothendieck as having provided a more general solution to part 1 in 1964. Myasuda 14:24, 9 September 2006 (UTC)

finite fields or algebraically closed fields of prime characteristic?[edit]

My knowledge of algebraic geometry is tiny, but it is usu. done over algebraically closed thus infinite fields, isn't it? So I wonder if the current 1st sentence with "...algebraic varieties over finite fields." might be incorrect. Rich 15:43, 23 August 2006 (UTC)

I've added a para of further explanation. Charles Matthews 08:35, 24 August 2006 (UTC)

I-adic or ℓ-adic?[edit]

Is it I-adic or ℓ-adic? Capital I is indistinguishable from lower-case l, but maybe there's some reason for preferring l over ℓ? Ditto for other articles mentioning l-adic. --Vaughan Pratt (talk) 22:20, 9 April 2011 (UTC)

I guess the traditional typography with ℓ really references the golfball typewriter. It still helps legibiility to use it, though. It is ℓ-adic everywhere in this article, at least. Charles Matthews (talk) 08:15, 10 April 2011 (UTC)

It is? IE, Firefox, and Chrome are all showing it to me as l-adic everywhere in this article. I'd fix it myself except that there must be so many articles needing that change that it would be better done by a bot. --Vaughan Pratt (talk) 19:50, 11 April 2011 (UTC)

No, what I meant is that it is not I-adic, as you wrote (no italic format). Charles Matthews (talk) 21:24, 11 April 2011 (UTC)

question on rational point in relation to curves over finite fields[edit]

Hi, I'm trying to rewrite rational point so that a college freshman can understand some of it. After I wrote the sentence: -"It should be pointed out that many algebraic curves contain no rational points, or just finitely many rational points."-, I put in the seemingly, but not, mickey mouse phrase: -"although they do contain infinitely many nonrational points"- Now I'm not sure that phrase is true for curves over finite fields, unless finite field is used loosely as meaning the completion of a finite field. I would like to keep it or a similar phrase in there to keep it understandable. Is it correct as it stands, or should we put in an exception for finite fields? Thank you.--Rich Peterson 18:04, 27 January 2012 (UTC) — Preceding unsigned comment added by 198.189.194.129 (talk) 198.189.194.129 (talk) 18:29, 27 January 2012 (UTC)

From any perspective you look at it, a curve always has infinitely many non-rational points.
If you take the viewpoint that a point is a map from the spectrum of a field into a space, then any non-empty scheme has infinitely many points: Take any point P, let k(P) be the residue field at P, and let {ki} be an infinite family of fields all isomorphic to k(P). These give an infinite family of points. But they're all trivially equivalent; a slightly less trivial example is to let ki = k(P)(x1, ..., xi) where the xs are indeterminates, because then all the points are distinct.
If you take the viewpoint that a point is a geometric point, then you're working over the algebraic closure of the field. Choose an affine open subscheme of the curve and embed it in affine space. If the curve had only finitely many points over an algebraically closed field, then this affine open subscheme would of course also be finite, so, by the Nullstellensatz, it would correspond to a finite intersection of maximal ideals in K[x1, ..., xn] (here K is an algebraically closed field containing the ground field and n is the dimension of the ambient affine space). But such an ideal would be zero-dimensional, which is a contradiction.
If you take the viewpoint that a point is a point of the underlying topological space, then you can apply the fact of the previous paragraph: Two non-equivalent geometric points map to distinct points of the topological space, and since there are infinitely many geometric points, the topological space has infinitely many points.
By the way, you should make an account! After you do, come join us at Wikipedia:Wikiproject Mathematics. Ozob (talk) 02:25, 28 January 2012 (UTC)

Lubkin ![edit]

Without Lubkin this article is incomplete. - 2601:4:3880:395:21E:C2FF:FEBB:A1D1 (talk) —Preceding undated comment added 03:18, 4 February 2014 (UTC)

Could you provide a reference for the facts you're interested in this article having? Or maybe, if you have the time, write a new section of the article yourself? Ozob (talk) 14:54, 4 February 2014 (UTC)