We all know that Prof.John Tate proposed a set of conjectures(along with Prof.Emil Artin) formally spread under the name of "Tate conjectures", they have a wide range of influence on various fields of studies, like for example recently Prof.Mathew Emerton in his magnificent answer quoted that Tate conjectures(in some sense) can be thought as an analogue of Birch and Swinnerton-Dyer conjectures and Prof.Emerton gave a short intuition into the subject by representing the how can one spread out elliptic curve into elliptic surface and also that how the conjecture can be restated in this case, and he writes that the equivalent formulation is " every Frobenius invariant element actually arises from a curve defined over $\mathbb F$ " .

But Prof.Emerton didn't had a need of elaborating the meaning behind the statement as he was trying to answer another question ( but he was kind in mentioning all these details instead of directly going into the point ) , but I have thought that the comparision could be found on internet and I read quite many articles but I couldn't find atleast one such article clearly stating the link between those conjectures, so I wanted to ask it under another question (as Math.SE mandatorily sees the effort of user, so I took much time reading myself and then wanted to ask as its out of my reach).

So my main doubt is that

" How can one think the B.S.D and Tate-conjecture imply each other ? "

Due to my past background I know that the rank-part of the B.S.D was intending to comment the cardinality of the group $E(\mathbb{Q})$ (set of rational points on $E$) by linking the algebraic part to analytic part, but it was clear in that case that if the curve has lot many rational points then the $N_p$ must be substantially large pushing the product to zero for range of large primes, but how can the link the frobenius-invariant to these things and how can one show that the statement is analogue of other.

To be sharp I am expecting an answer in giving the explanation that

How can one say that [$L(E,1)=0 \iff E(\mathbb{Q})$ is infinite]$\iff$[every Frobenius invariant element actually arises from a curve defined over $\mathbb F$]

I wanted to know the things happening under them, like I mean how can one compare each of them in deep.(comparing analogous terms present on each side)

P.S : If the person who answers this question has some time and energy left to answer, I will be happy to hear the implication chain of B.S.D . (Implication chain length is a new word introduced by me just for the purpose of explaining, [B.S.D for modular forms]$\iff$[B.S.D for rationals]$\iff$[Tate-conjectures]... , so complete implication chain in all possible directions is what I am intending to hear.

Thank you.

  • $\begingroup$ But I want to add something which may be a META-topic but I wanted to ask here, I have found that answers posted by Prof.Emerton surely appear as the lighthouses for the ones who wanted to learn about the overall gist of a complex topic, especially the previous answers to a question on hodge conjecture which is here and other answers were so beautiful, that made think that it would be very beautiful if someone neatly transform them (Answers) into expository articles [PDF files]....contd $\endgroup$
    – IDOK
    Commented Dec 4, 2011 at 10:02
  • $\begingroup$ Contd... and keep them in internet so that everyone could access it , and I even suggest that one can prepare a fantastic short notes collecting all answers from the great people who post answers here, I am looking for any volunteers to do it, I am ready to do it, but I neither don't have a web-page nor any space to post, but I would be happy to do any sort of help regarding that, let us wait for the comment of Prof.Emerton $\endgroup$
    – IDOK
    Commented Dec 4, 2011 at 10:07
  • 3
    $\begingroup$ "and keep them in internet so that everyone could access it" - this is what this site's for. No need for PDF conversions. $\endgroup$ Commented Dec 4, 2011 at 10:11
  • $\begingroup$ But PDF would be handy to print and read too, which one could not do it here @J.M. $\endgroup$
    – IDOK
    Commented Dec 4, 2011 at 10:12
  • $\begingroup$ You think? $\endgroup$ Commented Dec 4, 2011 at 10:14

1 Answer 1


In fact the equivalence of BSD and the Tate conjecture for elliptic curves over a function field is perhaps somewhat more subtle than I suggested in my previous answer. (As far as I understand, the subtelty lies in the aspects having to do with finiteness of the Shafarevich--Tate group (on the BSD side) and finiteness of the Brauer group (on the Tate conjecture side).)

As for why the two conjecture should be equivalent, you have to think about how the rational points of an elliptic curve $E$ over a function field $K(C)$ (for some curve $C$ over a finite field) are related to horizontal divisors of degree one on the corresponding model for $E$ as an elliptic surface $X \to C$. You then need to relate horizontal divisors of degree one to all divisors on $X$, and also understand how the $L$-series for $E$ over $K(C)$ relates to the $\zeta$-function (in the sense of the Weil conjectures) of $X$. Finally, you need to pass to Grothendieck's description of the $\zeta$-function in terms of the alternating product of the characteristic polynomials of Frobenius on the etale cohomology of $X$. The main contribution here will be from $H^2$ (since $X$ is a surface), and we are interested in the order of vanishing of the $L$-function at $1$, so it is the multiplicity of $1$ as a zero of the characteristic polynomial that matter, and this is (modulo the --- important but also difficult --- issue of semi-simplicity of the Frobenius action) the same as the dimension of the fixed part of Frobenius on $H^2$.

The above is just a sketch; I am not going to give more details here, since to flesh this out would be the topic of an advanced graduate course in arithmetic geometry, and I don't have time to write the notes for such a course!

  • $\begingroup$ That would be enough sir, Thanks a lot for your fantastic answer again $\endgroup$
    – IDOK
    Commented Dec 8, 2011 at 5:09

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