my question is that

already we know that the Birch and Swinnerton Dyer conjecture ,formally conjectures that the Hasse-weil L-function should have a zero at $s=1$ when curves have infinitely many rational points on it,

so my question is that imagine an elliptic curve $E/\mathbb{Q}$ which has rank $r>0$ and with $ \left|{E(\mathbb{Q})}\right|=\infty$ so we find that $L(E/\mathbb{Q},s)_{s=1}=0$

but i am interested in the zeroes on the line $s=1+it$, we know that above curve has got a zero at $s=1$ so are there any zeroes on the line $s=1+it$ ,if so tell me the cardinality of set of zeroes ,i mean whether there are finite zeroes or infinitely many zeroes

and if i get the answer there is a deep intuition behind the answer and properties of elliptic curves,

and may be someone can conjecture still more things knowing the zeroes there on the line

thank you, touch everyone's feet who helped me,by suggesting books,and resources and making me what i am today by studying privately

  • 1
    $\begingroup$ I imagine that the set of zeroes on the line $s = 1 + it$ will be infinite, just as with the Riemann $\zeta$-function. $\endgroup$
    – Emerton
    Jul 23, 2011 at 12:00
  • $\begingroup$ @Emerton:but the hasse-weil zeta function seems to be different right?? i recently read an article that tells , only the elliptic curves with Complex Multiplication,the L-function can be written as a Zeta function with Grossencharacters, i hope you dont misunderstand me,as if i am genius ,but i am learning things so please explain me,how did you get that imagination ??? $\endgroup$
    – Trust God
    Jul 23, 2011 at 15:19
  • $\begingroup$ @Emerton:but i always follow the quote "imagination is more important than knowledge "-albert Einstein fantastic imagination $\endgroup$
    – Trust God
    Jul 23, 2011 at 15:21
  • 6
    $\begingroup$ @trust god: I sincerely request you not to use sentences of the type : $\mathsf{touch everyone's feet}$ and so on. If you are thankful, to somebody then it would be worth saying thanks and some kind sentences, rather than using this type of language. $\endgroup$
    – C.S.
    Jul 23, 2011 at 19:16
  • 1
    $\begingroup$ Imagination won't get you anywhere if you don't already have a big, big storehouse of knowledge. $\endgroup$ Jul 24, 2011 at 5:03

2 Answers 2


The $L$-function has about $\displaystyle{\frac{T}{\pi} \log T \ }$ zeros in the strip with $0 < t < T$. See section 5.3 of Iwaniec and Kowalski's "Analytic Number Theory," in particular Theorem 5.8.

It should be possible, if it hasn't been done already, to show that a positive proportion of these zeros are on the critical line using Selberg's method. Hafner extended Selberg's method to various families of degree 2 $L$-functions in a series of papers in the 1980s.

  • 2
    $\begingroup$ Hafner did indeed prove the analogue of Selberg's theorem for L-functions of holomorphic cusp forms; the precise reference is "Zeros on the critical line for Dirichlet series attached to certain cusp forms", Mathematische Annalen 264, pages 9 to 37. $\endgroup$ Jul 24, 2011 at 4:59

+1 to Micah, read his answer first!

As an addendum, the $L$-function of an elliptic curve looks like (and is) an $L$-function of degree 2. It has a certain conductor, which is defined in terms of the primes of bad reduction and which affects the analytic behavior of the $L$-function.

But beyond that, there is unfortunately not a lot that is known from an analytic perspective for elliptic curve $L$-functions in particular. Once it is proved that they have analytic continuation and a functional equation, knowledge of where the $L$-function came from seems to be disappointingly useless for proving theorems. (With some exceptions, e.g., the value of the critical point; there is also the modularity theorem, etc. but I am not regarding that as "analytic".) There is a lot of general machinery, which is very well explained in Chapter 5 of Iwaniec and Kowalski for general $L$-functions in general. Beyond that, my impression (which could be mistaken) is that not a whole lot is known.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy