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I have read in the history of how Sir Swinnerton-Dyer and Prof. Bryan Birch, have found this conjecture,in that I have found a line like this,

...heuristically the value of the Hasse-Weil L-function in the infinite product at $s=1$ comes to be $L(E,1)=\prod_{p}\left(\frac{N_{p}}{p}\right)^{-1}$...

but I have two questions regarding that:

1) As we know that the infinite product makes sense only when $\Re(s)>3/2$ and if we plug $s=1$ it's meaningless ,and so it doesn't make any sense, my question is that how can one fix that the Hasse-Weil $L$-function (that infinite product) so that it makes sense at $s=1$. I don't want the answer to be "it's by analytical continuation", I want the infinite product to make sense at $s=1$ ,but that product $\prod_{p}\left(\frac1{1-a_{p}p^{-s}+p^{1-2s}}\right)$ will never make sense as it is a established theorem by Hasse, can we code a similar product which makes sense at $s=1$,and if I call that new infinite product to be $L^*(E,s)$ then my condition is that that new product should not harm the conjectural properties, to be clear, that new product is in such a way that BSD holds good for $L^*(E,s)$, I mean that function order of vanishing gives the rank at $s=1$ ,and the constant and everything ,and the only difference is the convergence ,the new product is well defined at $s=1$ and it produces $L^*(E,s)=\prod_{p}\left(\frac {N_{p}}{p}\right)^{-1}$ not heuristically but concretely,if we can't code a new product how does that first statement makes sense by substituting $s=1$ in $L(E,s)$

2) I studied that the Hasse-Minkowski theorem is not going to hold good for cubic curves,if so why does one consider the reduction modulo prime for elliptic curves?

thank you

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    Showing the Euler product makes sense for Re(s) > 3/2 does *not* necessarily mean the Euler product won't converge elsewhere. You're confusing an implicaton P ==> Q with an equivalence P <==> Q. Hasse showed the Euler product converges (absolutely) for Re(s) > 3/2 but he didn't prove anything about the product beyond that region. So your first question has some flawed premises.2011-06-21

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There are various accounts by Birch and Swinnterton-Dyer explaining the process by which they arrived at their conjecture. (I think one such is in a volume with the title "Computers and number theory", or something like that.) My memory (from having read these) is that they did begin by looking at the naive product $\prod_p (N_p/p)^{-1}$. Then, at some point, they learned that (in the CM case, which is all they explicitly studied, I think) they could use analytic continuation to compute $L(s,1)$, and so they shifted their attention from the naive product to the value of the $L$-function at $1$.

If I can find a reference for these accounts, I will update this answer to include it.

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    The products probably *do* converge at s = 1, but this lies deeper than GRH. See the top answer at the Math Overflow question http://mathoverflow.net/questions/25360/convergence-of-l-series/25417#25417 for a discussion of a result of Goldfeld on this topic. Birch and Swinnerton-Dyer indeed only numerically studied CM examples, for which the analytic continuation of the L-function was already known by work of Hecke.2011-06-21
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Let $f(x)=1+x+x^2+x^3+\dots$. What is $f(3)$? The series doesn't converge for $x=3$. But where the series does converge, it converges to $1/(1-x)$. If we take that as the definition of $f(x)$ then it's easy to get $f(3)$; $f(3)=-1/2$.

So: the sum makes no sense at $3$, but there's another formula (which is not a sum) which agrees with the sum wherever both make sense, and that other formula makes sense at $3$.

That's what happens with $L$-functions. There's another formula (which is not a product) which agrees with the product wherever both make sense, and which makes sense where the product doesn't.

And, like it or not, the process of constructing that other function is called analytic continuation.

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    @iyengar: You're missing the point. We already know that the infinite product has limited utility, so we try to find an alternative expression whose values agree with the infinite product at the product's domain of validity, but works also for values outside the original domain. Being able to analytically continue functions is routine in mathematics, much like making $x^n$ make sense not only for integer values of $n$.2011-05-02