Let $E\_1$ and $E\_2$ be two Elliptic curve defined over $\mathbb Q$ . Let $p$ be an fixed given odd prime of $\mathbb Q$ at which both the curves have good ordinary reduction. Moreover p-adic $L$-function of $E\_1$ and $E\_2$ are same . Does it mean that the complex $L$-function of $E\_1$ and $E\_2$ are also same ?
Is there some sufficient criteria on p-adic $L$-functions such that such that the $L$ function is determined? 



Does p-adic $L$- function determine the $L$ function
Hmmm. I am not so sure that the answer is *"No"*. In fact I would rather bet on *"Yes"*.


Of course, I totally agree that the characteristic ideal, i.e. the ideal generated by the $p$-adic $L$-function in $\Lambda$, is not enough to determine the elliptic curve. In particular there are plenty of curves for which the $p$-adic $L$-function is a unit in $\Lambda^{\times}$.


The $p$-adic $L$-function can be viewed as a measure $\mu$ on the Galois group of $F\_\infty/\mathbb{Q}$. It is build up from modular symbols of the form $\bigl[\frac{a}{p^k}\bigr]$ as $a$ and $k$ varies over all positive integers. Knowing the measure $\mu$ it is easy to extract the unit root $\alpha$ of the Frobenius at $p$ and hence the value of $a\_p$. Then it is not difficult to see from the definition of $\mu$ that one can compute all the modular symbols $\bigl[\frac{a}{p^k}\bigr]$. It is true that these values do not seem to carry the value of $a\_{\ell}$ for primes $\ell\neq p$ with them needed to reconstruct the complex $L$-function; we would need modular symbols with $\ell$ in the denominator and I can not see immediately how to get them from $\mu$.


Nevertheless, there are plenty of values of $a$ and $k$. And it would be a big surprise to me if there very by chance two elliptic curves such that all the values of the modular symbols $\bigl[\frac{a}{p^k}\bigr]$ would be equal. But I have no clue of how to prove this intuition.


So I ran through some examples. Let $F\_{\infty}$ be the cyclotomic $\mathbb{Z}\_p$-extension of $\mathbb{Q}$.
I picked a few elliptic curves of small conductor such that


* $E$ has good ordinary reduction at $p$.
* There are no torsion points in $E(F\_{\infty})$, simply by making sure that the $\ell$-adic Galois representation is surjective for all $\ell$.
* The Tamagawa numbers are all 1 for $E/F\_\infty$, by imposing that the Kodaira type at all bad places is $I\_1$.
* The curve is not anomalous at $p$, i.e. $a\_p \neq 1$. Actually, I fix $a\_p$.
* The Tate-Shafarevich group of $E/\mathbb{Q}$ is trivial.
* The rank of $E(F\_{\infty})$ is $0$. This will follow from the previous conditions if the rank of $E(\mathbb{Q})$ is $0$, since the $p$-adic $L$-function will be a unit.


Then I computed the $p$-adic $L$-functions $L\_p(T)$ for these with $T$ corresponding to $1+p$ under the cyclotomic character Gal$(F\_{\infty}/\mathbb{Q})$. By what I have imposed the leading term will be equal. For each $p^n$-th root of unity $\zeta$, the value of $L\_p(\zeta-1)$ is, up to a power of $\alpha$ which is the same for all my curves because I fixed $a\_p$, equal to the order of the Tate-Shafarevich group at the $n$-th level; at least if one believes the ($p$-adic version of the) Birch and Swinnerton-Dyer conjecture. From what I imposed, it is clear that the $p$-primary part will be trivial, but there may be different primes appearing in Sha for various curves. So there is no reason to believe that it would be easy to find two curves that have the same $p$-adic $L$-function in these family that I have chosen.


Here are some examples with $p=5$ and $a\_5=-1$.


139a1 $4 + 4 \cdot 5 + 4 \cdot 5^2 + O(5^5) + (1 + 4 \cdot 5 + O(5^2)) \cdot T + (3 + 5 + O(5^2)) \cdot T^2 + (3 + 2 \cdot 5 + O(5^2)) \cdot T^3 + (1 + 5 + O(5^2)) \cdot T^4 + O(T^5)$
141e1 $4 + 4 \cdot 5 + 4 \cdot 5^2 + O(5^5) + (4 + 3 \cdot 5 + O(5^2)) \cdot T + (3 \cdot 5 + O(5^2)) \cdot T^2 + (5 + O(5^2)) \cdot T^3 + (2 + 4 \cdot 5 + O(5^2)) \cdot T^4 + O(T^5) $
346a1 $ 4 + 4 \cdot 5 + 4 \cdot 5^2 + O(5^5) + (2 + 5 + O(5^2)) \cdot T + (4 \cdot 5 + O(5^2)) \cdot T^2 + O(5^2) \cdot T^3 + (1 + 2 \cdot 5 + O(5^2)) \cdot T^4 + O(T^5)$
906i1 $4 + 4 \cdot 5 + 4 \cdot 5^2 + O(5^5) + (3 + O(5^2)) \cdot T + (2 + 5 + O(5^2)) \cdot T^2 + (3 + 5 + O(5^2)) \cdot T^3 + (3 + 2 \cdot 5 + O(5^2)) \cdot T^4 + O(T^5)$


Finally a word why I believe the answer should be *Yes*. A big dream in the direction of BSD is to hope that there is a link between the $p$-adic and the complex $L$-function. Of course, we should believe that the order of vanishing at $s=1$ should be equal for instance (and this is only known when it is at most a simple zero). I would hope that such a link will be bijective. But that is a mere intuition and hence possibly wrong. But it also explains that the answer to this question might well be very difficult.


