Q(\sqrt{-l_0}) satisfies Heegner hypothesis for an Elliptic curve of conductor C implies C is a square

Trying to understand the proof of Corollary 2.3 in the following paper,

http://arxiv.org/pdf/1312.3884.pdf

I would like to be able to justify that the root number of the quadratic twist $E^{(-l_0)}$ is $-w_E$, where $w_E$ is the root number of $E$. To spare the readers some time, the set up is the following

$E$ is an elliptic curve over $\mathbb Q$ with conductor $C$ and $K=\mathbb{Q}(\sqrt{-l_0})$, where $l_0>3$ is a prime congruent to $3$ modulo $4$. It is given that $K$ satisfies Heegner hypothesis for $E$, namely every prime factor of $C$ splits in $K$. Does this imply that $C$ is a square or something? If yes, then I can prove my assertion about the root number.

• As far as I can tell, the argument uses Gross--Zagier: over $K$ one deduces from the fact that Heegner point has infinite order that the $L$-function has a simple zero (this is Gross--Zagier), thus the the sign in the fun'l equation over $K$ is $-1$. This is the product of the signs for each of $E$ and its twist. Earlier in the argument the sign for $E$ was shown to be $1$, and so the sign for the twist must be $-1$. As to why the root number over $K$ is always $-1$, you can look in GZ for a proof. Why should the Heegner hypothesis imply that $C$ is a square? E.g. if $C$ is prime, ... – tracing Feb 26 '15 at 0:41
• ... you can just choose $l_0$ such that $C$ splits in $K$. – tracing Feb 26 '15 at 0:42
• Thank you very much for your response, but as far as I understand, GZ just tells us that the $L'(E/K,1) \neq 0$. To be able to say that $L(E/K,s)$ vanishes at $s=1$, we need to look at something else. Is there something I miss? Thank you again – George Turcas Feb 26 '15 at 7:45
• In the context of GZ, the root number over $K$ is always $-1$. This is how one knowns that the $L$-function vanishes at $s = 1$, and why it makes sense to try to find a formula for its derivative (which is then given by the GZ formula). – tracing Feb 27 '15 at 0:59

It is a general fact that if $K$ is a number field, $K'/K$ is a quadratic extension, $E \rightarrow \mathrm{Spec}\, K$ is an elliptic curve, and $E' \rightarrow \mathrm{Spec}\, K'$ is its quadratic twist by $K'/K$, then the global root numbers are related by $$w(E_{K'}) = w(E)w(E'),$$ even though the corresponding formula for local root numbers fails (there is an extra term $(-1, K'/K)$ that disappears globally due to the product formula). This may be derived from standard properties of root numbers or, in your situation, from looking at $L$-functions (I can provide you with a reference, if you want).
Once you have the above formula, it remains to note that $w(E_{K'}) = -1$ because there is a single infinite place and all the "bad" finite places come in pairs (due to the Heegner hypothesis), so the corresponding local root numbers kill each other in the product.
• Thank you very much! I had the wrong impression that the character $\chi_{-l_0}(p)=-1$ if $p$ splits, and that's why I was desperately trying to see if $C$ happens to be a square. But $\chi_{-l_0}(p)$ in fact $1$ in this case. – George Turcas Feb 26 '15 at 7:52