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Let $E/Q$ be an elliptic curve of conductor $N$, $F$ an imaginary quadratic number field of discriminant $d$ with $d$ coprime to $N$, and $E^d/Q$ the quadratic twist of $E$ by $d$.

Let $p$ be prime where $E$ has semi-stable reduction, and let $w$ be a place of $F$ with $w\mid p$.

What is the Tamagawa number $c_w(E/K)$ of $E/K$ at $w$ in terms of the Tamagawa numbers $c_p(E/Q)$ and $c_p(E^d/Q)$ of $E/Q$ and $E^d/Q$ at $p$?

(Since the Tamagawa numbers should be the minimal discriminant exponents, I guess the answer is $c_w(E/K)=c_p(E/Q)=c_p(E^d/Q)$ if $w\nmid d$?)

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I assume $K$ is meant to be $F$ throughout the question.

There's a simple formula for the odd part of the Tamagawa numbers, which I'll denote by $c'$, at least when $p \neq 2$. Namely: $$c'_w(E/K) = c'_p(E/\mathbb{Q})c'_p(E^d/\mathbb{Q})$$ You can deduce this as follows, using Tate's algorithm (see the table in his Antwerp paper, but beware that he assumes algebraically closed residue field so you can only deduce things indirectly...) and some "basic" facts about $p$-adic uniformization. The most interesting case is when $E$ has split multiplicative reduction. Then the component group scheme is actually etale, and hence all components are $\mathbb{F}_p$-rational. The unramified quadratic twist will then "kill" the odd part of the component group (i.e. it will not be $\mathbb{F}_p$-rational). This implies the formula in this case. For a ramified quadratic twist, Tate's table shows that the component group has trivial odd part, again showing the formula. Other cases of bad reduction follow from this one. The case of good reduction is even easier, using the table.

A formula for the even part and the case $p = 2$ would be a bit more subtle, but you should be able to work something out with more work.

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  • $\begingroup$ That formula follows from representation theory, too: The $C_2=\operatorname{Gal}(K/\mathbb{Q})$-module $\Phi'(E/\mathbb{F}_w)$ splits as a $+1$-part, equal to $\Phi'(E/\mathbb{F}_p)$, and a $-1$ part, equal to $\Phi'(E^d/\mathbb{F}_p)$, as $\mathbb{Z}'[C_2]$-modules. Where primes indicate prime-to-2 everywhere and $\Phi$ are the groups of connected components each time. $\endgroup$
    – Chris Wuthrich
    Commented Sep 29, 2022 at 9:49
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    $\begingroup$ Agreed, but you do have to be careful if you try to prove the formula this way, e.g. if $E$ has good reduction, then $\Phi(E/\mathbb{F}_w)$ is trivial (including the even part) but $\Phi(E^d/\mathbb{F}_p)$ need not be. $\endgroup$
    – Ari Shnidman
    Commented Sep 29, 2022 at 12:17
  • $\begingroup$ Indeed you need to use the assumption that $d$ and $N$ are coprime otherwise the formula might not hold. $\endgroup$
    – Chris Wuthrich
    Commented Sep 29, 2022 at 13:10

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