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Let $r\geq 5$ be a prime. Suppose I have a specified hyperelliptic curve $C: y^2=f(x)$ defined over $\mathbb{Q}$, where $f\in\mathbb{Q}[x]$ has degree $r$. Note: the roots of $f$ are not rational but are defined over a certain extension of $\mathbb{Q}(\zeta_{r}+\zeta_{r}^{-1})$.

Consider the hyperelliptic curve $C': y^2=f'(x)$, where $f'(x)=f(x)(x-\alpha)$ for some fixed $\alpha\in\mathbb{Q}$.

I have found that the conductor exponent of $C$ at $r$ is equal to the conductor exponent of $C'$ at $r$. Is this to be expected since I am adjoining a rational root to $f$ i.e. is there an innate reason why this should happen?

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    $\begingroup$ At a glance it looks like the mod 2 representation will be related (maybe the same?). In many cases the conductor of the curve at odd primes is the same as that of the mod 2 representation, so that should explain your observation. On the other hand, they sometimes differ (there is a conductor drop from the 2-adic to the mod 2 representation) in which case your observation might not hold. So maybe a counter-example exists; how hard have you tried to look for one? $\endgroup$
    – Aurel
    Commented Jul 11 at 14:19
  • $\begingroup$ Thanks for your comment! I'm considering a fixed family of hyperelliptic curves - I think this is not necessarily the case in general? $\endgroup$
    – Maleeha
    Commented Jul 13 at 17:39
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    $\begingroup$ Indeed, it might be that there is no conductor drop in your family. $\endgroup$
    – Aurel
    Commented Jul 14 at 19:39

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