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$\DeclareMathOperator\Norm{Norm}$Suppose $E/\mathbb{Q}(j(E))$ is a CM elliptic curve and $d$ is a non-square. Let $E_d$ denote the twist of $E$ by $\mathbb{Q}(j(E))(\sqrt{d})$. I know if $d$ is relatively prime to the conductor of $E$, then we have $$N_{E_d} = d^2N_{E}$$ However, by computational investigations, there appear to be certain primes that always divide the norm of the conductor regardless of twist.

As a computational example: I will take a curve with CM by $K=\mathbb{Q}(\sqrt{-5})$. I take the curve with $j$-invariant $$j(E)= 5887918080 (31261995198\sqrt{5} - 69903946375)$$ (i just picked the one with smallest absolute norm). Taking the absolute norm of the conductor of $E$ gives the factorization $$\Norm(N_E)=5^2 * 7^4 * 11^4 * 19^2 * 41^2 * 47^4 * 199^2 $$ Taking $d=77$, I get the twisted curve $E_d$ which has conductor $$\Norm(N_{E_d}) =5^2 * 19^2 * 41^2 * 47^4 * 199^2 $$ No matter what integer I twist by, I cannot get rid of these factors. Do I need to twist by something in $\mathbb{Q}(j(E)) =\mathbb{Q}(\sqrt{5})$?

Other examples seem to indicate that there are primes of bad reduction common to all twists. For example for CM elliptic curves over $\mathbb{Q}$ with CM by $\mathbb{Q}(\sqrt{-q})$, where $q$ is an odd prime. $q$ is a prime of bad reduction among all curves with the same $j$-invariant.

My question: What is known about the primes of bad reduction common to all CM curves with the same $j$-invariant?

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These are exactly the primes where the Neron model of $E$ has fiber type, in Kodaira's table, anything other than $I_0$ and $I_0^*$. Here $I_0$ represents good reduction and $I_0^*$ represents a quadratic twist of good reduction by a ramified extension (quadratic twisting by an unramified extension preserves the reduction type).

(At least, this is true away from primes above 2, which are surely more complicated).

The fiber type can be computed by Tate's algorithm, but to tell whether the type is $I_0$ or $I_0^*$, you can simply perform a ramified quadratic twist and check whether either the curve or its quadratic twist has good reduction. The number you quadratic twist by doesn't matter, as long as it is $\pi$-adic valuation, where $\pi$ is the prime of bad reduction, is odd.

For $\pi$ in $\mathbb Q(\sqrt{-5})$ lying over a prime $p$ of $\mathbb Q$, unless $\pi$ is ramified (i.e. $\pi$ lies above $5$), then any number which has $p$-adic valuation odd will do the trick.

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  • $\begingroup$ @Rdrr One thing that might be happening is you could have $I_0^*$ reduction at what prime lying over $p$ and $I_0$ reduction at the other. Then twisting by $p$ will change the conductor but not the norm of the conductor. Twisting by a suitable $\alpha$ should fix this. $\endgroup$
    – Will Sawin
    Mar 13 at 15:15
  • $\begingroup$ I've tried running a version of Tate's Algorithm on $E_d$ as I specified. It says the factor $5$ is of type II, $47$ is of type III and $19,41$ and $19$ are of type $I_0^*$. I can twist by an element of norm 5 in $\mathbb{Q}(\sqrt{5})$ to get rid of the $5^4$ from the norm of the conductor, but I can't get rid of the $47^4$ which tate's algorithm says I should be able to. 47 remains prime in $\mathbb{Q}(\sqrt{5})$, and twisting by 47does nothing...any idea what's going on? $\endgroup$
    – Rdrr
    Mar 13 at 16:55
  • $\begingroup$ @Rdrr You can only get rid of factors of type III by doing a quartic twist, which you can only do if your curve has CM by $\mathbb Q(i)$. $\endgroup$
    – Will Sawin
    Mar 18 at 13:19
  • $\begingroup$ Do you have a reference for these claims? Does Tate mention this in his original paper? $\endgroup$
    – Rdrr
    Mar 21 at 16:51
  • $\begingroup$ @Rdrr If you look at where $I_0^*$ appears in Tate's algorithm, it occurs when $\pi$ divides $a_1$ and $a_2$, $\pi^2$ divides $a_3$ and $a_4$, and $\pi^3$ divides $a_6$. Outside characteristic 2 (which I need to assume for this to be correct, but that's OK since your curve has good reduction there), we can change variables so that $a_1=a_3=0$ and preserve this property. Then quadratic twisting by $\pi$ is just dividing $a_2$ by $\pi$, $a_4$ by $\pi^2$, and $a_6$ by $\pi^6$. $\endgroup$
    – Will Sawin
    Mar 21 at 17:03

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