Consider an elliptic curve $E: y^{2} = x^{3} + ax + b$. Then the quadratic twist by a squarefree $d$ is given by $E^{d} : dy^{2} = x^{3} + ax + b$. What is the relationship between the conductor of $E^{d}$ and $E$?
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OP, I think the conductor given in your comment is wrong. SAGE says that the conductor of $E^{19}$ is $2^4 \cdot 19^2 \cdot 53$. What David wrote is true in the case $d \equiv 1$ mod $4$ i.e. for $d=-19$ instead of $19$. You can imagine the twist by $19$ as a twist by $-19$ then by $-1$. $2$ appears in the conductor because of the twist by $-1$. I think this recent question is reusable here: writing down the minimal discriminant of an elliptic curve It seems that for $p \ge 5$ the twist by $\tilde{p} = \left( \frac{-1}{p} \right)p$ (which is always of the form $4k+1$) decreases the discriminant if and only if $p^6| \Delta$ and $p| c_4$. If $E$ has multiplicative reduction at $p$ then $E^{\tilde{p}}$ will have additive potentially multiplicative reduction therefore the exponent of $p$ in the conductor will change from $1$ to $2$. And if $E$ had additive, potentially multiplicative reduction at $p$, the exponent will change from $2$ to $1$. If $E$ has potentially good additive reduction at $p$ then the exponent of $p$ in the conductor can stay $2$ or decrease to $0$. In the latter case, the discriminant also decreases and we have $p^6 | \Delta$ and $p|c_4$. But there are examples where the discriminant decreases but the conductor does not e.g. Cremona 121A2 has $-11$-twist 121C1 where the discriminant changes from $-1 \cdot 11^{10}$ to $-1 \cdot 11^4$ but the conductor stays the same. (See e.g. http://www.lmfdb.org/EllipticCurve/Q/121.c2 and http://www.lmfdb.org/EllipticCurve/Q/121.a1 ) |
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You don't say what field $E$ is over -- is $\mathbb{Q}$? If so, then when $d$ is coprime to the conductor of $E$ the conductor of $E^{(d)}$ is $\Delta^2 N$, where $N$ is the conductor of $E$ and $\Delta$ is the discriminant of $\mathbb{Q}(\sqrt{d})$, which is either $|d|$ or $4|d|$ depending on whether $d$ is 1 or $\ne 1$ mod 4, as Tib points out. But if $d$ is not coprime to $N$ there is no simple formula -- the conductor can go down as well as up, as you can see by twisting by the same character twice! (Edited to fix the stupid mistake in the first version) |
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