Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $E$?

share|cite|improve this question

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. and )

share|cite|improve this answer

You don't say what field $E$ is over -- is $\mathbb{Q}$? I'll assume this is what you meant.

Let $\Delta$ be 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. Then if $\Delta$ is coprime to the conductor $N$ of $E$, the conductor of $E^{(d)}$ is $\Delta^2 N$. But if $\Delta$ 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, and edited again two years later to fix another stupid mistake in the second version)

share|cite|improve this answer
Ah yes, sorry I was working over $\mathbb{Q}$. But is this correct? For example, $E: y^{2} = x^{3} + 5x + 22$ has conductor 53, but a twist $E^{19}: y^{2} = x^{3} + 5\cdot 19^2 x + 22\cdot 19^3$ which has conductor $2^{3}\cdot 17\cdot 19^{2} \cdot 31$. – JHubert Oct 24 '12 at 11:01
For the record: I don't agree with your computation of the conductor of $E^{19}$; I get that its conductor is $2^4 \cdot 19^2 \cdot 53$, which is consistent with what I said above, since $\Delta = 2^2\cdot 19$ in this case. – David Loeffler Jul 9 '14 at 17:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.