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Let $f \in \mathbb{Z}[x]$ be a polynomial of odd degree $d \geq 5$, and consider the hyperelliptic curve $C$ with affine equation $y^2 = f(x)$. If $C'$ is another hyperelliptic curve such that for some $d \in \mathbb{Q}^\times$ we have that $C$ and $C'$ are isomorphic over $\mathbb{Q}(\sqrt{d})$, is it necessarily true that $C'$ is isomorphic to the curve with affine equation $dy^2 = f(x)$?

What I know: A simple argument involving the $j$-invariant says that the answer is yes for elliptic curves.

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  • $\begingroup$ The isomorphism is $\phi : \Bbb{Q}(C)(\sqrt{d})\to\Bbb{Q}(C')(\sqrt{d})$, with $\sqrt{d} \to - \sqrt{d}$ then $\phi$ becomes $\overline{\phi}$, I'd say one question is the genus of the subfield fixed by $\overline{\phi}\phi^{-1}$, if $\Bbb{Q}(C') = \Bbb{Q}(u)[v]/(dv^2 - f(u))$ that genus is $0$. $\endgroup$
    – reuns
    Aug 21, 2019 at 1:17

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In general the answer to the question is no, because a hyperelliptic curve $C_0/\mathbb{Q}$ whose automorphism group contains an element of order $2$ other than the hyperelliptic involution may have quadratic twists $C_1/\mathbb{Q}$ and $C_2/\mathbb{Q}$ that are both isomorphic to $C$ over $\mathbb{Q}(\sqrt{d})$ such that $C_0$, $C_1$, $C_2$ are all pairwise nonisomorphic over $\mathbb{Q}$.

For a concrete example, consider:

  • $C_0$: $y^2 = x^5 + x^4 + 2x^3 + x^2 + x$
  • $C_1$: $-3y^2 = x^5 + x^4 + 2x^3 + x^2 + x$
  • $C_2$: $y^2 = x^5 - 4x^3 - 7x^2 - 6x - 2$

which are all isomorphic over $\mathbb{Q}(\sqrt{-3})$ but pairwise nonisomorphic over $\mathbb{Q}$.

However, if we have a hyperelliptic curve $C/\mathbb{Q}$ that has no automorphisms other than the hyperelliptic involution $(x, y) \to (x, -y)$, then there will be a unique quadratic twist (up to $\mathbb{Q}$-isomorphism) associated to each quadratic field—this is actually the generic case, so the answer to the question is usually yes, but this fails in (infinitely many) special cases.

If you are curious as to how I found the example above, I searched the LMFDB for a genus $2$ curve over $\mathbb{Q}$ with automorphism group the Klein-$4$ group and at least one rational Weierstrass point (for a $y^2 = \text{quintic}(x)$ model):

http://www.lmfdb.org/Genus2Curve/Q/?num_rat_wpts=1-&aut_grp_id=%5B4%2C2%5D

Note that the LMFDB uses minimal equations, so the curves will often be written as $y^2 + h(x)y = g(x)$ when they have good reduction at $2$, but these can always be transformed to $y^2 = f(x) = 4g(x) + h(x)^2$ and made $y^2 = \text{quintic}(x)$ by moving the rational Weierstrass point to infinity.

The reason I looked for the Klein-$4$ group was that I knew I a curve with two automorphisms of order $2$ (that commute) in order to get two independent quadratic twists—this follows from the general fact that the twists of a curve are governed by the first cohomology "group" (a pointed set in the nonabelian case) of the automorphism group. For a proof of this fact, see Theorem 4.5.2 here:

http://www-math.mit.edu/~poonen/papers/Qpoints.pdf

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