Hyperelliptic Jacobians with (or without) CM

Let $$C$$ be a hyperelliptic curve $$y^2 = f(x)$$ defined over $$\mathbb{Q}$$, where $$f(x) \in \mathbb{Q} [x]$$ is a polynomial of degree $$n=5$$ or $$6$$, and $$J = Jac(C)$$ its Jacobian. I know Zarhin's result [Hyperelliptic Jacobians without complex multiplication], which states that if the Galois group $$G= Gal(f)$$ of $$f$$ is either the Symmetric group $$S_n$$ or the alternating group $$A_n$$, then the endomorphism ring of $$J$$ is $$\mathbb{Z}$$ (that is, $$J$$ has no complex multiplication).

My question is that whether there is a condition on $$G$$ under which $$J$$ has CM.

The reason why I ask this is the follwing; In [Wamelen, Examples of genus two CM curves defined over the rationals], Wamelen found 19 curve (of gunus 2) whose Jacobian has CM. In all of these examples, the Galois group of $$f$$ is either the cyclic group $$C_4$$ of order $$4$$ or the Frobenius group $$F_5$$ of order $$20$$. So my question is that;

Is there any example of $$C$$ without CM and with $$G = C_4$$ ?

The curve $$y^2 = x^5 + 2x^4 + 2x^3 +2x^2 + 2x +1$$ has $$G= C_4$$ and its Jacobian $$J$$ has endomorphism algebra $${\rm End}^0(J) \cong \mathbb{Q}$$.

For the curve $$y^2 = x^5 + x^4 + x^3 + x^2 + x$$ we have $$G= C_4$$ and its Jacobian $$J$$ has endomorphism algebra $${\rm End}^0(J) \cong \mathbb{Q} \times \mathbb{Q}$$.

As explained in Jedrzej's answer, it is possible to impose conditions on $$G$$ which restrict the possible endomorphism algebras. A relevant example here is: if $$C:y^2=f(x)$$ is a hyperelliptic curve with $$f(x) \in \mathbb{Q}[x]$$ of degree 5 or 6, and $$|G|$$ is odd, then the Jacobian of $$C$$ does not have CM. In fact, a stronger statement is true: if $$f(x)$$ has degree 5 or 6 with coefficients in some number field $$K$$, Galois group $$G$$ of odd order, and $$J$$ has CM, then $$K$$ must contain a real quadratic field.

Example of a (hyper)elliptic curve with $$G = C_4$$ and no CM:

For

$$C : y^2 = x^4 - x^3 + x^2 - x + 1$$

one has $$G = C_4$$ (see here). On the other hand, $$C$$ is the elliptic curve with label 200.b2 and has no complex multiplication.

(I searched for a polynomial with Galois group $$C_4$$ using lmfdb.org, then computed the conductor and j-invariant of the corresponding curve in Magma and then searched for the curve in lmfdb)

Comment: I believe that the philosophy is that the bigger the Galois group is, the smaller is the ring of automorphisms. In other words, it is possible to impose a comdition on the Galois group such that the endomorphism ring will be small. It seems unlikely that any condition on the Galois group will give you a large endomorphism ring.

• This does not answer the question, since the question was about curves of genus 2, not elliptic curves. – Michael Stoll Apr 8 '19 at 14:38
• Hmm, that's right. I'll try later to produce another counterexample. – Jędrzej Garnek Apr 8 '19 at 21:41