Torsion in the jacobian of a super elliptic curve Let $y^n = f(x)$ define a smooth projective curve $C$ over some field $k$ with $\deg f \geq n$ and odd and with $f(x)$ having no repeated roots. Let $J$ be the Jacobian of $C$ and $J[n]$ it's (geometric) n-torsion. Then is it true that the points $x= x_i, y=0$ generate the group $J[n]$ for the roots $x_i$ of $f(x)$?
Is there a similar explicit description if $f$ happens to have repeated roots? 
 A: As mentioned by the previous answers, this cannot be true for $n \ge 3$ by size considerations. When you identify the points $(x_i, 0)$ of $C$ inside its jacobian $J$, you are implicitly using some base-point. I will assume that $n$ and $\deg f$ are coprime, so that there is exactly one rational point at infinity (which I will denote by $\infty$). So you are considering the subgroup of $J[n]$ generated by the divisor classes
$$
[(x_i, 0) - \infty] \in J[n].
$$
Let $\zeta_n$ be a primitive $n$th root of unity and use $\zeta$ to denote both (i) the automorphism of $C$ defined by $(x, y) \mapsto (x, \zeta_n y)$ and (ii) the corresponding automorphism this induces on $J$. Note that each of the aforementioned divisor classes $[(x_i, 0) - \infty]$ are fixed by $\zeta$ and hence
$$
[(x_i, 0) - \infty] \in J[1 - \zeta].
$$
When $f$ is separable, it turns out that the $[(x_i, 0) - \infty]$ generate the subgroup $J[1 - \zeta]$ of $J[n]$; for a proof, see Proposition 3.2 of 
(1) Schaefer, Edward F. "Computing a Selmer group of a Jacobian using functions on the curve." Math. Ann 310 (1998): 447-471.
which is also available as arxiv:1507.08325. This proof might actually even work when $f$ is not assumed to be separable (I couldn't immediately see anything that breaks if the separability assumption is dropped).
When $n = p$ is prime, it turns out that $J[(1 - \zeta)^{p - 1}] = J[p]$; this is shown in Section 3 of (1). There has been some work on attempting to understand generators for/the field of definition of $J[(1 - \zeta)^{i}]$. Using my "division by $1 - \zeta$ formula" (arxiv:1810.07299), one can compute generators when $i \le 2$ and the field of definition for $i \le 3$ (this work does not assume that $n$ is prime, but when $n$ is not prime, it is not necessarily true anymore that $J[(1 - \zeta)^{i}] \subseteq J[n]$ for $i \le n - 1$). I also compute the field of definition of $J[(1 - \zeta)^{i}]$ for $i \le p$ in the specific case $y^p = x^q + 1$ (see arxiv:1910.14251, Theorem 3.6 5.1.6).
Additionally, the case $y^p = u^s (1 - u)$ for $1 \le s \le p - 2$ is studied in
(2) Greenberg, Ralph. "On the Jacobian variety of some algebraic curves." Compositio Mathematica 42.3 (1980): 345-359. 
(3) Tzermias, Pavlos. "Explicit rational functions on Fermat curves and a theorem of Greenberg." Compositio Mathematica 122.3 (2000): 337-345.
Letting $J_{p, s}$ be the jacobian of $y^p = u^s (1 - u)$ and $K$ be the cyclotomic field $K = \mathbf{Q}(\zeta_p)$, Tzermias's Theorem 1 in (3) (which is a combination of work by Greenberg (2), Gross-Rohrlich, Kurihara) classifies exactly "how much" of the $J_{p, s}[p^\infty]$ and $J_{p, s}[\ell^\infty]$ for $\ell \neq p$ is defined over $K$; the answer is that $J_{p, s}[p^\infty](K) = J_{p, s}[(1 - \zeta)^3]$ and $J_{p, s}[\ell^\infty](K) = \{ 0 \}$ except in the special cases $(\ell, p, s) \in \{ (2, 7, 2), (2, 7, 4) \}$. In addition, Tzermias in (3) computes generators for $J_{p, s}[(1 - \zeta)^i]$ for $i \le 3$.
A: Not in general. $J[n]$ has $n^{2g}$ elements, where $g$ is the genus. If $n$ and $\deg f$ are relatively prime, then $g=(n-1)(\deg f-1)/2$. The points you ask for generate a subgroup of order $n^{\deg f-1}$ (the upper bound is easy, for equality see my preprint). So for $n\geq 3$ $J[n]$ has too many points for your claim to hold.
A: You don't have enough points to generate the group if $n \ge 3$. Moreover, if $f(x)$ has rational roots, then all the points on the Jacobian you write down are rational, whereas $\mathbf{Q}(J[n])$ contains $\mathbf{Q}(\zeta_n) \ne \mathbf{Q}$ for $n > 2$.
