Halves of points of an odd degree hyperelliptic curve in its jacobian

Let $f(x)$ be a degree $(2g+1)$ monic polynomial with coefficients in an algebraically closed field $K$ with $char(K)\ne 2$ and without repeated roots. Let $\mathfrak{R}\subset K$ be the $(2g+1)$-element set of roots of $f(x)$. Let $\mathcal{C}: y^2=f(x)$ be an odd degree genus $g$ hyperelliptic curve over $K$. Let $J$ be the jacobian of $\mathcal{C}$ and $J[2]\subset J(K)$ the (sub)group of its points of order dividing $2$. We identify $\mathcal{C}$ with the image of its canonical embedding into $J$ (the infinite point of $\mathcal{C}$ goes to the identity element of $J$). Let $P=(a,b)\in \mathcal{C}(K)\subset J(K)$ and $M_{1/2,P}\subset J(K)$ the set of halves of $P$ in $J(K)$, which is $J[2]$-torsor. In a previous work we established an explicit bijection between $M_{1/2,P}$ and the set of collections of square roots $$\mathfrak{R}_{1/2,P}:=\{\mathfrak{r}: \mathfrak{R} \to K\mid \mathfrak{r}(\alpha)^2=a-\alpha \ \forall \alpha\in\mathfrak{R}; \ \prod_{\alpha\in\mathfrak{R}} \mathfrak{r}(\alpha)=-b\}.$$ The aim of this paper is to describe the induced action of $J[2]$ on $\mathfrak{R}_{1/2,P}$ (i.e., how signs of square roots $\mathfrak{r}(\alpha)=\sqrt{a-\alpha}$ should change)

Non-isogenous elliptic curves and hyperelliptic jacobians

Let $K$ be a field of characteristic different from $2$, $\bar{K}$ its algebraic closure. Let $n \ge 3$ be an odd prime such that $2$ is a primitive root modulo $n$. Let $f(x)$ and $h(x)$ be degree $n$ polynomials with coefficients in $K$ and without repeated roots. Let us consider genus $(n-1)/2$ hyperelliptic curves $C_f: y^2=f(x)$ and $C_h: y^2=h(x)$, and their jacobians $J(C_f)$ and $J(C_h)$, which are $(n-1)/2$-dimensional abelian varieties defined over $K$. Suppose that one of the polynomials is irreducible and the other reducible. We prove that if $J(C_f)$ and $J(C_h)$ are isogenous over $\bar{K}$ then both jacobians are abelian varieties of CM type with multiplication by the field of $n$th roots of $1$. We also discuss the case when both polynomials are irreducible while their splitting fields are linearly disjoint. In particular, we prove that if $char(K)=0$, the Galois group of one of the polynomials is doubly transitive and the Galois group of the other is a cyclic group of order $n$, then $J(C_f)$ and $J(C_h)$ are not isogenous over $\bar{K}$