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Let $K$ be an algebraically closed field of characteristic different from $2$, $g$ a positive integer, $f(x)in K[x]$ a degree $2g+1$ monic polynomial without repeated roots, $C_f: y^2=f(x)$ the corresponding genus g hyperelliptic curve over $K$, and $J$ the jacobian of $C_f$. We identify $C_f$ with the image of its canonical embedding into $J$ (the infinite point of $C_f$ goes to the zero of group law on $J$). It is known (arXiv:1809.03061 [math.AG]) that if $g>1$ then $C_f(K)$ does not contain torsion points, whose order lies between $3$ and $2g$. In this paper we study torsion points of order $2g+1$ on $C_f(K)$. Despite the striking difference between the cases of $g=1$ and $g> 1$, some of our results may be viewed as a generalization of well-known results about points of order $3$ on elliptic curves. E.g., if $p=2g+1$ is a prime that coincides with $char(K)$, then every odd degree genus $g$ hyperelliptic curve contains, at most, two points of order $p$. If $g$ is odd and $f(x)$ has real coefficients, then there are, at most, two real points of order $2g+1$ on $C_f$. If $f(x)$ has rational coefficients and $g<52$, then there are, at most, two rational points of order $2g+1$ on $C_f$. (However, there are exist genus $52$ hyperelliptic curves over the field of rational numbers that have, at least, four rational points of order 105.)
Let $C$ be a hyperelliptic curve of genus $g>1$ over an algebraically closed field $K$ of characteristic zero and $O$ one of the $(2g+2)$ Weierstrass points in $C(K)$. Let $J$ be the jacobian of $C$, which is a $g$-dimensional abelian variety over $K
Let $K$ be an algebraically closed field of characteristic different from 2, $g$ a positive integer, $f(x)$ a degree $(2g+1)$ polynomial with coefficients in $K$ and without multiple roots, $C:y^2=f(x)$ the corresponding genus $g$ hyperelliptic curve
In this article, we show that in each of four standard families of hyperelliptic curves, there is a density-$1$ subset of members with the property that their Jacobians have adelic Galois representation with image as large as possible. This result co
Let $C$ be a hyperelliptic curve of genus $g$ over the fraction field $K$ of a discrete valuation ring $R$. Assume that the residue field $k$ of $R$ is perfect and that $mathop{textrm{char}} k eq 2$. Assume that the Weierstrass points of $C$ are $K$
We describe a method to show that certain elliptic surfaces do not admit purely inseparable multisections (equivalently, that genus one curves over function fields admit no points over the perfect closure of the base field) and use it to show that an