Purely inseparable points on curves of higher genus
Let \(k\) be a field of characteristic \(p>0\) and \(K\) be a function field of one variable over \(k\) (that is: a field of transcendental degree one over \(k)\). Let \(C/K: f(x,y)=0\) be a plane curve with coefficients in \(K\). Denote by \(f^{(n)} (x,y)\) the \(n\)-th twist of \(f\) with the Frobenius map of \(K\) and let \(C^{(n)}\) be the plane curve defined by the affine equation \(f^{(n)} (x,y)=0\). If \((x,y)\) is a \(K\)-rational point of \(C^{(n-1)}\), that is a solution of \(f^{(n-1)} (x,y)=0\), then obviously \((x,y)^{(1)}: =(x^p,y^p)\) is a \(K\)-rational point of \(C^{(n)}\). Such a solution of \(f^{(n)}\) is called old. A new solution of \(f^{(n)}\) is a solution which is not the image of a solution of \(f^{(n-1)}\) by the Frobenius element. -- In the paper under review the author proves the following result: Theorem 1. Assume that the genus of the curve \(C\) (over \(\overline K)\) is at least 2 and suppose that infinitely many of the equations \(f^{(n)} (x,y)=0\) have new solutions. Then there is a change of variables (that is a birational map of curves) in some finite extension of \(K\) which sends \(f\) to a polynomial with coefficients in a finite field. Note that conversely, if \(f\) is a polynomial which has coefficients in some finite field, then infinitely many \(f^{(n)}\) will be equal to \(f\), hence have solutions which are not Frobenius twists if \(f\) does. Using the geometric Mordell conjecture, the author gets the following: Corollary 1. Assume that the genus of the curve \(C\) (over \(\overline K)\) is at least 2 and suppose it is not birational over any finite extension to a curve defined over a finite field. Then the set of solutions of \(f(x,y)=0\) in the field \(K^{1/p^\infty}: =\lim_n K^{1/p^n}\) is finite.
scientific article; zbMATH DE number 1117488