Simultaneous orderings in function fields
If \(R\) is a Dedekind domain, then a sequence \(u_0,u_1,u_2,\dots\) of elements of \(R\) is said to be a simultaneous ordering of \(R\) if for all prime ideals \(P\) of \(R\) and all \(x\in R\) one has \[ v_P\left(\prod_{k=0}^{n-1}(x-u_i)\right)\geq v_P\left(\prod_{k=0}^{n-1}(u_n-u_i)\right) \] for all \(n\). It has been proved by \textit{M. Bhargava} [Am. Math. Mon. 107, 783--799 (2000; Zbl 0987.05003)] that a sequence has this property if and only if the polynomials \(\prod_{i=0}^{n-1}(X-u_i)/(u_n-u_i)\) form a basis for the \(R\)-module consisting of all polynomials \(f\) over the field of fractions of \(R\), which map \(R\) into \(R\). It has been proved by \textit{M. Wood} [J. Number Theory 99, 36--56 (2003; Zbl 1076.13011)] that this cannot happen in the case when \(R\) is the ring of integers in an imaginary quadratic number field. The author considers quadratic extensions of the field \(F_q(T)\) (with odd \(q\)) having the form \(K=F_q(T)[Y]/(Y^2-D(T))\), where \(D(T)\) is a square-free polynomial over \(F_q\) with either odd degree, or even degree and a non-square leading coefficient, and shows that the ring of integers of \(K\) has a simultaneous ordering if and only if \(\deg D\leq 1\).
scientific article; zbMATH DE number 2183812