Purely simple Kronecker modules   $M$ , built from an algebraically closed field
 $M$ , built from an algebraically closed field   $K$ , arise from a triplet
 $K$ , arise from a triplet   $\left( m,\,h,\,\alpha \right)$  where
 $\left( m,\,h,\,\alpha \right)$  where   $m$  is a positive integer,
 $m$  is a positive integer,   $h:\,K\,\cup \,\{\infty \}\,\to \,\{\infty ,\,0,\,1,\,2,\,3,\,.\,.\,.\,\}$  is a height function, and
 $h:\,K\,\cup \,\{\infty \}\,\to \,\{\infty ,\,0,\,1,\,2,\,3,\,.\,.\,.\,\}$  is a height function, and   $\alpha$  is a
 $\alpha$  is a   $K$ -linear functional on the space
 $K$ -linear functional on the space   $K\left( X \right)$  of rational functions in one variable
 $K\left( X \right)$  of rational functions in one variable   $X$ . Every pair
 $X$ . Every pair   $\left( h,\,\alpha \right)$  comes with a polynomial
 $\left( h,\,\alpha \right)$  comes with a polynomial   $f$  in
 $f$  in   $K\left( X \right)\left[ Y\, \right]$  called the regulator. When the module
 $K\left( X \right)\left[ Y\, \right]$  called the regulator. When the module   $M$  admits non-trivial endomorphisms,
 $M$  admits non-trivial endomorphisms,   $f$  must be linear or quadratic in
 $f$  must be linear or quadratic in   $Y$ . In that case
 $Y$ . In that case   $M$  is purely simple if and only if
 $M$  is purely simple if and only if   $f$  is an irreducible quadratic. Then the
 $f$  is an irreducible quadratic. Then the   $K$ -algebra End
 $K$ -algebra End   $M$  embeds in the quadratic function field
 $M$  embeds in the quadratic function field   ${K\left( X \right)\left[ Y\, \right]}/{(f\,)}\;$ . For some height functions
 ${K\left( X \right)\left[ Y\, \right]}/{(f\,)}\;$ . For some height functions   $h$  of infinite support
 $h$  of infinite support   $I$ , the search for a functional
 $I$ , the search for a functional   $\alpha$  for which
 $\alpha$  for which   $\left( h,\,\alpha \right)$  has regulator
 $\left( h,\,\alpha \right)$  has regulator   $0$  comes down to having functions
 $0$  comes down to having functions   $\eta \,:\,I\,\to \,K$  such that no planar curve intersects the graph of
 $\eta \,:\,I\,\to \,K$  such that no planar curve intersects the graph of   $\eta$  on a cofinite subset. If
 $\eta$  on a cofinite subset. If   $K$  has characterictic not
 $K$  has characterictic not   $2$ , and the triplet
 $2$ , and the triplet   $\left( m,\,h,\,\alpha \right)$  gives a purely-simple Kronecker module
 $\left( m,\,h,\,\alpha \right)$  gives a purely-simple Kronecker module   $M$  having non-trivial endomorphisms, then
 $M$  having non-trivial endomorphisms, then   $h$  attains the value
 $h$  attains the value   $\infty$  at least once on
 $\infty$  at least once on   $\text{K}\,\cup \,\text{ }\!\!\{\!\!\text{ }\infty \text{ }\!\!\}\!\!\text{ }$  and
 $\text{K}\,\cup \,\text{ }\!\!\{\!\!\text{ }\infty \text{ }\!\!\}\!\!\text{ }$  and   $h$  is finite-valued at least twice on
 $h$  is finite-valued at least twice on   $\text{K}\,\cup \,\text{ }\!\!\{\!\!\text{ }\infty \text{ }\!\!\}\!\!\text{ }$ . Conversely all these
 $\text{K}\,\cup \,\text{ }\!\!\{\!\!\text{ }\infty \text{ }\!\!\}\!\!\text{ }$ . Conversely all these   $h$  form part of such triplets. The proof of this result hinges on the fact that a rational function
 $h$  form part of such triplets. The proof of this result hinges on the fact that a rational function   $r$  is a perfect square in
 $r$  is a perfect square in   $K\left( X \right)$  if and only if
 $K\left( X \right)$  if and only if   $r$  is a perfect square in the completions of
 $r$  is a perfect square in the completions of   $K\left( X \right)$  with respect to all of its valuations.
 $K\left( X \right)$  with respect to all of its valuations.