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