Polynomial ring in one variable/Integral closure and submersion/Example

Let ${}K$ be a field and consider ${}R=K[X]$. Since this is a principal ideal domain, the only interesting forcing algebras (if we are only interested in the local behavior around ${}(X)$) are of the form ${}K[X,T]/{\left(X^{n}T-X^{m}\right)}$. For ${}m\geq n$ this ${}K[X]$-algebra admits a section (corresponding to the fact that ${}X^{m}\in {\left(X^{n}\right)}$), and if ${}n\geq 1$ there exists an affine line over the maximal ideal ${}(X)$. So now assume ${}m<n$. If ${}m\geq 0$ then we have a hyperbola mapping to an affine line, with the fiber over ${}(X)$ being empty, corresponding to the fact that ${}1$ does not belong to the radical of ${}{\left(X^{n}\right)}$ for ${}n\geq 1$. So assume finally ${}1\leq m<n$. Then ${}X^{m}$ belongs to the radical of ${}{\left(X^{n}\right)}$, but not to its integral closure (which is the identical closure on a one-dimensional regular ring). We can write the forcing equation as ${}X^{n}T-X^{m}=X^{m}{\left(X^{n-m}T-1\right)}$. So the spectrum of the forcing algebra consists of a (thickened) line over ${}(X)$ and of a hyperbola. The forcing morphism is surjective, but it is not a submersion. For example, the preimage of ${}V(X)=\{(X)\}$ is a connected component hence open, but this single point is not open.