Power bases for unramified extensions of an affine genus 0 curve. $\newcommand{\Qbar}{\overline{\mathbb{Q}}}$
Let $R = $ Spec $\Qbar[x,\frac{1}{x-1},\frac{1}{x-2}]$, with fraction field $K$.
Let $f$ be integral over $R$.
Suppose the integral closure $\tilde{R}$ of $R$ in $K(f)$ is unramified over $R$. Must $\tilde{R} = R[f]$?
 A: Well I found a counterexample in the simplest example that could work.
Let $R = k[x]$, and let $f$ satisfy the polynomial $f^2 = x^2(x-1)$. This is precisely the equation of the nodal cubic, with node at $(x,f) = (0,0)$, and thus is not integrally closed.
Note that $R[f]$ is degree 2 over $R$, and ramified above $x = 0,x = 1$. The ramification above $x = 0$ is precisely the node.
On the other hand, to normalize, one adjoins $t := f/x$ to $R[f]$. Then $t^2 = x-1$, so the normalization is just $R[t]$ satisfying the relation $t^2 = x-1$. But note that over the $x$-line, this is only ramified at $x = 1$, so by inverting $x-1$, we find that $R[t]$ is unramified over $R[(x-1)^{-1}]$. Thus, we have the sequence
$$\text{Spec }R[t]\longrightarrow\text{Spec }R[f]\longrightarrow\text{Spec }R[x]$$
where the composite is ramified only above $x = 1$, but the second map is ramified above both $x = 0$ and $x = 1$. The "extra ramification" at $x = 0$ comes from the node, which is resolved when you take the integral closure, since normalization basically "splits the node in 2", so the answer is no because $R[t] \ne R[f]$.
