Normal ring between $k$ and $k[X]$ is singly generated. Let $k$ be a field and $A$ be a normal domain such that $k \subset A \subseteq k[X]$. Then $A=k[f]$ for some $f\in k[X]$.
My attempt to a possible solution:
By Lüroth's theorem, $\operatorname{Quot}(A)=k(t)$ where $t\in k(X)\setminus k$ is transcendental over $k$. The non-trivial valuation rings of $k(t)$ containing $k$ are $k[t^{-1}]_{(t^{-1})}$ and $k[t]_{(g(t))}$ where $g(t) \in k[t]$ irreducible. Now, $A$ is the intersection of all valuation rings of $k(t)$ containing it. But I don't know which of these contain $A$. How should I proceed?
 A: Fact: Let $k(y)$ be a function in field in one variable over $k$. Then all non-trivial valuation rings of $k(y)$ over $k$ are given by: $$R^y_h:=k[y]_{(h)}=\{\frac{f}{g}: f,g\in k[y], h\nmid g\}$$
where $h$ runs through all irreducible polynomials in $k[y]$ and $$R^y_\infty:=k[y^{-1}]_{(y^{-1})}=\{\frac{f}{g}:f,g\in k[y],\deg g\geq \deg f\}$$
Clearly we have $$k[y]=\bigcap_{h\ne\infty}R^y_h$$
and $$k=\bigcap_hR^y_h$$ where $h$ stands for either an irreducible polynomial in $k[y]$ or $\infty$.
Now let $k\subsetneq A\subseteq k[X]$. As you already noted we have $K:=\operatorname{Quot}(A)=k(t)$ for some $t\in k(X)\setminus k$. By taking $t^{-1}$ instead of $t$ we may assume wlog that $t\notin R^X_\infty$.
Claim: $A=\bigcap_{h\ne\infty} R^t_h$.
It then follows that $A=k[t]$ and we are done.
Proof of the claim: $A$ is normal, hence the intersection of the valuation rings containing it, thus it suffices to show that

*

*$A\subseteq R_h^t$ for all $h\ne\infty$ and

*$A\nsubseteq R^t_\infty$
For 1.: Let $k[t]\ni h\ne\infty$. By Chevalley's extension theorem we know that there is a valuation ring $R'=R^X_g$ of $k(X)$ such that $R^t_h=K\cap R'$. Since $t\in R^t_h$ and $t\notin R_\infty^X$ we get $g\ne\infty$. Hence $k[X]\subseteq R^X_g$ and therefore $A\subseteq K\cap k[X]\subseteq K\cap R^X_g=R^t_h$.
For 2.: If not we would have $A\subseteq \bigcap_hR^t_h=k$, a contradiction.
