Let $R$ be a $k$-algebra, $k$ is an algebraically closed field of characteristic zero. Assume that $R$ is an integral domain or a UFD (being a UFD makes things easier), $a$ is an algebraic element over $R$ and $S:=R[a]$ is also an integral domain.
Notice that if $R$ is a UFD, then $S$ is isomorphic to $\frac{R[T]}{(h)}$, where $h \in R[T]$ is the minimal polynomial of $a$ over $R$. Otherwise, $S$ is isomorphic to $\frac{R[T]}{(h_1,\ldots,h_m)}$, where $h_1,\ldots,h_m \in R[T]$, with probably $m > 1$.
Question 1: When $S$ is a normal ring, namely, integrally closed in its field of fractions?
I am not assuming that $R$ and $S$ have the same field of fractions; but I would appreciate also an answer for that case.
Example 1 (trivial): $R=\mathbb{C}$. Then $a \in R$ and $S=R$ is trivially normal.
Question 2 (special case): Take $R=k[x_1,\ldots,x_n]$, $n \geq 1$. Hopefully, results mentioned here may help in this special case.
Example 2: $n=1$, so $R=k[x]$, hence $S$ is isomorphic to $\frac{k[x][T]}{(h)}$, where $h \in k[x][T]$ is the minimal polynomial of $a$ over $k[x]$ ($h$ is irreducible). If, for example, $h=T^3-x^2$, then $S$ is isomorphic to $k[t^2,t^3]$ which is not normal.
In Question 2, I do not mind to concentrate on the special cases $n=1$ and $n=2$.
For $n=1$: When $\frac{k[x,y]}{(u)}$ is normal? where $u \in k[x,y]$ is irreducible.
For $n=2$: When $\frac{k[x,y,z]}{(v)}$ is normal? where $v \in k[x,y,z]$ is irreducible.
A partial answer, according to the above quoted question:
For $n=1$: If $u \in \mathcal{J}(u)$, then $S= \frac{k[x,y]}{(u+1)}$ is normal.
For $n=2$: If $v \in \mathcal{J}(v)$, then $S= \frac{k[x,y,z]}{(v+1)}$ is normal.
I have asked the above question in MSE.
Thank you very much!
