Normality condition for a DVR Let $R$ be a discrete valuation ring, with characteristic not equal to $2$. Let $R[x]$ be the ring of polymomials in one variable over $R$, and fix a non-zero $a\in R$. I must find necessary and sufficient conditions on the valuation of $R$ in order for $A:=R[x,\sqrt{ax}]$ to be normal.
Called $K:=\operatorname{Frac}R$, notice that $K(x)=R(x)$. It follows that $\operatorname{Frac}A$ can be expressed as $R(x)\oplus R(x)\sqrt{ax}$. So given a $t\in \operatorname{Frac}A$, write it as $r+s\sqrt{ax}$, with $r,s\in R (x)$. The minimal polynomial of $t$ over $R(x)$ is $X^2-2rX+(r^2-s^2ax)$. If $t$ is integral over $A$, it is integral over $R[x]$, and the coefficients $2r,r^2-s^2ax$ must be in $R[x]$. So I should find a condition on the valuation such that $2r,r^2-s^2ax\in R[x]$ if and only if $r,s\in R[x]$, for every $r,s\in R(x)$. For example here two conditions that seem sufficient are that $2$ is a unit and $a$ is square-free. Before trying to fill the details, I would like to know if this kind of reasoning can be useful or it takes nowhere.
Edit: I see now that in the statement in italics I took for granted that $R[x]$ is normal, but I don't know if it's true actually.
Second edit: the exercise asked also to compute the normalization of $A$ when it's not normal, i.e. when $a$ is not square-free. Again, if I take a $t\in\operatorname{Frac}A$ and I write it as $r+s\sqrt {ax}$, where $r,s$ in $R(x)$, I have that $2r\in R [x]$ and $r^2-s^2ax\in R[x]$. So $r\in R[x]$, while $s:=\frac fg$ (with $f,g\in R[x]$) must satisfy $g\in R$ (i.e. it is a constant polynomial) and $g^2$ divides $a$. So as a solution I'd say that the integral closure of $R[x,\sqrt{ax}]$ in its field of fractions $R(x)\oplus R(x)\sqrt{ax}$ consists of the elements of the form $\frac fg +\frac hb\sqrt{ax}$, where $f,g,h\in R[x]$ and $b\in R$ such that $b^2$ divides $a$.
 A: *

*For $K$ a field then $K[x]$ is integrally closed


*If $S$ is a DVR with discrete valuation $v$ and $K=Frac(S)$ then:

*

*let $Int(S[x])$ be the integral closure of $S[x]$ in $Frac(S[x])$.


*We have the discrete valuation $w(\sum_{j=0}^J c_j t^j) = \inf_j v(c_j)$ on $S[x]$, which extends to $Frac(S[x])$.


*let $O_w = \{ r\in Frac(S[x]):w(r)\ge 0\}$. It is a DVR, whence integrally closed, so $Int(S[x])\subset O_w$.


*Therefore, $Int(S[x]) \subset K[x]\cap O_w = S[x]$ ie. $S[x]$ is integrally closed.
Let $v'$ be the discrete valuation of $R$.

*

*If $v'(a)\ge 2$ then $a = \pi^2c$ with $c\in R,\pi \in R,v'(\pi)=1$ and $A=R[x,\sqrt{ax}]$ is not integrally closed since it doesn't contain $\sqrt{cx}$ which is a root of $T^2-cx\in A[T]_{monic}$.


*If $v'(a)=1$ then $S=R[\sqrt{a}]$ is a DVR with maximal ideal $(a)$.
$S[\sqrt{x}]$ is integrally closed and $S[\sqrt{x}]\cap Frac(A)= (A \oplus \sqrt{a} A) \cap Frac(A)=A$ so $A$ is integrally closed.


*If $v'(a)=0$ then $A= R[\sqrt{ax}]$ is integrally closed.
