Prove that $R[\sqrt{\pi}]$ is a DVR 
If $R$ is a DVR (discrete valuation ring) with uniformizer $\pi$, then prove that $R[\sqrt{\pi}]$ is a DVR.

How shall I begin? First do I have to find a candidate for the uniformizing element of $R[\sqrt{\pi}]$? What about $\sqrt{\pi}$? There are also $3$ conditions for a DVR:
$\bullet$ Noetherian property
$\bullet$ Integrally closed
$\bullet$ It must have exactly $1$ nonzero prime ideal
For the first property, if $R$ is noetherian, then so is $R[T]$ by Hilbert-Basis theorem and any quotient by an ideal is also noetherian. Then there is another theorem, which states: There is an isomorphism $R[T]/(f)\to R[\alpha]$ (provided $\alpha$ is an algebraic element in some extension of the field of fractions of $R$), is it useful?
Am I on the right track?
EDIT: After the hint of Mathmo123, one can write $a+b\sqrt{\pi}$, some element of $R[\sqrt{\pi}]$ as $u\pi^v+x\pi^y\sqrt{\pi}=u\left(\sqrt{\pi}\right)^{2v}+x\left(\sqrt{\pi}\right)^{2y+1}$, but required is I think to express it as $w\left(\sqrt{\pi}\right)^z$, or not?
 A: *

*$R[\sqrt{\pi}]$ Noetherian. Done.


*$\dim R[\sqrt{\pi}]=1$ follows from the integral extension $R\subset R[\sqrt{\pi}]$.


*$R[\sqrt{\pi}]$ integrally closed. The field of fractions of $R[\sqrt{\pi}]$ is $K[\sqrt{\pi}]$, where $K$ is the field of fractions of $R$. Every element $t\in K[\sqrt{\pi}]$ can be written as $t=a+b\sqrt{\pi}$ with $a,b\in K$. Obviously $t$ is a root of the polynomial $p_t(T)=T^2-2aT+a^2-b^2\pi$ which belongs to $K[T]$. By Gauss' Lemma we can conclude that $t$ is integral over $R[\sqrt{\pi}]$ if and only if $p_t$ has the coefficients in $R$. Thus we get $2a\in R$, and $a^2-b^2\pi\in R$.
If $2$ is invertible in $R$ it follows easily that $a,b\in R$, so $t\in R[\sqrt{\pi}]$. Unfortunately we can't count on this.
We then should content ourselves with $a^2-b^2\pi\in R$. If $a\in R$ or $b\in R$ we get $t\in R[\sqrt{\pi}]$. If $a\notin R$ and $b\notin R$, then $a=u\pi^{-m}$ and $b=v\pi^{-n}$ with $u,v\in R$ invertible, and $m,n\ge1$. We have $a^2-b^2\pi=u^2\pi^{-2m}-v^2\pi^{-2n+1}$. Now let's study the following cases:
$n>m$. Write $u^2\pi^{-2m}-v^2\pi^{-2n+1}=\pi^{-2n+1}(u^2\pi^{2(n-m)-1}-v^2)\in R$. But $u^2\pi^{2(n-m)-1}-v^2\in R\setminus(\pi)$, that is, it is invertible, so $\pi^{-2n+1}\in R$, a contradiction.
$n\le m$. Write $u^2\pi^{-2m}-v^2\pi^{-2n+1}=\pi^{-2m}(u^2-v^2\pi^{2(m-n)+1})\in R$. Similarly we get $\pi^{-2m}\in R$, a contradiction.

Edit. An alternative (and simpler) approach is to show that $R$ is a Noetherian local domain whose maximal ideal is principal. It is easy to show that the ideal generated by $\sqrt{\pi}$ is maximal. If $M$ is another maximal ideal, then $M\cap R=\pi R$. It follows that $\pi\in M$ and from $\sqrt{\pi}^2=\pi$ we get $\sqrt{\pi}\in M$, therefore $M=(\sqrt{\pi})$.
