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If $R$ be 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 ?

$\bf{EDIT}:$ 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 ?

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    $\begingroup$ You are on the right track. Have you been shown the equivalent definition of a DVR, that says R arises from a discrete valuation on its field of fractions? With this definition, the result is almost immediate $\endgroup$ – Mathmo123 Mar 1 '15 at 11:09
  • $\begingroup$ @Mathmo123 you mean: Every nonzero element of the field of fractions of a DVR $R$ with uniformizing element $\pi$ may be uniquely written as $u\pi^v$ with $u\in R^{\times}$ and $v\in\mathbb Z$ ? $\endgroup$ – inequal Mar 1 '15 at 11:12
  • $\begingroup$ That's more of a corollary of it. Another way of defining a DVR is to define a discrete valuation on its field of fractions, and set $R$ to be the elements with positive valuation. You can then deduce those properties, but if you haven't had this given as an equivalent definition, you should carry on with your current approach. $\endgroup$ – Mathmo123 Mar 1 '15 at 11:22
  • $\begingroup$ @Mathmo123 Yes we had it as a corollary of an equivalent definition, which is slightly different, Proposition 1.7 part (iv) here, but I didn't get also your formula can you explain it $\endgroup$ – inequal Mar 1 '15 at 11:41
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    $\begingroup$ You should use (ii) in that definition. You have that every non zero element of $R$ can be expressed as $u\pi^v$. Can you find a way of expressing every non zero element of $R[\pi]$ as $u\sqrt\pi^v$? $\endgroup$ – Mathmo123 Mar 1 '15 at 11:50
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  1. $R[\sqrt{\pi}]$ Noetherian. Done.

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

  3. $R[\sqrt{\pi}]$ 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 $u^2\pi^{2(n-m)-1}-v^2\in R-(\pi)$, 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.

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