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 ?

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.