TITLE: Showing a ring is DVR
QUESTION [5 upvotes]: Here $k$ is an algebraically closed field.

Consider $R=k[x,y,z]/(x^2+y^2+z^2-1)$ and $\mathfrak p=(x+iy,z-1)/(x^2+y^2+z^2-1)$ a prime ideal of $R$. I want to show that $R_{\mathfrak p}$ is a DVR.

I know that a ring $A$ is DVR iff $A$ is Noetherian, local, one dimensional and normal.
A ring is called a normal ring iff localisation at each of its prime ideals give integrally closed domains.
Clearly $R_{\mathfrak p}$ is a Noetherian local ring. Now $\dim R_{\mathfrak p}=\operatorname{ht}\mathfrak{p}=\dim R-\dim R/{\mathfrak{p}}$. Here $R/\mathfrak{p}\cong k[x]$. Hence $\dim R_{\mathfrak p}=1$. So it is enough to show that $R_{\mathfrak p}$ is a normal ring.
Clearly any integrally closed domain is a normal ring. So it is enough to show that this ring is integrally closed. I am stuck here. How do I show that $R_{\mathfrak p}$ is integrally closed?
Thank you in advance for the help.

REPLY [5 votes]: By Proposition 9.2 of Atiyah-Macdonald it suffices to show that the maximal ideal $\newcommand{\m}{\mathfrak{m}} \newcommand{\p}{\mathfrak{p}} \m := \mathfrak{p} R_\mathfrak{p}$ of $R_\mathfrak{p}$ is principal.  (The proposition actually contains many equivalent characterizations: see here for a few more.)
Note that $0 = x^2 + y^2 + z^2 - 1$ in $R_\mathfrak{p}$, so
$$
(x+iy)(x-iy) = x^2 + y^2 = -(z^2 - 1) = -(z+1)(z-1) \, .
$$
Since $x-iy \notin \p$ and $-(z+1) \notin \p$ then both are units in $R_\p$.  Thus $x+iy$ and $z-1$ are both generators for $\m$:
$$
\m = (x+iy,z-1) R_\mathfrak{p} = (x+iy)R_\mathfrak{p} = (z-1)R_\mathfrak{p}
$$
so $\m$ is principal, as desired.
(This is essentially the same idea as in user45765's comment.)