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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.

user26857
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  • I think hartshorne mentioned that normality is local property, if I am correct. It suffices to check on the maximal ideals of the local ring $R_p$. So looking at points, I guess? From noether normalization, you will get a polynomial ring. Then localize at the corresponding ideal. It should be a 1 variable polynomial ring at that point. Then it is a finite extension of integrally closed ring. Hence intergrally closed? Why finite extension of integrally closed ring integrally closed? – user45765 Sep 10 '17 at 21:37
  • I did not check this by hand but it looks reasonable to me by picture. And is $k$ here $C$ complex number. It would be better to have $char(k)\neq 2$ here. – user45765 Sep 10 '17 at 21:38
  • Yes normality is a local property as a ring $R$ is integrally closed iff $R_{\mathfrak m}$ is integrally closed for each of its maximal ideals $\mathfrak m$. Although it is only given that $k$ is algebraically closed you can assume $\text{char} (k)\neq 2$ if it simplies things. –  Sep 10 '17 at 21:43
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    This is my rough ideal. Reparametrize the surface by $\bar{w},w$, $w=(x+iy),\bar{w}=(x-iy)$. The prime ideal is rewritten as $(\bar{w}w+z^2-1)$. Then $k[x,y,z]=k[w,\bar{w},z]$. $(x+iy,z)$ ideal becomes $(w,z)$. So we can invert $\bar{w}$. Then the prime ideal becomes $(w+\frac{z^2-1}{\bar{w}})$. So the localized quotient ring becomes $k[\bar{w},z]_{(z-1)}$ which is still a UFD. So integrally closed. You can shift $z$ again. – user45765 Sep 10 '17 at 21:55

1 Answers1

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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.)

Viktor Vaughn
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  • Is it correct to say most non-singular varieties are normal? It seems that normal does imply non-singular. But there is an extra requirement on projective dimension. What is the relation between projective dimension and integrally closedness? – user45765 Sep 10 '17 at 23:04
  • There is a proof that nonsingular varieties are normal in these notes, but I haven't read it carefully. I don't think normal implies nonsingular: take the coordinate ring of the cone $z^2 = x^2 + y^2$. One can show that it is normal, but the cone has a singular point at the origin. I think the idea is that normality only ensures nonsingularity in codimension 1: a more precise statement of this is Serre's criterion. – Viktor Vaughn Sep 10 '17 at 23:30
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    I see. Thanks a lot. – user45765 Sep 10 '17 at 23:36