Let $k$ be a field of characteristic not $2$ and algebraically closed. I would like to show that $\operatorname{Spec} k[x,y,z]/(x^2 + y^2+z^2)$ is normal.
I would appreciate any help/hint. Thank you!
Asked
Active
Viewed 1,083 times
2
user26857
- 52,094
user211392
- 767
2 Answers
2
Normal is equivalent to showing that the ring $k[x,y,z]/(x^2+y^2+z^2)$ is integrally closed. Normal is also equivalent to the Serre conditions $R_1$ and $S_2$, which are that the singular locus is at least codimension 2 (this is $R_1$) and that a rational function on an open subset has no poles in codimension one, then it is in fact regular (this is $S_2$). Proving integrally closed is probably not too hard, but it will likely be somewhat long/tedious, although I'm sure someone on this site has a wonderful trick for it. I'll instead explain why this variety satisfies $R_1$ and $S_2$ and is therefore normal.
Since our variety $X\subset \mathbb{A}^3$ is a hypersurface ($x^2+y^2+z^2$ is irreducible), condition 2 is automatic (this works for any local complete intersection inside a smooth variety). So normal is equivalent to the singular locus being codimension 2 (in this case, zero-dimensional). We can compute the singular locus from the Jacobian. This matrix is $$\begin{pmatrix} 2x & 2y & 2z \end{pmatrix}$$
and since $\operatorname{char} k\neq 2$, this matrix is rank zero exactly when $x=y=z=0$, which is a single point. So we've shown that the singular locus is codimension two, and therefore this variety is normal.
-
Your "proof" would work for $k[x,y]/(x^{2}+y^{2})$, isn't it? – Bombyx mori Feb 09 '15 at 01:09
-
No, it would not. For curves, normal is equivalent to smooth by dimension considerations, but $k[x,y]/(x^2+y^2)$ is a non-smooth curve. – KReiser Feb 09 '15 at 01:11
-
This is clear, but your reasoning seems to work through (hypersurface, singular locus, etc) if OP did not see it. I think you need to add a word that it is irreducible because of Eisenstein's criterion or something. – Bombyx mori Feb 09 '15 at 01:13
-
I show that the singular locus is codimension two. I do not see how this would extend to the case you're talking about- there are no codimension two sets in a curve. – KReiser Feb 09 '15 at 01:16
-
In the case I am talking about, it does not work not because the singular locus is not of codimension 2 (the origin is singular and its orthgonal complement is the plane). But because $x^{2}+y^{2}=(x+iy)(x-iy)$ and it is not a hypersurface. So you need a proof for $n=3$ it is a hypersurface. – Bombyx mori Feb 09 '15 at 01:21
-
Ahhh, I see what you're after here. Yes, that is a good point. – KReiser Feb 09 '15 at 01:23
-
You mean $R_1$ and $S_2$, not $S_1$ and $S_2$. The condition $S_1$ has a different meaning (which is in any case subsumed by $S_2$). – tracing Feb 10 '15 at 02:32
-
@Bombyxmori: It is a general property that $k[x_1,\ldots,x_n]/f$ is is Cohen--Macaulay and pure of dimension $n=1$ (i.e. every irred. comp. has dimension $n-1$) for any $n$ and any non-zero $f$, and so is $S_i$ for all $i$. It has nothing to do with whether or not $f$ is irreducible. The point is that in the case when $n = 2$ and $f = x^2 + y^2$, the quotient is Cohen--Macaulay, but it is not $R_1$: the singular locus does not have codimension $2$. So KReiser's argument is complete as written. – tracing Feb 10 '15 at 02:37
-
@tracing: No, you do need a proof that the quotient is a domain. Otherwise the proof would collapse. See my answer below. You can also check Harthsorne, page 147. He has some hints as well. I should mention that Ravi Vakil's book is self-contained. So if Serre's criterion is not introduced before then it is not needed for this problem. – Bombyx mori Feb 10 '15 at 03:18
-
@Bombyxmori: Of course there are other approaches, but certainly Serre's criterion can be applied (although I agree it is not the approach Ravi had in mind at this point in his book), and there is no need to check you have a domain; I don't understand why you are insisting that there is. One way to think about it is that if $f = 0$ is a reducible hypersurface in $\mathbb A^3$, then the two components necessarily meet along a curve, not just a point, so that the singular locus will automatically lie in codim'n one, not codim'n two. Maybe you want normal to mean normal domain? If so, ... – tracing Feb 10 '15 at 03:22
-
... then I agree that you should check you have a domain. But normal does not normally mean normal domain, it means (on the ring side) a product of normal domains, or (on the scheme side) a scheme which is Zariski locally the Spec of a normal domain. And Serre's criterion applies to any Noetherian ring to determine if it is normal in this sense; there is never any need to explicitly consider whether the ring is a domain. If Serre's criterion applies (as it does here) then it automatically follows that the ring is a product of normal domains. – tracing Feb 10 '15 at 03:23
-
@tracing: No, I mean normal scheme. – Bombyx mori Feb 10 '15 at 03:23
-
@Bombyxmori: Then I don't understand your insistence on having to check that you have a domain. – tracing Feb 10 '15 at 03:24
-
@tracing: I do not think you follow my argument or understand what you wrote earlier. Since I have responded quite a bit I shall not respond any further as I have to teach tomorrow. – Bombyx mori Feb 10 '15 at 03:26
-
@Bombyxmori: I certainly understand what I wrote earlier, but you are correct that I don't understand your argument; indeed, I don't see any argument by you --- just a (false) assertion that KReiser's argument would apply to $k[x,y]/(x^2+y^2)$ (it wouldn't; this ring is not $R_1$), and references to Hartshorne and Vakil. If you don't want to continue the discussion, that's fine, and up to you. – tracing Feb 10 '15 at 03:30
-
@user26857: I finally had sometime off-teaching. You do not need to look up Eisenbud, this is in stack-project page: http://stacks.math.columbia.edu/tag/031O. – Bombyx mori Feb 16 '15 at 02:22
0
I think this is a standard HW problem in Hartshorne (somewhere in Chapter 2). A step by step guided proof based on other problems can be found in a series of problems in Ravi Vakil's book (page 162).
If you still cannot do it, there is in fact an identical post addressing your question posted a while ago. Hope this 25 cents helps.
Bombyx mori
- 19,638
- 6
- 52
- 112