I am looking to check whether the hypersurface in $A^{n}$ defined by $x_{1}^{2} + x_2^{2} + .... + x_n^{2} = 0$ is a normal variety.....In general, are there any nice sufficiency conditions to prove normality?
$\begingroup$
$\endgroup$
4
-
8$\begingroup$ Welcome to MathOverflow! A nice sufficient condition for normality is smoothness or (harder to check) just regularity. There is a more technical criterion for normality (due to Serre) and not very helpfully called "R1+S2" : you can read about it on page 183 of Matsumura's Commutative Ring Theory. $\endgroup$– Georges ElencwajgCommented Mar 30, 2011 at 23:55
-
2$\begingroup$ A quick and very minor comment. If the characteristic you are working in is 2, then the hypersurface is not normal. $\endgroup$– Karl SchwedeCommented Mar 31, 2011 at 15:00
-
2$\begingroup$ Take a look to Shafarevich's "Basic Algebraic Geometry 1: Varieties in Projective Space". This is exercise 5 in the section 5 of chapter 2, page 138. A detailed solution is given in the case $n=3$ some pages earlier. $\endgroup$– diveriettiCommented Apr 1, 2011 at 0:01
-
2$\begingroup$ @diverietti: I see that the first part of the question is indeed an exercise in Shafarevich. I couldn't see the detailed solution; could you give a more specific reference? $\endgroup$– Ravi VakilCommented Jun 22, 2013 at 19:57
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
3
Dear anonymous,
Here is an expansion of what Georges said in the comment. I will assume, as you wrote, that you are a beginner in AG but not in math. And please do not feel too bad about diverietti's comment, for this site to function well we do need to keep a certain standard. That's why it is a good idea to use your real name and state your background. People here would be a lot more accommodating if they know exactly where you come from.
As Georges wrote, normality is equivalent to two technical conditions: $R_1$ and $S_2$.
$R_1$ means ``regular in codimension one". In the case of your interest, which is a hypersurface $f \in \mathbb A^n$, it can be checked easily (I will assume you work over $\mathbb C$). Just take the ideal $J$ generated by all the partial derivatives of $f$ and let $d$ be the dimension of $\mathbb C[x_1,\cdots, x_n]/J$. As long as $n-d-1\geq 2$, your hypersurface will be $R_1$. In your particular case, $J = (x_1,\cdots, x_n)$ and $d=0$, so as long as $n\geq 3$ you will be OK. But this procedures works for any hypersurface, for example $x^3+y^5+z^7$.
The second condition $S_2$ is also known as ``Serre condition $S_2$". It is more technical to explain, and can actually be hard to check in general, but in this case, you are again in luck. Any hypersurface in $\mathbb A^n$ (for any $n$!) satisfies it.
So, in summary, your quadric hypersurface is normal as long as $n\geq 3$, but hopefully what I wrote will be helpful in other cases you might be interested in.
answered Mar 31, 2011 at 0:54
-
11$\begingroup$ @anonymous: In other words, a nice sufficient condition for normality (that you are looking for) is that if it is a hypersurface in something smooth and has a singular set of codimension at least 2, then it is normal. $\endgroup$ Commented Mar 31, 2011 at 2:33
-
$\begingroup$ @Sándor: Thank you for this clarifying comment. $\endgroup$ Commented Mar 31, 2011 at 3:36
-
$\begingroup$ Beware that in Hailong's great answer the codimension of the singular set has to be $\geq 3$ in $\mathbb A^3$, while in @Sándor's great comment it has to be $\geq 2$ in the hypersurface. These conditions are of course equivalent , but it might be not completely useless to point out this apparent discrepancy... $\endgroup$ Commented Mar 3, 2022 at 8:36
$\begingroup$
$\endgroup$
There is a simpler approach which works in this case, using the fact that this is a double-cover of affine $(n-1)$-space branched over a locus regular in codimension 1 (if the characteristic is not 2, as Karl Schwede pointed out). See Exercise 5.4.H (which gives a general useful tool) and Exercise 5.4.I(b) (which includes the question you ask) in the June 11, 2013 version of the notes available here.
(Also, welcome to mathoverflow!)
answered Jun 22, 2013 at 20:03
$\begingroup$
$\endgroup$
Given a Cohen-Macaulay affine ring $R$ over a perfect field, it satisfies $S_k$ for all $k$. Since $R$ is equidimensional, Jacobian criterion says that the singular locus is cut out by the Jacobian ideal $J$. It is then sufficient and necessary to have codim $J \ge 2$ in $R$ for $R$ to satisfy $R_1$.
Assuming characteristic is not 2, the coordinate ring $S = k[x_1,\cdots,x_n]/(x_1^2+\cdots+ x_r^2)$ is cut out by a non-zerodivisor and is thus Cohen-Macaulay. Its singular locus is cut out by $I = (x_1,\cdots, x_r)$ which has codim $r-1$ in $S$. Therefore $S$ is normal iff $r\ge 3$.
