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?

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    $\begingroup$ 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. $\endgroup$ Mar 30, 2011 at 23:55
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    $\begingroup$ A quick and very minor comment. If the characteristic you are working in is 2, then the hypersurface is not normal. $\endgroup$ Mar 31, 2011 at 15:00
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    $\begingroup$ 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. $\endgroup$
    – diverietti
    Apr 1, 2011 at 0:01
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    $\begingroup$ @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 Vakil
    Jun 22, 2013 at 19:57

3 Answers 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.

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    $\begingroup$ @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. $\endgroup$ Mar 31, 2011 at 2:33
  • $\begingroup$ @Sándor: Thank you for this clarifying comment. $\endgroup$ 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... $\endgroup$ Mar 3, 2022 at 8:36

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!)


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 $r\ge 3$.


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