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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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I sincerely that he first part of this question is a straightforward exercise, and the second one is really too vague... In other words, I don't think your question fits the standards of this site. – diverietti Mar 30 2011 at 18:06
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@diverietti 1.I have solved the first part now and I do not think it was entirely straightforward (for example, I had to look around a lot for certain commutative algebra facts, which were ultimately found in Zariski Samuel). 2.I am new here and do not have a proper idea of what the "standards" of this site are, or what you think they are.... 3.I am a beginner in Algebraic Geometry, but certainly not a beginner in math. I do not think the second question is "too" vague. 4.I don't think your condescending and thoroughly unhelpful comment "fits" the standards of normal decency. Thanks... – anonymous Mar 30 2011 at 23:04
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Dear anonymous, congratulations on your convincing and civil comment. For the record, I don't in the least share diverietti's point of view: welcome to MathOverflow! Back to mathematics: 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. – Georges Elencwajg Mar 30 2011 at 23:55
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A quick and very minor comment. If the characteristic you are working in is 2, then the hypersurface is not normal. – Karl Schwede Mar 31 2011 at 15:00
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Dear anonymous, nobody here (certainly not me) is doubting of your expertise in mathematics. Perhaps my comment seemed too rude, but it was really not my intention. When I talked about the standards of this site, of course I was not referring to a sort of "élite"... Just I wanted to point out that everything works better here if the question is as much precise as possible (if you are new here you can take a look to mathoverflow.net/howtoask). I don't think in any case that my comment was indecent. If you felt offended, I have no problem to kindly ask you to excuse me. – diverietti Mar 31 2011 at 17:57
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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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@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. – Sándor Kovács Mar 31 2011 at 2:33
@Sándor: Thank you for this clarifying comment. – Hailong Dao Mar 31 2011 at 3:36

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