Proving smoothness of a variety

I want to understand how smoothness is checked for algebraic varieties given as zero sets of functions.

Let me consider a "textbook" example to help fix ideas. Let the algebraic variety be given by these three polynomial equations,

$x_0x_3 -x_1x_2=0$

$x_1^2 -x_0x_2=0$

$x_2^2- x_1x_3=0$

Firstly I observe that if I assume $x_3$ and $x_2$ to be nonzero then the third equation follows from the first two.

Does that allow me to totally forget about the third equation?

Now I can write down the Jacobian matrix for the above set. I guess then I am supposed to check if the rank of that Jacobian evaluated on the common zeroset = or not to the number of functions.

But I can't figure out how this calculation has to be done. How do I determine the rank and further so on the common zero set?

I would be grateful if someone can walk me through the steps in this specific example.

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From a (basic) differential topology viewpoint a strategy could be to show that $0$ is a regular value of your polynomials (hence their zero-set is a smooth manifold) and then checking that the intersections are transversal. (Of course assuming a real or complex manifold) Unfortunately this does not work for your example... – Jose27 Sep 14 '11 at 7:05
Dear Anirbit, The third equation lets you solve for $x_2$, and eliminate it from the other equations, so that you are left with two equations in the three variables $x_0,x_1,x_3$. Now try solving these two equations just as you would have when you were in high school, and draw the resulting curve. Is it smooth? Regards, – Matt E Sep 14 '11 at 12:02
I predict you intended the last relation to be $x_2^2 - x_1 x_3$, because then this would be a standard example. Am I right? – David Speyer Sep 14 '11 at 21:13
Dear Anirbit, One difficulty with checking smoothness via Jacobians is that presumably you are interested in the variety cut out by your equations, not some possibly non-reduced scheme. For this, you need to compute the radical of the ideal generated by your equations before you move on to computing Jacobians. (E.g. $x = 0$ cuts out a smooth hyperplane, while $x^2 = 0$ cuts out the same hyperplane, but won't satisfy th e Jacobian condition.) Regards, – Matt E Sep 14 '11 at 21:41
@David Yes. I meant it to be $x_2^2$. Corrected it! – Anirbit Sep 17 '11 at 23:15

Taking Matt E's hint into account you could rewrite this as $x_0x_3-x_1^2x_3=0$ and $x_1^2-x_0x_1x_3=0$.

The Jacobian is $\left(\begin{matrix} x_3 & -2x_1x_3 & x_0-x_1^2 \\ -x_1x_3 & 2x_1-x_0x_3 & -x_0x_1\end{matrix}\right)$.

Since your equations aren't homogeneous, I'm assuming this isn't meant to be projective, so in particular $(0,0,0)$ is on your variety and also makes the Jacobian have $0$ in every entry so it isn't smooth.

Maybe you were hoping to see something more illuminating like how to check it has full rank when it actually does have full rank on the zero set?

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So is your conclusion that the variety I wrote down is NOT smooth? And yes, I was looking for a method to check that the rank version of thinking. I would think that one has to show that there exists a subset of the equations such that the Jacobian on the zero-set is of the same rank as the number of equations in that subset I guess thats what you are referring to as the Jacobian having a "full rank" on the zero set? – Anirbit Sep 14 '11 at 17:12
It is not smooth. If it were, then the Jacobian will have constant rank $n-r$ where $n$ is the number of variables and $r$ is the number of equations (so the rank is the dimension of the variety if it is a complete intersection). – Matt Sep 14 '11 at 17:57
@Anirbit: Dear Anirbit, You can easily figure out explicitly what variety you wrote down; it is reducible, a union of three one-dimensional components, two of which pass through the origin. Hence it is not smooth at the origin. Regards, – Matt E Sep 14 '11 at 21:42
@Matt E: As I understand it Anirbit actually doesn't care if this particular variety is smooth or not he (or she) is much more interested in how a general argument would go if one were to use the Jacobian criterion. – Matt Sep 14 '11 at 23:55
@Matt I had a typo in my equations. Sorry for the trouble. I can see Griffiths in his book saying that this variety is smooth. I can more or less understand his claim going through the Jacobian argument but I am not fully sure. May be you can elaborate. – Anirbit Sep 17 '11 at 23:17