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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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    @David Yes. I meant it to be $x_2^2$. Corrected it!2011-09-17

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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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    @Anirbit: Dear Anirbit, If you interpret the $x_i$ as homogeneous coordinates, so that this is the equation for a projective curve in $\mathbb P^3$, then yes, it is smooth --- it is precisely the rational cubic curve. If you interpret the $x_i$ as affine coordinates, then it is *not* smooth at the origin (the point where all $x_i = 0$). Regards,2011-09-18