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How do I show that on a projective hypersurface of homogenous equation F = 0, the singular points arise as common zeros of the partial derivatives dx_i(F)?

I have an assignment to calculate the singular points of the Steiner surface and it is not so hard using the above theorem. But I would like to understand why this fact holds in general.

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This is called the Jacobian criterion, and it is stated and proved in most introductions to algebraic geometry. You could consult---I will be unoriginal...---Hartshorne's book. –  Mariano Suárez-Alvarez Feb 26 '11 at 16:50

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As is often the case for such basic questions, it depends on your precise definition for "singular point" on an algebraic variety.

One standard definition is via the Jacobian condition: in an affine coordinate chart, the matrix of partial derivatives should have the largest possible rank: here, one. Your condition is simply a projectivized version of this basic condition: it is equivalent to their being an affine coordinate chart $f(x,y) = 0$ for which at least one of the two partial derivatives is non-vanishing. It may seem that it is easier for a projective curve $F(x,y,z) = 0$ to be smooth than an affine curve $f(x,y) = 0$ since in the projective case it is enough for any one of three partial derivatives to be nonzero, whereas in the affine case there are only two. However this is illusory, because of course we only care about the vanishing of the partial derivatives at points lying on the curve, and it turns out that in the projective case $\frac{\partial F}{\partial x}(P) = \frac{\partial F}{\partial y}(P) = \frac{\partial F}{\partial z}(P) = 0$ implies $F(P) = 0$, whereas this does not hold in the affine case.

The other common definition of nonsingularity of a variety at a point $P$ is that the local ring at that point is a regular local ring. The equivalence of this (more intrinsic) definition with the Jacobian condition is shown, for instance, in $\S I.5$ of Hartshorne's Algebraic Geometry.

(As an arithmetic geometer, I should point out that Hartshorne is rather unnecessarily working over an algebraically closed base field throughout his book -- despite spending a lot of time talking about "general schemes". This equivalence continues to hold over any perfect ground field $k$. For imperfect ground fields, smoothness is the stronger and more geometric condition: $X$ is smooth over $k$ iff the base extension of $X$ to an algebraic closure $\overline{k}$ is regular. But if $\overline{k}/k$ is not separable, funny things can happen...)

Much later on, in $\S III.10$, Hartshorne gives yet another equivalent condition for smoothness in terms of relative differentials. All three turn out to be important, but in my experience at least the Jacobian condition is by far the easiest to check computationally.

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I might be mistaken after a quick read, but it seems you first talk about "definition of singular point", but then actually give the definition of nonsingularity? –  Joachim Dec 20 '13 at 0:50
    
@Joachim: It seems so. I was expecting that it would be clear to the reader that a "singular point" is a point which is not nonsingular. –  Pete L. Clark Dec 20 '13 at 3:55
    
Yes, well, it was clear actually, i just wanted to made sure it wasnt a typo or something. So never mind, Pete! –  Joachim Dec 20 '13 at 21:54

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