Number of irreducible components of a variety

[GargleBlast42]

Hello,

I have the following problem:

I have an algebraic variety given as a zero locus of a set of polynomials. I know that there are points on this variety, which are singular (i.e. the dimension of the tangent space at these points is bigger than that of the variety). Now fixing one of these points, I would like to know, what is the number of irreducible components of the variety containing this point and what is their respective dimension. How can one do this?

 

[Eynstone]

This won't be an easy one : we must find out which components make the point singular and work backwards. The dimension of such a component depends on the variety at hand.

 

[GargleBlast42]

Well, for example, I have this variety in \mathbb{C}^9:
x_1^5+x_3^5+x_5^5+x_7^5+x_9^5=0
x_1^4+x_2 x_3^4+x_4 x_5^4+x_6 x_7^4+x_8 x_9^4=0
x_1^3+x_2^2 x_3^3+x_4^2 x_5^3+x_6^2 x_7^3+x_8^2 x_9^3=0
x_1^2+x_2^3 x_3^2+x_4^3 x_5^2+x_6^3 x_7^2+x_8^3 x_9^2=0
x_1+x_2^4 x_3+x_4^4 x_5+x_6^4 x_7+x_8^4 x_9=0
x_2^5+x_4^5+x_6^5+x_8^5+1=0
(this actually arose from an equation in \mathbb{CP}^9, where I took a patch in which one of the coordinates equals 1 (it's easy to see where the coordinate was in the original eqution))

Now in a generic point, the Jacobian of these polynomials has rank 6, and thus, one would expect that the dimension of the variety is 3 (provided that the polynomials form a radical ideal - is that the case here?). But now in a point like, e.g. x_1=0,x_3=0,x_5=0,x_6=0,x_8=0, the rank of the Jacobian is just 4 and thus, one would expect that these points are singular and that there are several irreducible components going through this points. How can one say how many they are and what their dimension is?