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Wednesday, November 11, 2015

UTM Ideals Varieties and Algorithm - Chapter 1 Section 2 Exercise 5

Problem:


Solution:

This is the variety V(x2,y,z+1)V(x2y). Therefore ,the answer is the single point (2,0,1) together with a parabola y=x2 extended to become a surface with arbitrary value z. The point is not on the surface itself so it stands out as a singular point.

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