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Tuesday, January 12, 2016

UTM Ideals Varieties and Algorithm - Chapter 1 Section 4 Exercise 9

Problem:


Solution:

The parameterization of the twisted cubic is (t,t2,t3), so all points in V must be of that form.

Now we test such points on the polynomial y2xz=(t2)2(t)(t3)=t4t4=0, so the polynomial vanishes on all points in the variety, so y2xzI(V).

For part (b), the key observation is that we have xz, so we multiply the second polynomial by x, the rest seems the just follow as:

(y)(yx2)(x)(zx3)=(y2xz)x2y+x4=(y2xz)x2(yx2).

So we just it back on the left hand side and get our answer as:

(x2+y)(yx2)(x)(zx3)=(y2xz).

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