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 y2−xz=(t2)2−(t)(t3)=t4−t4=0, so the polynomial vanishes on all points in the variety, so y2−xz∈I(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)(y−x2)−(x)(z−x3)=(y2−xz)−x2y+x4=(y2−xz)−x2(y−x2).
So we just it back on the left hand side and get our answer as:
(x2+y)(y−x2)−(x)(z−x3)=(y2−xz).
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 y2−xz=(t2)2−(t)(t3)=t4−t4=0, so the polynomial vanishes on all points in the variety, so y2−xz∈I(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)(y−x2)−(x)(z−x3)=(y2−xz)−x2y+x4=(y2−xz)−x2(y−x2).
So we just it back on the left hand side and get our answer as:
(x2+y)(y−x2)−(x)(z−x3)=(y2−xz).
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