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Saturday, January 9, 2016

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

Problem:


Solution:

For part (a), the second equation seems to be simpler and therefore we will start there. Assuming x0, we can write y=1x. Putting this back to the first equation gives:

x2+1x21=0.

This works if x0, but what if x=0? Turn out we do not need to worry about it because if x=0, then the second equation cannot possibly be satisfied. Envisioning part (b), let's convert it into a polynomial by multiplying through x2 and get:

x4x2+1=0

For part (b), we use the hint and get (xy1)(xy+1)=x2y21. To eliminate y2, we multiply the first equation by x2 so we get x2(x2+y21)=x4+x2y2x2.

Finally we get x4x2+1=x2(x2+y21)(xy1)(xy+1), therefore x4x2+1x2+y21,xy1

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