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Saturday, January 16, 2016

UTM Ideals Varieties and Algorithm - Chapter 1 Section 4 Exercise 15

Problem:


Solution:

For part (a), we show the set is closed from addition and multiplication.

Suppose f,g is in I(S), hk[x1,,xn] then for all xS, f(x)=g(x)=0.

Now (f+g)(x)=0 and also (hg)(x)=0, so f+g and hg are both in I(S).

Therefore I(S) is an ideal.

For part (b), we do exactly the same as in Exercise 10. Now we have f(x,y)=h(x,y)(xy)+r(x), f(a,a)=0r(a)=0, a1. so now we still have r(x)=0  and therefore the ideal is still xy.

For part (c), we already know polynomials that vanish on the whole integer grid is necessarily the zero polynomial, so I(Z) is simply {0}.

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