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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) $, $ h \in k[x_1, \cdots , x_n] $ then for all $ x \in S $, $ 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)(x - y) + r(x) $, $ f(a, a) = 0 \implies r(a) = 0 $, $ \forall a \neq 1 $. so now we still have $ r(x) = 0 $  and therefore the ideal is still $ \langle x  - y \rangle $.

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

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