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Monday, October 26, 2015

UTM Ideals Varieties and Algorithm - Chapter 1 Section 1 Exercise 6

Problem:


Solution:

Step 0: Induction hypothesis

Let $ S(n) $ be the statement that if $ f \in \mathbf{C}^n $ vanishes on every point of $ \mathbf{Z}^n $, then $ f $ is the zero polynomial.

Step 1: Induction basis

Suppose $ n = 1 $, then it is well known that if a polynomial has infinite number of zeroes, then it must be the zero polynomial.

Step 2: Induction

Suppose the statement $ S(n) $ is true for all $ n < k $, then when $ n = k $. For any fixed integers $ x_2 \cdots x_n $, $ f $ becomes a polynomial of 1 variable that vanish on all integers, so it must be a zero polynomial.

The coefficients of f (when interpreted as a polynomial in $ K[x_2, \cdots, x_n] $) must then be vanishing on all integers as well. Therefore, by the induction hypothesis, they must be zero polynomial as well.

Last but not least, a polynomial with all coefficients being zero polynomial is a zero polynomial itself, therefore we concluded $ S(k) $ is also true.

The presentation above is a little abstract, let show it in concrete terms.

Suppose we wanted to prove $ f(x, y) = axy + bx + cy $ is a constant polynomial, given this polynomial vanishes on all integers.

We write this polynomial in this form $ f(x, y) = (ax + c)y + bx $.

For any fixed x, say x = 2, we have $ g(y) = f(2, y) = (2a + c)y + 2b $, now this must be the zero polynomial because it vanishes on all integers, which means $ 2a + c $ and $ 2b $ are both 0.

But that must be true for all $ x $, not just $ x = 2 $, therefore we have
$ ax + c = 0 $ and $ bx = 0 $.

We can now show $ a, b, c $ must all be zero using the induction hypothesis now.

For part b, we replace the argument 'vanish on all integers' by 'vanish on more than $ M $ points and the proof proceed just like this above.

The key property is that, throughout the proof above, there will be no polynomial with degree larger than $ M $, therefore vanishing on a large enough grid is sufficient.



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