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.