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 $\{ 0 \}$.