UTM Ideals Varieties and Algorithm - Chapter 1 Section 4 Exercise 14

Problem:

Solution:

For the purpose of practicing, let's just prove Proposition 8.

For (i), if $V \subset W$, then for any polynomial $p \in I(W)$ must vanish in $W$ and therefore vanish in $V$, so $p \in I(V)$ and $I(W) \subset I(V)$.

On the other hand, if $I(W) \subset I(V)$. We know $W$ is a set of common zero for a set of polynomial $g_i$, so $\forall i$, $g_i \in I(W) \subset I(V)$, therefore $\forall i$, $\forall v \in V$, $g_i(v) = 0$, therefore $\forall v \in V$, $v \in W$.

To be honest, I cheated, I couldn't figure out the second part myself. The critical part I missed is the red part. Next time I should try harder before I cheat and maybe work backwards from the conclusion.

Part (a) is trivial though. We have part (i) already, so $V = W \implies V \subset W \text { and } W \subset V$, so $I(W) \subset I(V) \text{ and } I(V) \subset I(W)$, so we got the easy conclusion!

Part (b) is also trivial, it is simply (i) and not (ii) !