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) !
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) !
No comments:
Post a Comment