Let , for , and k an algebraically closed field, Prove that
i'll fix your latex for you..
V(I) = \{v \in V \ | \ g(v) = 0 \ \forall g \in G\}
and i think, and not .
and another thing, i think, it should be
(actually, i am assuming you want to prove a version of the weak nullstelensatz.. Ü)
now, the proof.. i'll prove one direction only.
(<=) Suppose .
By definition, . Since contains nonzero constant functions, say for a constant , then there is no such that . Thus, there is no such that for all . Hence, .
(=>) To prove this, Let , . show that contains a unit, since if it does, then is the entire (exclude the case that is empty.. it should be trivial that ).
or you can prove it by contrapostive: if I is a proper subset of k[x], show that V(I) is not empty. (in particular, let I be a maximal ideal of k[x])
can you do it?