Let , for , and k an algebraically closed field, Prove that

Printable View

- Oct 22nd 2008, 09:31 PMterr13Algebraic Geometry
Let , for , and k an algebraically closed field, Prove that

- Oct 23rd 2008, 07:49 AMkalagota
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?