Let $\displaystyle V(I) = {v \in V | g(v) = 0 \forall g \in G}$, for $\displaystyle I \subset k[x]$, and k an algebraically closed field, Prove that $\displaystyle V(I) = {0} \iff I=k[x]$
i'll fix your latex for you..
V(I) = \{v \in V \ | \ g(v) = 0 \ \forall g \in G\}
$\displaystyle V(I) = \{v \in V \ | \ g(v) = 0 \ \forall g \in G\}$
and i think, $\displaystyle V(I) = \emptyset \iff I=k[x]$ and not $\displaystyle V(I) =0$.
and another thing, i think, it should be $\displaystyle V(I) = \{v \in k \ | \ g(v) = 0 \ \forall g \in I\}$
(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 $\displaystyle I=k[x]$.
By definition, $\displaystyle V(I) = \{v \in k \ | \ g(v) = 0 \ \forall g \in I\}$. Since $\displaystyle I$ contains nonzero constant functions, say $\displaystyle p(x)=c$ for a constant $\displaystyle c$, then there is no $\displaystyle v\in k$ such that $\displaystyle p(v)=0$. Thus, there is no $\displaystyle v\in k$ such that $\displaystyle g(v)=0$ for all $\displaystyle g\in k[x]=I$. Hence, $\displaystyle V(I) = \emptyset$.
(=>) To prove this, Let $\displaystyle V(I) = \emptyset$, $\displaystyle I\subset k[x]$. show that $\displaystyle I$ contains a unit, since if it does, then $\displaystyle I$ is the entire $\displaystyle k[x]$ (exclude the case that $\displaystyle I$ is empty.. it should be trivial that $\displaystyle V(\emptyset)=k$).
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?