# Thread: Algebraic Geometry

1. ## Algebraic Geometry

Let $V(I) = {v \in V | g(v) = 0 \forall g \in G}$, for $I \subset k[x]$, and k an algebraically closed field, Prove that $V(I) = {0} \iff I=k[x]$

2. Originally Posted by terr13
Let $V(I) = {v \in V | g(v) = 0 \forall g \in G}$, for $I \subset k[x]$, and k an algebraically closed field, Prove that $V(I) = {0} \iff I=k[x]$
i'll fix your latex for you..

V(I) = \{v \in V \ | \ g(v) = 0 \ \forall g \in G\}

$V(I) = \{v \in V \ | \ g(v) = 0 \ \forall g \in G\}$

and i think, $V(I) = \emptyset \iff I=k[x]$ and not $V(I) =0$.

and another thing, i think, it should be $V(I) = \{v \in k \ | \ g(v) = 0 \ \forall g \in I\}$
(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 $I=k[x]$.

By definition, $V(I) = \{v \in k \ | \ g(v) = 0 \ \forall g \in I\}$. Since $I$ contains nonzero constant functions, say $p(x)=c$ for a constant $c$, then there is no $v\in k$ such that $p(v)=0$. Thus, there is no $v\in k$ such that $g(v)=0$ for all $g\in k[x]=I$. Hence, $V(I) = \emptyset$.

(=>) To prove this, Let $V(I) = \emptyset$, $I\subset k[x]$. show that $I$ contains a unit, since if it does, then $I$ is the entire $k[x]$ (exclude the case that $I$ is empty.. it should be trivial that $V(\emptyset)=k$).

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?