This is an exercises in Ideals, Varieties and Algorithms by Cox et al.
Let $V\subset \mathbb{C}^n$ be a nonempty variety. Let $\phi\in \mathbb{C}[V]$. Show that $\mathbf{V}_V(\phi)=\emptyset$ if and only if $\phi$ is invertible in $\mathbb{C}[V]$ (which means that there is some $\psi \in \mathbb{C}[V]$ such that $\phi\psi=[1]$ in $\mathbb{C}[V]$).
My attempt:
$$\mathbf{V}_V(\phi)=\{z\in V | \, \phi(z)=0\}=\emptyset$$
This means $\phi$ is never zero on $V$. I really don't have any idea from here. I think we can probably define $\psi(z)=\frac{1}{\phi(z)}$ for all $z\in V$. But how to prove it is a polynomial function in $\mathbb{C}[V]$?
This exercise also asks if it is still true when replacing $\mathbb{C}$ by $\mathbb{R}$. So I thought some examples may help. So I tried $V=V(z^2+1)$ and $\phi=z+2$. I still cannot find a polynomial function that can represent $\frac{1}{z+2}$ in $\mathbb{C}[V]$.
Any help would be appreciated!
Suppose $\phi$ is invertible. Then we have $\phi\psi=1$ for some $\psi$ and $(\phi)\supset (1)$. Now $\textbf{V}_V(\phi)\subset \textbf{V}_V(1)=\emptyset$. This proves one direction.
For the converse, suppose $\textbf{V}_V(\phi)=\emptyset$. We also have $\textbf{I}_V(\textbf{V}_V(\phi))=\textbf{I}_V(\emptyset)$. Now $1$ vanishes nowhere so that $1\in \textbf{I}_V(\emptyset)$. But this means $1\in \textbf{I}_V(\textbf{V}_V(\phi))$. Since $\mathbb{C}$ is algebraically closed Hilbert's Nullstellensatz says $\textbf{I}_V(\textbf{V}_V(\phi))=\sqrt{(\phi)}$. Since $1\in \sqrt{(\phi)}$, we have $1^n\in (\phi)$ for some $n$ (but it's all the same) so that $1\in (\phi)$. This means there is a $\psi$ such that $\phi\psi=1$ and $\phi$ is invertible.