Prove there exists a bijection between $spec_{max} ( \mathbb{C}[X] )$ and the points of an algebraic variety $V$ over $\mathbb{C}$

Question

I am looking for a model answer for the following question:

Prove that there is a bijection between points of an algebraic variety $V$ over $\mathbb{C}$, and $Spec_{max}(\mathbb{C}[V])$ by following the sequential steps:

1. For $x\in V$ define a map $\psi:\mathbb{C}[V]\to\mathbb{C}$ by $f\mapsto f(x)$. Prove that $\psi$ is a surjective ring homomorphism.
2. Prove that the kernel of $\psi$ is a maximal ideal of $\mathbb{C}[V]$. Denote this maximal ideal by $\mathfrak{m}_x$.
3. Define a map $\Psi:V\to Spec_{max}(\mathbb{C}[V])$ by $x\mapsto\mathfrak{m}_x$. Prove that $\Psi$ is injective.
4. Using Hilbert’s theorem prove that $\Psi$ is surjective.

I have been trying to understand this for a while and getting nowhere, and have no solution/hint resource. Part (ii) is quite straightforward: using the ring homomorphism theorem $f(A)\cong A/ker(f)\,(A=\mathbb{C}[V])$. As $f(A)$ is a field, $ker(f)$ must be maximal. For the rest I am at a bit of a loss.

Many thanks

Answer 1

1. The functions $\psi_x$ are surjective morphisms of rings, because: in $\mathbb{C}[V]$ one can find the equivalence classes of constant maps on $V$; and the remain reasoning is routine.
2. Since the image of any $\psi_x$ is a field, $\ker\psi_x=\mathfrak{m}_x$ is a maximal ideal of $\mathbb{C}[V]$.
3. Let $x\neq y\in V$, then $\psi_x\neq\psi_y\Rightarrow\mathfrak{m}_x\neq\mathfrak{m}_y$; in other words $\Psi$ is an injective map.
4. By Hilbert's Nullstellensatz: \begin{equation} \forall\mathfrak{m}\in Spec_{Max}\mathbb{C}[V],\,V(\mathfrak{m})=\{x\in V\mid\forall f\in\mathfrak{m},\,f(x)=0\}\neq\emptyset; \end{equation} by maximality of $\mathfrak{m}$, one can prove that there exists $x\in V$ such that $V(\mathfrak{m})=\{x\}$, that is $\mathfrak{m}=\mathfrak{m}_x$.