[proofplan]
We show that $V$ and $I$ are mutually inverse maps between the set of $k$-algebraic subsets of $\Omega^n$ and the set of radical ideals of $k[T_1,\dots,T_n]$. One composition is $V(I(X)) = X$, which follows from the definition of $k$-algebraic sets. The other composition is $I(V(\mathfrak{a})) = \mathfrak{a}$, which combines the [Strong Nullstellensatz](/theorems/2940) with the assumption that $\mathfrak{a}$ is radical. The order-reversing property is a direct consequence of the definitions.
[/proofplan]
[step:Verify that $I(X)$ is a radical ideal for every $k$-algebraic set $X$]
Let $X \subset \Omega^n$ be a $k$-algebraic set. The vanishing ideal is defined as
\begin{align*}
I(X) = \{f \in k[T_1,\dots,T_n] : f(\underline{x}) = 0 \text{ for all } \underline{x} \in X\}.
\end{align*}
This is an ideal of $k[T_1,\dots,T_n]$: if $f, g \in I(X)$ and $h \in k[T_1,\dots,T_n]$, then $(f + g)(\underline{x}) = f(\underline{x}) + g(\underline{x}) = 0$ and $(hf)(\underline{x}) = h(\underline{x})f(\underline{x}) = 0$ for all $\underline{x} \in X$.
We check $I(X)$ is radical. Suppose $f^\ell \in I(X)$ for some $\ell \geq 1$. Then for every $\underline{x} \in X$, $f(\underline{x})^\ell = f^\ell(\underline{x}) = 0$. Since $\Omega$ is a field (hence an integral domain), $f(\underline{x})^\ell = 0$ implies $f(\underline{x}) = 0$. Therefore $f \in I(X)$, and $I(X) = \sqrt{I(X)}$.
[/step]
[step:Prove $V(I(X)) = X$ for every $k$-algebraic set $X$]
Since $X$ is a $k$-algebraic set, there exists an ideal $\mathfrak{a} \trianglelefteq k[T_1,\dots,T_n]$ with $X = V(\mathfrak{a})$.
The inclusion $X \subset V(I(X))$ holds because every $\underline{x} \in X$ satisfies $f(\underline{x}) = 0$ for all $f \in I(X)$ (by definition of $I(X)$), so $\underline{x} \in V(I(X))$.
For the reverse inclusion, let $\underline{y} \in V(I(X))$, so $f(\underline{y}) = 0$ for all $f \in I(X)$. Since $X = V(\mathfrak{a})$, every $g \in \mathfrak{a}$ vanishes on $X$, so $\mathfrak{a} \subset I(X)$. In particular, $g(\underline{y}) = 0$ for all $g \in \mathfrak{a}$, giving $\underline{y} \in V(\mathfrak{a}) = X$.
Therefore $V(I(X)) = X$.
[guided]
The equality $V(I(X)) = X$ is the statement that a $k$-algebraic set is completely determined by its vanishing ideal. The inclusion $X \subset V(I(X))$ is tautological: if $\underline{x} \in X$, then every polynomial that vanishes on $X$ certainly vanishes at $\underline{x}$.
The key is the reverse inclusion. We use the fact that $X = V(\mathfrak{a})$ for some ideal $\mathfrak{a}$. Then $\mathfrak{a} \subset I(X)$, because every $g \in \mathfrak{a}$ vanishes at every point of $X = V(\mathfrak{a})$. Therefore $V(I(X)) \subset V(\mathfrak{a}) = X$ (since $V$ reverses inclusions: larger ideal $\implies$ smaller vanishing set). Combined with $X \subset V(I(X))$, we get $V(I(X)) = X$.
Note that this step does not use the Nullstellensatz — it uses only the definition of $V$ and $I$ and the fact that $X$ is an algebraic set.
[/guided]
[/step]
[step:Prove $I(V(\mathfrak{a})) = \mathfrak{a}$ for every radical ideal $\mathfrak{a}$ using the strong Nullstellensatz]
By the [Strong Nullstellensatz](/theorems/2940) (Part 2), for any ideal $\mathfrak{a} \trianglelefteq k[T_1,\dots,T_n]$:
\begin{align*}
I(V(\mathfrak{a})) = \sqrt{\mathfrak{a}}.
\end{align*}
If $\mathfrak{a}$ is a radical ideal, then $\sqrt{\mathfrak{a}} = \mathfrak{a}$ by definition. Therefore
\begin{align*}
I(V(\mathfrak{a})) = \mathfrak{a}.
\end{align*}
This is the step that requires the full strength of the Nullstellensatz: without it, we would only have $\mathfrak{a} \subset I(V(\mathfrak{a}))$, which is the easy direction.
[/step]
[step:Conclude the bijection and verify the order-reversing property]
The maps $V$ and $I$ satisfy:
- $V(I(X)) = X$ for every $k$-algebraic set $X$ (proved above),
- $I(V(\mathfrak{a})) = \mathfrak{a}$ for every radical ideal $\mathfrak{a}$ (by the strong Nullstellensatz and radicality).
Furthermore, $I(X)$ is a radical ideal for every $k$-algebraic set $X$ (proved in the first step), and $V(\mathfrak{a})$ is a $k$-algebraic set for every ideal $\mathfrak{a}$ (by definition). Therefore $V$ and $I$ are mutually inverse bijections between $k$-algebraic subsets of $\Omega^n$ and radical ideals of $k[T_1,\dots,T_n]$.
Both maps reverse inclusions: if $X_1 \subset X_2$, then every polynomial vanishing on $X_2$ also vanishes on $X_1$, so $I(X_2) \subset I(X_1)$. If $\mathfrak{a}_1 \subset \mathfrak{a}_2$, then every common zero of $\mathfrak{a}_2$ is a common zero of $\mathfrak{a}_1$, so $V(\mathfrak{a}_2) \subset V(\mathfrak{a}_1)$.
[/step]
