- It simply says that for all ideals $J$ in $\mathbb C[X_1, \dots, X_n]$, we have $I(V(J)) = sqrt J$
- Corollary: $I$ and $V$ are mutual inverses of inclusions between algebraic sets and radical ideals.

- Let $m$ be a maximal ideal.
- Let $K$ be the quotient ring $K \equiv \mathbb C[X_1, \dots, X_n] / m$.
- See that $K$ is a field because it is a ring quotiented by a maximal ideal.
- Consider the map $\alpha: \mathbb C[X_1, \dots, X_n] \rightarrow K$, or $\alpha : \mathbb C [X_1, \dots, X_n] \rightarrow \mathbb C[X_1, \dots, X_n] / m$ by sending elements into the quotient.
- We will show that $\alpha$ is an evaluation map, and $K = \mathbb C$. So we will get a function that evaluates polynomials at a given point, which will have a single point as a solution.
- Core idea: See that $\alpha(\mathbb C) = \mathbb C \subset K$. Hence $K$ is a field that contains $\mathbb C$. But $\mathbb C$ is algebraically closed, hence $K = C$.
- First see that $\mathbb C \subset K$, or that $\alpha$ preserves $\mathbb C$ [ie, $\alpha(\mathbb C) = \mathbb C$]. note that no complex number can be in $m$. If we had a complex number $z$ in $m$, then we would need to have $1 = 1/z \cdot z$ in $m$ (since an ideal is closed under multiplication by the full ring), which means $1 \in m$, due to which we get $m$ is the full ring. This can't be the case because $m$ is a proper maximal ideal.
- Hence, we have $\mathbb C \subseteq K$ or $K = \mathbb C$.
- Thus the map we have is $\alpha: \mathbb C[X_1, X_2, \dots, X_n] \rightarrow \mathbb C$.
- Define $z_i = \alpha(X_i)$. Now we get that $\alpha(\sum_{ij} a_{ij} X_i^j) = \sum_{ij} a_{ij} z_i^j$. That is, we have an evaluation map that sends $X_i \mapsto z_i$.
- CLAIM: The kernel of an evaluation map $\alpha$ is of the form $(X_1 - z_1, \dots, X_n - z_n)$.
- PROOF OF CLAIM:TODO
- The kernel is also $m$. Hence, $m = (X_1 - z_1, \dots, X_n - z_n)$, and point that corresponds to the maximal ideal is $(z_1, z_2, \dots, z_n)$.

- Suppose that wherever $f_1, \dots, f_m$ simultaneously vanish, then so does $g$. [that is, $g \in I(V(J))$where $J = (f_1, \dots, f_m)$].
- Then the polynomials $f_1, \dots, f_m, 1 - Yg$ have no common zeros where $Y$ is a new variable into the ring.
- Core idea of why they can't have common zeros: Contradiction. assume that $1 - Yg$, and all the $f_i$vanish at some point. Then we need $1 - Yg = 0$ which mean $Y = 1/g$, so $g$ cannot vanish, so $g \neq 0$. However, since all the $f_i$ vanish, $g$ also vanishes as $g \in (V(J))$. This is contradiction.
- Now by weak Nullstellensatz, the ideal $J = (f_1, \dots, f_m, (1-Y)g)$ cannot be contained in a maximal ideal (for then they would simultaneously vanish). Thus, $J = R$ and $1 \in J$.
- This means there are coefficients $c_i(Y, \vec x) \mathbb C[X_1, \dots , X_n, Y]$ such that

$1 = c_0(Y, \vec x) (1 - Yg(\vec x)) \sum_{i=1}^m c_i(Y, \vec x) f_i(\vec x)$

Since this holds when $\vec x, Y$ are arbitrary variables, it continues to hold
on substituting $Y = 1/g$, the coefficient $c_0(1-Yg) = c_0(1 - g/g) = c_0(1 - 1) = 0$ disappears. This gives:
$1 = \sum_{i=1}^m c_i (Y, \vec x) f_i(\vec x)$
since $Y = 1/g$, we can write $c_i(Y=1/g, \vec x) = n_i(\vec x)/g^r_i(\vec x)$. By clearing denominators, we get:
$1 = \sum{i=1}^m n_i(\vec x) f_i(\vec x)/ g^R(\vec x)$

This means that $g^R(\vec x) = \sum_{i=1}^m n_i(\vec x) f_i(\vec x)$
- We have $g \in I(V(J))$.
- We want to show that $g \in \sqrt{J}$ in $R$.
- This is the same as showing that $g \in \sqrt{0}$ in $R/J$. ( $J \mapsto 0$ in the quotient ring).
- If $g$ is nilpotent in $R/J$, then the $(R/J)_g$ becomes the trivial ring $\{ 0 \}$. [Intuitively, if $g$ is nilpotent and a unit, then we will have $g^n = 0$, that is unit raised to some power is 0, from which we can derive $1 = 0$].
- Localising at $g$ is the same as computing $R[Y]/(1 - Yg, J)$.
- But we have that $V(1 - Yg, J) = \emptyset$. Weak Nullstellensatz implies that $(1 - Yg, J) = (1)$.
- This means that $R[Y]/(1 - Yg,J) = R[Y]/(1) = \{ 0 \}$. Thus, $(R/J)_g$ has $g$ as nilpotent, or $g \in \sqrt J$ in $R$.

- Same statement: $I(V(J)) = \sqrt J$.
- $V(J)$ is the set of points on which $J$ vanihes. Evaluation is quotienting. So it's going to be set of prime ideals $p$ such that $J \xrightarrow{R/p} 0$. So $J \subset p$. This means that $V(J) = \{ p \text{prime ideal in } R, J \subseteq p \}$.
- $I(V(J))$ is the set of functions that vanish over every point in $V(J)$. The functions that vanish at $p \in V(J)$ are the elements of $p$. So the functions that vanish over all points is $I(V(J)) = \cap V(J)$.
- Unwrapping, this means that $I(V(J))$ is the intersection of all ideals in $V(J)$, which is the intersection of all primes that contains $J$, which is the radical of $J$.