= Solution
For an ideal $I\subseteq k[T_1,\ldots,T_n]$, define
$$
V(I)=\{a\in k^n:f(a)=0\text{ for every }f\in I\}.
$$
For $X\subseteq k^n$, define
$$
I(X)=\{f:f(a)=0\text{ for every }a\in X\},
$$
and recall that $\sqrt I=\{f:f^r\in I\text{ for some }r\geq1\}$ is the <radical of an ideal>.
For an <algebraically closed field> $k$, the <Weak Hilbert Nullstellensatz> says that every maximal ideal of $k[T_1,\ldots,T_n]$ is
$$
(T_1-a_1,\ldots,T_n-a_n)
$$
for a unique $a\in k^n$, equivalently every proper ideal has a common zero. The <Strong Hilbert Nullstellensatz> says
$$
\boxed{I(V(I))=\sqrt I}.
$$
To prove the weak form, let $\mathfrak m$ be maximal. The residue field
$$
K=k[T_1,\ldots,T_n]/\mathfrak m
$$
is a field finitely generated as a $k$-algebra. By the <Zariski lemma>, $K/k$ is finite algebraic; algebraic closedness gives $K=k$. If $a_i$ is the image of $T_i$, the quotient map is evaluation at $a=(a_1,\ldots,a_n)$ and its kernel is $(T_1-a_1,\ldots,T_n-a_n)$. This proves the assertion.
Solved by gpt-5.6-sol high.
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