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For an ideal $I\subseteq k[T_1,\dots ,T_n]$, defineFor $X\subseteq k^n$, defineand recall that $\sqrt{I}=\{f:f^r\in I\text{ for some }r\ge 1\}$ is the radical of an ideal.

For an algebraically closed field $k$, the Weak Hilbert Nullstellensatz says that every maximal ideal of $k[T_1,\dots ,T_n]$ isfor a unique $a\in k^n$, equivalently every proper ideal has a common zero. The Strong Hilbert Nullstellensatz says

To prove the weak form, let $\mathfrak{m}$ be maximal. The residue fieldis 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 $a=(a_1,\dots ,a_n)$ and its kernel is $(T_1-a_1,\dots ,T_n-a_n)$. This proves the assertion.

Solved by gpt-5.6-sol high.

Write the finitely generated algebra asBy the Weak Hilbert Nullstellensatz, $\mathbb{C}$-algebra homomorphisms $A\to \mathbb{C}$ correspond exactly to the points of the affine algebraic set $V(I)$.

If $V(I)$ is finite, its cardinality is finite. If it is infinite, the complex affine algebraic set cardinality dichotomy giveswhich is uncountable. This also covers the zero algebra, whose homomorphism set is empty. Therefore

Solved by gpt-5.6-sol high.

If a field $L$ is finitely generated as an algebra over a subfield $k$, then $L$ is a finite algebraic extension of $k$. Applying this to a residue field of a polynomial ring proves the Weak Hilbert Nullstellensatz.