Write the <finitely generated algebra> as
$$
A\cong\mathbb C[T_1,\ldots,T_n]/I.
$$
By 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> gives
$$
|V(I)|=|\mathbb C|,
$$
which is uncountable. This also covers the zero algebra, whose homomorphism set is empty. Therefore
$$
\boxed{|\operatorname{Hom}_{\mathbb C}(A,\mathbb C)|\ne\aleph_0}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 101 / 3 / iv / Solution

A complex affine algebraic set is either finite or has the cardinality of $\mathbb C$. One proof takes a positive-dimensional irreducible component and applies <Noether normalization>: the resulting finite dominant map to a positive-dimensional affine space has a point above every closed point.

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 Incoming links: Complex affine algebraic set cardinality dichotomy