Cardinality of $\operatorname{Hom}_{\mathbb{C}}(A,\mathbb{C})$ Question 
Prove there is no finite generated algebra $A$ over $\mathbb{C}$ such that the cardinality of $\operatorname{Hom}_{\mathbb{C}}(A,\mathbb{C})$ is exactly $\aleph_0$.
I need to prove it using commutative algebra tools such as Hilbert basis & Nullstellensatz theorem, Noether normalization theorem, etc... but can't figure out how.
Thank you!
 A: Hint: Noether normalization says there's an injective map $R=\Bbb C[x_1,\cdots,x_n]\to A$ which makes $A$ into a finite $R$-algebra (where $n=\dim A$). This induces a map on the hom-sets $\operatorname{Hom}_{\Bbb C}(A,\Bbb C)\to\operatorname{Hom}_{\Bbb C}(R,\Bbb C)$. What can you say about this map? What can you say about $\operatorname{Hom}_{\Bbb C}(R,\Bbb C)$?
More details under the spoiler, though I encourage you to make an effort without looking under the spoiler first.

 This map of hom-sets is surjective and finite-to-one. Why? Every element in $A$ satisfies some polynomial with coefficients from $R$, so if you know what happens to elements of $R$, then you know what happens with elements of $A$, except possibly up to some finite ambiguity. Now all you have to do is to argue that $\operatorname{Hom}_{\Bbb C}(R,\Bbb C)$ is finite or at least the cardinality of the continuum, which should not be hard using the Nullstellensatz once you remember that maximal ideals are in bijection with maps to a field.

