Characterized maximal ideal $\DeclareMathOperator\Alg{Alg}$Let $A$ be a commutative associative algebra with $1$ over $\mathbb{C}$. We define $\Alg(A,\mathbb{C}) $ to be the set of $\mathbb{C}$-algebra maps from $A$ to $\mathbb{C}$. What is the relationship between $\mathrm{Spec}(A)$ and $\Alg(A, \mathbb{C})$?
 A: What you have described is called the set of ``$\mathbb{C}$-points'' of the scheme $X = \mathrm{Spec}(A)$. In general for schemes $X$ and $T$ over a base $S$ we define the set of $T$-points of $X$ to be the set,
$$ X(T) = \mathrm{Hom}_S(T, X) $$
In particular, if $S = \mathrm{Spec}(R)$ for some ring $R$ and $T = \mathrm{Spec}(B)$ for some $R$-algebra $B$ then we write $X(B)$ for the set $X(\mathrm{Spec}(B))$.
Let's unwind what this means when $X$ is affine. Let $X = \mathrm{Spec}(A)$ where $A$ is an $R$-algebra. Then,
$$ X(B) = \mathrm{Hom}_S(\mathrm{Spec}(B), X) = \mathrm{Hom}_S(\mathrm{Spec}(B), \mathrm{Spec}(A)) = \mathrm{Hom}_{R-\mathrm{Alg}}(A,B) $$
Therefore the $B$-points of $X$ are exactly the $R$-algebra homomorphisms $A \to B$.
When $A$ is a finitely generated $\mathbb{C}$-algebra (this works for any algebraically closed field) then by the Nullstellensatz, every maximal ideal of $A$ has residue field $\mathbb{C}$ so the maximal ideals exactly correspond to homomorphisms $A \to \mathbb{C}$ so $X(\mathbb{C}) = \mathrm{Hom}_{\mathbb{C}}(A, \mathbb{C})$ is exactly the set of closed points of $\mathrm{Spec}(A)$.
