$\DeclareMathOperator\Alg{Alg}$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 $\Alg(A, \mathbb{C})$?
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$\begingroup$ Does some theorems about this question? We know that if f $\in Alg(A, \mathbb{C})$, then $ker f$ is a maximal ideal. What about else? $\endgroup$– user11090426Commented Nov 24, 2021 at 8:43
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$\begingroup$ What do you mean by "algebra"? the same as $\mathbf{Z}$-algebra? as $\mathbf{C}$-algebra? something else? $\endgroup$– YCorCommented Nov 24, 2021 at 8:43
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$\begingroup$ @YCor I have added it in the question, sorry. $\endgroup$– user11090426Commented Nov 24, 2021 at 8:45
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$\begingroup$ So you have an inclusion Alg(A,C)$\subset$Spec(A). The first correspond to quotients reduced to C. The second corresponds to quotients that are integral C-algebras. $\endgroup$– YCorCommented Nov 24, 2021 at 8:50
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$\begingroup$ What is the question (the last sentence does't make sense)? $\endgroup$– YCorCommented Nov 24, 2021 at 8:51
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1 Answer
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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 $\mathrm{Spec}(A)$.
