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Given two affine scheme $\text{Spec}(A)$ and $\text{Spec}(B)$ finite type over $\mathbb{C}$. For the purpose of this problem you can assume $\text{Spec}(A)$ is a Zariski open in some affine space $\mathbb{A}^n$. You can topologize the space of algebraic maps from $\text{Spec}(A)$ to $\text{Spec}(B)$ in two way:

  1. Consider it as a subspace of continuous maps with compact open topology. (Continuous maps from $\text{Spec}(A)^{an}$ to $\text{Spec}(B)^{an}$)
  2. The space of algebraic maps can be identified with the $\mathbb{C}$-algebra morphisms from $B$ to $A$. Assume $B$ is isomorphic to $\mathbb{C}[x_1,\ldots, x_n]/I$. This amounts to determining $n$ elements in $A$ as the images of the $x_i$'s in a way that they satisfy the relations coming from $I$. This identifies the space of maps as a subspace of $A^n$. Each $A$ has a topology induced from the space of complex value continuous functions on $\text{Spec}(A)^{an}$ (induced from the compact-open topology). With this topology note that $A$ is actually contractible. This induces a topology on the space of $\mathbb{C}$-algebra morphisms from $B$ to $A$.

Do these two topologies coincide?

My goal is to understand the space of regular maps from $\text{Spec}(A)$ to $\text{Spec}(B)$ with topology number 1. If number 2 is not the correct description I'd like to know the correct description of topology 1 on the space of $\mathbb{C}$-algebra maps from $B$ to $A$. I'd like to know when the homology of this space is finitely generated.

user127776
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  • An element of $A$ is not a complex-valued function on $\operatorname{Spec} A$, only on the closed points of $\operatorname{Spec} A$. – Eric Wofsey Sep 18 '21 at 22:42
  • @EricWofsey Yes true, but isn't that enough to identify it with a complex valued continuous function on the complex points. – user127776 Sep 18 '21 at 22:44
  • Also, what topologies are you using here (on $\mathbb{C}$, and on the closed points of $\operatorname{Spec} A$)? If you're using the Euclidean topology on $\mathbb{C}$ and the Zariski topology on $\operatorname{Spec} A$ then these functions are not continuous. – Eric Wofsey Sep 18 '21 at 22:45
  • (Also, if $A$ is not reduced, the map from $A$ to complex-valued functions on $\operatorname{Spec} A(\mathbb{C})$ is not injective.) – Eric Wofsey Sep 18 '21 at 22:49
  • @EricWofsey Are you implying that global sections of a complex affine scheme does not necessarily descent to a continuous function on the complex points? I find this a little bit strange, since global sections can be represented as restriction of a global section of the ambient $\mathbb{A}^n$ that the scheme lies in. These elements are all in polynomial form so I am confused why they cannot be continuous. – user127776 Sep 18 '21 at 22:58
  • Well it depends on what topologies you are using. A nonconstant polynomial map $\mathbb{C}\to\mathbb{C}$ will not be continuous if you're using the Euclidean topology on the codomain and the Zariski topology on the domain! – Eric Wofsey Sep 18 '21 at 22:59
  • No we use the analytic topology induced on the complex points to the analytic topology on the $\mathbb{C}$. – user127776 Sep 18 '21 at 23:01

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