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I have trouble understanding the proof in several pieces of the proposition below.

Proposition 3.2.7. Let $(S, \gamma)$ be a pointed connected noetherian scheme, $X_{1}$ and $X_{2}$ two etale covering spaces of S, $u: X_{1} \rightarrow X_{2} $ an $S$-morphism, and $X_{i}(\gamma)(i=1,2)$ the sets of geometric points of $ X_{i} $ lying above $\gamma$ . If the map $X_{1}(\gamma) \rightarrow X_{2}(\gamma)$ induced by $u$ is bijective, then $u$ is an isomorphism.

Proof. The image of any connected component of $X_{2}$ in $S$ is both open and closed. Since $S$ is connected, the image is $S$. Replacing $X_{2}$ by its connected components and $X_{1}$ by the inverse images of these components, we are reduced to the case where $X_{2}$ is connected. Note that $u$ is finite and etale. So $u_{*} \mathcal{O}_{X_{1}}$ is a locally free $\mathcal{O}_{X_{2}}$-module of constant finite rank. Let $s \in S$ be the image of $\gamma$ and let $x_{2} \in X_{2}$ be a point above $s$. Then $$X_{1} \times_{X_{2}} \operatorname{Spec} \mathcal{O}_{X_{2}, x_{2}} \cong \operatorname{Spec} A$$ for some $\mathcal{O}_{X_{2}, x_{2}} $-algebra $A$ which is free of finite rank as an $\mathcal{O}_{X_{2}, x_{2}} $-module. Since $X_{1}(\gamma) \rightarrow X_{2}(\gamma) $ is bijective, there is one and only one point $x_{1}$ in $X_{1}$ lying above $x_{2}$. So $A$ is a local ring. Since $u$ is etale, $\mathfrak{m}_{x_{2}}A$ is the maximal ideal of $A$ and $A / \mathfrak{m}_{x_{2}} A$ is finite separable over $\mathcal{O}_{X_{2}}$, ${x_{2}} / \mathfrak{m}_{x_{2}} $. Again because $X_{1}(\gamma) \rightarrow X_{2}(\gamma) $ is bijective, we must have $ \mathcal{O}_{X_{2}, x_{2}} / \mathfrak{m}_{x_{2}} \cong A / \mathfrak{m}_{2} A $. It follows that the rank of $u_{*} \mathcal{O}_{X_{1}} $ is 1. Let $x_{2}^{\prime} $ be an arbitrary point of $ X_{2} $ and let $A^{\prime} $ be an $\mathcal{O}_{X_{2}, x_{2}^{\prime}} $-algebra such that $$X_{1} \times_{X_{2}} \operatorname{Spec} \mathcal{O}_{X_{2}, x_{2}^{\prime}} \cong \operatorname{Spec} A^{\prime}.$$ Since $\operatorname{rank}\left(u_{*} \mathcal{O}_{X_{1}}\right)=1,$ $A^{\prime}$ is a free $\mathcal{O}_{X_{2}}, x_{2}^{\prime}-$module of rank 1. The homomorphism $\mathcal{O}_{X_{2}, x_{2}^{\prime}} / \mathfrak{m}_{x_{2}^{\prime}} \rightarrow A^{\prime} / \mathfrak{m}_{x_{2}^{\prime}} A $ is a nonzero homomorphism of one dimensional vectors spaces. It is necessarily surjective. By Nakayama's lemma, the homomorphism $\mathcal{O}_{X_{2}, x_{2}^{\prime}} \rightarrow A^{\prime} $ is also surjective. It is injective since it is faithfully flat. So we have $\mathcal{O}_{X_{2}, x_{2}^{\prime}} \cong A^{\prime} $. Hence $\mathcal{O}_{X_{2}} \cong u_{*} \mathcal{O}_{X_{1}} $, and $u$ is an isomorphism.

My questions are as follows:

  1. Why is the image of the connected component of $X_2$ closed? (This is in the first sentence in the proof.) Is a finite etale map a closed map?
  2. Why is the $X_{1} \times_{X_{2}} \operatorname {Spec} \mathcal{O}_{X_{2}, x_{2}} \cong \operatorname{Spec} A$ for some $A$?

When I read this book, I jump between chapters and sections and ignore many theorems. If you can tell me to notice some theorems before, I will also appreciate much.

KReiser
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Vingt et un
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1 Answers1

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Assuming "etale covering space" means finite etale map, both of these issues can be resolved by understanding the properties of finite morphisms.

  1. Finite morphisms are closed, so the image of any closed subset is again closed.

  2. Finite morphisms are by definition affine. Since affine morphisms are stable under base change, this means the map $X_1\times_{X_2} \operatorname{Spec} \mathcal{O}_{X_2,x_2}\to \operatorname{Spec} \mathcal{O}_{X_2,x_2}$ is affine. Since the target is affine, this means the source is affine as well.

Both of these things are probably more likely to be covered in a first course in algebraic geometry rather than one dealing with etale cohomology. While it's of course very reasonable to forget basic properies on occasion, if you find yourself consistently in the spot of having to reread some basics, you might need to make more of a concerted effort to review instead of pushing ahead. Your exact strategy will depend on your goals, consult your advisor for more personalized advice.

KReiser
  • 65,137