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Fixed a few typos and a sentence

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edit approved Dec 12, 2013 at 22:52

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I would like to ask one question (maybe very simple) about the proof of Proposition II 6.8 in Hartshorne's book Algebraic Geometry page 137. The question is about the proof of the finiteness of non-trivial morphismmorphisms between two projective non-singular curves. In his proof, let $f:\ X \to Y$ be a non-trivial morphism, i.e. $f(X) =Y$ and let $V = \mathrm{Spec} B$ be any open affine subset of $Y$. Denote by $A$ the integral closure of $B$ in $K(X)$, where $K(X)$ is the function field of $X$. Here we regard $B \subset K(Y) \subset K(X)$, then I understand $\mathrm{Spec} A$ is isomorphic to an open subset $U$ of $X$ by the definitiondefinitions of abstract non-singular curves and their topology introduced in ChpaterChapter I, section 6. But then the proof says clearly $\mathrm{Spec} A \cong f^{-1}(V)$. It is this argument that I do not understand. I can see $\mathrm{Spec} A \subset f^{-1}(V)$.

I would really appreciate if someone could explain this, i.e. why finding the inverse is the same as finding the integral closure in the functional field.

I would like to ask one question (maybe very simple) about the proof of Proposition II 6.8 in Hartshorne's book Algebraic Geometry page 137. The question is about the proof of the finiteness of non-trivial morphism between two projective non-singular curves. In his proof, let $f:\ X \to Y$ be a non-trivial morphism, i.e. $f(X) =Y$ and let $V = \mathrm{Spec} B$ be any open affine subset of $Y$. Denote by $A$ the integral closure of $B$ in $K(X)$, where $K(X)$ is the function field of $X$. Here we regard $B \subset K(Y) \subset K(X)$, then I understand $\mathrm{Spec} A$ is isomorphic to an open subset $U$ of $X$ by the definition of abstract non-singular curves and their topology introduced in Chpater I, section 6. But then the proof says clearly $\mathrm{Spec} A \cong f^{-1}(V)$. It is this argument that I do not understand. I can see $\mathrm{Spec} A \subset f^{-1}(V)$.

I really appreciate if someone could explain this, i.e. why finding the inverse is the same as finding the integral closure in the functional field.

I would like to ask one question (maybe very simple) about the proof of Proposition II 6.8 in Hartshorne's book Algebraic Geometry page 137. The question is about the proof of the finiteness of non-trivial morphisms between two projective non-singular curves. In his proof, let $f:\ X \to Y$ be a non-trivial morphism, i.e. $f(X) =Y$ and let $V = \mathrm{Spec} B$ be any open affine subset of $Y$. Denote by $A$ the integral closure of $B$ in $K(X)$, where $K(X)$ is the function field of $X$. Here we regard $B \subset K(Y) \subset K(X)$, then I understand $\mathrm{Spec} A$ is isomorphic to an open subset $U$ of $X$ by the definitions of abstract non-singular curves and their topology introduced in Chapter I, section 6. But then the proof says clearly $\mathrm{Spec} A \cong f^{-1}(V)$. It is this argument that I do not understand. I can see $\mathrm{Spec} A \subset f^{-1}(V)$.

I would really appreciate if someone could explain this, i.e. why finding the inverse is the same as finding the integral closure in the functional field.

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asked Dec 12, 2013 at 18:00

user41541

Proof of Hartshorne Proposition II 6.8

I would like to ask one question (maybe very simple) about the proof of Proposition II 6.8 in Hartshorne's book Algebraic Geometry page 137. The question is about the proof of the finiteness of non-trivial morphism between two projective non-singular curves. In his proof, let $f:\ X \to Y$ be a non-trivial morphism, i.e. $f(X) =Y$ and let $V = \mathrm{Spec} B$ be any open affine subset of $Y$. Denote by $A$ the integral closure of $B$ in $K(X)$, where $K(X)$ is the function field of $X$. Here we regard $B \subset K(Y) \subset K(X)$, then I understand $\mathrm{Spec} A$ is isomorphic to an open subset $U$ of $X$ by the definition of abstract non-singular curves and their topology introduced in Chpater I, section 6. But then the proof says clearly $\mathrm{Spec} A \cong f^{-1}(V)$. It is this argument that I do not understand. I can see $\mathrm{Spec} A \subset f^{-1}(V)$.

I really appreciate if someone could explain this, i.e. why finding the inverse is the same as finding the integral closure in the functional field.

ag.algebraic-geometry