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I have a question about a step in the proof of II.6.8 in Hartshorn's "Algebraic Geometry": How he concludes from the fact that $f(X)$ is closed in Y, proper over $k$ and irreducible that $f(X)=Y$ or a point.

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user267839
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1 Answers1

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If you agree that $f(X)$ is an closed and complete subvariety of $Y$, (the previous sentence), then we need only consider what the closed irreducible of subvarieties of $Y$ are. Since $Y$ is a curve, these are either single points, or an irreducible component. Since $Y$ is irreducible, this must be all of $Y$, and since $X$ is connected, case $(1)$ must be a single point.

Andres Mejia
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