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P23: Let $f:Y\rightarrow X$ be locally of finite-type. Prove that if $\Delta:Y\to Y\times Y$ is an open immersion, then $f$ is unramified.

$\newcommand{\spec}{\operatorname{spec}}$ ... we reduce the problem to the case of a morphism $f:Y\to \spec k$ where $k$ is an algebraically closed field. Let $y$ be a closed point of $Y$. Since $k$ is algebraically closed, there exists a section $g:\spec k\to Y$ whose image is $\{y\}$. (why?)

...$\{y\}$ is open in $Y$. Moreover, $\spec\mathcal{O}_y=\{y\}\to \spec k$ still has the property that $\spec\mathcal{O}_y\to \spec\mathcal{O}_y\otimes\mathcal{O}_y$ is an open immersion. ($\spec\mathcal{O}_y$ is not $V(y)$, why do we have $spec\mathcal{O}_y={y}$,and I cannot prove the rest of the statement.)

But $\mathcal{O}_y$ is an artin ring with residue field $k$ (why $k(y)=k$ ?) and $\spec\mathcal{O}_y\otimes\mathcal{O}_y$ has only one point. So $\mathcal{O}_y\otimes\mathcal{O}_y\to\mathcal{O}_y$ must be an isomorphism. (how can we characterize the struction of $\mathcal{O}_y\otimes\mathcal{O}_y$?)

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Why $k=k(y)$: since $Y\to \mathrm{Spec} k$ is locally of finite-type, if $y$ is a point of $Y$, then the local ring $\mathcal{O}_y$ is finite-type over $k$. In particular, if $y$ is a closed point, then $k(y)$ is finite type over $k$. But the only field of finite type over an algebraically closed $k$ is $k$ itself; this is one form of the Nullstellensatz. – Charles Rezk Dec 31 '10 at 3:34
thank you ,I forget it – Yingjin Bi Dec 31 '10 at 4:22
I had a go at improving the formatting, but I might have broken some of the LaTeX. If someone knows how to fix it please feel free to edit – Yemon Choi Dec 31 '10 at 5:02
thank you very much:Yemon Choi .thank you – Yingjin Bi Jan 1 '11 at 1:06

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