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Yemon Choi
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help needed with a proof in <etale cohomology> by milne ,please check it right

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.

P23:Let $f:Y\rightarrow X$ be locally of finite-type. prove:$\Delta:Y\rightarrow Y\times Y$ is an open immersion,then $f$ is unramified$\newcommand{\spec}{\operatorname{spec}}$ ....we we reduce the problem to the case of a morphisimmorphism $f:Y\rightarrow spec k$$f:Y\to \spec k$ where $k$ is an algebracallyalgebraically closed field.let Let $y$ be a closed point of $Y$.because Since $k$ is an algebracallyalgebraically closed  ,there exist there exists a section $g:spec k\rightarrow Y$$g:\spec k\to Y$ whose image is ${y}$$\{y\}$.(why ?)...(why?)

...${y}$$\{y\}$ is open in $Y$.moreover Moreover,$spec\mathcal{O}_y={y}\rightarrow spec k$ $\spec\mathcal{O}_y=\{y\}\to \spec k$ still has the property that $spac\mathcal{O}_y\rightarrow spac\mathcal{O}_y\otimes\mathcal{O}_y$$\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 have $spec\mathcal{O}_y={y}$,and I cannot prove the rest statement )($\spec\mathcal{O}_y$ 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(why $k(y)=k$ ?) and $spac\mathcal{O}_y\otimes\mathcal{O}_y$$\spec\mathcal{O}_y\otimes\mathcal{O}_y$ has only one point.and So $\mathcal{O}_y\otimes\mathcal{O}_y\rightarrow\mathcal{O}_y$$\mathcal{O}_y\otimes\mathcal{O}_y\to\mathcal{O}_y$ must be an isomorphism.(how can we characterize the struction of $\mathcal{O}_y\otimes\mathcal{O}_y$ )(how can we characterize the struction of $\mathcal{O}_y\otimes\mathcal{O}_y$?)

a proof in <etale cohomology> by milne ,please check it right

P23:Let $f:Y\rightarrow X$ be locally of finite-type. prove:$\Delta:Y\rightarrow Y\times Y$ is an open immersion,then $f$ is unramified ....we reduce the problem to the case of a morphisim $f:Y\rightarrow spec k$ where $k$ is an algebracally closed field.let $y$ be a closed point of $Y$.because $k$ is an algebracally closed  ,there exist a section $g:spec k\rightarrow Y$ whose image is ${y}$.(why ?)......${y}$ is open in $Y$.moreover ,$spec\mathcal{O}_y={y}\rightarrow spec k$ still has the property that $spac\mathcal{O}_y\rightarrow spac\mathcal{O}_y\otimes\mathcal{O}_y$ is an open immersion.($spec\mathcal{O}_y$ is not $V(y)$,why have $spec\mathcal{O}_y={y}$,and I cannot prove the rest statement ) But $\mathcal{O}_y$ is artin ring with residue field $k$(why $k(y)=k$ ?).and $spac\mathcal{O}_y\otimes\mathcal{O}_y$ has only one point.and $\mathcal{O}_y\otimes\mathcal{O}_y\rightarrow\mathcal{O}_y$ must be an isomorphism.(how can we characterize the struction of $\mathcal{O}_y\otimes\mathcal{O}_y$ )

help needed with a proof in <etale cohomology> by milne

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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P23:Let $f:Y\rightarrow X$ be locally of finite-type.  prove:the sheaf $\Omega_{Y/X}$ is zero,then $\Delta:Y\rightarrow Y\times Y$ is an open immersion,then $f$ is unramified ....we reduce the problem to the case of a morphisim $f:Y\rightarrow spec k$ where $k$ is an algebracally closed field.let $y$ be a closed point of $Y$.because $k$ is an algebracally closed ,there exist a section $g:spec k\rightarrow Y$ whose image is ${y}$.(why ?)......${y}$ is open in $Y$.moreover ,$spec\mathcal{O}_y={y}\rightarrow spec k$ still has the property that $spac\mathcal{O}_y\rightarrow spac\mathcal{O}_y\otimes\mathcal{O}_y$ is an open immersion.($spec\mathcal{O}_y$ is not $V(y)$,why have $spec\mathcal{O}_y={y}$,and I cannot prove the rest statement ) But $\mathcal{O}_y$ is artin ring with residue field $k$(why $k(y)=k$ ?).and $spac\mathcal{O}_y\otimes\mathcal{O}_y$ has only one point.and $\mathcal{O}_y\otimes\mathcal{O}_y\rightarrow\mathcal{O}_y$ must be an isomorphism.(how can we characterize the struction of $\mathcal{O}_y\otimes\mathcal{O}_y$ )

P23:Let $f:Y\rightarrow X$ be locally of finite-type.prove:the sheaf $\Omega_{Y/X}$ is zero,then $\Delta:Y\rightarrow Y\times Y$ is an open immersion ....we reduce the problem to the case of a morphisim $f:Y\rightarrow spec k$ where $k$ is an algebracally closed field.let $y$ be a closed point of $Y$.because $k$ is an algebracally closed ,there exist a section $g:spec k\rightarrow Y$ whose image is ${y}$.(why ?)......${y}$ is open in $Y$.moreover ,$spec\mathcal{O}_y={y}\rightarrow spec k$ still has the property that $spac\mathcal{O}_y\rightarrow spac\mathcal{O}_y\otimes\mathcal{O}_y$ is an open immersion.($spec\mathcal{O}_y$ is not $V(y)$,why have $spec\mathcal{O}_y={y}$,and I cannot prove the rest statement ) But $\mathcal{O}_y$ is artin ring with residue field $k$(why $k(y)=k$ ?).and $spac\mathcal{O}_y\otimes\mathcal{O}_y$ has only one point.and $\mathcal{O}_y\otimes\mathcal{O}_y\rightarrow\mathcal{O}_y$ must be an isomorphism.(how can we characterize the struction of $\mathcal{O}_y\otimes\mathcal{O}_y$ )

P23:Let $f:Y\rightarrow X$ be locally of finite-type.  prove:$\Delta:Y\rightarrow Y\times Y$ is an open immersion,then $f$ is unramified ....we reduce the problem to the case of a morphisim $f:Y\rightarrow spec k$ where $k$ is an algebracally closed field.let $y$ be a closed point of $Y$.because $k$ is an algebracally closed ,there exist a section $g:spec k\rightarrow Y$ whose image is ${y}$.(why ?)......${y}$ is open in $Y$.moreover ,$spec\mathcal{O}_y={y}\rightarrow spec k$ still has the property that $spac\mathcal{O}_y\rightarrow spac\mathcal{O}_y\otimes\mathcal{O}_y$ is an open immersion.($spec\mathcal{O}_y$ is not $V(y)$,why have $spec\mathcal{O}_y={y}$,and I cannot prove the rest statement ) But $\mathcal{O}_y$ is artin ring with residue field $k$(why $k(y)=k$ ?).and $spac\mathcal{O}_y\otimes\mathcal{O}_y$ has only one point.and $\mathcal{O}_y\otimes\mathcal{O}_y\rightarrow\mathcal{O}_y$ must be an isomorphism.(how can we characterize the struction of $\mathcal{O}_y\otimes\mathcal{O}_y$ )

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a proof in <etale cohomology> by milne ,please check it right

P23:Let $f:Y\rightarrow X$ be locally of finite-type.prove:the sheaf $\Omega_{Y/X}$ is zero,then $\Delta:Y\rightarrow Y\times Y$ is an open immersion ....we reduce the problem to the case of a morphisim $f:Y\rightarrow spec k$ where $k$ is an algebracally closed field.let $y$ be a closed point of $Y$.because $k$ is an algebracally closed ,there exist a section $g:spec k\rightarrow Y$ whose image is ${y}$.(why ?)......${y}$ is open in $Y$.moreover ,$spec\mathcal{O}_y={y}\rightarrow spec k$ still has the property that $spac\mathcal{O}_y\rightarrow spac\mathcal{O}_y\otimes\mathcal{O}_y$ is an open immersion.($spec\mathcal{O}_y$ is not $V(y)$,why have $spec\mathcal{O}_y={y}$,and I cannot prove the rest statement ) But $\mathcal{O}_y$ is artin ring with residue field $k$(why $k(y)=k$ ?).and $spac\mathcal{O}_y\otimes\mathcal{O}_y$ has only one point.and $\mathcal{O}_y\otimes\mathcal{O}_y\rightarrow\mathcal{O}_y$ must be an isomorphism.(how can we characterize the struction of $\mathcal{O}_y\otimes\mathcal{O}_y$ )