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 $\mathcal{O}_y\otimes\mathcal{O}_y$?)
