Question:Let $f:X \rightarrow Y$ be a morphism locally of finite presentation. If $f$ is smooth at all closed points of X, then is $f$ smooth? The definition of smooth is as follow: enter image description here
This question arise when I read the proof of flat morphisms locally of finite presentation with smooth geometoric fibers are smooth in the theorem 25.2.2 of Vakil's FOAG,(http://math.stanford.edu/~vakil/216blog/FOAGjun1113public.pdf).
I know a related post( Checking flat- and smoothness: enough to check on closed points?). But my situation is more generally.The smooth locus $X_{sm}$ of $X$ is open in $X$ and the set of all closed points of $X$ are contained in $X_{sm}$. But does any point of $X$ have a specialization that is a closed point of $X$?
I believe the following is true:
Proposition. Let $f\colon X \to Y$ be a morphism locally of finite presentation, and suppose that $X$ is quasi-compact. If $f$ is smooth at all closed points of $X$, then $f$ is smooth.
Proof. The locus $X_\mathrm{sm}$ on which $f$ is smooth is open in $X$. Moreover, every point in $X$ specializes to a closed point in $X$ by the proof of [Schwede, Prop. 4.1] since $X$ is quasi-compact. Thus, $X_\mathrm{sm} = X$. $\blacksquare$
On the other hand, if $X$ is not quasi-compact, then [Schwede, Thm. 4.6] shows that $X$ might not have any closed points at all!
EDIT. The condition "$X$ is quasi-compact" can be replaced by "$X$ is locally noetherian" since every point in a locally noetherian scheme specializes to a closed point by [Stacks, Tag 02IL].