I want to prove that the property of smoothness is stable under base change: if $X$ and $Y$ are two schemes over a scheme $S$ ($Y$ locally noetherian scheme) and $f:X\to Y$ smooth (that is finit for all $y\in Y$ the fiber $X_y$ is $k(y)$-smooth that is $X_y\times_{k(y)}\overline{k(y)}$ regular) and if $S$ is over $S'$ then $f_{S'}:X_{S'}\to Y_{S'}$ is smooth.
My idea: one has for all $y\in Y$, $X_y\to\operatorname{Spec}k(y)$ smooth ie $X_y\times_{k(y)}\overline{k(y)}=X\times_Y\overline{k(y)}$ regular and one wants to show for all $y'\in Y_{S'}=Y\times_S S'$ $(X_{S'})_{y'}\to\operatorname{Spec}k(y')$ smooth ie $(X\times_S S')\times_{Y'} k(y')\times_{k(y')} \overline{k(y')}=X\times_S S'\times_{Y'}\overline{k(y')}$ regular. I don't see how to do that.
Nota: I can't use that composition of smooth morphisms stay a smooth morphism because one prove that after using that fact.
If someone has an other approach with possibly another definition I take it too.