# Why does finitely presented imply quasi-separated ?

By the EGA definition, a morphism of schemes of finite presentation is required to be quasi-separated. As far as I can see, removing this requirement does not prevent from proving the basic properties such as stability of the notion under composition, products, etc. So my question is :

where exactly in proving important theorems involving morphisms of finite presentation is the quasi-separated assumption crucial ?

Note that a morphism of finite type is not required to be quasi-separated.

All kinds of examples and counter-examples will be appreciated.

Let me quickly try to explain why. The heart of the reduction to the noetherian case are theorems like the following: Let X over Spec A be a finitely presented scheme. Then there is a subring $A_0$ of $A$ which is a finitely generated $\mathbb{Z}$-algebra (and in particular noetherian) and an $A_0$-scheme $X_0$ of finite presentation such that $X$ arises from $X_0$ via the base-change $A_0\to A$. If $X$ is affine, this is pretty clear, as $X$ is definied by finitely many equations in an affine space over $A$. In order to pass from the affine case to the general case, it does NOT suffice to know that we can cover $X$ by finitely many affine pieces (which would be the assumption of quasi-compactness), but we also need that the glueing data for the affine pieces are somehow described by a finite number of equations. This is ensured by the assumption that the intersection of two affine pieces is quasi-compact which corresponds precisely to the assumption that $X$ is quasi-separated over A.
• Note also the trivial "converse" that since every f. type map $f_0:X_0 \rightarrow S_0$ to a noetherian $S_0$ is q−septd, any $f:X \rightarrow S$ arising from such $f_0$ by base change must be q-septd. So q-sep'tdness is not only sufficient for a f. type map to admit descent to a "noetherian" situation, but also necessary. For example, if $A$ is a ring and $U$ is a non-qc open in Spec($A$) then the f.type $A$-scheme gluing $X$ of Spec($A$) to itself along $U$ cannot descend to f.type scheme over noetherian ring; stronger than saying the EGA method of proof doesn't apply. Aug 26 '10 at 12:27