Algebraic space birational to a scheme

Let $S$ be a Noetherian scheme, let $Y$ be a scheme of finite type over $S$, and let $X$ be an algebraic space of finite type over $S$. Suppose that there is a morphism $f:Y \rightarrow X$ which is proper and birational. Must $X$ be a scheme?

• No, that is not true. In fact, the opposite is true if you add a few hypotheses (e.g., if S is a finite type scheme over a field or over an excellent DVR, and if the algebraic space $X$ is separated). This is often called "Chow's Lemma for algebraic spaces". There is a reference in Knutson's "Algebraic Spaces" . . . Commented May 29, 2018 at 20:52
• . . . I forget to look up the reference earlier. Here it is: Chow's Lemma, Theorem IV.3.1, p. 192, Donald Knutson, "Algebraic Spaces", LNM 203, Springer-Verlag, Berlin, 1971. Commented May 31, 2018 at 13:29
• Thank you very much. Can you post it as an answer so I will accept it.
– Rami
Commented Jun 1, 2018 at 21:42

For every separated, Noetherian scheme $S$, for every finitely presented, separated morphism of algebraic spaces, $X\to S$, there exists a morphism $f:Y\to X$ that is projective and birational such that $Y\to S$ is a quasi-projective morphism.