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Angelo
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Consider the fiber product $S \times_{\frak M} M$; its projection onto $S$ is smooth and surjective, hence open. Since $S$ is quasi-compact, there exists a quasi-compact open subset $U$ of $M$$S \times_{\frak M} M$ surjecting onto $S$. But sinceThe image of $U$ in $M$ is quasi-compact; let $V$ be a quasi-compact open subscheme of $M$ containing it. Since $S$ surjects onto $\frak M$, so does $U$. So $V \to \frak M$ is smooth and surjective, and $V$ is of finite type, so $\frak M$ is of finite type.

Consider the fiber product $S \times_{\frak M} M$; its projection onto $S$ is smooth and surjective. Since $S$ is quasi-compact, there exists a quasi-compact open subset $U$ of $M$ surjecting onto $S$. But since $S$ surjects onto $\frak M$, so does $U$, and $\frak M$ is of finite type.

Consider the fiber product $S \times_{\frak M} M$; its projection onto $S$ is smooth and surjective, hence open. Since $S$ is quasi-compact, there exists a quasi-compact open subset $U$ of $S \times_{\frak M} M$ surjecting onto $S$. The image of $U$ in $M$ is quasi-compact; let $V$ be a quasi-compact open subscheme of $M$ containing it. Since $S$ surjects onto $\frak M$, so does $U$. So $V \to \frak M$ is smooth and surjective, and $V$ is of finite type, so $\frak M$ is of finite type.

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Angelo
  • 27k
  • 6
  • 92
  • 112

Consider the fiber product $S \times_{\frak M} M$; its projection onto $S$ is smooth and surjective. Since $S$ is quasi-compact, there exists a quasi-compact open subset $U$ of $M$ surjecting onto $S$. But since $S$ surjects onto $\frak M$, so does $U$, and $\frak M$ is of finite type.