Lemma 66.19.1. Let $S$ be a scheme. Let $Y$ be a disjoint union of spectra of zero dimensional local rings over $S$. Let $f : X \to Y$ be a monomorphism of algebraic spaces over $S$. Then $f$ is representable, i.e., $X$ is a scheme.

Proof. This immediately reduces to the case $Y = \mathop{\mathrm{Spec}}(A)$ where $A$ is a zero dimensional local ring, i.e., $\mathop{\mathrm{Spec}}(A) = \{ \mathfrak m_ A\}$ is a singleton. If $X = \emptyset$, then there is nothing to prove. If not, choose a nonempty affine scheme $U = \mathop{\mathrm{Spec}}(B)$ and an étale morphism $U \to X$. As $|X|$ is a singleton (as a subset of $|Y|$, see Morphisms of Spaces, Lemma 65.10.9) we see that $U \to X$ is surjective. Note that $U \times _ X U = U \times _ Y U = \mathop{\mathrm{Spec}}(B \otimes _ A B)$. Thus we see that the ring maps $B \to B \otimes _ A B$ are étale. Since

$(B \otimes _ A B)/\mathfrak m_ A(B \otimes _ A B) = (B/\mathfrak m_ AB) \otimes _{A/\mathfrak m_ A} (B/\mathfrak m_ AB)$

we see that $B/\mathfrak m_ AB \to (B \otimes _ A B)/\mathfrak m_ A(B \otimes _ A B)$ is flat and in fact free of rank equal to the dimension of $B/\mathfrak m_ AB$ as a $A/\mathfrak m_ A$-vector space. Since $B \to B \otimes _ A B$ is étale, this can only happen if this dimension is finite (see for example Morphisms, Lemmas 29.55.9 and 29.55.10). Every prime of $B$ lies over $\mathfrak m_ A$ (the unique prime of $A$). Hence $\mathop{\mathrm{Spec}}(B) = \mathop{\mathrm{Spec}}(B/\mathfrak m_ A)$ as a topological space, and this space is a finite discrete set as $B/\mathfrak m_ A B$ is an Artinian ring, see Algebra, Lemmas 10.52.2 and 10.52.6. Hence all prime ideals of $B$ are maximal and $B = B_1 \times \ldots \times B_ n$ is a product of finitely many local rings of dimension zero, see Algebra, Lemma 10.52.5. Thus $B \to B \otimes _ A B$ is finite étale as all the local rings $B_ i$ are henselian by Algebra, Lemma 10.152.10. Thus $X$ is an affine scheme by Groupoids, Proposition 39.23.9. $\square$

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