Lemma 26.23.11. Let $k_1, \ldots , k_ n$ be fields. For any monomorphism of schemes $X \to \mathop{\mathrm{Spec}}(k_1 \times \ldots \times k_ n)$ there exists a subset $I \subset \{ 1, \ldots , n\}$ such that $X \cong \mathop{\mathrm{Spec}}(\prod _{i \in I} k_ i)$ as schemes over $\mathop{\mathrm{Spec}}(k_1 \times \ldots \times k_ n)$. More generally, if $X = \coprod _{i \in I} \mathop{\mathrm{Spec}}(k_ i)$ is a disjoint union of spectra of fields and $Y \to X$ is a monomorphism, then there exists a subset $J \subset I$ such that $Y = \coprod _{i \in J} \mathop{\mathrm{Spec}}(k_ i)$.

Proof. First reduce to the case $n = 1$ (or $\# I = 1$) by taking the inverse images of the open and closed subschemes $\mathop{\mathrm{Spec}}(k_ i)$. In this case $X$ has only one point hence is affine. The corresponding algebra problem is this: If $k \to R$ is an algebra map with $R \otimes _ k R \cong R$, then $R \cong k$ or $R = 0$. This holds for dimension reasons. See also Algebra, Lemma 10.107.8 $\square$

Comment #2580 by Matthew Emerton on

In the last line, should also allow the case $R = 0$, I guess.

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