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 $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (2)
Comment #2580 by Matthew Emerton on
Comment #2581 by Johan on
There are also: