Lemma 29.49.5. Let $X$ be a scheme with finitely many irreducible components $X_1, \ldots , X_ n$. If $\eta _ i \in X_ i$ is the generic point, then

\[ R(X) = \mathcal{O}_{X, \eta _1} \times \ldots \times \mathcal{O}_{X, \eta _ n} \]

If $X$ is reduced this is equal to $\prod \kappa (\eta _ i)$. If $X$ is integral then $R(X) = \mathcal{O}_{X, \eta } = \kappa (\eta )$ is a field.

**Proof.**
Let $U \subset X$ be an open dense subset. Then $U_ i = (U \cap X_ i) \setminus (\bigcup _{j \not= i} X_ j)$ is nonempty open as it contained $\eta _ i$, contained in $X_ i$, and $\bigcup U_ i \subset U \subset X$ is dense. Thus the identification in the lemma comes from the string of equalities

\begin{align*} R(X) & = \mathop{\mathrm{colim}}\nolimits _{U \subset X\text{ open dense}} \mathop{\mathrm{Mor}}\nolimits (U, \mathbf{A}^1_\mathbf {Z}) \\ & = \mathop{\mathrm{colim}}\nolimits _{U \subset X\text{ open dense}} \mathcal{O}_ X(U) \\ & = \mathop{\mathrm{colim}}\nolimits _{\eta _ i \in U_ i \subset X\text{ open}} \prod \mathcal{O}_ X(U_ i) \\ & = \prod \mathop{\mathrm{colim}}\nolimits _{\eta _ i \in U_ i \subset X\text{ open}} \mathcal{O}_ X(U_ i) \\ & = \prod \mathcal{O}_{X, \eta _ i} \end{align*}

where the second equality is Schemes, Example 26.15.2. The final statement follows from Algebra, Lemma 10.25.1.
$\square$

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