Proposition 33.31.1. Let X be a scheme. Let a : X \to \mathop{\mathrm{Spec}}(k_1) and b : X \to \mathop{\mathrm{Spec}}(k_2) be morphisms from X to spectra of fields. Assume a, b are locally of finite type, and X is reduced, and connected. Then we have k_1' = k_2', where k_ i' \subset \Gamma (X, \mathcal{O}_ X) is the integral closure of k_ i in \Gamma (X, \mathcal{O}_ X).
Proof. First, assume the lemma holds in case X is quasi-compact (we will do the quasi-compact case below). As X is locally of finite type over a field, it is locally Noetherian, see Morphisms, Lemma 29.15.6. In particular this means that it is locally connected, connected components of open subsets are open, and intersections of quasi-compact opens are quasi-compact, see Properties, Lemma 28.5.5, Topology, Lemma 5.7.11, Topology, Section 5.9, and Topology, Lemma 5.16.1. Pick an open covering X = \bigcup _{i \in I} U_ i such that each U_ i is quasi-compact and connected. For each i let K_ i \subset \mathcal{O}_ X(U_ i) be the integral closure of k_1 and of k_2. For each pair i, j \in I we decompose
U_ i \cap U_ j = \coprod U_{i, j, l}
into its finitely many connected components. Write K_{i, j, l} \subset \mathcal{O}(U_{i, j, l}) for the integral closure of k_1 and of k_2. By Lemma 33.28.4 the rings K_ i and K_{i, j, l} are fields. Now we claim that k_1' and k_2' both equal the kernel of the map
\prod K_ i \longrightarrow \prod K_{i, j, l}, \quad (x_ i)_ i \longmapsto x_ i|_{U_{i, j, l}} - x_ j|_{U_{i, j, l}}
which proves what we want. Namely, it is clear that k_1' is contained in this kernel. On the other hand, suppose that (x_ i)_ i is in the kernel. By the sheaf condition (x_ i)_ i corresponds to f \in \mathcal{O}(X). Pick some i_0 \in I and let P(T) \in k_1[T] be a monic polynomial with P(x_{i_0}) = 0. Then we claim that P(f) = 0 which proves that f \in k_1. To prove this we have to show that P(x_ i) = 0 for all i. Pick i \in I. As X is connected there exists a sequence i_0, i_1, \ldots , i_ n = i \in I such that U_{i_ t} \cap U_{i_{t + 1}} \not= \emptyset . Now this means that for each t there exists an l_ t such that x_{i_ t} and x_{i_{t + 1}} map to the same element of the field K_{i, j, l}. Hence if P(x_{i_ t}) = 0, then P(x_{i_{t + 1}}) = 0. By induction, starting with P(x_{i_0}) = 0 we deduce that P(x_ i) = 0 as desired.
To finish the proof of the lemma we prove the lemma under the additional hypothesis that X is quasi-compact. By Lemma 33.28.4 after replacing k_ i by k_ i' we may assume that k_ i is integrally closed in \Gamma (X, \mathcal{O}_ X). This implies that \mathcal{O}(X)^*/k_ i^* is a finitely generated abelian group, see Proposition 33.28.5. Let k_{12} = k_1 \cap k_2 as a subring of \mathcal{O}(X). Note that k_{12} is a field. Since
k_1^*/k_{12}^* \longrightarrow \mathcal{O}(X)^*/k_2^*
we see that k_1^*/k_{12}^* is a finitely generated abelian group as well. Hence there exist \alpha _1, \ldots , \alpha _ n \in k_1^* such that every element \lambda \in k_1 has the form
\lambda = c \alpha _1^{e_1} \ldots \alpha _ n^{e_ n}
for some e_ i \in \mathbf{Z} and c \in k_{12}. In particular, the ring map
k_{12}[x_1, \ldots , x_ n, \frac{1}{x_1 \ldots x_ n}] \longrightarrow k_1, \quad x_ i \longmapsto \alpha _ i
is surjective. By the Hilbert Nullstellensatz, Algebra, Theorem 10.34.1 we conclude that k_1 is a finite extension of k_{12}. In the same way we conclude that k_2 is a finite extension of k_{12}. In particular both k_1 and k_2 are contained in the integral closure k_{12}' of k_{12} in \Gamma (X, \mathcal{O}_ X). But since k_{12}' is a field by Lemma 33.28.4 and since we chose k_ i to be integrally closed in \Gamma (X, \mathcal{O}_ X) we conclude that k_1 = k_{12} = k_2 as desired. \square
Comments (0)