The Stacks project

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

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

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

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

for some $e_ i \in \mathbf{Z}$ and $c \in k_{12}$. In particular, the ring map

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. $\square$