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

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. $\square$

## 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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)