Lemma 33.28.4. Let $k$ be a field. Let $X$ be a scheme over $k$ which is connected and reduced. Then the integral closure of $k$ in $\Gamma (X, \mathcal{O}_ X)$ is a field.

**Proof.**
Let $k' \subset \Gamma (X, \mathcal{O}_ X)$ be the integral closure of $k$. Then $X \to \mathop{\mathrm{Spec}}(k)$ factors through $\mathop{\mathrm{Spec}}(k')$, see Schemes, Lemma 26.6.4. As $X$ is reduced we see that $k'$ has no nonzero nilpotent elements. As $k \to k'$ is integral we see that every prime ideal of $k'$ is both a maximal ideal and a minimal prime, and $\mathop{\mathrm{Spec}}(k')$ is totally disconnected, see Algebra, Lemmas 10.36.20 and 10.26.5. As $X$ is connected the morphism $X \to \mathop{\mathrm{Spec}}(k')$ is constant, say with image the point corresponding to $\mathfrak p \subset k'$. Then any $f \in k'$, $f \not\in \mathfrak p$ maps to an invertible element of $\mathcal{O}_ X$. By definition of $k'$ this then forces $f$ to be a unit of $k'$. Hence we see that $k'$ is local with maximal ideal $\mathfrak p$, see Algebra, Lemma 10.18.3. Since we've already seen that $k'$ is reduced this implies that $k'$ is a field, see Algebra, Lemma 10.25.1.
$\square$

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