Lemma 70.6.3. Let $S$ be a scheme. Let $X$ be a locally Noetherian integral algebraic space over $S$. Let $Z \subset X$ be a prime divisor and let $\xi \in |Z|$ be the generic point. Then the henselian local ring $\mathcal{O}_{X, \xi }^ h$ is a reduced $1$-dimensional Noetherian local ring and there is a canonical injective map

\[ R(X) \longrightarrow Q(\mathcal{O}_{X, \xi }^ h) \]

from the function field $R(X)$ of $X$ into the total ring of fractions.

**Proof.**
We will use the results of Decent Spaces, Section 66.11. Let $(U, u) \to (X, \xi )$ be an elementary étale neighbourhood. Observe that $U$ is locally Noetherian and reduced. Thus $\mathcal{O}_{U, u}$ is a $1$-dimensional (by our definition of prime divisors) reduced Noetherian ring. After replacing $U$ by an affine open neighbourhood of $u$ we may assume $U$ is Noetherian and affine. After replacing $U$ by a smaller open, we may assume every irreducible component of $U$ passes through $u$. Since $U \to X$ is open and $X$ irreducible, $U \to X$ is dominant. Hence we obtain a ring map $R(X) \to R(U)$ by composing rational maps, see Morphisms of Spaces, Section 65.47. Since $R(X)$ is a field, this map is injective. By our choice of $U$ we see that $R(U)$ is the total quotient ring $Q(\mathcal{O}_{U, u})$, see Morphisms, Lemma 29.49.5 and Algebra, Lemma 10.24.4.

At this point we have proved all the statements in the lemma with $\mathcal{O}_{U, u}$ in stead of $\mathcal{O}_{X, \xi }^ h$. However, $\mathcal{O}_{X, \xi }^ h$ is the henselization of $\mathcal{O}_{U, u}$. Thus $\mathcal{O}_{X, \xi }^ h$ is a $1$-dimensional reduced Noetherian ring, see More on Algebra, Lemmas 15.45.4, 15.45.7, and 15.45.3. Since $\mathcal{O}_{U, u} \to \mathcal{O}_{X, \xi }^ h$ is faithfully flat by More on Algebra, Lemma 15.45.1 it sends nonzerodivisors to nonzerodivisors. Therefore we obtain a canonical map $Q(\mathcal{O}_{U, u}) \to Q(\mathcal{O}_{X, \xi }^ h)$ and we obtain our map. We omit the verification that the map is independent of the choice of $(U, u) \to (X, x)$; a slightly better approach would be to first observe that $\mathop{\mathrm{colim}}\nolimits Q(\mathcal{O}_{U, u}) = Q(\mathcal{O}_{X, \xi }^ h)$.
$\square$

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