Lemma 41.20.5. Let $(A, I)$ be a henselian pair. Let $U \to \mathop{\mathrm{Spec}}(A)$ be a quasi-compact, separated, étale morphism such that $U \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A/I) \to \mathop{\mathrm{Spec}}(A/I)$ is finite. Then

\[ U = U_{fin} \amalg U_{away} \]

where $U_{fin} \to \mathop{\mathrm{Spec}}(A)$ is finite and $U_{away}$ has no points lying over $Z$.

**Proof.**
By Zariski's main theorem, the scheme $U$ is quasi-affine. In fact, we can find an open immersion $U \to T$ with $T$ affine and $T \to \mathop{\mathrm{Spec}}(A)$ finite, see More on Morphisms, Lemma 37.43.3. Write $Z = \mathop{\mathrm{Spec}}(A/I)$ and denote $U_ Z \to T_ Z$ the base change. Since $U_ Z \to Z$ is finite, we see that $U_ Z \to T_ Z$ is closed as well as open. Hence by More on Algebra, Lemma 15.11.6 we obtain a unique decomposition $T = T' \amalg T''$ with $T'_ Z = U_ Z$. Set $U_{fin} = U \cap T'$ and $U_{away} = U \cap T''$. Since $T'_ Z \subset U_ Z$ we see that all closed points of $T'$ are in $U$ hence $T' \subset U$, hence $U_{fin} = T'$, hence $U_{fin} \to \mathop{\mathrm{Spec}}(A)$ is finite. We omit the proof of uniqueness of the decomposition.
$\square$

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