Lemma 49.11.5. With notation as in Example 49.11.2 consider the open subscheme $U'_ d \subset X_ d$ over which $\pi _ d$ is étale. Then $U'_ d$ is a dense subset of the open $U_ d$ of Lemma 49.11.3.

**Proof.**
By exactly the same reasoning as in the proof of Lemma 49.11.3, using Morphisms, Lemma 29.36.17, there is a maximal open $U'_ d \subset X_ d$ over which $\pi _ d$ is étale. Moreover, since an étale morphism is syntomic, we see that $U'_ d \subset U_ d$. To finish the proof we have to show that $U'_ d \subset U_ d$ is dense. Let $u : \mathop{\mathrm{Spec}}(k) \to U_ d$ be a morphism where $k$ is a field. Let $B = k \otimes _{A_ d} B_ d$ as in the proof of Lemma 49.11.4. We will show there is a local domain $A'$ with residue field $k$ and a finite syntomic $A'$ algebra $B'$ with $B = k \otimes _{A'} B'$ whose generic fibre is étale. Exactly as in the previous paragraph this will determine a morphism $\mathop{\mathrm{Spec}}(A') \to U_ d$ which will map the generic point into $U'_ d$ and the closed point to $u$, thereby finishing the proof.

By Lemma 49.11.1 part (4) we can choose a presentation $B = k[x_1, \ldots , x_ n]/(f_1, \ldots , f_ n)$. Let $d'$ be the maximum total degree of the polynomials $f_1, \ldots , f_ n$. Let $Y_{n, d'} \to X_{n, d'}$ be as in Example 49.10.5. By construction there is a morphism $u' : \mathop{\mathrm{Spec}}(k) \to X_{n, d'}$ such that

Denote $A = \mathcal{O}_{X_{n, d'}, u'}^ h$ the henselization of the local ring of $X_{n, d'}$ at the image of $u'$. Then we can write

with $Z \to \mathop{\mathrm{Spec}}(A)$ finite and $W \to \mathop{\mathrm{Spec}}(A)$ having empty closed fibre, see Algebra, Lemma 10.153.3 part (13) or the discussion in More on Morphisms, Section 37.41. By Lemma 49.10.6 the local ring $A$ is regular (here we also use More on Algebra, Lemma 15.45.10) and the morphism $Z \to \mathop{\mathrm{Spec}}(A)$ is étale over the generic point of $\mathop{\mathrm{Spec}}(A)$ (because it is mapped to the generic point of $X_{d, n'}$). By construction $Z \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(k) \cong \mathop{\mathrm{Spec}}(B)$. This proves what we want except that the map from residue field of $A$ to $k$ may not be an isomorphism. By Algebra, Lemma 10.159.1 there exists a flat local ring map $A \to A'$ such that the residue field of $A'$ is $k$. If $A'$ isn't a domain, then we choose a minimal prime $\mathfrak p \subset A'$ (which lies over the unique minimal prime of $A$ by flatness) and we replace $A'$ by $A'/\mathfrak p$. Set $B'$ equal to the unique $A'$-algebra such that $Z \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A') = \mathop{\mathrm{Spec}}(B')$. This finishes the proof. $\square$

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