Lemma 49.10.6. With notation as in Example 49.10.5 the schemes $X_{n, d}$ and $Y_{n, d}$ are regular and irreducible, the morphism $Y_{n, d} \to X_{n, d}$ is locally quasi-finite and syntomic, and there is a dense open subscheme $V \subset Y_{n, d}$ such that $Y_{n, d} \to X_{n, d}$ restricts to an étale morphism $V \to X_{n, d}$.

**Proof.**
The scheme $X_{n, d}$ is the spectrum of the polynomial ring $A$. Hence $X_{n, d}$ is regular and irreducible. Since we can write

we see that the ring $B$ is isomorphic to the polynomial ring on $x_1, \ldots , x_ n$ and the elements $a_{i, E}$ with $E \not= (0, \ldots , 0)$. Hence $\mathop{\mathrm{Spec}}(B)$ is an irreducible and regular scheme and so is the open $Y_{n, d}$. The morphism $Y_{n, d} \to X_{n, d}$ is locally quasi-finite and syntomic by Lemma 49.10.1. To find $V$ it suffices to find a single point where $Y_{n, d} \to X_{n, d}$ is étale (the locus of points where a morphism is étale is open by definition). Thus it suffices to find a point of $X_{n, d}$ where the fibre of $Y_{n, d} \to X_{n, d}$ is nonempty and étale, see Morphisms, Lemma 29.36.15. We choose the point corresponding to the ring map $\chi : A \to \mathbf{Q}$ sending $f_ i$ to $1 + x_ i^ d$. Then

which is a nonzero étale algebra over $\mathbf{Q}$. $\square$

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