Lemma 37.39.7. With notation an assumptions as in Lemma 37.39.4 assume that $\varphi $ induces an isomorphism on completions. Then we can choose our diagram such that $f$ is étale.
Proof. We may assume $N \geq 2$ and we may replace $(T, t)$ with $(T', t')$ as in Lemma 37.39.6. Since $(V, v) \to (S, s)$ is an elementary étale neighbourhood, so is $(X \times _ S V, (x, v)) \to (X, x)$. Thus $\mathcal{O}_{X, x} \to \mathcal{O}_{X \times _ S V, (x, v)}$ induces an isomorphism on completions by More on Algebra, Lemma 15.43.9. We claim $\mathcal{O}_{X, x} \to \mathcal{O}_{W, w}$ induces an isomorphism on completions. Having proved this, Lemma 37.12.1 will show that $f$ is smooth at $w$ and of course $f$ is unramified at $u$ as well, so Morphisms, Lemma 29.36.5 tells us $f$ is étale at $w$.
First we use the commutativity in part (5) of Lemma 37.39.4 to see that for $i \leq N$ there is a commutative diagram
This implies that $f^\sharp _ w$ defines an isomorphism $\kappa (x) \to \kappa (w)$ on residue fields and an isomorphism $\mathfrak m_ x/\mathfrak m_ x^2 \to \mathfrak m_ w/\mathfrak m_ w^2$ on cotangent spaces. Hence $f^\sharp _ w$ defines a surjection $\mathcal{O}_{X, x}^\wedge \to \mathcal{O}_{W, w}^\wedge $ on complete local rings.
By Lemma 37.39.6 there is an isomorphism of $\text{Gr}_{\mathfrak m_ s}(\mathcal{O}_{(Y \times _{T, \tau } V, (y, v)})$ with $\text{Gr}_{\mathfrak m_ s}(\mathcal{O}_{Y_\sigma , y_\sigma })$. This follows by taking stalks of the isomorphism of conormal sheaves at the point $y$. Since our local rings are Noetherian taking associated graded with respect to $\mathfrak m_ s$ commutes with completion because completion with respect to an ideal is an exact functor on finite modules over Noetherian rings. This produces the right vertical isomorphism in the diagram of graded rings
We do not claim the diagram commutes. By the result of the previous paragraph the left arrow is surjective. The other three arrows are isomorphisms. It follows that the left arrow is a surjective map between isomorphic Noetherian rings. Hence it is an isomorphism by Algebra, Lemma 10.31.10 (you can argue this directly using Hilbert functions as well). In particular $\mathcal{O}_{X, x}^\wedge \to \mathcal{O}_{W, w}^\wedge $ must be injective as well as surjective which finishes the proof. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: