Fundamental group | Lovely little lemmas

It is well known that a finite étale morphism $f \colon X \to Y$ of schemes is étale locally given by a disjoint union of isomorphisms, i.e. there exists an étale cover $Y' \to Y$ such that the pullback $X' \to Y'$ is given by $X' = \coprod_{i=1}^n Y' \to Y'$ . Something similar is true for finite unramified morphisms:

Lemma. Let $f \colon X \to Y$ be a finite unramified¹ morphism of schemes. Then there exists an étale cover $Y' \to Y$ such that the pullback $X' \to Y'$ is given by $X' = \coprod_i Z_i \to Y'$ , where $Z_i \hookrightarrow Y'$ are closed immersions of finite presentation.

Proof. Let $y \in Y$ be a point, let $A = \mathcal O_{Y,y}^{\operatorname{sh}}$ be the strict henselisation of $Y$ at $y$ , and let $\Spec B \to \Spec A$ be the base change of $X \to Y$ along $\Spec A \to Y$ . Then $A \to B$ is unramified, so by Tag 04GL it splits as

$B = A_1 \times \ldots \times A_r \times C$

where $A \to A_i$ is surjective for each $i$ and no prime of $C$ lies above $\mathfrak m_y \subseteq A$ . But $A \to C$ is also finite, so by Tag 00GU the map $\Spec C \to \Spec A$ hits the maximal ideal if $\Spec C \neq \varnothing$ . Thus, we conclude that $C = 0$ , hence $B$ is a product of quotients of $A$ .

But $A$ is the colimit of $\mathcal O_{Y',y'}$ for $(Y',y') \to (Y,y)$ an étale neighbourhood inducing a separable extension $\kappa(y) \to \kappa(y')$ . Since $f$ is of finite presentation, each of the ideals $\ker(A \to A_i)$ and the projections $B \to A_i$ are defined over some étale neighbourhood $(Y',y') \to (Y,y)$ . Then the pullback $X' \to Y'$ is given by a finite disjoint union of closed immersions in $Y'$ .

Then $Y' \to Y$ might not be a covering, but since $y \in Y$ was arbitrary we can do this for each point separately and take a disjoint union. $\qedsymbol$

Remark. The number of $Z_i$ needed is locally bounded, but if $Y$ is not quasi-compact it might be infinite. For example, we can take $X \cong Y = \coprod_{i \in \N} \Spec k$ an infinite disjoint union of points, and $f \colon X \to Y$ such that the fibre over $y_i \in Y$ for $i \in \N$ has $i$ points.

Remark. In the étale case, we may actually take $Y' \to Y$ finite étale, by taking $Y'$ to be the Galois closure of $X \to Y$ , which exists in reasonable cases². For example, if $Y$ is normal, we may take $Y'$ to be the integral closure of $Y$ in the field extension corresponding to the Galois closure of $k(Y) \to k(X)$ . In general, if $Y$ is connected it follows from Tag 0BN2 that a suitable component of the $\deg(f)$ -fold fibre product of $X$ over $Y$ is a Galois closure $Y' \to Y$ of $X \to Y$ . If the connected components of $Y$ are open, apply this construction to each component.

In the unramified case, this is too much to hope for. For example, if $Y = \mathbf P^2_{\mathbf C}$ , then we may take $X$ to be a nontrivial finite étale cover of an elliptic curve $E \subseteq Y$ . This is finite and unramified, but does not split over any finite étale cover of $\mathbf P^2$ since there aren’t any. In fact, it cannot split over any connected étale cover $Y' \to \mathbf P^2$ whose image contains $E$ , since that implies the image only misses finitely many points (as $E$ is ample), which is again impossible since $\pi_1(\mathbf P^2 \setminus \{p_1,\ldots,p_r\}) = 0$ .

¹For the purposes of this post, unramified means in the sense of Grothendieck, i.e. including the finite presentation hypothesis. In Raynaud’s work on henselisations, this was weakened to finite type. See Tag 00US for definitions.

²I’m not sure what happens in general.

This is a cute proof that I ran into of the simple connectedness of $\mathbb P^1$ . It does not use Riemann–Hurwitz or differentials, and instead relies on a purely geometric argument.

Lemma. Let $k$ be an algebraically closed field. Then $\mathbb P^1_k$ is simply connected.

Proof. Let $f \colon C \to \mathbb P^1$ be a finite étale Galois cover with Galois group $G$ . We have to show that $f$ is an isomorphism. The diagonal $\Delta_{\mathbb P^1} \subseteq \mathbb P^1 \times \mathbb P^1$ is ample, so the same goes for the pullback $D = (f \times f)^* \Delta_{\mathbb P^1}$ to $C \times C$ [Hart, Exc. III.5.7(d)]. In particular, $D$ is connected [Hart, Cor. III.7.9].

But $D \cong C \times_{\mathbb P^1} C$ is isomorphic to $|G|$ copies of $C$ because the action

$\begin{align*} G \times C &\to C \times_{\mathbb P^1} C\\ (g,c) &\mapsto (gc,c) \end{align*}$

is an isomorphism. If $D$ is connected, this forces $|G| = 1$ , so $f$ is an isomorphism. $\qed$

The proof actually shows that if $X$ is a smooth projective variety such that $\Delta_X$ is a set-theoretic complete intersection of ample divisors, then $X$ is simply connected.

Example. For a smooth projective curve $C$ of genus $g \geq 1$ , the diagonal cannot be ample, as $\pi_1(C) \neq 0$ . We already knew this by computing the self-intersection $\Delta_C^2 = 2-2g \leq 0$ , ~~but the argument above is more elementary~~.