The Étale Fundamental GroupÉtale Morphisms

# 2 Étale Morphisms

We wish to define the notion of a “covering map” for arbitrary schemes. One way to think about covering maps is that they are local homeomorphisms plus some extra conditions. The notion of étale morphisms is the correct analogue of local homeomorphisms for schemes. In the next section, we will strengthen this to finite étale covers afterwards as an analogue of finite covering spaces, and prove that finite étale covers are exactly the “locally trivial” ones in a suitable sense.

We first give some examples of maps, and discuss why they should or should not be considered étale. Afterwards, we will write down the definition of étale.

Example

Let $X = \mathbb {P}^1$, and $f: X \to X$ the double cover given by squaring. Then this should not be étale, because it is ramified at $0$ and $\infty$, where there is only one geometric point in the preimage.

Algebraically, if we localize at the point $0$, then this is represented by the map $k[x]_{(x)} \to k[x]_{(x)}$ that sends $x$ to $x^2$, and in particular sends the maximal ideal $\mathfrak {m}$ to $\mathfrak {m}^2$. Thus, as an extension of discrete valuation rings, this is ramified.

Example

The inclusion of a proper closed subscheme should not be étale, because it is not a local homeomorphism. Alternatively, because it is not flat.

Example

The inclusion of an open subscheme is étale, because it is locally a homeomorphism, but should not be a étale cover.

Finally, for our theory to describe Galois theory proper, we want

Example

An extension of fields $\operatorname{Spec}K \to \operatorname{Spec}k$ is an étale cover iff it is a finite separable extension.

Based on the geometric examples, the following definition is entirely reasonable.

Definition (Étale morphism)

A morphism is étale if it is flat and unramified.

To make sense of such a definition, I must explain what it means to be unramified.
Theorem

Let $f: Y \to X$ be a morphism locally of finite type, $y \in Y$ and $x = f(y)$. Then the following are equivalent:

1. $\mathfrak {m}_y = f^*\mathfrak {m}_x$, and $k(y)$ is a finite separable extension of $k(x)$.

2. $\Omega _{Y/X}$ is trivial at $y$.

3. The diagonal morphism $\Delta _{Y/X}: Y \to Y \times _X Y$ is an open immersion in a neighbourhood of $y$.

If any (hence all) of these hold, we say $f$ is unramified at $y$. We say $f$ is unramified if it is unramified at all $y \in Y$.

Each of these are reasonable definitions of “unramified”.

Proof

• (1) $\Leftrightarrow$ (2): Assuming (1), we consider the diagram where the left-hand square is a pull-back by (1) and the right-hand square is a localization at a point in the total space. So they both preserve the differential. So $\Omega _{Y/X}$ is trivial at $y$ iff $\Omega _{\mathcal{O}_y/\mathcal{O}_x}$ is trivial, iff $\Omega _{k(y)/k(x)}$ is trivial by Nakayama, iff $k(y)/k(x)$ is finite separable.

To prove the converse, if $\Omega _{Y/X}$ is trivial at $y$, then pulling back shows that $\mathcal{O}_y/\mathfrak {m}_x \to k(x)$ has trivial Kähler differentials. So it suffices to show that if $A$ is a finite local $k$-algebra, then $\Omega _{A/k} = 0$ iff $A$ is a finite separable extension of $y$ (which implies $\mathfrak {m}_y = f^* \mathfrak {m}_x$ since $A/f^*\mathfrak {m}_x$ is a field, hence $f^*\mathfrak {m}_x$ is a maximal ideal).

By direct calculation, the result is true if $A = k[x]/f$ for some $f$, since the only interesting relation is $\mathrm{d} f = 0$. For general $A$, use that if $k \subseteq B \subseteq A$, then there is a surjection $\Omega _{A/k} \twoheadrightarrow \Omega _{B/k}$, and so taking $B$ to be the subalgebra generated by each $x \in A$, we see that every element is separable and invertible over $A$.

• (2) $\Rightarrow$ (3):

Restricted to affine patches, $\Delta _{Y/X}$ is a closed immersion, and $\Omega _{Y/X} = 0$ implies the (proper) ideal $I$ defining the image satisfies $I^2 = I$. By Nakayama, $I = 0$, so this is an open immersion as well. The converse is clear as well.

Proof

We also record the following immediate consequences:

Lemma
1. Open immersions are étale.

2. Pullback of étale maps are étale.

3. Composition of étale maps are étale.

4. Étale morphisms satisfy fpqc descent.

Recall that we say a property P of morphisms satisfies fpqc descent if whenever we have a pullback diagram if $f: X' \to X$ is fpqc (faithfully flat and quasi-compact, or fidélement plat et quasi-compact), then $p$ has property P iff $p'$ does.

We quickly look at some examples. It is clear from definition that

Example

An étale cover of a field is a disjoint union of finite separable extensions.

The following two theorems, which we shall not prove, characterize étale morphisms over smooth varieties, especially over $\mathbb {C}$. They are exactly what we would expect from the case of manifolds.

Theorem

Let $X$ be a smooth variety over an algebraically closed field, and $f: Y \to X$. Then $f$ is étale iff $Y$ is a smooth variety and the derivative $\mathrm{d} f_y: T_y Y \to Y_{f(y)}X$ is an isomorphism for all $y \in Y$.

Theorem (Riemann existence theorem)

Let $X$ be a smooth variety over $\mathbb {C}$. Then every finite covering space $X(\mathbb {C})$ has the structure of a smooth variety.

Example

The line with two origins is étale over $\mathbb {A}^1$ under the obvious projection map.

Returning to general theory, recall that covering spaces satisfy the unique lifting property. The same is true for étale maps if we assume, of course, that our morphism is separated.

Lemma

Let $f: Y \to X$ be a separated étale morphism. Then $\Delta : Y \to Y \times _X Y$ is an inclusion into a component.

Proof
It is true topologically since $\Delta$ is open (since $f$ is unramified) and closed (since $f$ is separated). It is true scheme-theoretically because $\Delta$ is locally an open immersion.
Proof

Corollary

Let $p: Y \to X$ be étale and separated, and $Z$ be a connected scheme over $X$. Suppose $z$ is a geometric point of $Z$ and $f_1, f_2: Z \to Y$ are morphisms such that $f_1(z) = f_2(z)$. Then $f_1 = f_2$.

Proof
Consider the map $(f_1, f_2): Z \to Y \times _X Y$. Since the diagonal in $Y \times _X Y$ is a component, the preimage under this map is open and closed. Since it is non-empty (it contains $z$), it must be everything. So it factors through the diagonal (scheme-theoretically, since $\Delta$ is a component), hence $f_1 = f_2$.
Proof