Example 41.8.1. Let k be a field. Unramified quasi-compact morphisms X \to \mathop{\mathrm{Spec}}(k) are affine. This is true because X has dimension 0 and is Noetherian, hence is a finite discrete set, and each point gives an affine open, so X is a finite disjoint union of affines hence affine. Noether normalization forces X to be the spectrum of a finite k-algebra A. This algebra is a product of finite separable field extensions of k. Thus, an unramified quasi-compact morphism to \mathop{\mathrm{Spec}}(k) corresponds to a finite number of finite separable field extensions of k. In particular, an unramified morphism with a connected source and a one point target is forced to be a finite separable field extension. As we will see later, X \to \mathop{\mathrm{Spec}}(k) is étale if and only if it is unramified. Thus, in this case at least, we obtain a very easy description of the étale topology of a scheme. Of course, the cohomology of this topology is another story.
41.8 Examples of unramified morphisms
Here are a few examples.
Example 41.8.2. Property (3) in Theorem 41.4.1 gives us a canonical source of examples for unramified morphisms. Fix a ring R and an integer n. Let I = (g_1, \ldots , g_ m) be an ideal in R[x_1, \ldots , x_ n]. Let \mathfrak q \subset R[x_1, \ldots , x_ n] be a prime. Assume I \subset \mathfrak q and that the matrix
has rank n. Then the morphism f : Z = \mathop{\mathrm{Spec}}(R[x_1, \ldots , x_ n]/I) \to \mathop{\mathrm{Spec}}(R) is unramified at the point x \in Z \subset \mathbf{A}^ n_ R corresponding to \mathfrak q. Clearly we must have m \geq n. In the extreme case m = n, i.e., the differential of the map \mathbf{A}^ n_ R \to \mathbf{A}^ n_ R defined by the g_ i's is an isomorphism of the tangent spaces, then f is also flat x and, hence, is an étale map (see Algebra, Definition 10.137.6, Lemma 10.137.7 and Example 10.137.8).
Example 41.8.3. Fix an extension of number fields L/K with rings of integers \mathcal{O}_ L and \mathcal{O}_ K. The injection K \to L defines a morphism f : \mathop{\mathrm{Spec}}(\mathcal{O}_ L) \to \mathop{\mathrm{Spec}}(\mathcal{O}_ K). As discussed above, the points where f is unramified in our sense correspond to the set of points where f is unramified in the conventional sense. In the conventional sense, the locus of ramification in \mathop{\mathrm{Spec}}(\mathcal{O}_ L) can be defined by vanishing set of the different; this is an ideal in \mathcal{O}_ L. In fact, the different is nothing but the annihilator of the module \Omega _{\mathcal{O}_ L/\mathcal{O}_ K}. Similarly, the discriminant is an ideal in \mathcal{O}_ K, namely it is the norm of the different. The vanishing set of the discriminant is precisely the set of points of K which ramify in L. Thus, denoting by X the complement of the closed subset defined by the different in \mathop{\mathrm{Spec}}(\mathcal{O}_ L), we obtain a morphism X \to \mathop{\mathrm{Spec}}(\mathcal{O}_ K) which is unramified. Furthermore, this morphism is also flat, as any local homomorphism of discrete valuation rings is flat, and hence this morphism is actually étale. If L/K is finite Galois, then denoting by Y the complement of the closed subset defined by the discriminant in \mathop{\mathrm{Spec}}(\mathcal{O}_ K), we see that we get even a finite étale morphism X \to Y. Thus, this is an example of a finite étale covering.
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