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 $X \to \mathop{\mathrm{Spec}}(\mathcal{O}_ L)$ 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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Comment #5912 by Jacksyn on