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.

## 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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (2)

Comment #5912 by Jacksyn on

Comment #6110 by Johan on