## 29.35 Unramified morphisms

We briefly discuss unramified morphisms before the (perhaps) more interesting class of étale morphisms. Recall that a ring map $R \to A$ is *unramified* if it is of finite type and $\Omega _{A/R} = 0$ (this is the definition of [Henselian]). A ring map $R \to A$ is called *G-unramified* if it is of finite presentation and $\Omega _{A/R} = 0$ (this is the definition of [EGA]). See Algebra, Definition 10.151.1.

Definition 29.35.1. Let $f : X \to S$ be a morphism of schemes.

We say that $f$ is *unramified at $x \in X$* if there exists an affine open neighbourhood $\mathop{\mathrm{Spec}}(A) = U \subset X$ of $x$ and affine open $\mathop{\mathrm{Spec}}(R) = V \subset S$ with $f(U) \subset V$ such that the induced ring map $R \to A$ is unramified.

We say that $f$ is *G-unramified at $x \in X$* if there exists an affine open neighbourhood $\mathop{\mathrm{Spec}}(A) = U \subset X$ of $x$ and affine open $\mathop{\mathrm{Spec}}(R) = V \subset S$ with $f(U) \subset V$ such that the induced ring map $R \to A$ is G-unramified.

We say that $f$ is *unramified* if it is unramified at every point of $X$.

We say that $f$ is *G-unramified* if it is G-unramified at every point of $X$.

Note that a G-unramified morphism is unramified. Hence any result for unramified morphisms implies the corresponding result for G-unramified morphisms. Moreover, if $S$ is locally Noetherian then there is no difference between G-unramified and unramified morphisms, see Lemma 29.35.6. A pleasing feature of this definition is that the set of points where a morphism is unramified (resp. G-unramified) is automatically open.

Lemma 29.35.2. Let $f : X \to S$ be a morphism of schemes. Then

$f$ is unramified if and only if $f$ is locally of finite type and $\Omega _{X/S} = 0$, and

$f$ is G-unramified if and only if $f$ is locally of finite presentation and $\Omega _{X/S} = 0$.

**Proof.**
By definition a ring map $R \to A$ is unramified (resp. G-unramified) if and only if it is of finite type (resp. finite presentation) and $\Omega _{A/R} = 0$. Hence the lemma follows directly from the definitions and Lemma 29.32.5.
$\square$

Note that there is no separation or quasi-compactness hypotheses in the definition. Hence the question of being unramified is local in nature on the source. Here is the precise result.

Lemma 29.35.3. Let $f : X \to S$ be a morphism of schemes. The following are equivalent

The morphism $f$ is unramified (resp. G-unramified).

For every affine open $U \subset X$, $V \subset S$ with $f(U) \subset V$ the ring map $\mathcal{O}_ S(V) \to \mathcal{O}_ X(U)$ is unramified (resp. G-unramified).

There exists an open covering $S = \bigcup _{j \in J} V_ j$ and open coverings $f^{-1}(V_ j) = \bigcup _{i \in I_ j} U_ i$ such that each of the morphisms $U_ i \to V_ j$, $j\in J, i\in I_ j$ is unramified (resp. G-unramified).

There exists an affine open covering $S = \bigcup _{j \in J} V_ j$ and affine open coverings $f^{-1}(V_ j) = \bigcup _{i \in I_ j} U_ i$ such that the ring map $\mathcal{O}_ S(V_ j) \to \mathcal{O}_ X(U_ i)$ is unramified (resp. G-unramified), for all $j\in J, i\in I_ j$.

Moreover, if $f$ is unramified (resp. G-unramified) then for any open subschemes $U \subset X$, $V \subset S$ with $f(U) \subset V$ the restriction $f|_ U : U \to V$ is unramified (resp. G-unramified).

**Proof.**
This follows from Lemma 29.14.3 if we show that the property “$R \to A$ is unramified” is local. We check conditions (a), (b) and (c) of Definition 29.14.1. These properties are proved in Algebra, Lemma 10.151.3.
$\square$

Lemma 29.35.4. The composition of two morphisms which are unramified is unramified. The same holds for G-unramified morphisms.

**Proof.**
The proof of Lemma 29.35.3 shows that being unramified (resp. G-unramified) is a local property of ring maps. Hence the first statement of the lemma follows from Lemma 29.14.5 combined with the fact that being unramified (resp. G-unramified) is a property of ring maps that is stable under composition, see Algebra, Lemma 10.151.3.
$\square$

Lemma 29.35.5. The base change of a morphism which is unramified is unramified. The same holds for G-unramified morphisms.

**Proof.**
The proof of Lemma 29.35.3 shows that being unramified (resp. G-unramified) is a local property of ring maps. Hence the lemma follows from Lemma 29.14.6 combined with the fact that being unramified (resp. G-unramified) is a property of ring maps that is stable under base change, see Algebra, Lemma 10.151.3.
$\square$

Lemma 29.35.6. Let $f : X \to S$ be a morphism of schemes. Assume $S$ is locally Noetherian. Then $f$ is unramified if and only if $f$ is G-unramified.

**Proof.**
Follows from the definitions and Lemma 29.21.9.
$\square$

Lemma 29.35.7. Any open immersion is G-unramified.

**Proof.**
This is true because an open immersion is a local isomorphism.
$\square$

Lemma 29.35.8. A closed immersion $i : Z \to X$ is unramified. It is G-unramified if and only if the associated quasi-coherent sheaf of ideals $\mathcal{I} = \mathop{\mathrm{Ker}}(\mathcal{O}_ X \to i_*\mathcal{O}_ Z)$ is of finite type (as an $\mathcal{O}_ X$-module).

**Proof.**
Follows from Lemma 29.21.7 and Algebra, Lemma 10.151.3.
$\square$

Lemma 29.35.9. An unramified morphism is locally of finite type. A G-unramified morphism is locally of finite presentation.

**Proof.**
An unramified ring map is of finite type by definition. A G-unramified ring map is of finite presentation by definition.
$\square$

Lemma 29.35.10. Let $f : X \to S$ be a morphism of schemes. If $f$ is unramified at $x$ then $f$ is quasi-finite at $x$. In particular, an unramified morphism is locally quasi-finite.

**Proof.**
See Algebra, Lemma 10.151.6.
$\square$

Lemma 29.35.11. Fibres of unramified morphisms.

Let $X$ be a scheme over a field $k$. The structure morphism $X \to \mathop{\mathrm{Spec}}(k)$ is unramified if and only if $X$ is a disjoint union of spectra of finite separable field extensions of $k$.

If $f : X \to S$ is an unramified morphism then for every $s \in S$ the fibre $X_ s$ is a disjoint union of spectra of finite separable field extensions of $\kappa (s)$.

**Proof.**
Part (2) follows from part (1) and Lemma 29.35.5. Let us prove part (1). We first use Algebra, Lemma 10.151.7. This lemma implies that if $X$ is a disjoint union of spectra of finite separable field extensions of $k$ then $X \to \mathop{\mathrm{Spec}}(k)$ is unramified. Conversely, suppose that $X \to \mathop{\mathrm{Spec}}(k)$ is unramified. By Algebra, Lemma 10.151.5 for every $x \in X$ the residue field extension $k \subset \kappa (x)$ is finite separable. Since $X \to \mathop{\mathrm{Spec}}(k)$ is locally quasi-finite (Lemma 29.35.10) we see that all points of $X$ are isolated closed points, see Lemma 29.20.6. Thus $X$ is a discrete space, in particular the disjoint union of the spectra of its local rings. By Algebra, Lemma 10.151.5 again these local rings are fields, and we win.
$\square$

The following lemma characterizes an unramified morphisms as morphisms locally of finite type with unramified fibres.

Lemma 29.35.12. Let $f : X \to S$ be a morphism of schemes.

If $f$ is unramified then for any $x \in X$ the field extension $\kappa (f(x)) \subset \kappa (x)$ is finite separable.

If $f$ is locally of finite type, and for every $s \in S$ the fibre $X_ s$ is a disjoint union of spectra of finite separable field extensions of $\kappa (s)$ then $f$ is unramified.

If $f$ is locally of finite presentation, and for every $s \in S$ the fibre $X_ s$ is a disjoint union of spectra of finite separable field extensions of $\kappa (s)$ then $f$ is G-unramified.

**Proof.**
Follows from Algebra, Lemmas 10.151.5 and 10.151.7.
$\square$

Here is a characterization of unramified morphisms in terms of the diagonal morphism.

Lemma 29.35.13. Let $f : X \to S$ be a morphism.

If $f$ is unramified, then the diagonal morphism $\Delta : X \to X \times _ S X$ is an open immersion.

If $f$ is locally of finite type and $\Delta $ is an open immersion, then $f$ is unramified.

If $f$ is locally of finite presentation and $\Delta $ is an open immersion, then $f$ is G-unramified.

**Proof.**
The first statement follows from Algebra, Lemma 10.151.4. The second statement from the fact that $\Omega _{X/S}$ is the conormal sheaf of the diagonal morphism (Lemma 29.32.7) and hence clearly zero if $\Delta $ is an open immersion.
$\square$

Lemma 29.35.14. Let $f : X \to S$ be a morphism of schemes. Let $x \in X$. Set $s = f(x)$. Assume $f$ is locally of finite type (resp. locally of finite presentation). The following are equivalent:

The morphism $f$ is unramified (resp. G-unramified) at $x$.

The fibre $X_ s$ is unramified over $\kappa (s)$ at $x$.

The $\mathcal{O}_{X, x}$-module $\Omega _{X/S, x}$ is zero.

The $\mathcal{O}_{X_ s, x}$-module $\Omega _{X_ s/s, x}$ is zero.

The $\kappa (x)$-vector space

\[ \Omega _{X_ s/s, x} \otimes _{\mathcal{O}_{X_ s, x}} \kappa (x) = \Omega _{X/S, x} \otimes _{\mathcal{O}_{X, x}} \kappa (x) \]

is zero.

We have $\mathfrak m_ s\mathcal{O}_{X, x} = \mathfrak m_ x$ and the field extension $\kappa (s) \subset \kappa (x)$ is finite separable.

**Proof.**
Note that if $f$ is unramified at $x$, then we see that $\Omega _{X/S} = 0$ in a neighbourhood of $x$ by the definitions and the results on modules of differentials in Section 29.32. Hence (1) implies (3) and the vanishing of the right hand vector space in (5). It also implies (2) because by Lemma 29.32.10 the module of differentials $\Omega _{X_ s/s}$ of the fibre $X_ s$ over $\kappa (s)$ is the pullback of the module of differentials $\Omega _{X/S}$ of $X$ over $S$. This fact on modules of differentials also implies the displayed equality of vector spaces in part (4). By Lemma 29.32.12 the modules $\Omega _{X/S, x}$ and $\Omega _{X_ s/s, x}$ are of finite type. Hence the modules $\Omega _{X/S, x}$ and $\Omega _{X_ s/s, x}$ are zero if and only if the corresponding $\kappa (x)$-vector space in (4) is zero by Nakayama's Lemma (Algebra, Lemma 10.20.1). This in particular shows that (3), (4) and (5) are equivalent. The support of $\Omega _{X/S}$ is closed in $X$, see Modules, Lemma 17.9.6. Assumption (3) implies that $x$ is not in the support. Hence $\Omega _{X/S}$ is zero in a neighbourhood of $x$, which implies (1). The equivalence of (1) and (3) applied to $X_ s \to s$ implies the equivalence of (2) and (4). At this point we have seen that (1) – (5) are equivalent.

Alternatively you can use Algebra, Lemma 10.151.3 to see the equivalence of (1) – (5) more directly.

The equivalence of (1) and (6) follows from Lemma 29.35.12. It also follows more directly from Algebra, Lemmas 10.151.5 and 10.151.7.
$\square$

Lemma 29.35.15. Let $f : X \to S$ be a morphism of schemes. Assume $f$ locally of finite type. Formation of the open set

\begin{align*} T & = \{ x \in X \mid X_{f(x)}\text{ is unramified over }\kappa (f(x))\text{ at }x\} \\ & = \{ x \in X \mid X\text{ is unramified over }S\text{ at }x\} \end{align*}

commutes with arbitrary base change: For any morphism $g : S' \to S$, consider the base change $f' : X' \to S'$ of $f$ and the projection $g' : X' \to X$. Then the corresponding set $T'$ for the morphism $f'$ is equal to $T' = (g')^{-1}(T)$. If $f$ is assumed locally of finite presentation then the same holds for the open set of points where $f$ is G-unramified.

**Proof.**
Let $s' \in S'$ be a point, and let $s = g(s')$. Then we have

\[ X'_{s'} = \mathop{\mathrm{Spec}}(\kappa (s')) \times _{\mathop{\mathrm{Spec}}(\kappa (s))} X_ s \]

In other words the fibres of the base change are the base changes of the fibres. In particular

\[ \Omega _{X_ s/s, x} \otimes _{\mathcal{O}_{X_ s, x}} \kappa (x') = \Omega _{X'_{s'}/s', x'} \otimes _{\mathcal{O}_{X'_{s'}, x'}} \kappa (x') \]

see Lemma 29.32.10. Whence $x' \in T'$ if and only if $x \in T$ by Lemma 29.35.14. The second part follows from the first because in that case $T$ is the (open) set of points where $f$ is G-unramified according to Lemma 29.35.14.
$\square$

Lemma 29.35.16. Let $f : X \to Y$ be a morphism of schemes over $S$.

If $X$ is unramified over $S$, then $f$ is unramified.

If $X$ is G-unramified over $S$ and $Y$ of finite type over $S$, then $f$ is G-unramified.

**Proof.**
Assume that $X$ is unramified over $S$. By Lemma 29.15.8 we see that $f$ is locally of finite type. By assumption we have $\Omega _{X/S} = 0$. Hence $\Omega _{X/Y} = 0$ by Lemma 29.32.9. Thus $f$ is unramified. If $X$ is G-unramified over $S$ and $Y$ of finite type over $S$, then by Lemma 29.21.11 we see that $f$ is locally of finite presentation and we conclude that $f$ is G-unramified.
$\square$

Lemma 29.35.17. Let $S$ be a scheme. Let $X$, $Y$ be schemes over $S$. Let $f, g : X \to Y$ be morphisms over $S$. Let $x \in X$. Assume that

the structure morphism $Y \to S$ is unramified,

$f(x) = g(x)$ in $Y$, say $y = f(x) = g(x)$, and

the induced maps $f^\sharp , g^\sharp : \kappa (y) \to \kappa (x)$ are equal.

Then there exists an open neighbourhood of $x$ in $X$ on which $f$ and $g$ are equal.

**Proof.**
Consider the morphism $(f, g) : X \to Y \times _ S Y$. By assumption (1) and Lemma 29.35.13 the inverse image of $\Delta _{Y/S}(Y)$ is open in $X$. And assumptions (2) and (3) imply that $x$ is in this open subset.
$\square$

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