## 10.151 Unramified ring maps

The definition of a G-unramified ring map is the one from EGA. The definition of an unramified ring map is the one from [Henselian].

Definition 10.151.1. Let $R \to S$ be a ring map.

We say $R \to S$ is *unramified* if $R \to S$ is of finite type and $\Omega _{S/R} = 0$.

We say $R \to S$ is *G-unramified* if $R \to S$ is of finite presentation and $\Omega _{S/R} = 0$.

Given a prime $\mathfrak q$ of $S$ we say that $S$ is *unramified at $\mathfrak q$* if there exists a $g \in S$, $g \not\in \mathfrak q$ such that $R \to S_ g$ is unramified.

Given a prime $\mathfrak q$ of $S$ we say that $S$ is *G-unramified at $\mathfrak q$* if there exists a $g \in S$, $g \not\in \mathfrak q$ such that $R \to S_ g$ is G-unramified.

Of course a G-unramified map is unramified.

Lemma 10.151.2. Let $R \to S$ be a ring map. The following are equivalent

$R \to S$ is formally unramified and of finite type, and

$R \to S$ is unramified.

Moreover, also the following are equivalent

$R \to S$ is formally unramified and of finite presentation, and

$R \to S$ is G-unramified.

**Proof.**
Follows from Lemma 10.148.2 and the definitions.
$\square$

Lemma 10.151.3. Properties of unramified and G-unramified ring maps.

The base change of an unramified ring map is unramified. The base change of a G-unramified ring map is G-unramified.

The composition of unramified ring maps is unramified. The composition of G-unramified ring maps is G-unramified.

Any principal localization $R \to R_ f$ is G-unramified and unramified.

If $I \subset R$ is an ideal, then $R \to R/I$ is unramified. If $I \subset R$ is a finitely generated ideal, then $R \to R/I$ is G-unramified.

An étale ring map is G-unramified and unramified.

If $R \to S$ is of finite type (resp. finite presentation), $\mathfrak q \subset S$ is a prime and $(\Omega _{S/R})_{\mathfrak q} = 0$, then $R \to S$ is unramified (resp. G-unramified) at $\mathfrak q$.

If $R \to S$ is of finite type (resp. finite presentation), $\mathfrak q \subset S$ is a prime and $\Omega _{S/R} \otimes _ S \kappa (\mathfrak q) = 0$, then $R \to S$ is unramified (resp. G-unramified) at $\mathfrak q$.

If $R \to S$ is of finite type (resp. finite presentation), $\mathfrak q \subset S$ is a prime lying over $\mathfrak p \subset R$ and $(\Omega _{S \otimes _ R \kappa (\mathfrak p)/\kappa (\mathfrak p)})_{\mathfrak q} = 0$, then $R \to S$ is unramified (resp. G-unramified) at $\mathfrak q$.

If $R \to S$ is of finite type (resp. presentation), $\mathfrak q \subset S$ is a prime lying over $\mathfrak p \subset R$ and $(\Omega _{S \otimes _ R \kappa (\mathfrak p)/\kappa (\mathfrak p)}) \otimes _{S \otimes _ R \kappa (\mathfrak p)} \kappa (\mathfrak q) = 0$, then $R \to S$ is unramified (resp. G-unramified) at $\mathfrak q$.

If $R \to S$ is a ring map, $g_1, \ldots , g_ m \in S$ generate the unit ideal and $R \to S_{g_ j}$ is unramified (resp. G-unramified) for $j = 1, \ldots , m$, then $R \to S$ is unramified (resp. G-unramified).

If $R \to S$ is a ring map which is unramified (resp. G-unramified) at every prime of $S$, then $R \to S$ is unramified (resp. G-unramified).

If $R \to S$ is G-unramified, then there exists a finite type $\mathbf{Z}$-algebra $R_0$ and a G-unramified ring map $R_0 \to S_0$ and a ring map $R_0 \to R$ such that $S = R \otimes _{R_0} S_0$.

If $R \to S$ is unramified, then there exists a finite type $\mathbf{Z}$-algebra $R_0$ and an unramified ring map $R_0 \to S_0$ and a ring map $R_0 \to R$ such that $S$ is a quotient of $R \otimes _{R_0} S_0$.

**Proof.**
We prove each point, in order.

Ad (1). Follows from Lemmas 10.131.12 and 10.14.2.

Ad (2). Follows from Lemmas 10.131.7 and 10.14.2.

Ad (3). Follows by direct computation of $\Omega _{R_ f/R}$ which we omit.

Ad (4). We have $\Omega _{(R/I)/R} = 0$, see Lemma 10.131.5, and the ring map $R \to R/I$ is of finite type. If $I$ is a finitely generated ideal then $R \to R/I$ is of finite presentation.

Ad (5). See discussion following Definition 10.143.1.

Ad (6). In this case $\Omega _{S/R}$ is a finite $S$-module (see Lemma 10.131.16) and hence there exists a $g \in S$, $g \not\in \mathfrak q$ such that $(\Omega _{S/R})_ g = 0$. By Lemma 10.131.8 this means that $\Omega _{S_ g/R} = 0$ and hence $R \to S_ g$ is unramified as desired.

Ad (7). Use Nakayama's lemma (Lemma 10.20.1) to see that the condition is equivalent to the condition of (6).

Ad (8) and (9). These are equivalent in the same manner that (6) and (7) are equivalent. Moreover $\Omega _{S \otimes _ R \kappa (\mathfrak p)/\kappa (\mathfrak p)} = \Omega _{S/R} \otimes _ S (S \otimes _ R \kappa (\mathfrak p))$ by Lemma 10.131.12. Hence we see that (9) is equivalent to (7) since the $\kappa (\mathfrak q)$ vector spaces in both are canonically isomorphic.

Ad (10). Follows from Lemmas 10.23.2 and 10.131.8.

Ad (11). Follows from (6) and (7) and the fact that the spectrum of $S$ is quasi-compact.

Ad (12). Write $S = R[x_1, \ldots , x_ n]/(g_1, \ldots , g_ m)$. As $\Omega _{S/R} = 0$ we can write

\[ \text{d}x_ i = \sum h_{ij}\text{d}g_ j + \sum a_{ijk}g_ j\text{d}x_ k \]

in $\Omega _{R[x_1, \ldots , x_ n]/R}$ for some $h_{ij}, a_{ijk} \in R[x_1, \ldots , x_ n]$. Choose a finitely generated $\mathbf{Z}$-subalgebra $R_0 \subset R$ containing all the coefficients of the polynomials $g_ i, h_{ij}, a_{ijk}$. Set $S_0 = R_0[x_1, \ldots , x_ n]/(g_1, \ldots , g_ m)$. This works.

Ad (13). Write $S = R[x_1, \ldots , x_ n]/I$. As $\Omega _{S/R} = 0$ we can write

\[ \text{d}x_ i = \sum h_{ij}\text{d}g_{ij} + \sum g'_{ik}\text{d}x_ k \]

in $\Omega _{R[x_1, \ldots , x_ n]/R}$ for some $h_{ij} \in R[x_1, \ldots , x_ n]$ and $g_{ij}, g'_{ik} \in I$. Choose a finitely generated $\mathbf{Z}$-subalgebra $R_0 \subset R$ containing all the coefficients of the polynomials $g_{ij}, h_{ij}, g'_{ik}$. Set $S_0 = R_0[x_1, \ldots , x_ n]/(g_{ij}, g'_{ik})$. This works.
$\square$

Lemma 10.151.4. Let $R \to S$ be a ring map. If $R \to S$ is unramified, then there exists an idempotent $e \in S \otimes _ R S$ such that $S \otimes _ R S \to S$ is isomorphic to $S \otimes _ R S \to (S \otimes _ R S)_ e$.

**Proof.**
Let $J = \mathop{\mathrm{Ker}}(S \otimes _ R S \to S)$. By assumption $J/J^2 = 0$, see Lemma 10.131.13. Since $S$ is of finite type over $R$ we see that $J$ is finitely generated, namely by $x_ i \otimes 1 - 1 \otimes x_ i$, where $x_ i$ generate $S$ over $R$. We win by Lemma 10.21.5.
$\square$

Lemma 10.151.5. Let $R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime lying over $\mathfrak p$ in $R$. If $S/R$ is unramified at $\mathfrak q$ then

we have $\mathfrak p S_{\mathfrak q} = \mathfrak qS_{\mathfrak q}$ is the maximal ideal of the local ring $S_{\mathfrak q}$, and

the field extension $\kappa (\mathfrak p) \subset \kappa (\mathfrak q)$ is finite separable.

**Proof.**
We may first replace $S$ by $S_ g$ for some $g \in S$, $g \not\in \mathfrak q$ and assume that $R \to S$ is unramified. The base change $S \otimes _ R \kappa (\mathfrak p)$ is unramified over $\kappa (\mathfrak p)$ by Lemma 10.151.3. By Lemma 10.140.3 it is smooth hence étale over $\kappa (\mathfrak p)$. Hence we see that $S \otimes _ R \kappa (\mathfrak p) = (R \setminus \mathfrak p)^{-1} S/\mathfrak pS$ is a product of finite separable field extensions of $\kappa (\mathfrak p)$ by Lemma 10.143.4. This implies the lemma.
$\square$

Lemma 10.151.6. Let $R \to S$ be a finite type ring map. Let $\mathfrak q$ be a prime of $S$. If $R \to S$ is unramified at $\mathfrak q$ then $R \to S$ is quasi-finite at $\mathfrak q$. In particular, an unramified ring map is quasi-finite.

**Proof.**
An unramified ring map is of finite type. Thus it is clear that the second statement follows from the first. To see the first statement apply the characterization of Lemma 10.122.2 part (2) using Lemma 10.151.5.
$\square$

Lemma 10.151.7. Let $R \to S$ be a ring map. Let $\mathfrak q$ be a prime of $S$ lying over a prime $\mathfrak p$ of $R$. If

$R \to S$ is of finite type,

$\mathfrak p S_{\mathfrak q}$ is the maximal ideal of the local ring $S_{\mathfrak q}$, and

the field extension $\kappa (\mathfrak p) \subset \kappa (\mathfrak q)$ is finite separable,

then $R \to S$ is unramified at $\mathfrak q$.

**Proof.**
By Lemma 10.151.3 (8) it suffices to show that $\Omega _{S \otimes _ R \kappa (\mathfrak p) / \kappa (\mathfrak p)}$ is zero when localized at $\mathfrak q$. Hence we may replace $S$ by $S \otimes _ R \kappa (\mathfrak p)$ and $R$ by $\kappa (\mathfrak p)$. In other words, we may assume that $R = k$ is a field and $S$ is a finite type $k$-algebra. In this case the hypotheses imply that $S_{\mathfrak q} \cong \kappa (\mathfrak q)$. Thus $(\Omega _{S/k})_{\mathfrak q} = \Omega _{S_\mathfrak q/k} = \Omega _{\kappa (\mathfrak q)/k}$ is zero as desired (the first equality is Lemma 10.131.8).
$\square$

Lemma 10.151.8. Let $R \to S$ be a ring map. The following are equivalent

$R \to S$ is étale,

$R \to S$ is flat and G-unramified, and

$R \to S$ is flat, unramified, and of finite presentation.

**Proof.**
Parts (2) and (3) are equivalent by definition. The implication (1) $\Rightarrow $ (3) follows from the fact that étale ring maps are of finite presentation, Lemma 10.143.3 (flatness of étale maps), and Lemma 10.151.3 (étale maps are unramified). Conversely, the characterization of étale ring maps in Lemma 10.143.7 and the structure of unramified ring maps in Lemma 10.151.5 shows that (3) implies (1). (This uses that $R \to S$ is étale if $R \to S$ is étale at every prime $\mathfrak q \subset S$, see Lemma 10.143.3.)
$\square$

Lemma 10.151.9. Let $k$ be a field. Let

\[ \varphi : k[x_1, \ldots , x_ n] \to A, \quad x_ i \longmapsto a_ i \]

be a finite type ring map. Then $\varphi $ is étale if and only if we have the following two conditions: (a) the local rings of $A$ at maximal ideals have dimension $n$, and (b) the elements $\text{d}(a_1), \ldots , \text{d}(a_ n)$ generate $\Omega _{A/k}$ as an $A$-module.

**Proof.**
Assume (a) and (b). Condition (b) implies that $\Omega _{A/k[x_1, \ldots , x_ n]} = 0$ and hence $\varphi $ is unramified. Thus it suffices to prove that $\varphi $ is flat, see Lemma 10.151.8. Let $\mathfrak m \subset A$ be a maximal ideal. Set $X = \mathop{\mathrm{Spec}}(A)$ and denote $x \in X$ the closed point corresponding to $\mathfrak m$. Then $\dim (A_\mathfrak m)$ is $\dim _ x X$, see Lemma 10.114.6. Thus by Lemma 10.140.3 we see that if (a) and (b) hold, then $A_\mathfrak m$ is a regular local ring for every maximal ideal $\mathfrak m$. Then $k[x_1, \ldots , x_ n]_{\varphi ^{-1}(\mathfrak m)} \to A_\mathfrak m$ is flat by Lemma 10.128.1 (and the fact that a regular local ring is CM, see Lemma 10.106.3). Thus $\varphi $ is flat by Lemma 10.39.18.

Assume $\varphi $ is étale. Then $\Omega _{A/k[x_1, \ldots , x_ n]} = 0$ and hence (b) holds. On the other hand, étale ring maps are flat (Lemma 10.143.3) and quasi-finite (Lemma 10.143.6). Hence for every maximal ideal $\mathfrak m$ of $A$ we my apply Lemma 10.112.7 to $k[x_1, \ldots , x_ n]_{\varphi ^{-1}(\mathfrak m)} \to A_\mathfrak m$ to see that $\dim (A_\mathfrak m) = n$ and hence (a) holds.
$\square$

## Comments (0)