
Lemma 10.147.3. Properties of unramified and G-unramified ring maps.

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

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

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

4. 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.

5. An étale ring map is G-unramified and unramified.

6. 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 $\mathfrak q$.

7. 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 $\mathfrak q$.

8. 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 $\mathfrak q$.

9. 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 $\mathfrak q$.

10. 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).

11. 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).

12. 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 $S = R \otimes _{R_0} S_0$.

13. 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 10.130.12 and 10.13.2.

Ad (2). Follows from Lemmas 10.130.7 and 10.13.2.

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 10.130.5, 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 10.141.1.

Ad (6). In this case $\Omega _{S/R}$ is a finite $S$-module (see Lemma 10.130.16) and hence there exists a $g \in S$, $g \not\in \mathfrak q$ such that $(\Omega _{S/R})_ g = 0$. By Lemma 10.130.8 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 10.19.1) 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 10.130.12. 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 10.22.2 and 10.130.8.

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. $\square$

## Comments (2)

Comment #1975 by Matthieu Romagny on

Typo in proof : Ad (8) & (9)

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