
## 10.147 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 .

Definition 10.147.1. Let $R \to S$ be a ring map.

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

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

3. 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.

4. 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 10.147.2. Let $R \to S$ be a ring map. The following are equivalent

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

2. $R \to S$ is unramified.

Moreover, also the following are equivalent

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

2. $R \to S$ is G-unramified.

Proof. Follows from Lemma 10.144.2 and the definitions. $\square$

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. $\square$

Lemma 10.147.4. 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 10.130.13. 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 10.20.5. $\square$

Lemma 10.147.5. 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

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

2. 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 10.147.3. By Lemma 10.138.3 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 10.141.4. This implies the lemma. $\square$

Lemma 10.147.6. 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 10.121.2 part (2) using Lemma 10.147.5. $\square$

Lemma 10.147.7. 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

1. $R \to S$ is of finite type,

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

3. 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 10.147.3 (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 $S_{\mathfrak q} \cong \kappa (\mathfrak q)$ and hence $S = \kappa (\mathfrak q) \times S'$ (see Lemma 10.121.1). Hence $(\Omega _{S/k})_{\mathfrak q} = \Omega _{\kappa (\mathfrak q)/k}$ which is zero as desired. $\square$

Lemma 10.147.8. Let $R \to S$ be a ring map. The following are equivalent

1. $R \to S$ is étale,

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

3. $R \to S$ 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 10.141.3 (flatness of étale maps), and Lemma 10.147.3 (étale maps are unramified). Conversely, the characterization of étale ring maps in Lemma 10.141.7 and the structure of unramified ring maps in Lemma 10.147.5 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 10.141.3.) $\square$

Proposition 10.147.9. Let $R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime. If $R \to S$ is unramified at $\mathfrak q$, then there exist

1. a $g \in S$, $g \not\in \mathfrak q$,

2. a standard étale ring map $R \to S'$, and

3. a surjective $R$-algebra map $S' \to S_ g$.

Proof. This proof is the “same” as the proof of Proposition 10.141.16. The proof is a little roundabout and there may be ways to shorten it.

Step 1. By Definition 10.147.1 there exists a $g \in S$, $g \not\in \mathfrak q$ such that $R \to S_ g$ is unramified. Thus we may assume that $S$ is unramified over $R$.

Step 2. By Lemma 10.147.3 there exists an unramified ring map $R_0 \to S_0$ with $R_0$ of finite type over $\mathbf{Z}$, and a ring map $R_0 \to R$ such that $S$ is a quotient of $R \otimes _{R_0} S_0$. Denote $\mathfrak q_0$ the prime of $S_0$ corresponding to $\mathfrak q$. If we show the result for $(R_0 \to S_0, \mathfrak q_0)$ then the result follows for $(R \to S, \mathfrak q)$ by base change. Hence we may assume that $R$ is Noetherian.

Step 3. Note that $R \to S$ is quasi-finite by Lemma 10.147.6. By Lemma 10.122.14 there exists a finite ring map $R \to S'$, an $R$-algebra map $S' \to S$, an element $g' \in S'$ such that $g' \not\in \mathfrak q$ such that $S' \to S$ induces an isomorphism $S'_{g'} \cong S_{g'}$. (Note that $S'$ may not be unramified over $R$.) Thus we may assume that (a) $R$ is Noetherian, (b) $R \to S$ is finite and (c) $R \to S$ is unramified at $\mathfrak q$ (but no longer necessarily unramified at all primes).

Step 4. Let $\mathfrak p \subset R$ be the prime corresponding to $\mathfrak q$. Consider the fibre ring $S \otimes _ R \kappa (\mathfrak p)$. This is a finite algebra over $\kappa (\mathfrak p)$. Hence it is Artinian (see Lemma 10.52.2) and so a finite product of local rings

$S \otimes _ R \kappa (\mathfrak p) = \prod \nolimits _{i = 1}^ n A_ i$

see Proposition 10.59.6. One of the factors, say $A_1$, is the local ring $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}$ which is isomorphic to $\kappa (\mathfrak q)$, see Lemma 10.147.5. The other factors correspond to the other primes, say $\mathfrak q_2, \ldots , \mathfrak q_ n$ of $S$ lying over $\mathfrak p$.

Step 5. We may choose a nonzero element $\alpha \in \kappa (\mathfrak q)$ which generates the finite separable field extension $\kappa (\mathfrak p) \subset \kappa (\mathfrak q)$ (so even if the field extension is trivial we do not allow $\alpha = 0$). Note that for any $\lambda \in \kappa (\mathfrak p)^*$ the element $\lambda \alpha$ also generates $\kappa (\mathfrak q)$ over $\kappa (\mathfrak p)$. Consider the element

$\overline{t} = (\alpha , 0, \ldots , 0) \in \prod \nolimits _{i = 1}^ n A_ i = S \otimes _ R \kappa (\mathfrak p).$

After possibly replacing $\alpha$ by $\lambda \alpha$ as above we may assume that $\overline{t}$ is the image of $t \in S$. Let $I \subset R[x]$ be the kernel of the $R$-algebra map $R[x] \to S$ which maps $x$ to $t$. Set $S' = R[x]/I$, so $S' \subset S$. Here is a diagram

$\xymatrix{ R[x] \ar[r] & S' \ar[r] & S \\ R \ar[u] \ar[ru] \ar[rru] & & }$

By construction the primes $\mathfrak q_ j$, $j \geq 2$ of $S$ all lie over the prime $(\mathfrak p, x)$ of $R[x]$, whereas the prime $\mathfrak q$ lies over a different prime of $R[x]$ because $\alpha \not= 0$.

Step 6. Denote $\mathfrak q' \subset S'$ the prime of $S'$ corresponding to $\mathfrak q$. By the above $\mathfrak q$ is the only prime of $S$ lying over $\mathfrak q'$. Thus we see that $S_{\mathfrak q} = S_{\mathfrak q'}$, see Lemma 10.40.11 (we have going up for $S' \to S$ by Lemma 10.35.22 since $S' \to S$ is finite as $R \to S$ is finite). It follows that $S'_{\mathfrak q'} \to S_{\mathfrak q}$ is finite and injective as the localization of the finite injective ring map $S' \to S$. Consider the maps of local rings

$R_{\mathfrak p} \to S'_{\mathfrak q'} \to S_{\mathfrak q}$

The second map is finite and injective. We have $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q} = \kappa (\mathfrak q)$, see Lemma 10.147.5. Hence a fortiori $S_{\mathfrak q}/\mathfrak q'S_{\mathfrak q} = \kappa (\mathfrak q)$. Since

$\kappa (\mathfrak p) \subset \kappa (\mathfrak q') \subset \kappa (\mathfrak q)$

and since $\alpha$ is in the image of $\kappa (\mathfrak q')$ in $\kappa (\mathfrak q)$ we conclude that $\kappa (\mathfrak q') = \kappa (\mathfrak q)$. Hence by Nakayama's Lemma 10.19.1 applied to the $S'_{\mathfrak q'}$-module map $S'_{\mathfrak q'} \to S_{\mathfrak q}$, the map $S'_{\mathfrak q'} \to S_{\mathfrak q}$ is surjective. In other words, $S'_{\mathfrak q'} \cong S_{\mathfrak q}$.

Step 7. By Lemma 10.125.7 there exist $g \in S$, $g \not\in \mathfrak q$ and $g' \in S'$, $g' \not\in \mathfrak q'$ such that $S'_{g'} \cong S_ g$. As $R$ is Noetherian the ring $S'$ is finite over $R$ because it is an $R$-submodule of the finite $R$-module $S$. Hence after replacing $S$ by $S'$ we may assume that (a) $R$ is Noetherian, (b) $S$ finite over $R$, (c) $S$ is unramified over $R$ at $\mathfrak q$, and (d) $S = R[x]/I$.

Step 8. Consider the ring $S \otimes _ R \kappa (\mathfrak p) = \kappa (\mathfrak p)[x]/\overline{I}$ where $\overline{I} = I \cdot \kappa (\mathfrak p)[x]$ is the ideal generated by $I$ in $\kappa (\mathfrak p)[x]$. As $\kappa (\mathfrak p)[x]$ is a PID we know that $\overline{I} = (\overline{h})$ for some monic $\overline{h} \in \kappa (\mathfrak p)$. After replacing $\overline{h}$ by $\lambda \cdot \overline{h}$ for some $\lambda \in \kappa (\mathfrak p)$ we may assume that $\overline{h}$ is the image of some $h \in R[x]$. (The problem is that we do not know if we may choose $h$ monic.) Also, as in Step 4 we know that $S \otimes _ R \kappa (\mathfrak p) = A_1 \times \ldots \times A_ n$ with $A_1 = \kappa (\mathfrak q)$ a finite separable extension of $\kappa (\mathfrak p)$ and $A_2, \ldots , A_ n$ local. This implies that

$\overline{h} = \overline{h}_1 \overline{h}_2^{e_2} \ldots \overline{h}_ n^{e_ n}$

for certain pairwise coprime irreducible monic polynomials $\overline{h}_ i \in \kappa (\mathfrak p)[x]$ and certain $e_2, \ldots , e_ n \geq 1$. Here the numbering is chosen so that $A_ i = \kappa (\mathfrak p)[x]/(\overline{h}_ i^{e_ i})$ as $\kappa (\mathfrak p)[x]$-algebras. Note that $\overline{h}_1$ is the minimal polynomial of $\alpha \in \kappa (\mathfrak q)$ and hence is a separable polynomial (its derivative is prime to itself).

Step 9. Let $m \in I$ be a monic element; such an element exists because the ring extension $R \to R[x]/I$ is finite hence integral. Denote $\overline{m}$ the image in $\kappa (\mathfrak p)[x]$. We may factor

$\overline{m} = \overline{k} \overline{h}_1^{d_1} \overline{h}_2^{d_2} \ldots \overline{h}_ n^{d_ n}$

for some $d_1 \geq 1$, $d_ j \geq e_ j$, $j = 2, \ldots , n$ and $\overline{k} \in \kappa (\mathfrak p)[x]$ prime to all the $\overline{h}_ i$. Set $f = m^ l + h$ where $l \deg (m) > \deg (h)$, and $l \geq 2$. Then $f$ is monic as a polynomial over $R$. Also, the image $\overline{f}$ of $f$ in $\kappa (\mathfrak p)[x]$ factors as

$\overline{f} = \overline{h}_1 \overline{h}_2^{e_2} \ldots \overline{h}_ n^{e_ n} + \overline{k}^ l \overline{h}_1^{ld_1} \overline{h}_2^{ld_2} \ldots \overline{h}_ n^{ld_ n} = \overline{h}_1(\overline{h}_2^{e_2} \ldots \overline{h}_ n^{e_ n} + \overline{k}^ l \overline{h}_1^{ld_1 - 1} \overline{h}_2^{ld_2} \ldots \overline{h}_ n^{ld_ n}) = \overline{h}_1 \overline{w}$

with $\overline{w}$ a polynomial relatively prime to $\overline{h}_1$. Set $g = f'$ (the derivative with respect to $x$).

Step 10. The ring map $R[x] \to S = R[x]/I$ has the properties: (1) it maps $f$ to zero, and (2) it maps $g$ to an element of $S \setminus \mathfrak q$. The first assertion is clear since $f$ is an element of $I$. For the second assertion we just have to show that $g$ does not map to zero in $\kappa (\mathfrak q) = \kappa (\mathfrak p)[x]/(\overline{h}_1)$. The image of $g$ in $\kappa (\mathfrak p)[x]$ is the derivative of $\overline{f}$. Thus (2) is clear because

$\overline{g} = \frac{\text{d}\overline{f}}{\text{d}x} = \overline{w}\frac{\text{d}\overline{h}_1}{\text{d}x} + \overline{h}_1\frac{\text{d}\overline{w}}{\text{d}x},$

$\overline{w}$ is prime to $\overline{h}_1$ and $\overline{h}_1$ is separable.

Step 11. We conclude that $\varphi : R[x]/(f) \to S$ is a surjective ring map, $R[x]_ g/(f)$ is étale over $R$ (because it is standard étale, see Lemma 10.141.14) and $\varphi (g) \not\in \mathfrak q$. Thus the map $(R[x]/(f))_ g \to S_{\varphi (g)}$ is the desired surjection. $\square$

Lemma 10.147.10. Let $R \to S$ be a ring map. Let $\mathfrak q$ be a prime of $S$ lying over $\mathfrak p \subset R$. Assume that $R \to S$ is of finite type and unramified at $\mathfrak q$. Then there exist

1. an étale ring map $R \to R'$,

2. a prime $\mathfrak p' \subset R'$ lying over $\mathfrak p$.

3. a product decomposition

$R' \otimes _ R S = A \times B$

with the following properties

1. $R' \to A$ is surjective, and

2. $\mathfrak p'A$ is a prime of $A$ lying over $\mathfrak p'$ and over $\mathfrak q$.

Proof. We may replace $(R \to S, \mathfrak p, \mathfrak q)$ with any base change $(R' \to R'\otimes _ R S, \mathfrak p', \mathfrak q')$ by an étale ring map $R \to R'$ with a prime $\mathfrak p'$ lying over $\mathfrak p$, and a choice of $\mathfrak q'$ lying over both $\mathfrak q$ and $\mathfrak p'$. Note also that given $R \to R'$ and $\mathfrak p'$ a suitable $\mathfrak q'$ can always be found.

The assumption that $R \to S$ is of finite type means that we may apply Lemma 10.141.23. Thus we may assume that $S = A_1 \times \ldots \times A_ n \times B$, that each $R \to A_ i$ is finite with exactly one prime $\mathfrak r_ i$ lying over $\mathfrak p$ such that $\kappa (\mathfrak p) \subset \kappa (\mathfrak r_ i)$ is purely inseparable and that $R \to B$ is not quasi-finite at any prime lying over $\mathfrak p$. Then clearly $\mathfrak q = \mathfrak r_ i$ for some $i$, since an unramified morphism is quasi-finite (see Lemma 10.147.6). Say $\mathfrak q = \mathfrak r_1$. By Lemma 10.147.5 we see that $\kappa (\mathfrak p) \subset \kappa (\mathfrak r_1)$ is separable hence the trivial field extension, and that $\mathfrak p(A_1)_{\mathfrak r_1}$ is the maximal ideal. Also, by Lemma 10.40.11 (which applies to $R \to A_1$ because a finite ring map satisfies going up by Lemma 10.35.22) we have $(A_1)_{\mathfrak r_1} = (A_1)_{\mathfrak p}$. It follows from Nakayama's Lemma 10.19.1 that the map of local rings $R_{\mathfrak p} \to (A_1)_{\mathfrak p} = (A_1)_{\mathfrak r_1}$ is surjective. Since $A_1$ is finite over $R$ we see that there exists a $f \in R$, $f \not\in \mathfrak p$ such that $R_ f \to (A_1)_ f$ is surjective. After replacing $R$ by $R_ f$ we win. $\square$

Lemma 10.147.11. Let $R \to S$ be a ring map. Let $\mathfrak p$ be a prime of $R$. If $R \to S$ is unramified then there exist

1. an étale ring map $R \to R'$,

2. a prime $\mathfrak p' \subset R'$ lying over $\mathfrak p$.

3. a product decomposition

$R' \otimes _ R S = A_1 \times \ldots \times A_ n \times B$

with the following properties

1. $R' \to A_ i$ is surjective,

2. $\mathfrak p'A_ i$ is a prime of $A_ i$ lying over $\mathfrak p'$, and

3. there is no prime of $B$ lying over $\mathfrak p'$.

Proof. We may apply Lemma 10.141.23. Thus, after an étale base change, we may assume that $S = A_1 \times \ldots \times A_ n \times B$, that each $R \to A_ i$ is finite with exactly one prime $\mathfrak r_ i$ lying over $\mathfrak p$ such that $\kappa (\mathfrak p) \subset \kappa (\mathfrak r_ i)$ is purely inseparable, and that $R \to B$ is not quasi-finite at any prime lying over $\mathfrak p$. Since $R \to S$ is quasi-finite (see Lemma 10.147.6) we see there is no prime of $B$ lying over $\mathfrak p$. By Lemma 10.147.5 we see that $\kappa (\mathfrak p) \subset \kappa (\mathfrak r_ i)$ is separable hence the trivial field extension, and that $\mathfrak p(A_ i)_{\mathfrak r_ i}$ is the maximal ideal. Also, by Lemma 10.40.11 (which applies to $R \to A_ i$ because a finite ring map satisfies going up by Lemma 10.35.22) we have $(A_ i)_{\mathfrak r_ i} = (A_ i)_{\mathfrak p}$. It follows from Nakayama's Lemma 10.19.1 that the map of local rings $R_{\mathfrak p} \to (A_ i)_{\mathfrak p} = (A_ i)_{\mathfrak r_ i}$ is surjective. Since $A_ i$ is finite over $R$ we see that there exists a $f \in R$, $f \not\in \mathfrak p$ such that $R_ f \to (A_ i)_ f$ is surjective. After replacing $R$ by $R_ f$ we win. $\square$

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