**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.
$\square$

## Comments (2)

Comment #3258 by Dario Weißmann on

Comment #3354 by Johan on