## Tag `00U0`

## 10.141. Étale ring maps

An étale ring map is a smooth ring map whose relative dimension is equal to zero. This is the same as the following slightly more direct definition.

Definition 10.141.1. Let $R \to S$ be a ring map. We say $R \to S$ is

étaleif it is of finite presentation and the naive cotangent complex $\mathop{N\!L}\nolimits_{S/R}$ is quasi-isomorphic to zero. Given a prime $\mathfrak q$ of $S$ we say that $R \to S$ isétale at $\mathfrak q$if there exists a $g \in S$, $g \not \in \mathfrak q$ such that $R \to S_g$ is étale.In particular we see that $\Omega_{S/R} = 0$ if $S$ is étale over $R$. If $R \to S$ is smooth, then $R \to S$ is étale if and only if $\Omega_{S/R} = 0$. From our results on smooth ring maps we automatically get a whole host of results for étale maps. We summarize these in Lemma 10.141.3 below. But before we do so we prove that

anyétale ring map is standard smooth.Lemma 10.141.2. Any étale ring map is standard smooth. More precisely, if $R \to S$ is étale, then there exists a presentation $S = R[x_1, \ldots, x_n]/(f_1, \ldots, f_n)$ such that the image of $\det(\partial f_j/\partial x_i)$ is invertible in $S$.

Proof.Let $R \to S$ be étale. Choose a presentation $S = R[x_1, \ldots, x_n]/I$. As $R \to S$ is étale we know that $$ \text{d} : I/I^2 \longrightarrow \bigoplus\nolimits_{i = 1, \ldots, n} S\text{d}x_i $$ is an isomorphism, in particular $I/I^2$ is a free $S$-module. Thus by Lemma 10.134.6 we may assume (after possibly changing the presentation), that $I = (f_1, \ldots, f_c)$ such that the classes $f_i \bmod I^2$ form a basis of $I/I^2$. It follows immediately from the fact that the displayed map above is an isomorphism that $c = n$ and that $\det(\partial f_j/\partial x_i)$ is invertible in $S$. $\square$Lemma 10.141.3. Results on étale ring maps.

- The ring map $R \to R_f$ is étale for any ring $R$ and any $f \in R$.
- Compositions of étale ring maps are étale.
- A base change of an étale ring map is étale.
- The property of being étale is local: Given a ring map $R \to S$ and elements $g_1, \ldots, g_m \in S$ which generate the unit ideal such that $R \to S_{g_j}$ is étale for $j = 1, \ldots, m$ then $R \to S$ is étale.
- Given $R \to S$ of finite presentation, and a flat ring map $R \to R'$, set $S' = R' \otimes_R S$. The set of primes where $R' \to S'$ is étale is the inverse image via $\mathop{\mathrm{Spec}}(S') \to \mathop{\mathrm{Spec}}(S)$ of the set of primes where $R \to S$ is étale.
- An étale ring map is syntomic, in particular flat.
- If $S$ is finite type over a field $k$, then $S$ is étale over $k$ if and only if $\Omega_{S/k} = 0$.
- Any étale ring map $R \to S$ is the base change of an étale ring map $R_0 \to S_0$ with $R_0$ of finite type over $\mathbf{Z}$.
- Let $A = \mathop{\mathrm{colim}}\nolimits A_i$ be a filtered colimit of rings. Let $A \to B$ be an étale ring map. Then there exists an étale ring map $A_i \to B_i$ for some $i$ such that $B \cong A \otimes_{A_i} B_i$.
- Let $A$ be a ring. Let $S$ be a multiplicative subset of $A$. Let $S^{-1}A \to B'$ be étale. Then there exists an étale ring map $A \to B$ such that $B' \cong S^{-1}B$.

Proof.In each case we use the corresponding result for smooth ring maps with a small argument added to show that $\Omega_{S/R}$ is zero.Proof of (1). The ring map $R \to R_f$ is smooth and $\Omega_{R_f/R} = 0$.

Proof of (2). The composition $A \to C$ of smooth maps $A \to B$ and $B \to C$ is smooth, see Lemma 10.135.14. By Lemma 10.130.7 we see that $\Omega_{C/A}$ is zero as both $\Omega_{C/B}$ and $\Omega_{B/A}$ are zero.

Proof of (3). Let $R \to S$ be étale and $R \to R'$ be arbitrary. Then $R' \to S' = R' \otimes_R S$ is smooth, see Lemma 10.135.4. Since $\Omega_{S'/R'} = S' \otimes_S \Omega_{S/R}$ by Lemma 10.130.12 we conclude that $\Omega_{S'/R'} = 0$. Hence $R' \to S'$ is étale.

Proof of (4). Assume the hypotheses of (4). By Lemma 10.135.13 we see that $R \to S$ is smooth. We are also given that $\Omega_{S_{g_i}/R} = (\Omega_{S/R})_{g_i} = 0$ for all $i$. Then $\Omega_{S/R} = 0$, see Lemma 10.23.2.

Proof of (5). The result for smooth maps is Lemma 10.135.17. In the proof of that lemma we used that $\mathop{N\!L}\nolimits_{S/R} \otimes_S S'$ is homotopy equivalent to $\mathop{N\!L}\nolimits_{S'/R'}$. This reduces us to showing that if $M$ is a finitely presented $S$-module the set of primes $\mathfrak q'$ of $S'$ such that $(M \otimes_S S')_{\mathfrak q'} = 0$ is the inverse image of the set of primes $\mathfrak q$ of $S$ such that $M_{\mathfrak q} = 0$. This follows from Lemma 10.39.6.

Proof of (6). Follows directly from the corresponding result for smooth ring maps (Lemma 10.135.10).

Proof of (7). Follows from Lemma 10.138.3 and the definitions.

Proof of (8). Lemma 10.136.14 gives the result for smooth ring maps. The resulting smooth ring map $R_0 \to S_0$ satisfies the hypotheses of Lemma 10.129.8, and hence we may replace $S_0$ by the factor of relative dimension $0$ over $R_0$.

Proof of (9). Follows from (8) since $R_0 \to A$ will factor through $A_i$ for some $i$ by Lemma 10.126.3.

Proof of (10). Follows from (9), (1), and (2) since $S^{-1}A$ is a filtered colimit of principal localizations of $A$. $\square$

Next we work out in more detail what it means to be étale over a field.

Lemma 10.141.4. Let $k$ be a field. A ring map $k \to S$ is étale if and only if $S$ is isomorphic as a $k$-algebra to a finite product of finite separable extensions of $k$.

Proof.If $k \to k'$ is a finite separable field extension then we can write $k' = k(\alpha) \cong k[x]/(f)$. Here $f$ is the minimal polynomial of the element $\alpha$. Since $k'$ is separable over $k$ we have $\gcd(f, f') = 1$. This implies that $\text{d} : k'\cdot f \to k' \cdot \text{d}x$ is an isomorphism. Hence $k \to k'$ is étale.Conversely, suppose that $k \to S$ is étale. Let $\overline{k}$ be an algebraic closure of $k$. Then $S \otimes_k \overline{k}$ is étale over $\overline{k}$. Suppose we have the result over $\overline{k}$. Then $S \otimes_k \overline{k}$ is reduced and hence $S$ is reduced. Also, $S \otimes_k \overline{k}$ is finite over $\overline{k}$ and hence $S$ is finite over $k$. Hence $S$ is a finite product $S = \prod k_i$ of fields, see Lemma 10.52.2 and Proposition 10.59.6. The result over $\overline{k}$ means $S \otimes_k \overline{k}$ is isomorphic to a finite product of copies of $\overline{k}$, which implies that each $k \subset k_i$ is finite separable, see for example Lemmas 10.43.1 and 10.43.3. Thus we have reduced to the case $k = \overline{k}$. In this case Lemma 10.138.2 (combined with $\Omega_{S/k} = 0$) we see that $S_{\mathfrak m} \cong k$ for all maximal ideals $\mathfrak m \subset S$. This implies the result because $S$ is the product of the localizations at its maximal ideals by Lemma 10.52.2 and Proposition 10.59.6 again. $\square$

Lemma 10.141.5. 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 étale 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.First we may replace $S$ by $S_g$ for some $g \in S$, $g \not \in \mathfrak q$ and assume that $R \to S$ is étale. Then the lemma follows from Lemma 10.141.4 by unwinding the fact that $S \otimes_R \kappa(\mathfrak p)$ is étale over $\kappa(\mathfrak p)$. $\square$Lemma 10.141.6. An étale ring map is quasi-finite.

Proof.Let $R \to S$ be an étale ring map. By definition $R \to S$ is of finite type. For any prime $\mathfrak p \subset R$ the fibre ring $S \otimes_R \kappa(\mathfrak p)$ is étale over $\kappa(\mathfrak p)$ and hence a finite products of fields finite separable over $\kappa(\mathfrak p)$, in particular finite over $\kappa(\mathfrak p)$. Thus $R \to S$ is quasi-finite by Lemma 10.121.4. $\square$Lemma 10.141.7. 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 presentation,
- $R_{\mathfrak p} \to S_{\mathfrak q}$ is flat
- $\mathfrak p S_{\mathfrak q}$ 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 étale at $\mathfrak q$.

Proof.Apply Lemma 10.121.2 to find a $g \in S$, $g \not \in \mathfrak q$ such that $\mathfrak q$ is the only prime of $S_g$ lying over $\mathfrak p$. We may and do replace $S$ by $S_g$. Then $S \otimes_R \kappa(\mathfrak p)$ has a unique prime, hence is a local ring, hence is equal to $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q} \cong \kappa(\mathfrak q)$. By Lemma 10.135.16 there exists a $g \in S$, $g \not \in \mathfrak q$ such that $R \to S_g$ is smooth. Replace $S$ by $S_g$ again we may assume that $R \to S$ is smooth. By Lemma 10.135.10 we may even assume that $R \to S$ is standard smooth, say $S = R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$. Since $S \otimes_R \kappa(\mathfrak p) = \kappa(\mathfrak q)$ has dimension $0$ we conclude that $n = c$, i.e., if $R \to S$ is étale. $\square$Here is a completely new phenomenon.

Lemma 10.141.8. Let $R \to S$ and $R \to S'$ be étale. Then any $R$-algebra map $S' \to S$ is étale.

Proof.First of all we note that $S' \to S$ is of finite presentation by Lemma 10.6.2. Let $\mathfrak q \subset S$ be a prime ideal lying over the primes $\mathfrak q' \subset S'$ and $\mathfrak p \subset R$. By Lemma 10.141.5 the ring map $S'_{\mathfrak q'}/\mathfrak p S'_{\mathfrak q'} \to S_{\mathfrak q}/\mathfrak p S_{\mathfrak q}$ is a map finite separable extensions of $\kappa(\mathfrak p)$. In particular it is flat. Hence by Lemma 10.127.8 we see that $S'_{\mathfrak q'} \to S_{\mathfrak q}$ is flat. Thus $S' \to S$ is flat. Moreover, the above also shows that $\mathfrak q'S_{\mathfrak q}$ is the maximal ideal of $S_{\mathfrak q}$ and that the residue field extension of $S'_{\mathfrak q'} \to S_{\mathfrak q}$ is finite separable. Hence from Lemma 10.141.7 we conclude that $S' \to S$ is étale at $\mathfrak q$. Since being étale is local (see Lemma 10.141.3) we win. $\square$Lemma 10.141.9. Let $\varphi : R \to S$ be a ring map. If $R \to S$ is surjective, flat and finitely presented then there exist an idempotent $e \in R$ such that $S = R_e$.

First proof.Let $I$ be the kernel of $\varphi$. We have that $I$ is finitely generated by Lemma 10.6.3 since $\varphi$ is of finite presentation. Moreover, since $S$ is flat over $R$, tensoring the exact sequence $0 \to I \to R \to S \to 0$ over $R$ with $S$ gives $I/I^2 = 0$. Now we conclude by Lemma 10.20.5. $\square$

Second proof.Since $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is a homeomorphism onto a closed subset (see Lemma 10.16.7) and is open (see Proposition 10.40.8) we see that the image is $D(e)$ for some idempotent $e \in R$ (see Lemma 10.20.3). Thus $R_e \to S$ induces a bijection on spectra. Now this map induces an isomorphism on all local rings for example by Lemmas 10.77.4 and 10.19.1. Then it follows that $R_e \to S$ is also injective, for example see Lemma 10.23.1. $\square$Lemma 10.141.10. Let $R$ be a ring and let $I \subset R$ be an ideal. Let $R/I \to \overline{S}$ be an étale ring map. Then there exists an étale ring map $R \to S$ such that $\overline{S} \cong S/IS$ as $R/I$-algebras.

Proof.By Lemma 10.141.2 we can write $\overline{S} = (R/I)[x_1, \ldots, x_n]/(\overline{f}_1, \ldots, \overline{f}_n)$ as in Definition 10.135.6 with $\overline{\Delta} = \det(\frac{\partial \overline{f}_i}{\partial x_j})_{i, j = 1, \ldots, n}$ invertible in $\overline{S}$. Just take some lifts $f_i$ and set $S = R[x_1, \ldots, x_n, x_{n+1}]/(f_1, \ldots, f_n, x_{n + 1}\Delta - 1)$ where $\Delta = \det(\frac{\partial f_i}{\partial x_j})_{i, j = 1, \ldots, n}$ as in Example 10.135.8. This proves the lemma. $\square$Lemma 10.141.11. Consider a commutative diagram $$ \xymatrix{ 0 \ar[r] & J \ar[r] & B' \ar[r] & B \ar[r] & 0 \\ 0 \ar[r] & I \ar[r] \ar[u] & A' \ar[r] \ar[u] & A \ar[r] \ar[u] & 0 } $$ with exact rows where $B' \to B$ and $A' \to A$ are surjective ring maps whose kernels are ideals of square zero. If $A \to B$ is étale, and $J = I \otimes_A B$, then $A' \to B'$ is étale.

Proof.By Lemma 10.141.10 there exists an étale ring map $A' \to C$ such that $C/IC = B$. Then $A' \to C$ is formally smooth (by Proposition 10.136.13) hence we get an $A'$-algebra map $\varphi : C \to B'$. Since $A' \to C$ is flat we have $I \otimes_A B = I \otimes_A C/IC = IC$. Hence the assumption that $J = I \otimes_A B$ implies that $\varphi$ induces an isomorphism $IC \to J$ and an isomorphism $C/IC \to B'/IB'$, whence $\varphi$ is an isomorphism. $\square$Example 10.141.12. Let $n , m \geq 1$ be integers. Consider the ring map \begin{eqnarray*} R = \mathbf{Z}[a_1, \ldots, a_{n + m}] & \longrightarrow & S = \mathbf{Z}[b_1, \ldots, b_n, c_1, \ldots, c_m] \\ a_1 & \longmapsto & b_1 + c_1 \\ a_2 & \longmapsto & b_2 + b_1 c_1 + c_2 \\ \ldots & \ldots & \ldots \\ a_{n + m} & \longmapsto & b_n c_m \end{eqnarray*} of Example 10.134.7. Write symbolically $$ S = R[b_1, \ldots, c_m]/(\{a_k(b_i, c_j) - a_k\}_{k = 1, \ldots, n + m}) $$ where for example $a_1(b_i, c_j) = b_1 + c_1$. The matrix of partial derivatives is $$ \left( \begin{matrix} 1 & c_1 & \ldots & c_m & 0 & \ldots & 0 \\ 0 & 1 & c_1 & \ldots & c_m & \ldots & 0 \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ 0 & \ldots & 0 & 1 & c_1 & \ldots & c_m \\ 1 & b_1 & \ldots & b_n & 0 & \ldots & 0 \\ 0 & 1 & b_1 & \ldots & b_n & \ldots & 0 \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ 0 & \ldots & 0 & 1 & b_1 & \ldots & b_n \\ \end{matrix} \right) $$ The determinant $\Delta$ of this matrix is better known as the

resultantof the polynomials $g = x^n + b_1 x^{n - 1} + \ldots + b_n$ and $h = x^m + c_1 x^{m - 1} + \ldots + c_m$, and the matrix above is known as theSylvester matrixassociated to $g, h$. In a formula $\Delta = \text{Res}_x(g, h)$. The Sylvester matrix is the transpose of the matrix of the linear map \begin{eqnarray*} S[x]_{< m} \oplus S[x]_{< n} & \longrightarrow & S[x]_{< n + m} \\ a \oplus b & \longmapsto & ag + bh \end{eqnarray*} Let $\mathfrak q \subset S$ be any prime. By the above the following are equivalent:

- $R \to S$ is étale at $\mathfrak q$,
- $\Delta = \text{Res}_x(g, h) \not \in \mathfrak q$,
- the images $\overline{g}, \overline{h} \in \kappa(\mathfrak q)[x]$ of the polynomials $g, h$ are relatively prime in $\kappa(\mathfrak q)[x]$.
The equivalence of (2) and (3) holds because the image of the Sylvester matrix in $\text{Mat}(n + m, \kappa(\mathfrak q))$ has a kernel if and only if the polynomials $\overline{g}, \overline{h}$ have a factor in common. We conclude that the ring map $$ R \longrightarrow S[\frac{1}{\Delta}] = S[\frac{1}{\text{Res}_x(g, h)}] $$ is étale.

Lemma 10.141.2 tells us that it does not really make sense to define a standard étale morphism to be a standard smooth morphism of relative dimension $0$. As a model for an étale morphism we take the example given by a finite separable extension $k \subset k'$ of fields. Namely, we can always find an element $\alpha \in k'$ such that $k' = k(\alpha)$ and such that the minimal polynomial $f(x) \in k[x]$ of $\alpha$ has derivative $f'$ which is relatively prime to $f$.

Definition 10.141.13. Let $R$ be a ring. Let $g , f \in R[x]$. Assume that $f$ is monic and the derivative $f'$ is invertible in the localization $R[x]_g/(f)$. In this case the ring map $R \to R[x]_g/(f)$ is said to be

standard étale.Lemma 10.141.14. Let $R \to R[x]_g/(f)$ be standard étale.

- The ring map $R \to R[x]_g/(f)$ is étale.
- For any ring map $R \to R'$ the base change $R' \to R'[x]_g/(f)$ of the standard étale ring map $R \to R[x]_g/(f)$ is standard étale.
- Any principal localization of $R[x]_g/(f)$ is standard étale over $R$.
- A composition of standard étale maps is
notstandard étale in general.

Proof.Omitted. Here is an example for (4). The ring map $\mathbf{F}_2 \to \mathbf{F}_{2^2}$ is standard étale. The ring map $\mathbf{F}_{2^2} \to \mathbf{F}_{2^2} \times \mathbf{F}_{2^2} \times \mathbf{F}_{2^2} \times \mathbf{F}_{2^2}$ is standard étale. But the ring map $\mathbf{F}_2 \to \mathbf{F}_{2^2} \times \mathbf{F}_{2^2} \times \mathbf{F}_{2^2} \times \mathbf{F}_{2^2}$ is not standard étale. $\square$Standard étale morphisms are a convenient way to produce étale maps. Here is an example.

Lemma 10.141.15. Let $R$ be a ring. Let $\mathfrak p$ be a prime of $R$. Let $\kappa(\mathfrak p) \subset L$ be a finite separable field extension. There exists an étale ring map $R \to R'$ together with a prime $\mathfrak p'$ lying over $\mathfrak p$ such that the field extension $\kappa(\mathfrak p) \subset \kappa(\mathfrak p')$ is isomorphic to $\kappa(\mathfrak p) \subset L$.

Proof.By the theorem of the primitive element we may write $L = \kappa(\mathfrak p)[\alpha]$. Let $\overline{f} \in \kappa(\mathfrak p)[x]$ denote the minimal polynomial for $\alpha$ (in particular this is monic). After replacing $\alpha$ by $c\alpha$ for some $c \in R$, $c\not \in \mathfrak p$ we may assume all the coefficients of $\overline{f}$ are in the image of $R \to \kappa(\mathfrak p)$ (verification omitted). Thus we can find a monic polynomial $f \in R[x]$ which maps to $\overline{f}$ in $\kappa(\mathfrak p)[x]$. Since $\kappa(\mathfrak p) \subset L$ is separable, we see that $\gcd(\overline{f}, \overline{f}') = 1$. Hence there is an element $\gamma \in L$ such that $\overline{f}'(\alpha) \gamma = 1$. Thus we get a $R$-algebra map \begin{eqnarray*} R[x, 1/f']/(f) & \longrightarrow & L \\ x & \longmapsto & \alpha \\ 1/f' & \longmapsto & \gamma \end{eqnarray*} The left hand side is a standard étale algebra $R'$ over $R$ and the kernel of the ring map gives the desired prime. $\square$Proposition 10.141.16. Let $R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime. If $R \to S$ is étale at $\mathfrak q$, then there exists a $g \in S$, $g \not \in \mathfrak q$ such that $R \to S_g$ is standard étale.

Proof.The following proof is a little roundabout and there may be ways to shorten it.Step 1. By Definition 10.141.1 there exists a $g \in S$, $g \not \in \mathfrak q$ such that $R \to S_g$ is étale. Thus we may assume that $S$ is étale over $R$.

Step 2. By Lemma 10.141.3 there exists an étale 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 $R = R \otimes_{R_0} S_0$. 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 10.141.6. By Lemma 10.122.15 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 of course $S'$ is not étale over $R$ in general.) Thus we may assume that (a) $R$ is Noetherian, (b) $R \to S$ is finite and (c) $R \to S$ is étale at $\mathfrak q$ (but no longer necessarily étale 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 10.141.5. 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 10.141.5. 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 étale 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 $\overline{h} \in \kappa(\mathfrak p)[x]$. 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 $h \in I \subset R[x]$. (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 $\varphi(g) \not \in \mathfrak q$. Pick an element $g' \in R[x]/(f)$ such that also $\varphi(g') \not \in \mathfrak q$ and $S_{\varphi(g')}$ is étale over $R$ (which exists since $S$ is étale over $R$ at $\mathfrak q$). Then the ring map $R[x]_{gg'}/(f) \to S_{\varphi(gg')}$ is a surjective map of étale algebras over $R$. Hence it is étale by Lemma 10.141.8. Hence it is a localization by Lemma 10.141.9. Thus a localization of $S$ at an element not in $\mathfrak q$ is isomorphic to a localization of a standard étale algebra over $R$ which is what we wanted to show. $\square$

The following two lemmas say that the étale topology is coarser than the topology generated by Zariski coverings and finite flat morphisms. They should be skipped on a first reading.

Lemma 10.141.17. Let $R \to S$ be a standard étale morphism. There exists a ring map $R \to S'$ with the following properties

- $R \to S'$ is finite, finitely presented, and flat (in other words $S'$ is finite projective as an $R$-module),
- $\mathop{\mathrm{Spec}}(S') \to \mathop{\mathrm{Spec}}(R)$ is surjective,
- for every prime $\mathfrak q \subset S$, lying over $\mathfrak p \subset R$ and every prime $\mathfrak q' \subset S'$ lying over $\mathfrak p$ there exists a $g' \in S'$, $g' \not \in \mathfrak q'$ such that the ring map $R \to S'_{g'}$ factors through a map $\varphi : S \to S'_{g'}$ with $\varphi^{-1}(\mathfrak q'S'_{g'}) = \mathfrak q$.

Proof.Let $S = R[x]_g/(f)$ be a presentation of $S$ as in Definition 10.141.13. Write $f = x^n + a_1 x^{n - 1} + \ldots + a_n$ with $a_i \in R$. By Lemma 10.134.9 there exists a finite locally free and faithfully flat ring map $R \to S'$ such that $f = \prod (x - \alpha_i)$ for certain $\alpha_i \in S'$. Hence $R \to S'$ satisfies conditions (1), (2). Let $\mathfrak q \subset R[x]/(f)$ be a prime ideal with $g \not \in \mathfrak q$ (i.e., it corresponds to a prime of $S$). Let $\mathfrak p = R \cap \mathfrak q$ and let $\mathfrak q' \subset S'$ be a prime lying over $\mathfrak p$. Note that there are $n$ maps of $R$-algebras \begin{eqnarray*} \varphi_i : R[x]/(f) & \longrightarrow & S' \\ x & \longmapsto & \alpha_i \end{eqnarray*} To finish the proof we have to show that for some $i$ we have (a) the image of $\varphi_i(g)$ in $\kappa(\mathfrak q')$ is not zero, and (b) $\varphi_i^{-1}(\mathfrak q') = \mathfrak q$. Because then we can just take $g' = \varphi_i(g)$, and $\varphi = \varphi_i$ for that $i$.Let $\overline{f}$ denote the image of $f$ in $\kappa(\mathfrak p)[x]$. Note that as a point of $\mathop{\mathrm{Spec}}(\kappa(\mathfrak p)[x]/(\overline{f}))$ the prime $\mathfrak q$ corresponds to an irreducible factor $f_1$ of $\overline{f}$. Moreover, $g \not \in \mathfrak q$ means that $f_1$ does not divide the image $\overline{g}$ of $g$ in $\kappa(\mathfrak p)[x]$. Denote $\overline{\alpha}_1, \ldots, \overline{\alpha}_n$ the images of $\alpha_1, \ldots, \alpha_n$ in $\kappa(\mathfrak q')$. Note that the polynomial $\overline{f}$ splits completely in $\kappa(\mathfrak q')[x]$, namely $$ \overline{f} = \prod\nolimits_i (x - \overline{\alpha}_i) $$ Moreover $\varphi_i(g)$ reduces to $\overline{g}(\overline{\alpha}_i)$. It follows we may pick $i$ such that $f_1(\overline{\alpha}_i) = 0$ and $\overline{g}(\overline{\alpha}_i) \not = 0$. For this $i$ properties (a) and (b) hold. Some details omitted. $\square$

Lemma 10.141.18. Let $R \to S$ be a ring map. Assume that

- $R \to S$ is étale, and
- $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is surjective.
Then there exists a ring map $R \to S'$ such that

- $R \to S'$ is finite, finitely presented, and flat (in other words it is finite projective as an $R$-module),
- $\mathop{\mathrm{Spec}}(S') \to \mathop{\mathrm{Spec}}(R)$ is surjective,
- for every prime $\mathfrak q' \subset S'$ there exists a $g' \in S'$, $g' \not \in \mathfrak q'$ such that the ring map $R \to S'_{g'}$ factors as $R \to S \to S'_{g'}$.

Proof.By Proposition 10.141.16 and the quasi-compactness of $\mathop{\mathrm{Spec}}(S)$ (see Lemma 10.16.10) we can find $g_1, \ldots, g_n \in S$ generating the unit ideal of $S$ such that each $R \to S_{g_i}$ is standard étale. If we prove the lemma for the ring map $R \to \prod_{i = 1, \ldots, n} S_{g_i}$ then the lemma follows for the ring map $R \to S$. Hence we may assume that $S = \prod_{i = 1, \ldots, n} S_i$ is a finite product of standard étale morphisms.For each $i$ choose a ring map $R \to S_i'$ as in Lemma 10.141.17 adapted to the standard étale morphism $R \to S_i$. Set $S' = S_1' \otimes_R \ldots \otimes_R S_n'$; we will use the $R$-algebra maps $S_i' \to S'$ without further mention below. We claim this works. Properties (1) and (2) are immediate. For property (3) suppose that $\mathfrak q' \subset S'$ is a prime. Denote $\mathfrak p$ its image in $\mathop{\mathrm{Spec}}(R)$. Choose $i \in \{1, \ldots, n\}$ such that $\mathfrak p$ is in the image of $\mathop{\mathrm{Spec}}(S_i) \to \mathop{\mathrm{Spec}}(R)$; this is possible by assumption. Set $\mathfrak q_i' \subset S_i'$ the image of $\mathfrak q'$ in the spectrum of $S_i'$. By construction of $S'_i$ there exists a $g'_i \in S_i'$ such that $R \to (S_i')_{g_i'}$ factors as $R \to S_i \to (S_i')_{g_i'}$. Hence also $R \to S'_{g_i'}$ factors as $$ R \to S_i \to (S_i')_{g_i'} \to S'_{g_i'} $$ as desired. $\square$

Lemma 10.141.19. Let $R$ be a ring. Let $f \in R[x]$ be a monic polynomial. Let $\mathfrak p$ be a prime of $R$. Let $f \bmod \mathfrak p = \overline{g} \overline{h}$ be a factorization of the image of $f$ in $\kappa(\mathfrak p)[x]$. If $\gcd(\overline{g}, \overline{h}) = 1$, then there exist

- an étale ring map $R \to R'$,
- a prime $\mathfrak p' \subset R'$ lying over $\mathfrak p$, and
- a factorization $f = g h$ in $R'[x]$
such that

- $\kappa(\mathfrak p) = \kappa(\mathfrak p')$,
- $\overline{g} = g \bmod \mathfrak p'$, $\overline{h} = h \bmod \mathfrak p'$, and
- the polynomials $g, h$ generate the unit ideal in $R'[x]$.

Proof.Suppose $\overline{g} = \overline{b}_0 x^n + \overline{b}_1 x^{n - 1} + \ldots + \overline{b}_n$, and $\overline{h} = \overline{c}_0 x^m + \overline{c}_1 x^{m - 1} + \ldots + \overline{c}_m$ with $\overline{b}_0, \overline{c}_0 \in \kappa(\mathfrak p)$ nonzero. After localizing $R$ at some element of $R$ not contained in $\mathfrak p$ we may assume $\overline{b}_0$ is the image of an invertible element $b_0 \in R$. Replacing $\overline{g}$ by $\overline{g}/b_0$ and $\overline{h}$ by $b_0\overline{h}$ we reduce to the case where $\overline{g}$, $\overline{h}$ are monic (verification omitted). Say $\overline{g} = x^n + \overline{b}_1 x^{n - 1} + \ldots + \overline{b}_n$, and $\overline{h} = x^m + \overline{c}_1 x^{m - 1} + \ldots + \overline{c}_m$. Write $f = x^{n + m} + a_1 x^{n - 1} + \ldots + a_{n + m}$. Consider the fibre product $$ R' = R \otimes_{\mathbf{Z}[a_1, \ldots, a_{n + m}]} \mathbf{Z}[b_1, \ldots, b_n, c_1, \ldots, c_m] $$ where the map $\mathbf{Z}[a_k] \to \mathbf{Z}[b_i, c_j]$ is as in Examples 10.134.7 and 10.141.12. By construction there is an $R$-algebra map $$ R' = R \otimes_{\mathbf{Z}[a_1, \ldots, a_{n + m}]} \mathbf{Z}[b_1, \ldots, b_n, c_1, \ldots, c_m] \longrightarrow \kappa(\mathfrak p) $$ which maps $b_i$ to $\overline{b}_i$ and $c_j$ to $\overline{c}_j$. Denote $\mathfrak p' \subset R'$ the kernel of this map. Since by assumption the polynomials $\overline{g}, \overline{h}$ are relatively prime we see that the element $\Delta = \text{Res}_x(g, h) \in \mathbf{Z}[b_i, c_j]$ (see Example 10.141.12) does not map to zero in $\kappa(\mathfrak p)$ under the displayed map. We conclude that $R \to R'$ is étale at $\mathfrak p'$. In fact a solution to the problem posed in the lemma is the ring map $R \to R'[1/\Delta]$ and the prime $\mathfrak p' R'[1/\Delta]$. Because $\text{Res}_x(f, g)$ is invertible in this ring the Sylvester matrix is invertible over $R'$ and hence $1 = a g + b h$ for some $a, b \in R'[x]$ see Example 10.141.12. $\square$The following lemmas say roughly that after an étale extension a quasi-finite ring map becomes finite. To help interpret the results recall that the locus where a finite type ring map is quasi-finite is open (see Lemma 10.122.14) and that formation of this locus commutes with arbitrary base change (see Lemma 10.121.8).

Lemma 10.141.20. Let $R \to S' \to S$ be ring maps. Let $\mathfrak p \subset R$ be a prime. Let $g \in S'$ be an element. Assume

- $R \to S'$ is integral,
- $R \to S$ is finite type,
- $S'_g \cong S_g$, and
- $g$ invertible in $S' \otimes_R \kappa(\mathfrak p)$.
Then there exists a $f \in R$, $f \not \in \mathfrak p$ such that $R_f \to S_f$ is finite.

Proof.By assumption the image $T$ of $V(g) \subset \mathop{\mathrm{Spec}}(S')$ under the morphism $\mathop{\mathrm{Spec}}(S') \to \mathop{\mathrm{Spec}}(R)$ does not contain $\mathfrak p$. By Section 10.40 especially, Lemma 10.40.6 we see $T$ is closed. Pick $f \in R$, $f \not \in \mathfrak p$ such that $T \cap D(f) = \emptyset$. Then we see that $g$ becomes invertible in $S'_f$. Hence $S'_f \cong S_f$. Thus $S_f$ is both of finite type and integral over $R_f$, hence finite. $\square$Lemma 10.141.21. Let $R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime lying over the prime $\mathfrak p \subset R$. Assume $R \to S$ finite type and quasi-finite at $\mathfrak q$. Then there exists

- an étale ring map $R \to R'$,
- a prime $\mathfrak p' \subset R'$ lying over $\mathfrak p$,
- a product decomposition $$ R' \otimes_R S = A \times B $$
with the following properties

- $\kappa(\mathfrak p) = \kappa(\mathfrak p')$,
- $R' \to A$ is finite,
- $A$ has exactly one prime $\mathfrak r$ lying over $\mathfrak p'$, and
- $\mathfrak r$ lies over $\mathfrak q$.

Proof.Let $S' \subset S$ be the integral closure of $R$ in $S$. Let $\mathfrak q' = S' \cap \mathfrak q$. By Zariski's Main Theorem 10.122.13 there exists a $g \in S'$, $g \not \in \mathfrak q'$ such that $S'_g \cong S_g$. Consider the fibre rings $F = S \otimes_R \kappa(\mathfrak p)$ and $F' = S' \otimes_R \kappa(\mathfrak p)$. Denote $\overline{\mathfrak q}'$ the prime of $F'$ corresponding to $\mathfrak q'$. Since $F'$ is integral over $\kappa(\mathfrak p)$ we see that $\overline{\mathfrak q}'$ is a closed point of $\mathop{\mathrm{Spec}}(F')$, see Lemma 10.35.19. Note that $\mathfrak q$ defines an isolated closed point $\overline{\mathfrak q}$ of $\mathop{\mathrm{Spec}}(F)$ (see Definition 10.121.3). Since $S'_g \cong S_g$ we have $F'_g \cong F_g$, so $\overline{\mathfrak q}$ and $\overline{\mathfrak q}'$ have isomorphic open neighbourhoods in $\mathop{\mathrm{Spec}}(F)$ and $\mathop{\mathrm{Spec}}(F')$. We conclude the set $\{\overline{\mathfrak q}'\} \subset \mathop{\mathrm{Spec}}(F')$ is open. Combined with $\mathfrak q'$ being closed (shown above) we conclude that $\overline{\mathfrak q}'$ defines an isolated closed point of $\mathop{\mathrm{Spec}}(F')$ as well.An additional small remark is that under the map $\mathop{\mathrm{Spec}}(F) \to \mathop{\mathrm{Spec}}(F')$ the point $\overline{\mathfrak q}$ is the only point mapping to $\overline{\mathfrak q}'$. This follows from the discussion above.

By Lemma 10.22.3 we may write $F' = F'_1 \times F'_2$ with $\mathop{\mathrm{Spec}}(F'_1) = \{\overline{\mathfrak q}'\}$. Since $F' = S' \otimes_R \kappa(\mathfrak p)$, there exists an $s' \in S'$ which maps to the element $(r, 0) \in F'_1 \times F'_2 = F'$ for some $r \in R$, $r \not \in \mathfrak p$. In fact, what we will use about $s'$ is that it is an element of $S'$, not contained in $\mathfrak q'$, and contained in any other prime lying over $\mathfrak p$.

Let $f(x) \in R[x]$ be a monic polynomial such that $f(s') = 0$. Denote $\overline{f} \in \kappa(\mathfrak p)[x]$ the image. We can factor it as $\overline{f} = x^e \overline{h}$ where $\overline{h}(0) \not = 0$. By Lemma 10.141.19 we can find an étale ring extension $R \to R'$, a prime $\mathfrak p'$ lying over $\mathfrak p$, and a factorization $f = h i$ in $R'[x]$ such that $\kappa(\mathfrak p) = \kappa(\mathfrak p')$, $\overline{h} = h \bmod \mathfrak p'$, $x^e = i \bmod \mathfrak p'$, and we can write $a h + b i = 1$ in $R'[x]$ (for suitable $a, b$).

Consider the elements $h(s'), i(s') \in R' \otimes_R S'$. By construction we have $h(s')i(s') = f(s') = 0$. On the other hand they generate the unit ideal since $a(s')h(s') + b(s')i(s') = 1$. Thus we see that $R' \otimes_R S'$ is the product of the localizations at these elements: $$ R' \otimes_R S' = (R' \otimes_R S')_{h(s')} \times (R' \otimes_R S')_{i(s')} = S'_1 \times S'_2 $$ Moreover this product decomposition is compatible with the product decomposition we found for the fibre ring $F'$; this comes from our choice of $s', h$ which guarantee that $\overline{\mathfrak q}'$ is the only prime of $F'$ which does not contain the image of $h(s')$ in $F'$. Here we use that the fibre ring of $R'\otimes_R S'$ over $R'$ at $\mathfrak p'$ is the same as $F'$ due to the fact that $\kappa(\mathfrak p) = \kappa(\mathfrak p')$. It follows that $S'_1$ has exactly one prime, say $\mathfrak r'$, lying over $\mathfrak p'$ and that this prime lies over $\mathfrak q$. Hence the element $g \in S'$ maps to an element of $S'_1$ not contained in $\mathfrak r'$.

The base change $R'\otimes_R S$ inherits a similar product decomposition $$ R' \otimes_R S = (R' \otimes_R S)_{h(s')} \times (R' \otimes_R S)_{i(s')} = S_1 \times S_2 $$ It follows from the above that $S_1$ has exactly one prime, say $\mathfrak r$, lying over $\mathfrak p'$ (consider the fibre ring as above), and that this prime lies over $\mathfrak q$.

Now we may apply Lemma 10.141.20 to the ring maps $R' \to S'_1 \to S_1$, the prime $\mathfrak p'$ and the element $g$ to see that after replacing $R'$ by a principal localization we can assume that $S_1$ is finite over $R'$ as desired. $\square$

Lemma 10.141.22. Let $R \to S$ be a ring map. Let $\mathfrak p \subset R$ be a prime. Assume $R \to S$ finite type. Then there exists

- an étale ring map $R \to R'$,
- a prime $\mathfrak p' \subset R'$ lying over $\mathfrak p$,
- a product decomposition $$ R' \otimes_R S = A_1 \times \ldots \times A_n \times B $$
with the following properties

- we have $\kappa(\mathfrak p) = \kappa(\mathfrak p')$,
- each $A_i$ is finite over $R'$,
- each $A_i$ has exactly one prime $\mathfrak r_i$ lying over $\mathfrak p'$, and
- $R' \to B$ not quasi-finite at any prime lying over $\mathfrak p'$.

Proof.Denote $F = S \otimes_R \kappa(\mathfrak p)$ the fibre ring of $S/R$ at the prime $\mathfrak p$. As $F$ is of finite type over $\kappa(\mathfrak p)$ it is Noetherian and hence $\mathop{\mathrm{Spec}}(F)$ has finitely many isolated closed points. If there are no isolated closed points, i.e., no primes $\mathfrak q$ of $S$ over $\mathfrak p$ such that $S/R$ is quasi-finite at $\mathfrak q$, then the lemma holds. If there exists at least one such prime $\mathfrak q$, then we may apply Lemma 10.141.21. This gives a diagram $$ \xymatrix{ S \ar[r] & R'\otimes_R S \ar@{=}[r] & A_1 \times B' \\ R \ar[r] \ar[u] & R' \ar[u] \ar[ru] } $$ as in said lemma. Since the residue fields at $\mathfrak p$ and $\mathfrak p'$ are the same, the fibre rings of $S/R$ and $(A \times B)/R'$ are the same. Hence, by induction on the number of isolated closed points of the fibre we may assume that the lemma holds for $R' \to B$ and $\mathfrak p'$. Thus we get an étale ring map $R' \to R''$, a prime $\mathfrak p'' \subset R''$ and a decomposition $$ R'' \otimes_{R'} B' = A_2 \times \ldots \times A_n \times B $$ We omit the verification that the ring map $R \to R''$, the prime $\mathfrak p''$ and the resulting decomposition $$ R'' \otimes_R S = (R'' \otimes_{R'} A_1) \times A_2 \times \ldots \times A_n \times B $$ is a solution to the problem posed in the lemma. $\square$Lemma 10.141.23. Let $R \to S$ be a ring map. Let $\mathfrak p \subset R$ be a prime. Assume $R \to S$ finite type. Then there exists

- an étale ring map $R \to R'$,
- a prime $\mathfrak p' \subset R'$ lying over $\mathfrak p$,
- a product decomposition $$ R' \otimes_R S = A_1 \times \ldots \times A_n \times B $$
with the following properties

- each $A_i$ is finite over $R'$,
- each $A_i$ has exactly one prime $\mathfrak r_i$ lying over $\mathfrak p'$,
- the finite field extensions $\kappa(\mathfrak p') \subset \kappa(\mathfrak r_i)$ are purely inseparable, and
- $R' \to B$ not quasi-finite at any prime lying over $\mathfrak p'$.

Proof.The strategy of the proof is to make two étale ring extensions: first we control the residue fields, then we apply Lemma 10.141.22.Denote $F = S \otimes_R \kappa(\mathfrak p)$ the fibre ring of $S/R$ at the prime $\mathfrak p$. As in the proof of Lemma 10.141.22 there are finitely may primes, say $\mathfrak q_1, \ldots, \mathfrak q_n$ of $S$ lying over $R$ at which the ring map $R \to S$ is quasi-finite. Let $\kappa(\mathfrak p) \subset L_i \subset \kappa(\mathfrak q_i)$ be the subfield such that $\kappa(\mathfrak p) \subset L_i$ is separable, and the field extension $L_i \subset \kappa(\mathfrak q_i)$ is purely inseparable. Let $\kappa(\mathfrak p) \subset L$ be a finite Galois extension into which $L_i$ embeds for $i = 1, \ldots, n$. By Lemma 10.141.15 we can find an étale ring extension $R \to R'$ together with a prime $\mathfrak p'$ lying over $\mathfrak p$ such that the field extension $\kappa(\mathfrak p) \subset \kappa(\mathfrak p')$ is isomorphic to $\kappa(\mathfrak p) \subset L$. Thus the fibre ring of $R' \otimes_R S$ at $\mathfrak p'$ is isomorphic to $F \otimes_{\kappa(\mathfrak p)} L$. The primes lying over $\mathfrak q_i$ correspond to primes of $\kappa(\mathfrak q_i) \otimes_{\kappa(\mathfrak p)} L$ which is a product of fields purely inseparable over $L$ by our choice of $L$ and elementary field theory. These are also the only primes over $\mathfrak p'$ at which $R' \to R' \otimes_R S$ is quasi-finite, by Lemma 10.121.8. Hence after replacing $R$ by $R'$, $\mathfrak p$ by $\mathfrak p'$, and $S$ by $R' \otimes_R S$ we may assume that for all primes $\mathfrak q$ lying over $\mathfrak p$ for which $S/R$ is quasi-finite the field extensions $\kappa(\mathfrak p) \subset \kappa(\mathfrak q)$ are purely inseparable.

Next apply Lemma 10.141.22. The result is what we want since the field extensions do not change under this étale ring extension. $\square$

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 37175–38407 (see updates for more information).

```
\section{\'Etale ring maps}
\label{section-etale}
\noindent
An \'etale ring map is a smooth ring map whose relative dimension
is equal to zero. This is the same as the following slightly more
direct definition.
\begin{definition}
\label{definition-etale}
Let $R \to S$ be a ring map. We say $R \to S$ is {\it \'etale} if it is
of finite presentation and the naive cotangent complex
$\NL_{S/R}$ is quasi-isomorphic to zero. Given a prime $\mathfrak q$
of $S$ we say that $R \to S$ is {\it \'etale at $\mathfrak q$}
if there exists a $g \in S$, $g \not \in \mathfrak q$ such that
$R \to S_g$ is \'etale.
\end{definition}
\noindent
In particular we see that $\Omega_{S/R} = 0$ if $S$ is \'etale over $R$.
If $R \to S$ is smooth,
then $R \to S$ is \'etale if and only if $\Omega_{S/R} = 0$.
From our results on smooth ring maps we automatically get a whole host
of results for \'etale maps. We summarize these in Lemma \ref{lemma-etale}
below. But before we do so we prove that {\it any} \'etale ring map is
standard smooth.
\begin{lemma}
\label{lemma-etale-standard-smooth}
Any \'etale ring map is standard smooth. More precisely, if
$R \to S$ is \'etale, then there exists a presentation
$S = R[x_1, \ldots, x_n]/(f_1, \ldots, f_n)$ such that
the image of $\det(\partial f_j/\partial x_i)$ is invertible in $S$.
\end{lemma}
\begin{proof}
Let $R \to S$ be \'etale. Choose a presentation $S = R[x_1, \ldots, x_n]/I$.
As $R \to S$ is \'etale we know that
$$
\text{d} :
I/I^2
\longrightarrow
\bigoplus\nolimits_{i = 1, \ldots, n} S\text{d}x_i
$$
is an isomorphism, in particular $I/I^2$ is a free $S$-module.
Thus by Lemma \ref{lemma-huber} we may assume (after possibly changing
the presentation), that $I = (f_1, \ldots, f_c)$ such that the classes
$f_i \bmod I^2$ form a basis of $I/I^2$. It follows immediately from
the fact that the displayed map above is an isomorphism that $c = n$ and
that $\det(\partial f_j/\partial x_i)$ is invertible in $S$.
\end{proof}
\begin{lemma}
\label{lemma-etale}
Results on \'etale ring maps.
\begin{enumerate}
\item The ring map $R \to R_f$ is \'etale for any ring $R$ and any $f \in R$.
\item Compositions of \'etale ring maps are \'etale.
\item A base change of an \'etale ring map is \'etale.
\item The property of being \'etale is local: Given a ring map
$R \to S$ and elements $g_1, \ldots, g_m \in S$ which generate the unit ideal
such that $R \to S_{g_j}$ is \'etale for $j = 1, \ldots, m$ then
$R \to S$ is \'etale.
\item Given $R \to S$ of finite presentation, and a flat ring map
$R \to R'$, set $S' = R' \otimes_R S$. The set of primes where $R' \to S'$
is \'etale is the inverse image via $\Spec(S') \to \Spec(S)$
of the set of primes where $R \to S$ is \'etale.
\item An \'etale ring map is syntomic, in particular flat.
\item If $S$ is finite type over a field $k$, then $S$ is \'etale over
$k$ if and only if $\Omega_{S/k} = 0$.
\item Any \'etale ring map $R \to S$ is the base change of an \'etale
ring map $R_0 \to S_0$ with $R_0$ of finite type over $\mathbf{Z}$.
\item Let $A = \colim A_i$ be a filtered colimit of rings.
Let $A \to B$ be an \'etale ring map. Then there exists an \'etale ring
map $A_i \to B_i$ for some $i$ such that $B \cong A \otimes_{A_i} B_i$.
\item Let $A$ be a ring. Let $S$ be a multiplicative subset of $A$.
Let $S^{-1}A \to B'$ be \'etale. Then there exists an \'etale ring map
$A \to B$ such that $B' \cong S^{-1}B$.
\end{enumerate}
\end{lemma}
\begin{proof}
In each case we use the corresponding result for smooth ring maps with
a small argument added to show that $\Omega_{S/R}$ is zero.
\medskip\noindent
Proof of (1). The ring map $R \to R_f$ is smooth and $\Omega_{R_f/R} = 0$.
\medskip\noindent
Proof of (2). The composition $A \to C$ of smooth maps $A \to B$ and
$B \to C$ is smooth, see Lemma \ref{lemma-compose-smooth}. By
Lemma \ref{lemma-exact-sequence-differentials} we see that
$\Omega_{C/A}$ is zero as both $\Omega_{C/B}$ and $\Omega_{B/A}$ are zero.
\medskip\noindent
Proof of (3). Let $R \to S$ be \'etale and $R \to R'$ be arbitrary.
Then $R' \to S' = R' \otimes_R S$ is smooth, see
Lemma \ref{lemma-base-change-smooth}. Since
$\Omega_{S'/R'} = S' \otimes_S \Omega_{S/R}$ by
Lemma \ref{lemma-differentials-base-change}
we conclude that $\Omega_{S'/R'} = 0$. Hence $R' \to S'$ is \'etale.
\medskip\noindent
Proof of (4). Assume the hypotheses of (4). By
Lemma \ref{lemma-locally-smooth} we see that $R \to S$ is smooth.
We are also given that $\Omega_{S_{g_i}/R} = (\Omega_{S/R})_{g_i} = 0$
for all $i$. Then $\Omega_{S/R} = 0$, see Lemma \ref{lemma-cover}.
\medskip\noindent
Proof of (5). The result for smooth maps is
Lemma \ref{lemma-flat-base-change-locus-smooth}.
In the proof of that lemma we used that $\NL_{S/R} \otimes_S S'$
is homotopy equivalent to $\NL_{S'/R'}$.
This reduces us to showing that if $M$ is a finitely presented
$S$-module the set of primes $\mathfrak q'$ of $S'$
such that $(M \otimes_S S')_{\mathfrak q'} = 0$ is the inverse
image of the set of primes $\mathfrak q$ of $S$ such that
$M_{\mathfrak q} = 0$. This follows from Lemma \ref{lemma-support-base-change}.
\medskip\noindent
Proof of (6). Follows directly from the corresponding result for
smooth ring maps (Lemma \ref{lemma-smooth-syntomic}).
\medskip\noindent
Proof of (7). Follows from Lemma \ref{lemma-characterize-smooth-over-field}
and the definitions.
\medskip\noindent
Proof of (8). Lemma \ref{lemma-finite-presentation-fs-Noetherian}
gives the result for smooth ring maps. The resulting smooth ring map
$R_0 \to S_0$ satisfies the
hypotheses of Lemma \ref{lemma-relative-dimension-CM}, and hence we may
replace $S_0$ by the factor of relative dimension $0$ over $R_0$.
\medskip\noindent
Proof of (9). Follows from (8) since $R_0 \to A$ will factor through
$A_i$ for some $i$ by Lemma \ref{lemma-characterize-finite-presentation}.
\medskip\noindent
Proof of (10). Follows from (9), (1), and (2) since $S^{-1}A$ is a
filtered colimit of principal localizations of $A$.
\end{proof}
\noindent
Next we work out in more detail what it means to be \'etale
over a field.
\begin{lemma}
\label{lemma-etale-over-field}
Let $k$ be a field. A ring map $k \to S$ is \'etale if and only if $S$
is isomorphic as a $k$-algebra to a finite product of finite separable
extensions of $k$.
\end{lemma}
\begin{proof}
If $k \to k'$ is a finite separable field extension then we can
write $k' = k(\alpha) \cong k[x]/(f)$. Here $f$ is the minimal
polynomial of the element $\alpha$. Since $k'$ is separable over $k$
we have $\gcd(f, f') = 1$. This
implies that $\text{d} : k'\cdot f \to k' \cdot \text{d}x$
is an isomorphism. Hence $k \to k'$ is \'etale.
\medskip\noindent
Conversely, suppose that $k \to S$ is \'etale. Let $\overline{k}$
be an algebraic closure of $k$. Then $S \otimes_k \overline{k}$
is \'etale over $\overline{k}$. Suppose we have the result over $\overline{k}$.
Then $S \otimes_k \overline{k}$ is reduced and hence $S$ is reduced.
Also, $S \otimes_k \overline{k}$ is finite over $\overline{k}$
and hence $S$ is finite over $k$. Hence $S$ is a finite product
$S = \prod k_i$
of fields, see
Lemma \ref{lemma-finite-dimensional-algebra}
and
Proposition \ref{proposition-dimension-zero-ring}.
The result over $\overline{k}$ means $S \otimes_k \overline{k}$
is isomorphic to a finite product of copies of $\overline{k}$, which
implies that each $k \subset k_i$ is finite separable, see for example
Lemmas \ref{lemma-characterize-separable-field-extensions} and
\ref{lemma-geometrically-reduced-finite-purely-inseparable-extension}.
Thus we have reduced to the case $k = \overline{k}$.
In this case
Lemma \ref{lemma-characterize-smooth-kbar}
(combined with $\Omega_{S/k} = 0$) we see that $S_{\mathfrak m} \cong k$
for all maximal ideals $\mathfrak m \subset S$. This implies the result
because $S$ is the product of the localizations at its maximal ideals by
Lemma \ref{lemma-finite-dimensional-algebra}
and
Proposition \ref{proposition-dimension-zero-ring}
again.
\end{proof}
\begin{lemma}
\label{lemma-etale-at-prime}
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 \'etale at $\mathfrak q$ then
\begin{enumerate}
\item we have $\mathfrak p S_{\mathfrak q} = \mathfrak qS_{\mathfrak q}$
is the maximal ideal of the local ring $S_{\mathfrak q}$, and
\item the field extension $\kappa(\mathfrak p) \subset \kappa(\mathfrak q)$
is finite separable.
\end{enumerate}
\end{lemma}
\begin{proof}
First we may replace $S$ by $S_g$ for some $g \in S$, $g \not \in \mathfrak q$
and assume that $R \to S$ is \'etale. Then the lemma follows from
Lemma \ref{lemma-etale-over-field} by unwinding the
fact that $S \otimes_R \kappa(\mathfrak p)$ is \'etale over
$\kappa(\mathfrak p)$.
\end{proof}
\begin{lemma}
\label{lemma-etale-quasi-finite}
An \'etale ring map is quasi-finite.
\end{lemma}
\begin{proof}
Let $R \to S$ be an \'etale ring map. By definition $R \to S$ is of finite type.
For any prime $\mathfrak p \subset R$ the fibre ring
$S \otimes_R \kappa(\mathfrak p)$ is \'etale over $\kappa(\mathfrak p)$
and hence a finite products of fields finite separable over
$\kappa(\mathfrak p)$, in particular finite over $\kappa(\mathfrak p)$.
Thus $R \to S$ is quasi-finite by Lemma \ref{lemma-quasi-finite}.
\end{proof}
\begin{lemma}
\label{lemma-characterize-etale}
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
\begin{enumerate}
\item $R \to S$ is of finite presentation,
\item $R_{\mathfrak p} \to S_{\mathfrak q}$ is flat
\item $\mathfrak p S_{\mathfrak q}$ is the maximal ideal
of the local ring $S_{\mathfrak q}$, and
\item the field extension $\kappa(\mathfrak p) \subset \kappa(\mathfrak q)$
is finite separable,
\end{enumerate}
then $R \to S$ is \'etale at $\mathfrak q$.
\end{lemma}
\begin{proof}
Apply
Lemma \ref{lemma-isolated-point-fibre}
to find a $g \in S$, $g \not \in \mathfrak q$ such that
$\mathfrak q$ is the only prime of $S_g$ lying over $\mathfrak p$.
We may and do replace $S$ by $S_g$. Then
$S \otimes_R \kappa(\mathfrak p)$ has a unique prime, hence is a
local ring, hence is equal to
$S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}
\cong \kappa(\mathfrak q)$.
By Lemma \ref{lemma-flat-fibre-smooth}
there exists a $g \in S$, $g \not \in \mathfrak q$
such that $R \to S_g$ is smooth. Replace $S$ by $S_g$ again we may
assume that $R \to S$ is smooth. By
Lemma \ref{lemma-smooth-syntomic} we may even assume that
$R \to S$ is standard smooth, say $S = R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$.
Since $S \otimes_R \kappa(\mathfrak p) = \kappa(\mathfrak q)$
has dimension $0$ we conclude that $n = c$, i.e.,
if $R \to S$ is \'etale.
\end{proof}
\noindent
Here is a completely new phenomenon.
\begin{lemma}
\label{lemma-map-between-etale}
Let $R \to S$ and $R \to S'$ be \'etale.
Then any $R$-algebra map $S' \to S$ is \'etale.
\end{lemma}
\begin{proof}
First of all we note that $S' \to S$ is of finite presentation by
Lemma \ref{lemma-compose-finite-type}.
Let $\mathfrak q \subset S$ be a prime ideal lying over the primes
$\mathfrak q' \subset S'$ and $\mathfrak p \subset R$.
By Lemma \ref{lemma-etale-at-prime} the ring map
$S'_{\mathfrak q'}/\mathfrak p S'_{\mathfrak q'} \to
S_{\mathfrak q}/\mathfrak p S_{\mathfrak q}$
is a map finite separable extensions of $\kappa(\mathfrak p)$.
In particular it is flat. Hence by
Lemma \ref{lemma-criterion-flatness-fibre} we see that
$S'_{\mathfrak q'} \to S_{\mathfrak q}$ is flat. Thus $S' \to S$
is flat. Moreover, the above also shows that $\mathfrak q'S_{\mathfrak q}$
is the maximal ideal of $S_{\mathfrak q}$ and that the residue
field extension of $S'_{\mathfrak q'} \to S_{\mathfrak q}$ is
finite separable. Hence from Lemma \ref{lemma-characterize-etale}
we conclude that $S' \to S$ is \'etale at $\mathfrak q$. Since
being \'etale is local (see Lemma \ref{lemma-etale}) we win.
\end{proof}
\begin{lemma}
\label{lemma-surjective-flat-finitely-presented}
Let $\varphi : R \to S$ be a ring map. If $R \to S$ is surjective, flat and
finitely presented then there exist an idempotent $e \in R$ such that
$S = R_e$.
\end{lemma}
\begin{proof}[First proof]
Let $I$ be the kernel of $\varphi$.
We have that $I$ is finitely generated by
Lemma \ref{lemma-finite-presentation-independent}
since $\varphi$ is of finite presentation.
Moreover, since $S$ is flat over $R$, tensoring the exact sequence
$0 \to I \to R \to S \to 0$ over $R$ with $S$
gives $I/I^2 = 0$. Now we conclude by
Lemma \ref{lemma-ideal-is-squared-union-connected}.
\end{proof}
\begin{proof}[Second proof]
Since $\Spec(S) \to \Spec(R)$ is a homeomorphism
onto a closed subset (see Lemma \ref{lemma-spec-closed}) and
is open (see Proposition \ref{proposition-fppf-open}) we see that
the image is $D(e)$ for some idempotent $e \in R$ (see
Lemma \ref{lemma-disjoint-decomposition}). Thus $R_e \to S$
induces a bijection on spectra. Now this map induces an isomorphism
on all local rings for example by
Lemmas \ref{lemma-finite-flat-local} and \ref{lemma-NAK}.
Then it follows that $R_e \to S$ is also injective, for example
see Lemma \ref{lemma-characterize-zero-local}.
\end{proof}
\begin{lemma}
\label{lemma-lift-etale}
Let $R$ be a ring and let $I \subset R$ be an ideal.
Let $R/I \to \overline{S}$ be an \'etale ring map.
Then there exists an \'etale ring map
$R \to S$ such that $\overline{S} \cong S/IS$ as $R/I$-algebras.
\end{lemma}
\begin{proof}
By Lemma \ref{lemma-etale-standard-smooth} we can write
$\overline{S} =
(R/I)[x_1, \ldots, x_n]/(\overline{f}_1, \ldots, \overline{f}_n)$
as in Definition \ref{definition-standard-smooth} with
$\overline{\Delta} =
\det(\frac{\partial \overline{f}_i}{\partial x_j})_{i, j = 1, \ldots, n}$
invertible in $\overline{S}$. Just take some lifts $f_i$ and set
$S = R[x_1, \ldots, x_n, x_{n+1}]/(f_1, \ldots, f_n, x_{n + 1}\Delta - 1)$
where $\Delta = \det(\frac{\partial f_i}{\partial x_j})_{i, j = 1, \ldots, n}$
as in Example \ref{example-make-standard-smooth}.
This proves the lemma.
\end{proof}
\begin{lemma}
\label{lemma-lift-etale-infinitesimal}
Consider a commutative diagram
$$
\xymatrix{
0 \ar[r] &
J \ar[r] &
B' \ar[r] &
B \ar[r] & 0 \\
0 \ar[r] &
I \ar[r] \ar[u] &
A' \ar[r] \ar[u] &
A \ar[r] \ar[u] & 0
}
$$
with exact rows where $B' \to B$ and $A' \to A$ are surjective ring maps
whose kernels are ideals of square zero. If $A \to B$ is \'etale,
and $J = I \otimes_A B$, then $A' \to B'$ is \'etale.
\end{lemma}
\begin{proof}
By
Lemma \ref{lemma-lift-etale}
there exists an \'etale ring map $A' \to C$ such that $C/IC = B$.
Then $A' \to C$ is formally smooth (by
Proposition \ref{proposition-smooth-formally-smooth})
hence we get an $A'$-algebra map $\varphi : C \to B'$.
Since $A' \to C$ is flat we have $I \otimes_A B = I \otimes_A C/IC = IC$.
Hence the assumption that $J = I \otimes_A B$ implies that
$\varphi$ induces an isomorphism $IC \to J$ and an isomorphism
$C/IC \to B'/IB'$, whence $\varphi$ is an isomorphism.
\end{proof}
\begin{example}
\label{example-factor-polynomials-etale}
Let $n , m \geq 1$ be integers. Consider the ring map
\begin{eqnarray*}
R = \mathbf{Z}[a_1, \ldots, a_{n + m}]
& \longrightarrow &
S = \mathbf{Z}[b_1, \ldots, b_n, c_1, \ldots, c_m] \\
a_1 & \longmapsto & b_1 + c_1 \\
a_2 & \longmapsto & b_2 + b_1 c_1 + c_2 \\
\ldots & \ldots & \ldots \\
a_{n + m} & \longmapsto & b_n c_m
\end{eqnarray*}
of Example \ref{example-factor-polynomials}.
Write symbolically
$$
S = R[b_1, \ldots, c_m]/(\{a_k(b_i, c_j) - a_k\}_{k = 1, \ldots, n + m})
$$
where for example $a_1(b_i, c_j) = b_1 + c_1$.
The matrix of partial derivatives is
$$
\left(
\begin{matrix}
1 & c_1 & \ldots & c_m & 0 & \ldots & 0 \\
0 & 1 & c_1 & \ldots & c_m & \ldots & 0 \\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
0 & \ldots & 0 & 1 & c_1 & \ldots & c_m \\
1 & b_1 & \ldots & b_n & 0 & \ldots & 0 \\
0 & 1 & b_1 & \ldots & b_n & \ldots & 0 \\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
0 & \ldots & 0 & 1 & b_1 & \ldots & b_n \\
\end{matrix}
\right)
$$
The determinant $\Delta$ of this matrix is better known as the
{\it resultant} of the polynomials $g = x^n + b_1 x^{n - 1} + \ldots + b_n$
and $h = x^m + c_1 x^{m - 1} + \ldots + c_m$, and the matrix above
is known as the {\it Sylvester matrix} associated to $g, h$.
In a formula $\Delta = \text{Res}_x(g, h)$. The Sylvester matrix
is the transpose of the matrix of the linear map
\begin{eqnarray*}
S[x]_{< m} \oplus S[x]_{< n} & \longrightarrow & S[x]_{< n + m} \\
a \oplus b & \longmapsto & ag + bh
\end{eqnarray*}
Let $\mathfrak q \subset S$ be any prime. By the above the
following are equivalent:
\begin{enumerate}
\item $R \to S$ is \'etale at $\mathfrak q$,
\item $\Delta = \text{Res}_x(g, h) \not \in \mathfrak q$,
\item the images $\overline{g}, \overline{h} \in \kappa(\mathfrak q)[x]$
of the polynomials $g, h$ are relatively prime in $\kappa(\mathfrak q)[x]$.
\end{enumerate}
The equivalence of (2) and (3) holds because the image of the
Sylvester matrix in $\text{Mat}(n + m, \kappa(\mathfrak q))$
has a kernel if and only if the polynomials $\overline{g}, \overline{h}$
have a factor in common. We conclude that the ring map
$$
R \longrightarrow S[\frac{1}{\Delta}] = S[\frac{1}{\text{Res}_x(g, h)}]
$$
is \'etale.
\end{example}
\noindent
Lemma \ref{lemma-etale-standard-smooth} tells us that it does not really
make sense to define a standard \'etale morphism to be
a standard smooth morphism of relative dimension $0$.
As a model for an \'etale morphism we take the example given
by a finite separable extension $k \subset k'$ of fields.
Namely, we can always find an element $\alpha \in k'$ such
that $k' = k(\alpha)$ and such that the minimal polynomial
$f(x) \in k[x]$ of $\alpha$ has derivative $f'$ which is
relatively prime to $f$.
\begin{definition}
\label{definition-standard-etale}
Let $R$ be a ring. Let $g , f \in R[x]$.
Assume that $f$ is monic and the derivative $f'$ is invertible in
the localization $R[x]_g/(f)$.
In this case the ring map $R \to R[x]_g/(f)$ is said to be
{\it standard \'etale}.
\end{definition}
\begin{lemma}
\label{lemma-standard-etale}
Let $R \to R[x]_g/(f)$ be standard \'etale.
\begin{enumerate}
\item The ring map $R \to R[x]_g/(f)$ is \'etale.
\item For any ring map $R \to R'$ the base change $R' \to R'[x]_g/(f)$
of the standard \'etale ring map $R \to R[x]_g/(f)$ is standard \'etale.
\item Any principal localization of $R[x]_g/(f)$ is standard \'etale over $R$.
\item A composition of standard \'etale maps is {\bf not} standard \'etale
in general.
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted. Here is an example for (4).
The ring map $\mathbf{F}_2 \to \mathbf{F}_{2^2}$ is standard \'etale.
The ring map
$\mathbf{F}_{2^2} \to \mathbf{F}_{2^2} \times \mathbf{F}_{2^2}
\times \mathbf{F}_{2^2} \times \mathbf{F}_{2^2}$ is standard \'etale.
But the ring map
$\mathbf{F}_2 \to \mathbf{F}_{2^2} \times \mathbf{F}_{2^2}
\times \mathbf{F}_{2^2} \times \mathbf{F}_{2^2}$ is not standard \'etale.
\end{proof}
\noindent
Standard \'etale morphisms are a convenient way to produce \'etale maps.
Here is an example.
\begin{lemma}
\label{lemma-make-etale-map-prescribed-residue-field}
Let $R$ be a ring.
Let $\mathfrak p$ be a prime of $R$.
Let $\kappa(\mathfrak p) \subset L$ be a finite separable field extension.
There exists an \'etale ring map $R \to R'$ together with a prime $\mathfrak p'$
lying over $\mathfrak p$ such that the field extension
$\kappa(\mathfrak p) \subset \kappa(\mathfrak p')$ is isomorphic
to $\kappa(\mathfrak p) \subset L$.
\end{lemma}
\begin{proof}
By the theorem of the primitive element we may write
$L = \kappa(\mathfrak p)[\alpha]$. Let
$\overline{f} \in \kappa(\mathfrak p)[x]$
denote the minimal polynomial for $\alpha$ (in particular this is monic).
After replacing $\alpha$ by $c\alpha$ for some $c \in R$,
$c\not \in \mathfrak p$ we may assume all the coefficients
of $\overline{f}$ are in the image of $R \to \kappa(\mathfrak p)$
(verification omitted). Thus we can find a monic polynomial
$f \in R[x]$ which maps to $\overline{f}$ in $\kappa(\mathfrak p)[x]$.
Since $\kappa(\mathfrak p) \subset L$ is separable, we see
that $\gcd(\overline{f}, \overline{f}') = 1$.
Hence there is an element $\gamma \in L$ such that
$\overline{f}'(\alpha) \gamma = 1$. Thus we get a $R$-algebra map
\begin{eqnarray*}
R[x, 1/f']/(f) & \longrightarrow & L \\
x & \longmapsto & \alpha \\
1/f' & \longmapsto & \gamma
\end{eqnarray*}
The left hand side is a standard \'etale algebra $R'$ over $R$
and the kernel of the ring map gives the desired prime.
\end{proof}
\begin{proposition}
\label{proposition-etale-locally-standard}
Let $R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime.
If $R \to S$ is \'etale at $\mathfrak q$, then there exists
a $g \in S$, $g \not \in \mathfrak q$ such that $R \to S_g$
is standard \'etale.
\end{proposition}
\begin{proof}
The following proof is a little roundabout and there may be ways to
shorten it.
\medskip\noindent
Step 1. By Definition \ref{definition-etale}
there exists a $g \in S$, $g \not \in \mathfrak q$
such that $R \to S_g$ is \'etale. Thus we may assume that $S$ is \'etale
over $R$.
\medskip\noindent
Step 2. By Lemma \ref{lemma-etale} there exists an \'etale 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 $R = R \otimes_{R_0} S_0$. 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.
\medskip\noindent
Step 3.
Note that $R \to S$ is quasi-finite by Lemma \ref{lemma-etale-quasi-finite}.
By Lemma \ref{lemma-quasi-finite-open-integral-closure}
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 of course $S'$ is not \'etale over $R$ in general.)
Thus we may assume that (a) $R$ is Noetherian, (b) $R \to S$ is finite
and (c) $R \to S$ is \'etale at $\mathfrak q$
(but no longer necessarily \'etale at all primes).
\medskip\noindent
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 \ref{lemma-finite-dimensional-algebra}) and
so a finite product of local rings
$$
S \otimes_R \kappa(\mathfrak p) = \prod\nolimits_{i = 1}^n A_i
$$
see Proposition \ref{proposition-dimension-zero-ring}. 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 \ref{lemma-etale-at-prime}. The other factors correspond to
the other primes, say $\mathfrak q_2, \ldots, \mathfrak q_n$ of
$S$ lying over $\mathfrak p$.
\medskip\noindent
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 $\alpha \not = 0$.
\medskip\noindent
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 \ref{lemma-unique-prime-over-localize-below} (we have
going up for $S' \to S$ by Lemma \ref{lemma-integral-going-up}
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 \ref{lemma-etale-at-prime}.
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 \ref{lemma-NAK} 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,
$S'_{\mathfrak q'} \cong S_{\mathfrak q}$.
\medskip\noindent
Step 7. By Lemma \ref{lemma-isomorphic-local-rings} 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 \'etale over $R$ at $\mathfrak q$, and (d) $S = R[x]/I$.
\medskip\noindent
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
$\overline{h} \in \kappa(\mathfrak p)[x]$. 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 $h \in I \subset R[x]$.
(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).
\medskip\noindent
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 $x$).
\medskip\noindent
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.
\medskip\noindent
Step 11.
We conclude that $\varphi : R[x]/(f) \to S$ is a surjective ring map,
$R[x]_g/(f)$ is \'etale over $R$ (because it is standard \'etale,
see Lemma \ref{lemma-standard-etale}) and $\varphi(g) \not \in \mathfrak q$.
Pick an element $g' \in R[x]/(f)$ such that
also $\varphi(g') \not \in \mathfrak q$ and $S_{\varphi(g')}$
is \'etale over $R$ (which exists since $S$ is \'etale over $R$ at
$\mathfrak q$). Then the ring map
$R[x]_{gg'}/(f) \to S_{\varphi(gg')}$ is a surjective map of \'etale
algebras over $R$. Hence it is \'etale by Lemma \ref{lemma-map-between-etale}.
Hence it is a localization by
Lemma \ref{lemma-surjective-flat-finitely-presented}.
Thus a localization of $S$ at an element not in $\mathfrak q$ is
isomorphic to a localization of a standard \'etale algebra over $R$
which is what we wanted to show.
\end{proof}
\noindent
The following two lemmas say that the \'etale topology is coarser than the
topology generated by Zariski coverings and finite flat morphisms.
They should be skipped on a first reading.
\begin{lemma}
\label{lemma-standard-etale-finite-flat-Zariski}
Let $R \to S$ be a standard \'etale morphism.
There exists a ring map $R \to S'$ with the following properties
\begin{enumerate}
\item $R \to S'$ is finite, finitely presented, and flat
(in other words $S'$ is finite projective as an $R$-module),
\item $\Spec(S') \to \Spec(R)$ is surjective,
\item for every prime $\mathfrak q \subset S$, lying over
$\mathfrak p \subset R$ and every prime
$\mathfrak q' \subset S'$ lying over $\mathfrak p$ there exists
a $g' \in S'$, $g' \not \in \mathfrak q'$
such that the ring map $R \to S'_{g'}$ factors
through a map $\varphi : S \to S'_{g'}$ with
$\varphi^{-1}(\mathfrak q'S'_{g'}) = \mathfrak q$.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $S = R[x]_g/(f)$ be a presentation of $S$ as in
Definition \ref{definition-standard-etale}.
Write $f = x^n + a_1 x^{n - 1} + \ldots + a_n$ with $a_i \in R$.
By Lemma \ref{lemma-adjoin-roots} there exists a finite locally free
and faithfully flat ring map $R \to S'$ such that $f = \prod (x - \alpha_i)$
for certain $\alpha_i \in S'$. Hence $R \to S'$ satisfies conditions (1), (2).
Let $\mathfrak q \subset R[x]/(f)$ be a prime ideal with
$g \not \in \mathfrak q$ (i.e., it corresponds to a prime of $S$).
Let $\mathfrak p = R \cap \mathfrak q$ and let
$\mathfrak q' \subset S'$ be a prime lying over $\mathfrak p$.
Note that there are
$n$ maps of $R$-algebras
\begin{eqnarray*}
\varphi_i : R[x]/(f) & \longrightarrow & S' \\
x & \longmapsto & \alpha_i
\end{eqnarray*}
To finish the proof we have to show that for some $i$ we have
(a) the image of $\varphi_i(g)$ in $\kappa(\mathfrak q')$ is not zero,
and (b) $\varphi_i^{-1}(\mathfrak q') = \mathfrak q$.
Because then we can just take $g' = \varphi_i(g)$, and
$\varphi = \varphi_i$ for that $i$.
\medskip\noindent
Let $\overline{f}$ denote the image of $f$ in $\kappa(\mathfrak p)[x]$.
Note that as a point of $\Spec(\kappa(\mathfrak p)[x]/(\overline{f}))$
the prime $\mathfrak q$ corresponds to an irreducible factor
$f_1$ of $\overline{f}$. Moreover, $g \not \in \mathfrak q$ means
that $f_1$ does not divide the image $\overline{g}$ of $g$ in
$\kappa(\mathfrak p)[x]$.
Denote $\overline{\alpha}_1, \ldots, \overline{\alpha}_n$ the images
of $\alpha_1, \ldots, \alpha_n$ in $\kappa(\mathfrak q')$.
Note that the polynomial $\overline{f}$ splits completely
in $\kappa(\mathfrak q')[x]$, namely
$$
\overline{f} = \prod\nolimits_i (x - \overline{\alpha}_i)
$$
Moreover $\varphi_i(g)$ reduces to $\overline{g}(\overline{\alpha}_i)$.
It follows we may pick $i$ such that $f_1(\overline{\alpha}_i) = 0$ and
$\overline{g}(\overline{\alpha}_i) \not = 0$.
For this $i$ properties (a) and (b) hold. Some details omitted.
\end{proof}
\begin{lemma}
\label{lemma-etale-finite-flat-zariski}
Let $R \to S$ be a ring map.
Assume that
\begin{enumerate}
\item $R \to S$ is \'etale, and
\item $\Spec(S) \to \Spec(R)$ is surjective.
\end{enumerate}
Then there exists a ring map $R \to S'$ such that
\begin{enumerate}
\item $R \to S'$ is finite, finitely presented, and flat
(in other words it is finite projective as an $R$-module),
\item $\Spec(S') \to \Spec(R)$ is surjective,
\item for every prime $\mathfrak q' \subset S'$ there exists a
$g' \in S'$, $g' \not \in \mathfrak q'$ such that
the ring map $R \to S'_{g'}$ factors as $R \to S \to S'_{g'}$.
\end{enumerate}
\end{lemma}
\begin{proof}
By Proposition \ref{proposition-etale-locally-standard} and
the quasi-compactness of $\Spec(S)$ (see Lemma \ref{lemma-quasi-compact})
we can find $g_1, \ldots, g_n \in S$ generating the unit ideal
of $S$ such that each $R \to S_{g_i}$ is standard \'etale.
If we prove the lemma for the ring map $R \to \prod_{i = 1, \ldots, n} S_{g_i}$
then the lemma follows for the ring map $R \to S$.
Hence we may assume that $S = \prod_{i = 1, \ldots, n} S_i$
is a finite product of standard \'etale morphisms.
\medskip\noindent
For each $i$ choose a ring map $R \to S_i'$ as in
Lemma \ref{lemma-standard-etale-finite-flat-Zariski}
adapted to the standard \'etale morphism $R \to S_i$.
Set $S' = S_1' \otimes_R \ldots \otimes_R S_n'$; we will use
the $R$-algebra maps $S_i' \to S'$ without further mention below.
We claim this works. Properties (1) and (2) are immediate.
For property (3) suppose that $\mathfrak q' \subset S'$ is a prime.
Denote $\mathfrak p$ its image in $\Spec(R)$.
Choose $i \in \{1, \ldots, n\}$ such that $\mathfrak p$
is in the image of $\Spec(S_i) \to \Spec(R)$; this is
possible by assumption. Set $\mathfrak q_i' \subset S_i'$
the image of $\mathfrak q'$ in the spectrum of $S_i'$.
By construction of $S'_i$ there exists a $g'_i \in S_i'$
such that $R \to (S_i')_{g_i'}$ factors as
$R \to S_i \to (S_i')_{g_i'}$. Hence also
$R \to S'_{g_i'}$ factors as
$$
R \to S_i \to (S_i')_{g_i'} \to S'_{g_i'}
$$
as desired.
\end{proof}
\begin{lemma}
\label{lemma-factor-mod-lift-etale}
Let $R$ be a ring. Let $f \in R[x]$ be a monic polynomial. Let $\mathfrak p$
be a prime of $R$. Let $f \bmod \mathfrak p = \overline{g} \overline{h}$
be a factorization of the image of $f$ in $\kappa(\mathfrak p)[x]$.
If $\gcd(\overline{g}, \overline{h}) = 1$, then there exist
\begin{enumerate}
\item an \'etale ring map $R \to R'$,
\item a prime $\mathfrak p' \subset R'$ lying over $\mathfrak p$, and
\item a factorization $f = g h$ in $R'[x]$
\end{enumerate}
such that
\begin{enumerate}
\item $\kappa(\mathfrak p) = \kappa(\mathfrak p')$,
\item $\overline{g} = g \bmod \mathfrak p'$,
$\overline{h} = h \bmod \mathfrak p'$, and
\item the polynomials $g, h$ generate the unit ideal in $R'[x]$.
\end{enumerate}
\end{lemma}
\begin{proof}
Suppose
$\overline{g} = \overline{b}_0 x^n + \overline{b}_1 x^{n - 1} + \ldots
+ \overline{b}_n$, and
$\overline{h} = \overline{c}_0 x^m + \overline{c}_1 x^{m - 1} + \ldots
+ \overline{c}_m$ with $\overline{b}_0, \overline{c}_0 \in \kappa(\mathfrak p)$
nonzero. After localizing $R$ at some element of $R$ not contained in
$\mathfrak p$ we may assume $\overline{b}_0$ is the
image of an invertible element $b_0 \in R$. Replacing
$\overline{g}$ by $\overline{g}/b_0$ and
$\overline{h}$ by $b_0\overline{h}$ we reduce to the case where
$\overline{g}$, $\overline{h}$ are monic (verification omitted).
Say $\overline{g} = x^n + \overline{b}_1 x^{n - 1} + \ldots + \overline{b}_n$,
and $\overline{h} = x^m + \overline{c}_1 x^{m - 1} + \ldots + \overline{c}_m$.
Write $f = x^{n + m} + a_1 x^{n - 1} + \ldots + a_{n + m}$.
Consider the fibre product
$$
R' = R \otimes_{\mathbf{Z}[a_1, \ldots, a_{n + m}]}
\mathbf{Z}[b_1, \ldots, b_n, c_1, \ldots, c_m]
$$
where the map $\mathbf{Z}[a_k] \to \mathbf{Z}[b_i, c_j]$
is as in Examples \ref{example-factor-polynomials} and
\ref{example-factor-polynomials-etale}. By construction there
is an $R$-algebra map
$$
R' = R \otimes_{\mathbf{Z}[a_1, \ldots, a_{n + m}]}
\mathbf{Z}[b_1, \ldots, b_n, c_1, \ldots, c_m]
\longrightarrow
\kappa(\mathfrak p)
$$
which maps $b_i$ to $\overline{b}_i$ and $c_j$ to $\overline{c}_j$.
Denote $\mathfrak p' \subset R'$ the kernel of this map.
Since by assumption the polynomials $\overline{g}, \overline{h}$
are relatively prime we see that the element
$\Delta = \text{Res}_x(g, h) \in \mathbf{Z}[b_i, c_j]$
(see Example \ref{example-factor-polynomials-etale})
does not map to zero in $\kappa(\mathfrak p)$ under the displayed map.
We conclude that $R \to R'$ is \'etale at $\mathfrak p'$.
In fact a solution to the problem posed in the lemma is
the ring map $R \to R'[1/\Delta]$ and the prime
$\mathfrak p' R'[1/\Delta]$. Because $\text{Res}_x(f, g)$ is
invertible in this ring the Sylvester matrix is invertible over
$R'$ and hence $1 = a g + b h$ for some $a, b \in R'[x]$
see Example \ref{example-factor-polynomials-etale}.
\end{proof}
\noindent
The following lemmas say roughly that after an \'etale extension
a quasi-finite ring map becomes finite.
To help interpret the results recall that the locus where a
finite type ring map is quasi-finite is open
(see Lemma \ref{lemma-quasi-finite-open}) and that formation of
this locus commutes with arbitrary base change
(see Lemma \ref{lemma-quasi-finite-base-change}).
\begin{lemma}
\label{lemma-produce-finite}
Let $R \to S' \to S$ be ring maps.
Let $\mathfrak p \subset R$ be a prime.
Let $g \in S'$ be an element.
Assume
\begin{enumerate}
\item $R \to S'$ is integral,
\item $R \to S$ is finite type,
\item $S'_g \cong S_g$, and
\item $g$ invertible in $S' \otimes_R \kappa(\mathfrak p)$.
\end{enumerate}
Then there exists a $f \in R$, $f \not \in \mathfrak p$ such
that $R_f \to S_f$ is finite.
\end{lemma}
\begin{proof}
By assumption the image $T$ of $V(g) \subset \Spec(S')$ under
the morphism $\Spec(S') \to \Spec(R)$ does not
contain $\mathfrak p$. By Section \ref{section-going-up}
especially, Lemma \ref{lemma-going-up-closed} we see $T$ is closed.
Pick $f \in R$, $f \not \in \mathfrak p$ such that
$T \cap D(f) = \emptyset$. Then we see that $g$ becomes invertible
in $S'_f$. Hence $S'_f \cong S_f$. Thus $S_f$ is both of finite type
and integral over $R_f$, hence finite.
\end{proof}
\begin{lemma}
\label{lemma-etale-makes-quasi-finite-finite-one-prime}
Let $R \to S$ be a ring map.
Let $\mathfrak q \subset S$ be a prime lying over
the prime $\mathfrak p \subset R$.
Assume $R \to S$ finite type and quasi-finite at $\mathfrak q$.
Then there exists
\begin{enumerate}
\item an \'etale ring map $R \to R'$,
\item a prime $\mathfrak p' \subset R'$ lying over $\mathfrak p$,
\item a product decomposition
$$
R' \otimes_R S = A \times B
$$
\end{enumerate}
with the following properties
\begin{enumerate}
\item $\kappa(\mathfrak p) = \kappa(\mathfrak p')$,
\item $R' \to A$ is finite,
\item $A$ has exactly one prime $\mathfrak r$ lying over $\mathfrak p'$, and
\item $\mathfrak r$ lies over $\mathfrak q$.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $S' \subset S$ be the integral closure of $R$ in $S$.
Let $\mathfrak q' = S' \cap \mathfrak q$.
By Zariski's Main Theorem \ref{theorem-main-theorem}
there exists a $g \in S'$, $g \not \in \mathfrak q'$ such
that $S'_g \cong S_g$. Consider the fibre rings
$F = S \otimes_R \kappa(\mathfrak p)$ and
$F' = S' \otimes_R \kappa(\mathfrak p)$. Denote $\overline{\mathfrak q}'$
the prime of $F'$ corresponding to $\mathfrak q'$. Since
$F'$ is integral over $\kappa(\mathfrak p)$ we see
that $\overline{\mathfrak q}'$ is a closed point of
$\Spec(F')$, see Lemma \ref{lemma-integral-over-field}.
Note that $\mathfrak q$ defines an isolated closed point
$\overline{\mathfrak q}$ of
$\Spec(F)$ (see Definition \ref{definition-quasi-finite}).
Since $S'_g \cong S_g$ we have $F'_g \cong F_g$,
so $\overline{\mathfrak q}$ and $\overline{\mathfrak q}'$
have isomorphic open neighbourhoods in $\Spec(F)$
and $\Spec(F')$. We conclude the set
$\{\overline{\mathfrak q}'\} \subset \Spec(F')$ is
open. Combined with $\mathfrak q'$ being closed (shown above)
we conclude that $\overline{\mathfrak q}'$ defines
an isolated closed point of $\Spec(F')$ as well.
\medskip\noindent
An additional small remark is that under the map
$\Spec(F) \to \Spec(F')$ the point $\overline{\mathfrak q}$
is the only point mapping to $\overline{\mathfrak q}'$. This follows
from the discussion above.
\medskip\noindent
By Lemma \ref{lemma-disjoint-implies-product} we may write
$F' = F'_1 \times F'_2$ with
$\Spec(F'_1) = \{\overline{\mathfrak q}'\}$.
Since $F' = S' \otimes_R \kappa(\mathfrak p)$, there
exists an $s' \in S'$ which maps to the element
$(r, 0) \in F'_1 \times F'_2 = F'$ for some $r \in R$, $r \not \in \mathfrak p$.
In fact, what we will use about $s'$ is that it is an element of $S'$,
not contained in $\mathfrak q'$, and contained in any other prime
lying over $\mathfrak p$.
\medskip\noindent
Let $f(x) \in R[x]$ be a monic polynomial such that $f(s') = 0$.
Denote $\overline{f} \in \kappa(\mathfrak p)[x]$ the image.
We can factor it as $\overline{f} = x^e \overline{h}$ where
$\overline{h}(0) \not = 0$. By Lemma \ref{lemma-factor-mod-lift-etale}
we can find an \'etale ring extension $R \to R'$,
a prime $\mathfrak p'$ lying over $\mathfrak p$, and
a factorization $f = h i$ in $R'[x]$ such that
$\kappa(\mathfrak p) = \kappa(\mathfrak p')$,
$\overline{h} = h \bmod \mathfrak p'$,
$x^e = i \bmod \mathfrak p'$, and
we can write $a h + b i = 1$ in $R'[x]$ (for suitable $a, b$).
\medskip\noindent
Consider the elements $h(s'), i(s') \in R' \otimes_R S'$.
By construction we have $h(s')i(s') = f(s') = 0$. On the other
hand they generate the unit ideal since $a(s')h(s') + b(s')i(s') = 1$.
Thus we see that $R' \otimes_R S'$ is the product of the
localizations at these elements:
$$
R' \otimes_R S'
=
(R' \otimes_R S')_{h(s')}
\times
(R' \otimes_R S')_{i(s')}
=
S'_1 \times S'_2
$$
Moreover this product decomposition is compatible with the product
decomposition we found for the fibre ring $F'$; this comes from our
choice of $s', h$ which guarantee that $\overline{\mathfrak q}'$
is the only prime of $F'$ which does not contain the image of $h(s')$
in $F'$. Here we use that the fibre ring of $R'\otimes_R S'$ over $R'$ at
$\mathfrak p'$ is the same as $F'$ due to the fact that
$\kappa(\mathfrak p) = \kappa(\mathfrak p')$.
It follows that $S'_1$ has exactly
one prime, say $\mathfrak r'$,
lying over $\mathfrak p'$ and
that this prime lies over $\mathfrak q$.
Hence the element $g \in S'$ maps to an element of $S'_1$ not contained
in $\mathfrak r'$.
\medskip\noindent
The base change $R'\otimes_R S$ inherits a similar product decomposition
$$
R' \otimes_R S
=
(R' \otimes_R S)_{h(s')}
\times
(R' \otimes_R S)_{i(s')}
=
S_1 \times S_2
$$
It follows from the above that $S_1$ has exactly
one prime, say $\mathfrak r$,
lying over $\mathfrak p'$ (consider the fibre ring as above),
and that this prime lies over $\mathfrak q$.
\medskip\noindent
Now we may apply Lemma \ref{lemma-produce-finite} to the ring maps
$R' \to S'_1 \to S_1$, the prime $\mathfrak p'$ and
the element $g$ to see that after replacing $R'$ by
a principal localization we can assume that $S_1$ is
finite over $R'$ as desired.
\end{proof}
\begin{lemma}
\label{lemma-etale-makes-quasi-finite-finite}
Let $R \to S$ be a ring map.
Let $\mathfrak p \subset R$ be a prime.
Assume $R \to S$ finite type.
Then there exists
\begin{enumerate}
\item an \'etale ring map $R \to R'$,
\item a prime $\mathfrak p' \subset R'$ lying over $\mathfrak p$,
\item a product decomposition
$$
R' \otimes_R S = A_1 \times \ldots \times A_n \times B
$$
\end{enumerate}
with the following properties
\begin{enumerate}
\item we have $\kappa(\mathfrak p) = \kappa(\mathfrak p')$,
\item each $A_i$ is finite over $R'$,
\item each $A_i$ has exactly one prime $\mathfrak r_i$ lying over
$\mathfrak p'$, and
\item $R' \to B$ not quasi-finite at any prime lying over $\mathfrak p'$.
\end{enumerate}
\end{lemma}
\begin{proof}
Denote $F = S \otimes_R \kappa(\mathfrak p)$ the fibre ring of $S/R$
at the prime $\mathfrak p$. As $F$ is of finite type over $\kappa(\mathfrak p)$
it is Noetherian and hence $\Spec(F)$ has finitely many isolated closed
points. If there are no isolated closed points,
i.e., no primes $\mathfrak q$ of $S$ over $\mathfrak p$ such that
$S/R$ is quasi-finite at $\mathfrak q$, then the lemma holds.
If there exists at least one such prime $\mathfrak q$, then
we may apply Lemma \ref{lemma-etale-makes-quasi-finite-finite-one-prime}.
This gives a diagram
$$
\xymatrix{
S \ar[r] & R'\otimes_R S \ar@{=}[r] & A_1 \times B' \\
R \ar[r] \ar[u] & R' \ar[u] \ar[ru]
}
$$
as in said lemma. Since the residue fields at $\mathfrak p$ and $\mathfrak p'$
are the same, the fibre rings of $S/R$ and $(A \times B)/R'$
are the same. Hence, by induction on the number of isolated closed points
of the fibre we may assume that the lemma holds for
$R' \to B$ and $\mathfrak p'$. Thus we get an \'etale ring
map $R' \to R''$, a prime $\mathfrak p'' \subset R''$ and
a decomposition
$$
R'' \otimes_{R'} B' = A_2 \times \ldots \times A_n \times B
$$
We omit the verification that the ring map $R \to R''$, the
prime $\mathfrak p''$ and the resulting decomposition
$$
R'' \otimes_R S = (R'' \otimes_{R'} A_1) \times
A_2 \times \ldots \times A_n \times B
$$
is a solution to the problem posed in the lemma.
\end{proof}
\begin{lemma}
\label{lemma-etale-makes-quasi-finite-finite-variant}
Let $R \to S$ be a ring map.
Let $\mathfrak p \subset R$ be a prime.
Assume $R \to S$ finite type.
Then there exists
\begin{enumerate}
\item an \'etale ring map $R \to R'$,
\item a prime $\mathfrak p' \subset R'$ lying over $\mathfrak p$,
\item a product decomposition
$$
R' \otimes_R S = A_1 \times \ldots \times A_n \times B
$$
\end{enumerate}
with the following properties
\begin{enumerate}
\item each $A_i$ is finite over $R'$,
\item each $A_i$ has exactly one prime $\mathfrak r_i$ lying over
$\mathfrak p'$,
\item the finite field extensions
$\kappa(\mathfrak p') \subset \kappa(\mathfrak r_i)$
are purely inseparable, and
\item $R' \to B$ not quasi-finite at any prime lying over $\mathfrak p'$.
\end{enumerate}
\end{lemma}
\begin{proof}
The strategy of the proof is to make two \'etale ring
extensions: first we control the residue fields, then we
apply Lemma \ref{lemma-etale-makes-quasi-finite-finite}.
\medskip\noindent
Denote $F = S \otimes_R \kappa(\mathfrak p)$ the fibre ring of $S/R$
at the prime $\mathfrak p$.
As in the proof of Lemma \ref{lemma-etale-makes-quasi-finite-finite}
there are finitely may primes, say
$\mathfrak q_1, \ldots, \mathfrak q_n$ of $S$ lying over
$R$ at which the ring map $R \to S$ is quasi-finite.
Let $\kappa(\mathfrak p) \subset L_i \subset \kappa(\mathfrak q_i)$
be the subfield such that $\kappa(\mathfrak p) \subset L_i$
is separable, and the field extension $L_i \subset \kappa(\mathfrak q_i)$
is purely inseparable. Let $\kappa(\mathfrak p) \subset L$
be a finite Galois extension into which $L_i$ embeds for $i = 1, \ldots, n$.
By Lemma \ref{lemma-make-etale-map-prescribed-residue-field}
we can find an \'etale ring extension
$R \to R'$ together with a prime $\mathfrak p'$ lying over $\mathfrak p$
such that the field extension
$\kappa(\mathfrak p) \subset \kappa(\mathfrak p')$ is isomorphic
to $\kappa(\mathfrak p) \subset L$.
Thus the fibre ring of $R' \otimes_R S$ at $\mathfrak p'$ is
isomorphic to $F \otimes_{\kappa(\mathfrak p)} L$.
The primes lying over $\mathfrak q_i$ correspond to primes
of $\kappa(\mathfrak q_i) \otimes_{\kappa(\mathfrak p)} L$
which is a product of fields purely inseparable over
$L$ by our choice of $L$ and elementary field theory.
These are also the only primes over $\mathfrak p'$
at which $R' \to R' \otimes_R S$ is quasi-finite, by
Lemma \ref{lemma-quasi-finite-base-change}.
Hence after replacing $R$ by $R'$, $\mathfrak p$ by $\mathfrak p'$,
and $S$ by $R' \otimes_R S$ we may assume that for all
primes $\mathfrak q$ lying over $\mathfrak p$
for which $S/R$ is quasi-finite the field extensions
$\kappa(\mathfrak p) \subset \kappa(\mathfrak q)$
are purely inseparable.
\medskip\noindent
Next apply Lemma \ref{lemma-etale-makes-quasi-finite-finite}.
The result is what we want since the field extensions do not
change under this \'etale ring extension.
\end{proof}
```

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