The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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 étale if 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 $R \to S$ is étale 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 any étale 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.

  1. The ring map $R \to R_ f$ is étale for any ring $R$ and any $f \in R$.

  2. Compositions of étale ring maps are étale.

  3. A base change of an étale ring map is étale.

  4. 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.

  5. 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.

  6. An étale ring map is syntomic, in particular flat.

  7. If $S$ is finite type over a field $k$, then $S$ is étale over $k$ if and only if $\Omega _{S/k} = 0$.

  8. 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 $\mathbf{Z}$.

  9. 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 $B \cong A \otimes _{A_ i} B_ i$.

  10. 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 10.22.2.

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

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

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

Proof. 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 $\kappa (\mathfrak p)$. $\square$

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

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

  2. $R_{\mathfrak p} \to S_{\mathfrak q}$ is flat

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

  4. 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., $R \to S$ 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 10.22.1. $\square$

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 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 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:

  1. $R \to S$ is étale at $\mathfrak q$,

  2. $\Delta = \text{Res}_ x(g, h) \not\in \mathfrak q$,

  3. 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.

  1. The ring map $R \to R[x]_ g/(f)$ is étale.

  2. 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.

  3. Any principal localization of $R[x]_ g/(f)$ is standard étale over $R$.

  4. A composition of standard étale maps is not standard 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 10.122.14 there exists a finite ring map $R \to S'$, an $R$-algebra map $S' \to S$, an element $g' \in S'$ such that $g' \not\in \mathfrak q$ such that $S' \to S$ induces an isomorphism $S'_{g'} \cong S_{g'}$. (Note that 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

  1. $R \to S'$ is finite, finitely presented, and flat (in other words $S'$ is finite projective as an $R$-module),

  2. $\mathop{\mathrm{Spec}}(S') \to \mathop{\mathrm{Spec}}(R)$ is surjective,

  3. 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

  1. $R \to S$ is étale, and

  2. $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is surjective.

Then there exists a ring map $R \to S'$ such that

  1. $R \to S'$ is finite, finitely presented, and flat (in other words it is finite projective as an $R$-module),

  2. $\mathop{\mathrm{Spec}}(S') \to \mathop{\mathrm{Spec}}(R)$ is surjective,

  3. 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

  1. an étale ring map $R \to R'$,

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

  3. a factorization $f = g h$ in $R'[x]$

such that

  1. $\kappa (\mathfrak p) = \kappa (\mathfrak p')$,

  2. $\overline{g} = g \bmod \mathfrak p'$, $\overline{h} = h \bmod \mathfrak p'$, and

  3. 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 10.122.13) 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

  1. $R \to S'$ is integral,

  2. $R \to S$ is finite type,

  3. $S'_ g \cong S_ g$, and

  4. $g$ 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

  1. an étale ring map $R \to R'$,

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

  3. a product decomposition

    \[ R' \otimes _ R S = A \times B \]

with the following properties

  1. $\kappa (\mathfrak p) = \kappa (\mathfrak p')$,

  2. $R' \to A$ is finite,

  3. $A$ has exactly one prime $\mathfrak r$ lying over $\mathfrak p'$, and

  4. $\mathfrak r$ 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 10.122.12 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 10.23.3 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

  1. an étale ring map $R \to R'$,

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

  3. a product decomposition

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

with the following properties

  1. we have $\kappa (\mathfrak p) = \kappa (\mathfrak p')$,

  2. each $A_ i$ is finite over $R'$,

  3. each $A_ i$ has exactly one prime $\mathfrak r_ i$ lying over $\mathfrak p'$, and

  4. $R' \to B$ 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

  1. an étale ring map $R \to R'$,

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

  3. a product decomposition

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

with the following properties

  1. each $A_ i$ is finite over $R'$,

  2. each $A_ i$ has exactly one prime $\mathfrak r_ i$ lying over $\mathfrak p'$,

  3. the finite field extensions $\kappa (\mathfrak p') \subset \kappa (\mathfrak r_ i)$ are purely inseparable, and

  4. $R' \to B$ 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. $\square$


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