## 10.144 Local structure of étale ring maps

Lemma 10.143.2 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 $k'/k$ 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 10.144.1. 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.

In Proposition 10.144.4 we show that every étale ring map is locally standard étale.

Lemma 10.144.2. 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 10.144.3. Let $R$ be a ring. Let $\mathfrak p$ be a prime of $R$. Let $L/\kappa (\mathfrak p)$ 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 $\kappa (\mathfrak p')/\kappa (\mathfrak p)$ 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 10.144.4. 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 10.143.1 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 10.143.3 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 10.143.6. By Lemma 10.123.14 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 10.53.2) and so a finite product of local rings

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

see Proposition 10.60.7. 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 10.143.5. 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 $\kappa (\mathfrak q)/\kappa (\mathfrak p)$ (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 10.41.11 (we have going up for $S' \to S$ by Lemma 10.36.22 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 10.143.5. 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 10.20.1 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 10.126.7 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 10.144.2) 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 10.143.8. Hence it is a localization by Lemma 10.143.9. 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 10.144.5. 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 10.144.1. Write $f = x^ n + a_1 x^{n - 1} + \ldots + a_ n$ with $a_ i \in R$. By Lemma 10.136.9 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 10.144.6. 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 10.144.4 and the quasi-compactness of $\mathop{\mathrm{Spec}}(S)$ (see Lemma 10.17.10) 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 10.144.5 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. $\square$

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