The Stacks project

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 10.136.14 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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0G1A. Beware of the difference between the letter 'O' and the digit '0'.