Definition 10.143.1. 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.

## 10.143 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.

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 10.143.3 below. But before we do so we prove that *any* étale ring map is standard smooth.

Lemma 10.143.2. 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

is an isomorphism, in particular $I/I^2$ is a free $S$-module. Thus by Lemma 10.136.6 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 10.143.3. 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 $B' \cong S^{-1}B$.

Let $A$ be a ring. Let $B = B' \times B''$ be a product of $A$-algebras. Then $B$ is étale over $A$ if and only if both $B'$ and $B''$ are étale over $A$.

**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 10.137.14. By Lemma 10.131.7 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 10.137.4. Since $\Omega _{S'/R'} = S' \otimes _ S \Omega _{S/R}$ by Lemma 10.131.12 we conclude that $\Omega _{S'/R'} = 0$. Hence $R' \to S'$ is étale.

Proof of (4). Assume the hypotheses of (4). By Lemma 10.137.13 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 10.137.18. 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 10.40.6.

Proof of (6). Follows directly from the corresponding result for smooth ring maps (Lemma 10.137.10).

Proof of (7). Follows from Lemma 10.140.3 and the definitions.

Proof of (8). Lemma 10.138.14 gives the result for smooth ring maps. The resulting smooth ring map $R_0 \to S_0$ satisfies the hypotheses of Lemma 10.130.8, 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 10.127.3.

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

Proof of (11). Use Lemma 10.137.15 to see the result for smoothness and then use that $\Omega _{B/A}$ is zero if and only if both $\Omega _{B'/A}$ and $\Omega _{B''/A}$ are zero. $\square$

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

Lemma 10.143.4. 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.**
We are going to use without further mention: if $S = S_1 \times \ldots \times S_ n$ is a finite product of $k$-algebras, then $S$ is étale over $k$ if and only if each $S_ i$ is étale over $k$. See Lemma 10.143.3 part (11).

If $k'/k$ 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. Thus if $S$ is a finite product of finite separable extensions of $k$, then $S$ is étale over $k$.

Conversely, suppose that $k \to S$ is étale. Then $S$ is smooth over $k$ and $\Omega _{S/k} = 0$. By Lemma 10.140.3 we see that $\dim _\mathfrak m \mathop{\mathrm{Spec}}(S) = 0$ for every maximal ideal $\mathfrak m$ of $S$. Thus $\dim (S) = 0$. By Proposition 10.60.7 we find that $S$ is a finite product of Artinian local rings. By the already used Lemma 10.140.3 these local rings are fields. Hence we may assume $S = k'$ is a field. By the Hilbert Nullstellensatz (Theorem 10.34.1) we see that the extension $k'/k$ is finite. The smoothness of $k \to k'$ implies by Lemma 10.140.9 that $k'/k$ is a separable extension and the proof is complete. $\square$

Lemma 10.143.5. 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 $\kappa (\mathfrak q)/\kappa (\mathfrak p)$ 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 10.143.4 by unwinding the fact that $S \otimes _ R \kappa (\mathfrak p)$ is étale over $\kappa (\mathfrak p)$.
$\square$

Lemma 10.143.6. 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 10.122.4.
$\square$

Lemma 10.143.7. 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 $\kappa (\mathfrak q)/\kappa (\mathfrak p)$ is finite separable,

then $R \to S$ is étale at $\mathfrak q$.

**Proof.**
Apply Lemma 10.122.2 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 10.137.17 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 10.137.10 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 10.143.8. 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 10.143.5 the ring map $S'_{\mathfrak q'}/\mathfrak p S'_{\mathfrak q'} \to S_{\mathfrak q}/\mathfrak p S_{\mathfrak q}$ is a map of finite separable extensions of $\kappa (\mathfrak p)$. In particular it is flat. Hence by Lemma 10.128.8 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 10.143.7 we conclude that $S' \to S$ is étale at $\mathfrak q$. Since being étale is local (see Lemma 10.143.3) we win.
$\square$

Lemma 10.143.9. 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 10.21.5.
$\square$

**Second proof.**
Since $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is a homeomorphism onto a closed subset (see Lemma 10.17.7) and is open (see Proposition 10.41.8) we see that the image is $D(e)$ for some idempotent $e \in R$ (see Lemma 10.21.3). Thus $R_ e \to S$ induces a bijection on spectra. Now this map induces an isomorphism on all local rings for example by Lemmas 10.78.5 and 10.20.1. Then it follows that $R_ e \to S$ is also injective, for example see Lemma 10.23.1.
$\square$

Lemma 10.143.10. 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 10.143.2 we can write $\overline{S} = (R/I)[x_1, \ldots , x_ n]/(\overline{f}_1, \ldots , \overline{f}_ n)$ as in Definition 10.137.6 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 10.137.8. This proves the lemma.
$\square$

Lemma 10.143.11. Consider a commutative diagram

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 10.143.10 there exists an étale ring map $A' \to C$ such that $C/IC = B$. Then $A' \to C$ is formally smooth (by Proposition 10.138.13) 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 10.143.12. Let $n , m \geq 1$ be integers. Consider the ring map

of Example 10.136.7. Write symbolically

where for example $a_1(b_ i, c_ j) = b_1 + c_1$. The matrix of partial derivatives is

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

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

is étale.

Lemma 10.143.13. 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 $f = x^{n + m} + a_1 x^{n + m - 1} + \ldots + a_{n + m}$. Consider the fibre product

where the map $\mathbf{Z}[a_ k] \to \mathbf{Z}[b_ i, c_ j]$ is as in Examples 10.136.7 and 10.143.12. By construction there is an $R$-algebra map

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 10.143.12) 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 $R'[1/\Delta ]$ and hence $1 = a g + b h$ for some $a, b \in R'[1/\Delta ][x]$ see Example 10.143.12. $\square$

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