The Stacks project

Proof. Write $f(x + y) - f(x) = f'(x)y + g(x, y) y^2$ in $R[x, y]$ (this is possible as one sees by expanding $f(x + y)$; details omitted). Then we see that $0 = f(b) - f(a) = f(a + (b - a)) - f(a) = f'(a)(b - a) + c (b - a)^2$ for some $c \in R$. By assumption $f'(a)$ is a unit in $R$. Hence $(b - a)(1 + f'(a)^{-1}c(b - a)) = 0$. By assumption $b - a \in \mathfrak m$, hence $1 + f'(a)^{-1}c(b - a)$ is a unit in $R$. Hence $b - a = 0$ in $R$. $\square$

Proof. Here is a list of the easier implications:

1. 2$\Rightarrow $1 because in (2) we consider all polynomials and in (1) only monic ones,

2. 5$\Rightarrow $3 because in (5) we consider all polynomials and in (3) only monic ones,

3. 6$\Rightarrow $4 because in (6) we consider all polynomials and in (4) only monic ones,

4. 4$\Rightarrow $3 is obvious,

5. 6$\Rightarrow $5 is obvious,

6. 8$\Rightarrow $7 is obvious,

7. 10$\Rightarrow $9 is obvious,

8. 11$\Leftrightarrow $12 by definition of being quasi-finite at a prime,

9. 11$\Rightarrow $13 by definition of being quasi-finite,

Proof of 1$\Rightarrow $8. Assume (1). Let $R \to S$ be étale, and let $\mathfrak q \subset S$ be a prime ideal such that $\kappa (\mathfrak q) \cong \kappa $. By Proposition 10.144.4 we can find a $g \in S$, $g \not\in \mathfrak q$ such that $R \to S_ g$ is standard étale. After replacing $S$ by $S_ g$ we may assume that $S = R[t]_ g/(f)$ is standard étale (details omitted). Since the prime $\mathfrak q$ has residue field $\kappa $ it corresponds to a root $a_0$ of $\overline{f}$ which is not a root of $\overline{g}$. By definition of a standard étale algebra this also means that $\overline{f'}(a_0) \not= 0$. Since also $f$ is monic by definition of a standard étale algebra again we may use that $R$ is henselian to conclude that there exists an $a \in R$ with $a_0 = \overline{a}$ such that $f(a) = 0$. This implies that $g(a)$ is a unit of $R$ and we obtain the desired map $\tau : S = R[t]_ g/(f) \to R$ by the rule $t \mapsto a$. By construction $\tau ^{-1}(\mathfrak m) = \mathfrak q$. By Lemma 10.153.2 the map $\tau $ is unique. This proves (8) holds.

Proof of 7$\Rightarrow $8. (This is really unimportant and should be skipped.) Assume (7) holds and assume $R \to S$ is étale. Let $\mathfrak q_1, \ldots , \mathfrak q_ r$ be the other primes of $S$ lying over $\mathfrak m$. Then we can find a $g \in S$, $g \not\in \mathfrak q$ and $g \in \mathfrak q_ i$ for $i = 1, \ldots , r$. Namely, we can argue that $\bigcap _{i=1}^{r} \mathfrak {q}_{i} \not\subset \mathfrak {q}$ since otherwise $\mathfrak {q}_{i} \subset \mathfrak {q}$ for some $i$, but this cannot happen as the fiber of an étale morphism is discrete (use Lemma 10.143.4 for example). Apply (7) to the étale ring map $R \to S_ g$ and the prime $\mathfrak qS_ g$. This gives a retraction $\tau _ g : S_ g \to R$ such that the composition $\tau : S \to S_ g \to R$ has the property $\tau ^{-1}(\mathfrak m) = \mathfrak q$. Details omitted.

Proof of 8$\Rightarrow $11. Assume (8) and let $R \to S$ be a finite type ring map. Apply Lemma 10.145.3. We find an étale ring map $R \to R'$ and a prime $\mathfrak m' \subset R'$ lying over $\mathfrak m$ with $\kappa = \kappa (\mathfrak m')$ such that $R' \otimes _ R S = A' \times B'$ with $A'$ finite over $R'$ and $B'$ not quasi-finite over $R'$ at any prime lying over $\mathfrak m'$. Apply (8) to get a retraction $\tau : R' \to R$ with $\mathfrak m = \tau ^{-1}(\mathfrak m')$. Then use that

which gives a decomposition as in (11).

Proof of 8$\Rightarrow $10. Assume (8) and let $R \to S$ be a finite ring map. Apply Lemma 10.145.3. We find an étale ring map $R \to R'$ and a prime $\mathfrak m' \subset R'$ lying over $\mathfrak m$ with $\kappa = \kappa (\mathfrak m')$ such that $R' \otimes _ R S = A'_1 \times \ldots \times A'_ n \times B'$ with $A'_ i$ finite over $R'$ having exactly one prime over $\mathfrak m'$ and $B'$ not quasi-finite over $R'$ at any prime lying over $\mathfrak m'$. Apply (8) to get a retraction $\tau : R' \to R$ with $\mathfrak m' = \tau ^{-1}(\mathfrak m)$. Then we obtain

The factor $B$ is finite over $R$ but $R \to B$ is not quasi-finite at any prime lying over $\mathfrak m$. Hence $B = 0$. The factors $A_ i$ are finite $R$-algebras having exactly one prime lying over $\mathfrak m$, hence they are local rings. This proves that $S$ is a finite product of local rings.

Proof of 9$\Rightarrow $10. This holds because if $S$ is finite over the local ring $R$, then it has at most finitely many maximal ideals. Namely, by going up for $R \to S$ the maximal ideals of $S$ all lie over $\mathfrak m$, and $S/\mathfrak mS$ is Artinian hence has finitely many primes.

Proof of 10$\Rightarrow $1. Assume (10). Let $f \in R[T]$ be a monic polynomial and $a_0 \in \kappa $ a simple root of $\overline{f}$. Then $S = R[T]/(f)$ is a finite $R$-algebra. Applying (10) we get $S = A_1 \times \ldots \times A_ r$ is a finite product of local $R$-algebras. In particular we see that $S/\mathfrak mS = \prod A_ i/\mathfrak mA_ i$ is the decomposition of $\kappa [T]/(\overline{f})$ as a product of local rings. This means that one of the factors, say $A_1/\mathfrak mA_1$ is the quotient $\kappa [T]/(\overline{f}) \to \kappa [T]/(T - a_0)$. Since $A_1$ is a summand of the finite free $R$-module $S$ it is a finite free $R$-module itself. As $A_1/\mathfrak mA_1$ is a $\kappa $-vector space of dimension 1 we see that $A_1 \cong R$ as an $R$-module. Clearly this means that $R \to A_1$ is an isomorphism. Let $a \in R$ be the image of $T$ under the map $R[T] \to S \to A_1 \to R$. Then $f(a) = 0$ and $\overline{a} = a_0$ as desired.

Proof of 13$\Rightarrow $1. Assume (13). Let $f \in R[T]$ be a monic polynomial and $a_0 \in \kappa $ a simple root of $\overline{f}$. Then $S_1 = R[T]/(f)$ is a finite $R$-algebra. Let $g \in R[T]$ be any element such that $\overline{g} = \overline{f}/(T - a_0)$. Then $S = (S_1)_ g$ is a quasi-finite $R$-algebra such that $S \otimes _ R \kappa \cong \kappa [T]_{\overline{g}}/(\overline{f}) \cong \kappa [T]/(T - a_0) \cong \kappa $. Applying (13) to $S$ we get $S = A \times B$ with $A$ finite over $R$ and $B \otimes _ R \kappa = 0$. In particular we see that $\kappa \cong S/\mathfrak mS = A/\mathfrak mA$. Since $A$ is a summand of the flat $R$-algebra $S$ we see that it is finite flat, hence free over $R$. As $A/\mathfrak mA$ is a $\kappa $-vector space of dimension 1 we see that $A \cong R$ as an $R$-module. Clearly this means that $R \to A$ is an isomorphism. Let $a \in R$ be the image of $T$ under the map $R[T] \to S \to A \to R$. Then $f(a) = 0$ and $\overline{a} = a_0$ as desired.

Proof of 8$\Rightarrow $2. Assume (8). Let $f \in R[T]$ be any polynomial and let $a_0 \in \kappa $ be a simple root. Then the algebra $S = R[T]_{f'}/(f)$ is étale over $R$. Let $\mathfrak q \subset S$ be the prime generated by $\mathfrak m$ and $T - b$ where $b \in R$ is any element such that $\overline{b} = a_0$. Apply (8) to $S$ and $\mathfrak q$ to get $\tau : S \to R$. Then the image $\tau (T) = a \in R$ works in (2).

At this point we see that (1), (2), (7), (8), (9), (10), (11), (12), (13) are all equivalent. The weakest assertion of (3), (4), (5) and (6) is (3) and the strongest is (6). Hence we still have to prove that (3) implies (1) and (1) implies (6).

Proof of 3$\Rightarrow $1. Assume (3). Let $f \in R[T]$ be monic and let $a_0 \in \kappa $ be a simple root of $\overline{f}$. This gives a factorization $\overline{f} = (T - a_0)h_0$ with $h_0(a_0) \not= 0$, so $\gcd (T - a_0, h_0) = 1$. Apply (3) to get a factorization $f = gh$ with $\overline{g} = T - a_0$ and $\overline{h} = h_0$. Set $S = R[T]/(f)$ which is a finite free $R$-algebra. We will write $g$, $h$ also for the images of $g$ and $h$ in $S$. Then $gS + hS = S$ by Nakayama's Lemma 10.20.1 as the equality holds modulo $\mathfrak m$. Since $gh = f = 0$ in $S$ this also implies that $gS \cap hS = 0$. Hence by the Chinese Remainder theorem we obtain $S = S/(g) \times S/(h)$. This implies that $A = S/(g)$ is a summand of a finite free $R$-module, hence finite free. Moreover, the rank of $A$ is $1$ as $A/\mathfrak mA = \kappa [T]/(T - a_0)$. Thus the map $R \to A$ is an isomorphism. Setting $a \in R$ equal to the image of $T$ under the maps $R[T] \to S \to A \to R$ gives an element of $R$ with $f(a) = 0$ and $\overline{a} = a_0$.

Proof of 1$\Rightarrow $6. Assume (1) or equivalently all of (1), (2), (7), (8), (9), (10), (11), (12), (13). Let $f \in R[T]$ be a polynomial. Suppose that $\overline{f} = g_0h_0$ is a factorization with $\gcd (g_0, h_0) = 1$. We may and do assume that $g_0$ is monic. Consider $S = R[T]/(f)$. Because we have the factorization we see that the coefficients of $f$ generate the unit ideal in $R$. This implies that $S$ has finite fibres over $R$, hence is quasi-finite over $R$. It also implies that $S$ is flat over $R$ by Lemma 10.128.5. Combining (13) and (10) we may write $S = A_1 \times \ldots \times A_ n \times B$ where each $A_ i$ is local and finite over $R$, and $B \otimes _ R \kappa = 0$. After reordering the factors $A_1, \ldots , A_ n$ we may assume that

as quotients of $\kappa [T]$. The finite flat $R$-algebra $A = A_1 \times \ldots \times A_ r$ is free as an $R$-module, see Lemma 10.78.5. Its rank is $\deg _ T(g_0)$. Let $g \in R[T]$ be the characteristic polynomial of the $R$-linear operator $T : A \to A$. Then $g$ is a monic polynomial of degree $\deg _ T(g) = \deg _ T(g_0)$ and moreover $\overline{g} = g_0$. By Cayley-Hamilton (Lemma 10.16.1) we see that $g(T_ A) = 0$ where $T_ A$ indicates the image of $T$ in $A$. Hence we obtain a well defined surjective map $R[T]/(g) \to A$ which is an isomorphism by Nakayama's Lemma 10.20.1. The map $R[T] \to A$ factors through $R[T]/(f)$ by construction hence we may write $f = gh$ for some $h$. This finishes the proof. $\square$

Proof. Part (2) implies part (1) since $S$ as in part (1) is a finite product of its localizations at the primes lying over $\mathfrak m$ by Lemma 10.153.3 part (10). Part (2) also follows from Lemma 10.153.3 part (10) since any finite $S$-algebra is also a finite $R$-algebra (of course any finite ring map between local rings is local).

Let $R \to S$ and $\mathfrak q$ be as in (3). Write $S = A \times B$ with $A$ finite over $R$ and $B$ not quasi-finite over $R$ at any prime lying over $\mathfrak m$, see Lemma 10.153.3 part (11). Hence $S_\mathfrak q$ is a localization of $A$ at a maximal ideal and we deduce (3) from (1). Part (4) follows from part (3). $\square$

Proof. This is a combination of parts (11) and (10) of Lemma 10.153.3. $\square$

Proof. First write $S = A_1 \times \ldots \times A_ n \times B$ as in Lemma 10.153.5. The field extension $\kappa (\mathfrak m_{A_ i})/\kappa $ is finite and $\kappa $ is separably algebraically closed, hence it is finite purely inseparable. $\square$

Proof. Denote $\mathcal{C} \to \mathcal{D}$ the functor of categories of the statement. Suppose that $R \to S$ is finite étale. Then we may write

with $A_ i$ local and finite étale over $S$, use either Lemma 10.153.5 or Lemma 10.153.3 part (10). In particular $A_ i/\mathfrak mA_ i$ is a finite separable field extension of $\kappa $, see Lemma 10.143.5. Thus we see that every object of $\mathcal{C}$ and $\mathcal{D}$ decomposes canonically into irreducible pieces which correspond via the given functor. Next, suppose that $S_1$, $S_2$ are finite étale over $R$ such that $\kappa _1 = S_1/\mathfrak mS_1$ and $\kappa _2 = S_2/\mathfrak mS_2$ are fields (finite separable over $\kappa $). Then $S_1 \otimes _ R S_2$ is finite étale over $R$ and we may write

as before. Then we see that $\mathop{\mathrm{Hom}}\nolimits _ R(S_1, S_2)$ is identified with the set of indices $i \in \{ 1, \ldots , n\} $ such that $S_2 \to A_ i$ is an isomorphism. To see this use that given any $R$-algebra map $\varphi : S_1 \to S_2$ the map $\varphi \times 1 : S_1 \otimes _ R S_2 \to S_2$ is surjective, and hence is equal to projection onto one of the factors $A_ i$. But in exactly the same way we see that $\mathop{\mathrm{Hom}}\nolimits _\kappa (\kappa _1, \kappa _2)$ is identified with the set of indices $i \in \{ 1, \ldots , n\} $ such that $\kappa _2 \to A_ i/\mathfrak mA_ i$ is an isomorphism. By the discussion above these sets of indices match, and we conclude that our functor is fully faithful. Finally, let $\kappa '/\kappa $ be a finite separable field extension. By Lemma 10.144.3 there exists an étale ring map $R \to S$ and a prime $\mathfrak q$ of $S$ lying over $\mathfrak m$ such that $\kappa \subset \kappa (\mathfrak q)$ is isomorphic to the given extension. By part (1) we may write $S = A_1 \times \ldots \times A_ n \times B$. Since $R \to S$ is quasi-finite we see that there exists no prime of $B$ over $\mathfrak m$. Hence $S_{\mathfrak q}$ is equal to $A_ i$ for some $i$. Hence $R \to A_ i$ is finite étale and produces the given residue field extension. Thus the functor is essentially surjective and we win. $\square$

Proof. First write $S = A_1 \times \ldots \times A_ n \times B$ as in Lemma 10.153.5. Now we see that $R \to A_ i$ is finite unramified and $A_ i$ local. Hence the maximal ideal of $A_ i$ is $\mathfrak mA_ i$ and its residue field $A_ i / \mathfrak m A_ i$ is a finite separable extension of $\kappa $, see Lemma 10.151.5. However, the condition that $R$ is strictly henselian means that $\kappa $ is separably algebraically closed, so $\kappa = A_ i / \mathfrak m A_ i$. By Nakayama's Lemma 10.20.1 we conclude that $R \to A_ i$ is surjective as desired. $\square$

Proof. Let $f \in R[T]$ be monic. Denote $f_ n \in R/\mathfrak m^{n + 1}[T]$ the image. Denote $f'_ n$ the derivative of $f_ n$ with respect to $T$. Let $a_0 \in \kappa $ be a simple root of $f_0$. We lift this to a solution of $f$ over $R$ inductively as follows: Suppose given $a_ n \in R/\mathfrak m^{n + 1}$ such that $a_ n \bmod \mathfrak m = a_0$ and $f_ n(a_ n) = 0$. Pick any element $b \in R/\mathfrak m^{n + 2}$ such that $a_ n = b \bmod \mathfrak m^{n + 1}$. Then $f_{n + 1}(b) \in \mathfrak m^{n + 1}/\mathfrak m^{n + 2}$. Set

(Newton's method). This makes sense as $f'_{n + 1}(b) \in R/\mathfrak m^{n + 2}$ is invertible by the condition on $a_0$. Then we compute $f_{n + 1}(a_{n + 1}) = f_{n + 1}(b) - f_{n + 1}(b) = 0$ in $R/\mathfrak m^{n + 2}$. Since the system of elements $a_ n \in R/\mathfrak m^{n + 1}$ so constructed is compatible we get an element $a \in \mathop{\mathrm{lim}}\nolimits R/\mathfrak m^ n = R$ (here we use that $R$ is complete). Moreover, $f(a) = 0$ since it maps to zero in each $R/\mathfrak m^ n$. Finally $\overline{a} = a_0$ and we win. $\square$

Proof. Let $R \to S$ be a finite ring map. By Lemma 10.153.3 it suffices to show that $S$ is a product of local rings. By Lemma 10.36.21 $S$ has finitely many primes $\mathfrak m_1, \ldots , \mathfrak m_ r$ which all lie over $\mathfrak m$. There are no inclusions among these primes, see Lemma 10.36.20, hence they are all maximal. Every element of $\mathfrak m_1 \cap \ldots \cap \mathfrak m_ r$ is nilpotent by Lemma 10.17.2. It follows $S$ is the product of the localizations of $S$ at the primes $\mathfrak m_ i$ by Lemma 10.53.5. $\square$

Proof. Consider $A \otimes _ R S$. This is an étale algebra over $S$, see Lemma 10.143.3. Moreover, the kernel

of the map using the map given in (3) is a prime ideal lying over $\mathfrak m_ S$ with residue field equal to the residue field of $S$. Hence by Lemma 10.153.3 there exists a unique splitting $\tau : A \otimes _ R S \to S$ with $\tau ^{-1}(\mathfrak m_ S) = \mathfrak q'$. Set $f$ equal to the composition $A \to A \otimes _ R S \to S$. $\square$

Proof. The first solution set is canonically isomorphic to the set

As $R$ is henselian the map $R \to R/\mathfrak m_ R$ induces a bijection between this set and the set of solutions in the residue field $R/\mathfrak m_ R$, see Lemma 10.153.3. The same is true for $S$. Now since $R[x_1, \ldots , x_ n]/(P_1, \ldots , P_ n)$ is étale over $R$ and $R/\mathfrak m_ R$ is separably algebraically closed we see that $R/\mathfrak m_ R[x_1, \ldots , x_ n]/(\overline{P}_1, \ldots , \overline{P}_ n)$ is a finite product of copies of $R/\mathfrak m_ R$ where $\overline{P}_ i$ is the image of $P_ i$ in $R/\mathfrak m_ R[x_1, \ldots , x_ n]$. Hence the tensor product

is also a finite product of copies of $S/\mathfrak m_ S$ with the same index set. This proves the lemma. $\square$

Proof. Let $M$ be a countably generated and Mittag-Leffler $R$-module. We claim that for any element $x \in M$ there exists a direct sum decomposition $M = N \oplus K$ with $x \in N$, the module $N$ finitely presented, and $K$ Mittag-Leffler.

Suppose the claim is true. Choose generators $x_1, x_2, x_3, \ldots $ of $M$. By the claim we can inductively find direct sum decompositions

with $N_ i$ finitely presented, $x_1, \ldots , x_ n \in N_1 \oplus \ldots \oplus N_ n$, and $K_ n$ Mittag-Leffler. Repeating ad infinitum we see that $M = \bigoplus N_ i$.

We still have to prove the claim. Let $x \in M$. By Lemma 10.92.2 there exists an endomorphism $\alpha : M \to M$ such that $\alpha $ factors through a finitely presented module, and $\alpha (x) = x$. Say $\alpha $ factors as

Set $a = \pi \circ \alpha \circ i : P \to P$, so $i \circ a \circ \pi = \alpha ^3$. By Lemma 10.16.2 there exists a monic polynomial $P \in R[T]$ such that $P(a) = 0$. Note that this implies formally that $\alpha ^2 P(\alpha ) = 0$. Hence we may think of $M$ as a module over $R[T]/(T^2P)$. Assume that $x \not= 0$. Then $\alpha (x) = x$ implies that $0 = \alpha ^2P(\alpha )x = P(1)x$ hence $P(1) = 0$ in $R/I$ where $I = \{ r \in R \mid rx = 0\} $ is the annihilator of $x$. As $x \not= 0$ we see $I \subset \mathfrak m_ R$, hence $1$ is a root of $\overline{P} = P \bmod \mathfrak m_ R \in R/\mathfrak m_ R[T]$. As $R$ is henselian we can find a factorization

for some $Q_1, Q_2 \in R[T]$ with $Q_2 = (T - 1)^ e \bmod \mathfrak m_ R R[T]$ and $Q_1(1) \not= 0 \bmod \mathfrak m_ R$, see Lemma 10.153.3. Let $N = \mathop{\mathrm{Im}}(\alpha ^2Q_1(\alpha ) : M \to M)$ and $K = \mathop{\mathrm{Im}}(Q_2(\alpha ) : M \to M)$. As $T^2Q_1$ and $Q_2$ generate the unit ideal of $R[T]$ we get a direct sum decomposition $M = N \oplus K$. Moreover, $Q_2$ acts as zero on $N$ and $T^2Q_1$ acts as zero on $K$. Note that $N$ is a quotient of $P$ hence is finitely generated. Also $x \in N$ because $\alpha ^2Q_1(\alpha )x = Q_1(1)x$ and $Q_1(1)$ is a unit in $R$. By Lemma 10.89.10 the modules $N$ and $K$ are Mittag-Leffler. Finally, the finitely generated module $N$ is finitely presented as a finitely generated Mittag-Leffler module is finitely presented, see Example 10.91.1 part (1). $\square$