**Proof.**
Here is a list of the easier implications:

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

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

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

4$\Rightarrow $3 is obvious,

6$\Rightarrow $5 is obvious,

8$\Rightarrow $7 is obvious,

10$\Rightarrow $9 is obvious,

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

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. 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 $\tau ^{-1}(\mathfrak q) = \mathfrak m$. 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 section $\tau _ g : S_ g \to R$ such that the composition $\tau : S \to S_ g \to R$ has the property $\tau ^{-1}(\mathfrak m) = \mathfrak q$. Minor 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 section $\tau : R' \to R$ with $\mathfrak m = \tau ^{-1}(\mathfrak m')$. Then use that

\[ S = (S \otimes _ R R') \otimes _{R', \tau } R = (A' \times B') \otimes _{R', \tau } R = (A' \otimes _{R', \tau } R) \times (B' \otimes _{R', \tau } R) \]

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 section $\tau : R' \to R$ with $\mathfrak m' = \tau ^{-1}(\mathfrak m)$. Then we obtain

\begin{align*} S & = (S \otimes _ R R') \otimes _{R', \tau } R \\ & = (A'_1 \times \ldots \times A'_ n \times B') \otimes _{R', \tau } R \\ & = (A'_1 \otimes _{R', \tau } R) \times \ldots \times (A'_1 \otimes _{R', \tau } R) \times (B' \otimes _{R', \tau } R) \\ & = A_1 \times \ldots \times A_ n \times B \end{align*}

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

\[ \kappa [T]/(g_0) = A_1/\mathfrak m A_1 \times \ldots \times A_ r/\mathfrak mA_ r, \ \kappa [T]/(h_0) = A_{r + 1}/\mathfrak mA_{r + 1} \times \ldots \times A_ n/\mathfrak mA_ n \]

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.
$\square$

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