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Chapter 10: Commutative Algebra > Section 10.148: Henselian local rings

Characterizations of henselian local rings

Lemma 10.148.3. Let $(R, \mathfrak m, \kappa)$ be a local ring. The following are equivalent

  1. $R$ is henselian,
  2. for every $f \in R[T]$ and every root $a_0 \in \kappa$ of $\overline{f}$ such that $\overline{f'}(a_0) \not = 0$ there exists an $a \in R$ such that $f(a) = 0$ and $a_0 = \overline{a}$,
  3. for any monic $f \in R[T]$ and any factorization $\overline{f} = g_0 h_0$ with $\gcd(g_0, h_0) = 1$ there exists a factorization $f = gh$ in $R[T]$ such that $g_0 = \overline{g}$ and $h_0 = \overline{h}$,
  4. for any monic $f \in R[T]$ and any factorization $\overline{f} = g_0 h_0$ with $\gcd(g_0, h_0) = 1$ there exists a factorization $f = gh$ in $R[T]$ such that $g_0 = \overline{g}$ and $h_0 = \overline{h}$ and moreover $\deg_T(g) = \deg_T(g_0)$,
  5. for any $f \in R[T]$ and any factorization $\overline{f} = g_0 h_0$ with $\gcd(g_0, h_0) = 1$ there exists a factorization $f = gh$ in $R[T]$ such that $g_0 = \overline{g}$ and $h_0 = \overline{h}$,
  6. for any $f \in R[T]$ and any factorization $\overline{f} = g_0 h_0$ with $\gcd(g_0, h_0) = 1$ there exists a factorization $f = gh$ in $R[T]$ such that $g_0 = \overline{g}$ and $h_0 = \overline{h}$ and moreover $\deg_T(g) = \deg_T(g_0)$,
  7. for any étale ring map $R \to S$ and prime $\mathfrak q$ of $S$ lying over $\mathfrak m$ with $\kappa = \kappa(\mathfrak q)$ there exists a section $\tau : S \to R$ of $R \to S$,
  8. for any étale ring map $R \to S$ and prime $\mathfrak q$ of $S$ lying over $\mathfrak m$ with $\kappa = \kappa(\mathfrak q)$ there exists a section $\tau : S \to R$ of $R \to S$ with $\mathfrak q = \tau^{-1}(\mathfrak m)$,
  9. any finite $R$-algebra is a product of local rings,
  10. any finite $R$-algebra is a finite product of local rings,
  11. any finite type $R$-algebra $S$ can be written as $A \times B$ with $R \to A$ finite and $R \to B$ not quasi-finite at any prime lying over $\mathfrak m$,
  12. any finite type $R$-algebra $S$ can be written as $A \times B$ with $R \to A$ finite such that each irreducible component of $\mathop{\rm Spec}(B \otimes_R \kappa)$ has dimension $\geq 1$, and
  13. any quasi-finite $R$-algebra $S$ can be written as $S = A \times B$ with $R \to A$ finite such that $B \otimes_R \kappa = 0$.

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 10.141.16 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 10.141.4 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 $\tau^{-1}(\mathfrak q) = \mathfrak m$. Minor details omitted.

Proof of 8$\Rightarrow$11. Assume (8) and let $R \to S$ be a finite type ring map. Apply Lemma 10.141.22. 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 10.141.22. 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 10.19.1 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 10.98.2. 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 10.77.4. 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 10.15.1) 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 10.19.1. 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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    \begin{lemma}
    \label{lemma-characterize-henselian}
    \begin{slogan}
    Characterizations of henselian local rings
    \end{slogan}
    Let $(R, \mathfrak m, \kappa)$ be a local ring.
    The following are equivalent
    \begin{enumerate}
    \item $R$ is henselian,
    \item for every $f \in R[T]$ and every root $a_0 \in \kappa$
    of $\overline{f}$ such that $\overline{f'}(a_0) \not = 0$
    there exists an $a \in R$ such that $f(a) = 0$ and
    $a_0 = \overline{a}$,
    \item for any monic $f \in R[T]$ and any factorization
    $\overline{f} = g_0 h_0$ with $\gcd(g_0, h_0) = 1$ there
    exists a factorization $f = gh$ in $R[T]$ such that
    $g_0 = \overline{g}$ and $h_0 = \overline{h}$,
    \item for any monic $f \in R[T]$ and any factorization
    $\overline{f} = g_0 h_0$ with $\gcd(g_0, h_0) = 1$ there
    exists a factorization $f = gh$ in $R[T]$ such that
    $g_0 = \overline{g}$ and $h_0 = \overline{h}$ and moreover
    $\deg_T(g) = \deg_T(g_0)$,
    \item for any $f \in R[T]$ and any factorization
    $\overline{f} = g_0 h_0$ with $\gcd(g_0, h_0) = 1$ there
    exists a factorization $f = gh$ in $R[T]$ such that
    $g_0 = \overline{g}$ and $h_0 = \overline{h}$,
    \item for any $f \in R[T]$ and any factorization
    $\overline{f} = g_0 h_0$ with $\gcd(g_0, h_0) = 1$ there
    exists a factorization $f = gh$ in $R[T]$ such that
    $g_0 = \overline{g}$ and $h_0 = \overline{h}$ and
    moreover $\deg_T(g) = \deg_T(g_0)$,
    \item for any \'etale ring map $R \to S$ and prime $\mathfrak q$ of $S$
    lying over $\mathfrak m$ with $\kappa = \kappa(\mathfrak q)$
    there exists a section $\tau : S \to R$ of $R \to S$,
    \item for any \'etale ring map $R \to S$ and prime $\mathfrak q$ of $S$
    lying over $\mathfrak m$ with $\kappa = \kappa(\mathfrak q)$
    there exists a section $\tau : S \to R$ of $R \to S$ with
    $\mathfrak q = \tau^{-1}(\mathfrak m)$,
    \item any finite $R$-algebra is a product of local rings,
    \item any finite $R$-algebra is a finite product of local rings,
    \item any finite type $R$-algebra $S$ can be written as
    $A \times B$ with $R \to A$ finite
    and $R \to B$ not quasi-finite at any prime lying over $\mathfrak m$,
    \item any finite type $R$-algebra $S$ can be written as
    $A \times B$ with $R \to A$ finite
    such that each irreducible component of $\Spec(B \otimes_R \kappa)$
    has dimension $\geq 1$, and
    \item any quasi-finite $R$-algebra $S$ can be written as
    $S = A \times B$ with $R \to A$ finite such that $B \otimes_R \kappa = 0$.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    Here is a list of the easier implications:
    \begin{enumerate}
    \item[2$\Rightarrow$1] because in (2) we consider all polynomials and
    in (1) only monic ones,
    \item[5$\Rightarrow$3] because in (5) we consider all polynomials and
    in (3) only monic ones,
    \item[6$\Rightarrow$4] because in (6) we consider all polynomials and
    in (4) only monic ones,
    \item[4$\Rightarrow$3] is obvious,
    \item[6$\Rightarrow$5] is obvious,
    \item[8$\Rightarrow$7] is obvious,
    \item[10$\Rightarrow$9] is obvious,
    \item[11$\Leftrightarrow$12] by definition of being quasi-finite at a prime,
    \item[11$\Rightarrow$13] by definition of being quasi-finite,
    \end{enumerate}
    
    \noindent
    Proof of 1$\Rightarrow$8. Assume (1).
    Let $R \to S$ be \'etale, and let $\mathfrak q \subset S$
    be a prime ideal such that $\kappa(\mathfrak q) \cong \kappa$. By
    Proposition \ref{proposition-etale-locally-standard}
    we can find a $g \in S$, $g \not \in \mathfrak q$ such that
    $R \to S_g$ is standard \'etale. After replacing $S$ by $S_g$ we may assume
    that $S = R[t]_g/(f)$ is standard \'etale. 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 \'etale algebra this also means that
    $\overline{f'}(a_0) \not = 0$.
    Since also $f$ is monic by definition of a standard \'etale 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.
    
    \medskip\noindent
    Proof of 7$\Rightarrow$8. (This is really unimportant and should be
    skipped.) Assume (7) holds and assume $R \to S$ is \'etale.
    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
    \'etale morphism is discrete (use Lemma \ref{lemma-etale-over-field}
    for example).
    Apply (7) to the \'etale 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 $\tau^{-1}(\mathfrak q) = \mathfrak m$.
    Minor details omitted.
    
    \medskip\noindent
    Proof of 8$\Rightarrow$11. Assume (8) and let $R \to S$ be a finite type
    ring map. Apply
    Lemma \ref{lemma-etale-makes-quasi-finite-finite}.
    We find an \'etale 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).
    
    \medskip\noindent
    Proof of 8$\Rightarrow$10. Assume (8) and let $R \to S$ be a finite
    ring map. Apply
    Lemma \ref{lemma-etale-makes-quasi-finite-finite}.
    We find an \'etale 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.
    
    \medskip\noindent
    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.
    
    \medskip\noindent
    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.
    
    \medskip\noindent
    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.
    
    \medskip\noindent
    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 \'etale 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).
    
    \medskip\noindent
    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).
    
    \medskip\noindent
    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 \ref{lemma-NAK}
    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 $\overline{a} = a_0$.
    
    \medskip\noindent
    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 \ref{lemma-grothendieck}.
    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 \ref{lemma-finite-flat-local}.
    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 \ref{lemma-charpoly})
    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 \ref{lemma-NAK}. 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.
    \end{proof}

    Comments (4)

    Comment #1449 by Johan Commelin (site) on May 7, 2015 a 9:20 am UTC

    Suggested slogan: Equivalent definitions of henselian local ring

    Comment #2870 by Ko Aoki on October 5, 2017 a 3:39 pm UTC

    Possibly wrong reference: Proof of 1 $\Rightarrow$ 6 refers to the lemma [00MF] (https://stacks.math.columbia.edu/tag/00MF) in order to prove $R \to S$ is flat, but how can it be used?

    Comment #2932 by Johan (site) on October 10, 2017 a 2:02 am UTC

    It can be used because $R \to R[T]$ is flat and hence if we divide by the nonzerodivisor $f$ the quotient $S = R[T]/(f)$ is flat over $R$ by the flatness mentioned in Lemma 00MF. To really apply the lemma you look at the localizations at the maximal ideals of $R[T]$.

    Comment #2951 by Ko Aoki on October 11, 2017 a 12:45 pm UTC

    Thank you for teaching me. I got it. I just thought it's wrong as 00MF has a noetherian hypothesis.

    There are also 2 comments on Section 10.148: Commutative Algebra.

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