Complete local rings are Henselian by Newton's method

Lemma 10.153.9. Let $(R, \mathfrak m, \kappa )$ be a complete local ring, see Definition 10.160.1. Then $R$ is henselian.

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

$a_{n + 1} = b - f_{n + 1}(b)/f'_{n + 1}(b)$

(Newton's method). This makes sense as $f'_{n + 1}(b) \in R/\mathfrak m^{n + 1}$ 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. $\square$

Comment #5780 by Rohrbach on

Suggested slogan: complete local rings are Henselian.

Comment #5786 by on

Thanks. I threw in an extra word or two, see this commit. Please complain if you don't like it.

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