The Stacks project

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 $f'_{n + 1}(b) \in R/\mathfrak m^{n + 2}$ 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. $\square$


Comments (4)

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.

Comment #9371 by Hsueh-Yung on

A typo just after the displayed formula: it should be .

There are also:

  • 6 comment(s) on Section 10.153: Henselian local rings

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 04GM. Beware of the difference between the letter 'O' and the digit '0'.