Lemma 10.153.2 (06RR)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.153: Henselian local rings
Lemma 10.153.2 (cite)

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Lemma 10.153.2. Let $(R, \mathfrak m, \kappa )$ be a local ring. Let $f \in R[T]$ . Let $a, b \in R$ such that $f(a) = f(b) = 0$ , $a = b \bmod \mathfrak m$ , and $f'(a) \not\in \mathfrak m$ . Then $a = b$ .

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 $R$ . $\square$

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