Corollary 1

Let $A$ be a ring and $m$ an ideal of $A$ such that the ordered pair $(A, m)$ satisfies Hensel's conditions. Let $f \in A\{X\}$, $a \in A$ and write $e = f'(a)$. If $f(a) \equiv 0$ (mod. $e^2 m$), then there exists $b \in A$ such that $f(b) = 0$ and $b \equiv a$ (mod. $em$). If b' is another element of $\mathbf{A}$ such that $f(b') = 0$ and $b' \equiv a \pmod{\mathfrak{m}}$, then $e(b - b') = 0$. In particular, $b$ is unique if $e$ is not a divisor of zero in $\mathbf{A}$.

Let $f(a) = e^2 c$ where $c \in \mathfrak{m}$; formula (20) for $n = 1$ gives

and it is therefore sufficient to take $y = -c$, $b = a + eh(-c)$. Moreover if $b = a + ex, b' = a + ex', x \in \mathfrak{m}, x' \in m, f(b) = f(b') = 0$, we deduce from (19) that $e^2(g(x) - g(x')) = 0$. As $g(\mathbf{X}) - g(\mathbf{Y}) = (\mathbf{X} - \mathbf{Y})u(\mathbf{X}, \mathbf{Y})$, where $u$ is restricted and $u(0, 0) = 1$, $g(x) - g(x') = (x - x')v$, where $v \in \mathbf{A}$ is invertible, for, since $m$ is closed, $v - 1 = u(x, x') - 1 \in \mathfrak{m}$ and $m$ is contained in the Jacobson radical of $\mathbf{A}$; whence the relation $e(b - b') = 0$.