Let $X$ be an algebraic variety, and $ \mathcal{O}_{X,x} $ its local ring at a point $x \in X$, and $ \mathfrak{m}_x $ its maximal ideal. Let set $ k_x = \mathcal{O}_{X,x} / \mathfrak{m}_x $ the residue field and $ \epsilon_x $ the morphism of algebras : $ \mathcal{O}_{X,x} \to k_x $, the evaluation in $x$ definded by : $ \epsilon_x (f) = f(x) $.
As we can say : $ \mathcal{O}_{X,x} \simeq \mathfrak{m}_x \oplus \mathcal{O}_{X,x} / \mathfrak{m}_x = \mathfrak{m}_x \oplus k(x) = \mathfrak{m}_x \oplus k.1 $
It's follows that : $ \mathcal{O}_{X,x}^* \simeq \mathfrak{m}_{x}^* \oplus k^* \simeq \mathfrak{m}_{X}^* \oplus k^k $ and : $ \mathfrak{m}_{x}* = \{ f \in \mathcal{O}_{X,x}^* \ / \ f \not \in k^k \ \} = \{ \ f \in \mathcal{O}_{X,x}^* \ / \ f(1) = 0 \ \} $
In my book which i'm learning actually, the author say without explanation, that : $ ( \mathfrak{m}_x / \mathfrak{m}_{x}^2 )^* = \{ \ f \in \mathcal{O}_{X,x}^* \ / \ f(1 ) = 0 = f ( \mathfrak{m}_{x}^{2} ) \ \} $
Could you explain me in detail, why please?
Thanks in advance for your help.