Proof explanation, singular point of an affine variety Let $K$ be an algebraically closed field, and let $X$ be an affine variety, i.e. an irreducible closed subset of $A_K^n$, and let $P=(a_1,..,a_n)$ be a point of $X$. Let $\alpha_P$ be the ideal of $P$, i.e. $\alpha_P=(x_1-a_1,..,x_n-a_n)$. Then $\alpha_p$ is a maximal ideal in $K[x_1,..,x_n]$.
Let $\theta\colon K[x_1,\dots,x_n]\to K^n$ defined by $\theta(f)=(\frac{\partial f}{\partial x_1}(P),..,\frac{\partial f}{\partial x_n}(P)).$
Then $\theta: \frac{\alpha_P}{\alpha_P^2}\to K^n$ is an isomorphism of $K$ vector spaces.
Let $X=Z(f_1,..,f_t)$, with $f_i \in  K[x_1,..,x_n]$, so $I(X)=(f_1,..,f_t)$ is a prime ideal of $K[x_1,..,x_n]$, and $I(X)\subseteq \alpha_P$.
Let $O_{X,p}$ be the local ring of $X$ at $P$: it consists of all regular functions defined in an open neighborhood of $P$, but it doesn't matter the exact neighborhood. I know that $O_{X,p}\simeq A(X)_{MP}$ where $A(X)=K[x_1,..,x_n]/I(X)$ and $M_p=\alpha_P/I(X)$. Let $m_P$ be the maximal ideal of the local ring $O_{X,p}$.
Let $J(P)=(\frac{\partial f_i}{\partial x_j}(P))$. Then i know that $rankJ(P)=dim_K(\frac{I(X)+\alpha_P^2}{\alpha_P^2})$

I want to show that dim$_K(m_P/m_P^2)=n-rank J(P)$     $\;\;$ [1]

My book says that $$\frac{m_p}{m_P^2}\simeq\frac{\alpha_P}{\alpha_P^2+I(X)} \;\;\; [2]$$ from wich deduces the result [1], but i can't see [2].
 A: I wish I can help you. From the book Silverman, the arithmetic of elliptic curves:
In page 4, Definition(Jacobian criterion): Let $V$ be a variety and $f_1,...,f_m\in k[X]$ a set of generators for $I(V)$. Then $V$ is nonsingular at $p$ if the matrix $$J=\left( \frac{\partial f_i}{\partial x_j}(p)\right)_{m\ge i \ge1, n\ge j \ge1}$$ has rank $n-dim(V)$.
In page 5, Let be $k[V]$ is a coordinate ring and $m_p=\{f\in k[V]; f(p)=0 \}$ is a maximal ideal, since $k[V]/m_p\cong k$. Then $m_p/m_p^2$ is a finite dimensional $k$ vector space. 
Lemma: Let $V$ be a variety. A point $p\in V$ is nonsingular if and only if $$\dim_k m_p/m_p²=\dim V $$

So we have $\dim_k m_p/m_p^2=dimV=n-$rank$J$ 

In page 17, Lemma: Let $C$ be a curve (projective variety of dimenision one) and $p\in C$ a smooth point. Then $k[C]_p$ is a DVR, so $\dim_k m_p/m_p^2=1$ 
In the book Daniel Bump, Algebraic geometry: In page 42, We have speical case: Let $X=\Bbb A^d$ and $x=(0,..,0)$, then $$O_x=\left\{\frac{f}{g}; f,g\in k[x_1 ,...x_d]; g(x)\ne0 \right\}$$
and $$m_x=\left\{\frac{f}{g}; f,g\in k[x_1 ,...x_d]; f(x)=0 \right\}$$ then 
$$\dim\left(m_x^n / m_x^{n+1}\right)= \begin{pmatrix} d+n-1\\n \end{pmatrix} $$
and 
$$\dim\left(O_x / m_x^{n}\right)= \begin{pmatrix} d+n-1\\n-1 \end{pmatrix} $$
