# 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].

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 $$\dim\left(O_x / m_x^{n}\right)= \begin{pmatrix} d+n-1\\n-1 \end{pmatrix}$$