Dimension of $m/m^2$ over $k$ where $m=m_P(V)$ and $V=V(x^2-y^3,y^2-z^3)$ Consider the variety $V=V(x^2-y^3,y^2-z^3)$ and coordinate ring $\Gamma(V)=\frac{k[x,y,z]}{I(V)}$. Let $k(V)$ be the field of fractions of $\Gamma(V)$. Let $P=(0,0,0)$. An element $f\in k(V)$ with $f=\frac{a}{b}$ and $b(P)\not=0$ is called a rational function on $V$ defined at $P$. Let $\mathscr O_P(V)$ be the set of rational function on $V$ defined at $P$. Now $m:=\{f\in \mathscr O_P(V):f(P)=0\}$ is a maximal ideal of the local ring $\mathscr O_P(V)$.
My question is the following. What is the dimension of $\frac{m}{m^2}$ over $k$ i.e. $dim_k\bigg(\frac{m}{m^2}\bigg)$. What I can guess is that $x+m^2,y+m^2,z+m^2$ will be a generating set of the vector space $\frac{m}{m^2}$. Now how do I show they are linearly independent also.
 A: The quantity ${\rm dim}_k (m / m^2)$ has a nice geometric interpretation as the dimension of the tangent space to $V$ at $P$.
If $n$ is the dimension of the affine space $\mathbb A^n$ that we're working in, then the dimension of the tangent space to $V$ at $P$ can also be defined as
$$ {\rm dim}(T_P V) := n - {\rm rank} \left[ \left. \frac{\partial f_i}{\partial x_j}\right\vert_P\right],$$
where $I = (f_1, \dots, f_2) \subset k[x_1, \dots, x_n]$ is the ideal defining the variety $V$, and $\left[ \left. \frac{\partial f_i}{\partial x_j}\right\vert_P\right]$ is the Jacobian matrix at the point $P$.
The equivalence between ${\rm dim}_k (m / m^2)$ and $\left( n - {\rm rank} \left[ \left. \frac{\partial f_i}{\partial x_j}\right\vert_P\right]\right)$ is proved in Hartshorne 5.1. Or see  this answer for a sketch.
It's usually the case that the Jacobian definition is easier to compute with.
In our case, we have
$$ f_1 = x^2 - y^3, \ \ \ f_2 = y^2 - z^3,$$
so the Jacobian is
$$ \begin{bmatrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} & \frac{\partial f_1}{\partial z} \\ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} & \frac{\partial f_2}{\partial z} \end{bmatrix}= \begin{bmatrix} 2x & -3y^2 & 0 \\ 0 & 2y & - 3z^2 \end{bmatrix}. $$
Evaluating this Jacobian matrix at $P = (0,0,0)$, we find get the zero matrix,
$$ \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix},$$
which has rank zero. Hence the dimension of the tangent space at $P$ is equal to $3$.
