Let $W=V(y^{2}-x^{3}) \subseteq \mathbb{A}^{2}$ and $k$ algebraically closed. Clearly the dimension of the tangent space at the origin is $2$. I want to compute this using the definition the fact that $\operatorname{dim} T_{(0,0)}W=\operatorname{dim}_{k} \mathfrak{m}/\mathfrak{m}^{2}$. Where $\mathfrak{m}$ is the maximal ideal of the local ring $\mathcal{O}_{(0,0),W}$.
OK according to Hartshorne the local ring is isomorphic to the localization of the corresponding coordinate ring localized at the ideal $(x,y)$ right?
So we need to compute the maximal ideal of $k[x,y]/(y^{2}-x^{3})$ localized at $(x,y)$. I know that in general given a ring $A$ then if $\mathfrak{p} \in \operatorname{Spec}(A)$ we have $\mathfrak{p}A_{p}=\{\frac{g}{h}: g,h \in P\}$ is the unique maximal ideal. However I don't see how to simplify things here. How to compute this?
Best Answer
If $M=(x,y)\subset k[X,Y]/(Y^2-X^3)=k[x,y]\stackrel{\text {def}}{=}A \;$, the maximal ideal you are interested in is $\mathfrak m=MA_M \subset A_M$ the maximal ideal of the local ring $A_M=\mathcal O_{(0,0),W}$.
The $k$-vector space $\mathfrak m/\mathfrak m^2$ is generated by $\bar x$ and $\bar y$ and all you have to check is that they are linearly independent.
A linear dependence relation $q\bar x+r \bar y=0$ ($q,r\in k$) means $qx+ry \in \mathfrak m^2=(x^2,xy,y^2)$.
Lifting to actual polynomials this means $$qX+rY\in (X^2,XY,Y^2)+(Y^2-X^3)=(X^2,XY,Y^2)\subset k[X,Y]$$ and implies $q=r=0$.
So $\bar x$ and $\bar y$ are linearly independent in $\mathfrak m/\mathfrak m^2$ and $dim_k(\mathfrak m/\mathfrak m^2)=2$