Zariski tangent space.

I'd like to investigate in two examples of Zariski tangent space: Let $X$ be $Spec(\mathbb{C}[x,y]/(y-x^2))$ with structure sheaf $\mathcal{O}_X$. Let $p=(x-a,y-a^2)$, How can calculate $T_pX$? We know that cotangent space is $\mathfrak{m}_{p,X}/\mathfrak{m}^2_{p,X}$.  If we take $Y=Spec(\mathbb{C}[x,y]/(xy))$, then how can I calculate $T_qY$ is $q=(x,y)$?

• The Zariski tangent space is (isomorphic to) the dual space of the Zariski cotangent space. – Zhen Lin Oct 8 '13 at 14:56

In general, the tangent space at a point $P_0\in C=V(f)\subset \mathbb A^2$ is given by $$T_PC=V(\nabla_Pf\cdot (P-P_0))\subset\mathbb A^2,$$ where $P=(x,y)$ is the generic point of $\mathbb A^2$.
• Let $f=y-x^2$. Then, the gradient of $f$ is the vector $(-2x\,\,1)$. So if $P_0=(a,a^2)$, we get \begin{align} T_{P_0}X&=V(\nabla_{P_0}f\cdot (x-a\,\,\,y-a^2))=V((-2a\,\,1)\cdot (x-a\,\,\,y-a^2))\\\notag &=V(-2ax+2a^2+y-a^2)=V(y-2ax+a^2)\subset \mathbb A^2. \end{align} For instance, the tangent space (line!) at $(1,1)$ is $y=2x-1$, as one would expect.
• Let $g=xy$. Then, the gradient of $g$ is the vector $(y\,\,x)$. If $Q=(0,0)$ is the origin, then $\nabla_Q\,g=(y\,\,x)|_Q=(0\,\,0)$ and, as before, $$T_QY=V((0\,\,0)\cdot (x\,\,y)^T)=V(0)=\mathbb A^2.$$ Now the tangent space is the whole of $\mathbb A^2$ because $Q=(0,0)$ is a singular point of $Y=V(g)$.