Let $k$ be an algebraically closed field. We identify the space $M_{2}(k)$ of $2 \times 2$ with $\mathbb{A}^{4}(k)$ with coordinates a,b,c,d. Let $X$ be the set of all matrices $A$ in $M_{2}(k)$ such that $A^{2}=0$.
Show that $X$ is isomorphic to the affine cone of the projective variety $W=V(x^{2}-yz) \subseteq \mathbb{P}^{2}$ and find the singular points of $X$. Finally compute the Zariski tangent space at such point.
My work:
Doing the computations shows that $X=V(ad-bc,a+d)$. Now note that the affine cone of $W$ is $V(x^{2}-yz) \subseteq \mathbb{A}^{3}$.
So let $f: X \rightarrow V(x^{2}-yz) \subseteq \mathbb{A}^{3}$ be given by $(x,y,z,w) \mapsto (x,-y,z)$ and let $g: V(x^{2}-yz) \subseteq \mathbb{A}^{3} \rightarrow X$ be given by $g(x,y,z)=(x,-y,z,-x)$.
(Unless I did something wrong) $g$ is the inverse of $f$ so $f$ is an isomorphism.
Now observe that $V(x^{2}-yz) \subseteq \mathbb{A}^{3}$ has only one singular point, namely the origin $(0,0,0)$ thus via the above isomorphism we see that $(0,0,0,0)$ is the only singular point of $X$.
The part I'm not sure is how to compute the Zariski tangent space we first need to find a set of generators for $I(X)$ no? or is there an easier way?
EDIT: Using software one can check that $I(X)=(xyz-x^{2}w+yzw-xw^{2})$ so the tangent space is $\mathbb{A}^{4}$ because all partials of $xyz-x^{2}w+yzw-xw^{2}$ vanish at $(0,0,0,0)$. However is it possible to solve this question without calculating $I(X)$?