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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 $I(X)$?

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up vote 2 down vote accepted

a) The proof that $X$ is isomorphic to $V(x^2-yz)\subset \mathbb A^3$ is perfect: congratulations!
Let me call $V$ that variety $V(x^2-yz)\subset \mathbb A^3$

b) You haven't given a proof, but it is true that $V$ has $O=(0,0,0)$ as its only singularity.
A possible proof proof is to use the Jacobian criterion:
Letting $F(x,y,z)=x^2-yz$ we have $Jac(F)=(2x,-z,-y)$ and the singularities of the hypersurface $V$ are given by the system $x^2-yz=2x=-z=-y=0$ so that the only singularity of $V$ is indeed $O$.

c) The Zariski tangent space $T_O(V)$ is the subspace of $k^3$ determined by the equation $Jac_O(F)(t)=0$, where $Jac_O(F)$ is seen as a linear form on $k^3$ and $t=(t_1,t_2,t_3)$ as a vector in $k^3$.
Since $Jac_O(F)=0$ is the zero linear form here, it follows that $T_O(V)=k^3$.
Translated back in terms of the original matrices, this means that the Zariski tangent space of your variety $X$ of nilpotent matrices is the hyperplane $a+d=0$ of $M_2(k)$.

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Thanks! I can see that the tangent space of $V(x^{2}-yz) \subseteq \mathbb{A}^{3}$ is the whole $\mathbb{A}^{3}$. How did you translate this to $X$ to conclude that the tangent space consists of the hyperplane $a+d=0$? – user10 Jun 21 '12 at 22:44
is it simply because if $g: X \cong V$ is an isomorphism of varieties then the tangent spaces $T_{(0,0,0,0)}(X) \cong T_{g((0,0,0)}(V)$ are isomorphic as vector spaces? Now since the tangent space of $V$ is $\mathbb{A}^{3}$ then the points are of the form $(x,y,z) \in k^{3}$. Therefore the points of the tangent space of $X$ at the origin are of the form $g(x,y,z)$ where $g$ is the above isomorphism, so points of the form $(x,-y,z,-x)$ where $(x,y,z) \in k^{3}$ and since the first and last coordinate add $0$ then we characterize this set as a+d=0. – user10 Jun 21 '12 at 23:07
Dear user10, yes your argument in the comment is completely correct.In other words you have an isomorphism $F:V(a+d)\to k^3: (a,b,c,d)\mapsto (a,-b,c)$ restricting to your $f:X\to V$, so that the Zariski tangent space $k^3$ corresponds under $F$ to $F^{-1}(k^3)=V(a+d)$. – Georges Elencwajg Jun 22 '12 at 8:13

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