What is the fastest way to compute the dimension of the Zariski tangent space at the origin of $\operatorname{Spec} A$, where $A$ is the ring
$$A = k[w, x, y, z] /(wz – xy)?$$
We know it suffices to compute the dimension of the cotangent space $$\mathfrak m / \mathfrak m^2$$
where $\mathfrak m$ is the unique maximal ideal of $A_{(w, x, y, z)}$. Is there a nice way to quickly find this maximal ideal and the resulting dimension?
We know the maximal ideal of $A_{(w, x, y, z)}$ corresponds to the maximal ideal in $A$ contained in $(w, x, y, z)$ which corresponds to the maximal ideal of $k[w, x, y, z]$ containing $(wz – xy)$ contained in $(w, x, y, z)$, which should be just $(w, x, y, z)$, which feels morally wrong to me somehow. What am I missing?
Best Answer
The tangent space of $V(I)$ at a point $p$ is the kernel of the Jacobian matrix with entries evaluated at $p$ (see here for more details if you need them). This makes computing the dimension of the tangent space fairly straightforward: pick generators for $I$, take derivatives, evaluate at $p$, and apply linear algebra to compute the nullity of the matrix. In this case, $I$ has a single generator $wz-xy$, and the Jacobian matrix is $$\begin{pmatrix} -y & -x & w & z \end{pmatrix}$$ which when evaluated at the maximal ideal $(x,y,z,w)$ is the zero matrix. So the tangent space is 4-dimensional at $(x,y,z,w)$. (For the connection between this and the taylor-series approach mentioned in the comments, recall that taking the derivative and evaluating at a point is exactly what you do to get the coefficient of the linear term of the taylor series of a function at that point.)
If you're looking for a more hands-on calculation, you can pick vector space bases of everything in sight and see what happens. $(x,y,z,w)/(x,y,z,w)^2$ has a vector space basis $\{x,y,z,w\}$, while $((x,y,z,w)/(wz-xy))/((x,y,z,w)^2/(wz-xy))$ has the same vector space basis.