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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?

Johnny Apple
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    The fastest way is to use Macaulay2, in case what you are looking for is just an answer. – Aurelio Oct 09 '20 at 15:55
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    @Aurelio maybe in general, but for this specific example that's not true: you can just look at it! The Taylor expansion of the equation at the point in question has no linear terms and so the tangent space is 4-dimensional. – KReiser Oct 09 '20 at 16:55
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    @KReiser I completely forgot you can just kill the nonlinear terms of wz - xy! May I ask what is actually incorrect about my solution, and how I would compute the dimension from there, just for completeness? – Johnny Apple Oct 09 '20 at 17:07
  • I don't see anything incorrect in your post, but I don't see a full solution either - just an argument about what the maximal ideal is. If you've computed the maximal ideal, then you just need to take the quotient - is that what you need help with? – KReiser Oct 09 '20 at 17:11
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    @KReiser I am just having trouble seeing why modding out by the square doesn't reduce the dimension. I see it intuitively, just not rigorously. I am also having trouble understanding how your answer applies to this same curve applied to other points besides the origin. What is it exactly that we are removing the nonlinear terms of in those cases? – Johnny Apple Oct 09 '20 at 17:14
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    You're removing the nonlinear terms of the Taylor expansion of the function at that point. So think about the Taylor expansion of $wz-xy$ at, say, $(1,0,0,0)$: this is $(wz-(x-1)y)+(-y)$, and so there's a linear term left hanging around, just as there's something in the Jacobian matrix which doesn't vanish after evaluation here. – KReiser Oct 09 '20 at 17:59
  • @KReiser I can get a four dimensional tangent space indeed with Macaulay2. What I do, technically, is to look at the normal sheaf of a point. – Aurelio Oct 09 '20 at 18:35

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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.

KReiser
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