We know in differential geometry, given a $C^k$ manifold for $k>1$, the tangent space at a point in this manifold is parametrized by curves passing through this point modulo certain equivalence relation. The tangent space is given by the velocity vectors to these curves at this point.


Intuitively, I would think this should be true even for Zariski topology in the case $X$ is a smooth projective variety. In particular, we could take a tangent vector $t$ and associate to this the subscheme $Y\_t$ of $X$ obtained by intersecting all some subschemes of $X$ which has $t$ as an element in its tangent space. The question is whether we can somehow ensure that $Y\_t$ is $1$-dimensional? 



Tangent space in Algebraic geometry and Differential geometry
If p is a nonsingular point, then we can define a tangent vector as an equivalence class of (nonsingular) curves, just as in the differentiable case. In fact, being nonsingular is equivalent to every tangent vector being tangent to a curve.
In this case, the intersection of all curves in the equivalence class is a zero-dimensional closed subscheme with a one-dimensional tangent space, isomorphic to the spectrum of the dual numbers described above by Martin Brandenburg.


Warning: I'm thinking of schemes of finite type over an algebraically closed field K, and by point I mean a K-valued point. 


