# Motivation behind the definition of Zariski tangent space

Intuitively, I think of tangent space at a point as the set of all points lying in the tangent plane passing through that point.

Here is the definition of the Zariski tangent space:

Let X be an algebraic variety and $p \in X$. The tangent space of $X$ at point $p$ is defined as $$T_pX= \operatorname{Der}_k(O_{X,p}.k).$$

How does the above definition match with my intuition? Or more specifically, can someone give a one-one correspondence between $T_pX$ and the set of all points lying in the tangent plane passing through $p$?

• This definition (or something very close) is fairly standard in differential geometry. It might be worth opening up a differential geometry book and seeing why this definition is equivalent to more intuitive definitions of the tangent space of a smooth manifold. Feb 11, 2013 at 15:55
• You can also write down a formal definition for your "intuitive" tangent space and then prove that the vector spaces you get are isomorphic. It's not too hard, though there is a trick involved. Feb 11, 2013 at 16:00
• What will be the formal definition of my intuitive tangent space? Feb 11, 2013 at 16:11
• I believe it's in Shafarevich, but I don't know for sure. Just take $X$ to be an affine variety embedded in affine space and write down the obvious thing associated with your mental picture. Feb 11, 2013 at 18:31
• @ZhenLin what is this trick you mention? Sep 2, 2014 at 12:03

The tangent space at $p$ is the space of all directions in which you can take a directional derivative at $p$. Whatever "directional derivative" means, it should only depend on the germ at $p$, so it's a function on $\mathcal{O}_{X, p}$. And it should be linear and obey the Leibniz rule, so it's a derivation. These conditions turn out to be enough to give a notion of tangent space that agrees with intuition (e.g. you can compute the Zariski tangent space to a variety cut out by various polynomials and it will be the thing you think it is).

• great answer but his view of tangent spaces is the extrinsic view and he asked if there is a 1-to-1 correspondence or an isomorphism between his view and directional derivatives at a point, it would be easier to just define basis vectors to a point on say a surface in the elementary differential geometry way, as partial derivatives of a position vector, and then show that the vector space will still work even if we forgot about the position vector and worked just with the partial derivative operators as the basis. Apr 17, 2020 at 20:36

First use GAGA to pass to complex manifolds. Then use base change to pass to a real manifold....Now, given any smooth embedding (use the Whitney embedding theorem) of your manifold in Euclidean space, assume that the tangent surface at p exists and does everything you want it to. The one-to-one correspondence you ask for, which was not given by the other answer, is:

Given a vector starting at p and lying in the tangent surface, consider the geodesic line lying in your manifold, passing through p, and whose tangent vector in the usual Euclidean sense is the given vector. Problem: to define a derivation of the space of germs of functions on your manifold at p. Hint: First extend your function to the entire ambient space, smoothly. Then use the ordinary directional derivative.
Remark: this is not canonical, but you didn't ask for a canonical isomorphism.

Second remark: you didn't specify whether the field has characteristic zero or not...I assume it does. Third remark: what you wrote down isn't actually the definition of the Zariski tangent space. The definition of the Zariski tangent space is $m_p/m_p^2$ where $m_p$ is the ideal of all algebraic functions which vanish at p.

• There is no such thing as "the" definition of the Zariski tangent space. Any definition provably equivalent to any other can be used. Feb 27, 2013 at 4:58

There is a one-to-one correspondence, here is a simple example from elementary differential geometry :

let $$\Gamma:\mathbb{R}\to \mathbb{R}^2$$ be a curve in $$\mathbb{R}^2$$, the derivative $$\frac{d\Gamma}{dt}|_p$$ of $$\Gamma$$ at some point $$p$$ on the curve is a vector tangent to the curve at this point, so it can be the basis vector for the tangent space at $$p$$ by spanning it so : $$T_p\Gamma=\left\{ v\frac{d\Gamma}{dt}\Big{|}_p,~v\in\mathbb{R} \right\}$$
now we can forget about $$\Gamma$$ and define it using only the operator $$\frac{d}{dt}|_p$$ and nothing will be effected: $$T'_p\Gamma=\left\{ v\frac{d}{dt}\Big{|}_p,~v\in\mathbb{R} \right\}$$

you can see that the old and the new $$T_p\Gamma$$ are isomorphic so you can use any one of them as the tangent space