Why is $\mathfrak{m}/\mathfrak{m}^{2}$ the cotangent space to a variety?

Question

Let $X$ be a variety (or just a scheme) then for a point $p \in X$ with local ring $\mathcal{O}_{p}$ at $p$, we define the cotangent space at $p$ to be the $\kappa(p)$-vector space $\mathfrak{m}/\mathfrak{m}^{2}$, where $\mathfrak{m}$ is the maximal ideal of $\mathcal{O}_{p}$ and $\kappa(p)$ is the residue field at $p$.

But why? I cannot see the intuition behind this. The vector space $\mathfrak{m}/\mathfrak{m}^{2}$ is just the space of functions vanishing at $p$, modulo higher order terms. So in other words, first order approximations of functions vanishing at $p$.

All of my intuition tells me this is precisely what a tangent space is. So why is this the cotangent space? Why do we not then call the space of derivations the cotangent space?

I was tempted to say this was just a matter of convention, but most algebraic geometry books claim that the cotangent space is more natural, and suggest that this is a quirk of algebraic geometry as opposed to differential geometry. So this suggests it is not just a convention but that there is some real meaningful difference.

I am sorry if this is an extraordinarily basic question, but I find I am just going about using the definition without actually really knowing what or why I am doing it.

Answer 1

Your intuition is not correct: a tangent vector at a point is something you can take a directional derivative along. This leads naturally to the identification of the tangent space with the derivations acting on local functions defined at our point by sending a direction to the derivative along that direction (one can check that this is reversible). In other words, if $X$ is a $C^\infty$ manifold with a point $p\in X$, then $T_pX=\operatorname{Der}(\mathcal{O}_{X,p},\Bbb R)$.

If we do the same thing for a scheme, we get $T_pX=\operatorname{Der}(\mathcal{O}_{X,p},\kappa(p))$, and this is the dual of $\mathfrak{m}/\mathfrak{m}^2$ because a derivation can only see the linear term. Voila, the Zariski cotangent space!

Answer 2

Question: "I am sorry if this is an extraordinarily basic question, but I find I am just going about using the definition without actually really knowing what or why I am doing it."

Answer: This is a much asked question by students: To "mod out" by the square of a maximal ideal corresponds to "taking derivatives".

Example: If $A:=k[x,y]$ and $I:=(x-a,y-b)$ a maximal ideal with $p:=(a,b)\in k^2$, and $k$ a field, there is a canonical projection map

$$T^1: A \rightarrow A/I^2.$$

Any polynomial $f(x,y)\in A$ may be written as

$$f(x,y)=f(a,b)+ \partial_x(f)(p)(x-a)+\partial_y(f)(p)(y-b) +\cdots \in A.$$

You get a map

$$D: A \rightarrow I/I^2$$

defined by

$$D(f):=T^1(f)-f(p)=\partial_x(f)(p)(x-a)+\partial_y(f)(p)(y-b) \in k\{(x-a),(y-b)\}.$$

Hence the map $D$ takes the derivative of $f(x,y)$ at $p=(a,b)$.

Note: You find an elementary and more detailed answer to your question below. You also find a discussion of the tangent cone that may provide intuition. It explains why it is natural to define the tangent space as the dual of the cotangent space. Note that if $k \rightarrow A$ is any $k$-algebra and $\mathfrak{m} \subseteq A$ is a maximal ideal with residue field $k$, it follows the fiber

$$\Omega^1_{A/k}\otimes_A A/\mathfrak{m}\cong \mathfrak{m}/\mathfrak{m}^2$$

is the cotangent space. Hence

"the cotangent space is the fiber of the cotangent module"

$\Omega^1_{A/k}$ at $k$-rational points.

Example: If $J \subseteq A\otimes_k A$ is the ideal of the diagonal, you may define $\Omega^1_{A/k}:=J/J^2$. The map $D$ defined above is the composed map

$$d: A \rightarrow J/J^2 \rightarrow J/J^2\otimes_A A/I \cong I/I^2$$

where $d: A\rightarrow \Omega^1_{A/k}:=J/J^2$ is the universal derivation:

$$d(f):=1\otimes f - f\otimes 1.$$

When $A$ is a regular $k$-algebra of finite type it follows $\Omega^1_{A/k}$ is a finite rank projective $A$-module: It is the "cotangent bundle" of $A/k$.

If $A_{\mathfrak{m}}$ is the local ring of $A$ at a maximal ideal $\mathfrak{m}$, there is an isomorphism of $A/\mathfrak{m}$-vector spaces

$$\mathfrak{m}_{\mathfrak{m}}/\mathfrak{m}_{\mathfrak{m}}^2 \cong \mathfrak{m}/\mathfrak{m}^2,$$

hence the cotangent space can be defined in terms of the local ring of $A$ at $\mathfrak{m}$. The local ring is intrinsic and does not depend on a "choice of coordinates" and one wants to define the cotangent and tangent space independently of choice of coordinates. If you choose a closed embedding

$$\phi: Spec(A) \subseteq \mathbb{A}^n$$

into affine $n$-space (equivalently: You choose a set of coordinates of $A$), you may define the "embedded tangent space $E(x)$ of $Spec(A)$ at $x$" using the embedding. Then you must prove that $E(x)$ is independent of choice of embedding $\phi$. When using the local ring, you avoid this problem. See the attached links for more information:

$aX + bY$ is an element of $M^2$ if and only if the line $aX + bY = 0$ is tangent to $W$ at $(0, 0)$

About the definition of tangent space and tangent cone.

Dimension of the tangent space in algebraic geometry