Understand Cotangent Space as an Equivalence Class

Solution 1:

It is obvious that $\mathcal I_p^2 \subset \mathcal I_p$. Well, it is an easy exercise: simply use the definition of $\mathcal I_p$ to check that each element of $\mathcal I_p^2$ is in $\mathcal I_p$. Hint: 1) trivial: $a \cdot 0 = 0$ $\forall a \in \mathbb{R}$; 2) less trivial: products of smooth functions are smooth.

Straightforward verifications show the further necessary algebraic properties (being subspaces, ideals...) as the commenters have pointed out.

An important property of the space $\mathcal I_p^2$ is that each element of it has the vanishing differential at point $p$: $\forall f \in \mathcal I_p^2 \; \mathrm{d}f(p)=0$. Hint: use the product rule.

And the crucial fact: the quotient $\mathcal I_p / \mathcal I_p^2$ is a finite dimensional vector space. For a proof one uses the Taylor theorem (in the form that is also known as Hadamard's lemma).

The beauty of this construction is that we obtain a naturally defined finite dimensional vector space at each point of a smooth manifold without using coordinates!

It turns out that elements of this space can act on tangent vectors viewed at an appropriate angle (e.g. as derivations). The duality between vectors and covectors emerges immediately.

What is more important to get is that there is the notion of a cotangent space (of a smooth manifold at a point), and there are various definitions. The notion is something to visualize, the definitions being tools that allow us to work efficiently with the problems. Without a context everything will look unjustified and clumsy.

I hope that my remarks can help. For more details please see this and this notes. Actually, Wikipedia's articles on the tangent and the cotangent spaces are quite helpful too.